diff options
Diffstat (limited to '3864')
131 files changed, 5580 insertions, 0 deletions
diff --git a/3864/CH10/EX10.1/Ex10_1.sce b/3864/CH10/EX10.1/Ex10_1.sce new file mode 100644 index 000000000..b17b14525 --- /dev/null +++ b/3864/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,61 @@ +clear +// +// + +//Initilization of Variables + +P_e=300 //N/mm**2 //Elastic Limit in tension +FOS=3 //Factor of safety +mu=0.3 //Poissons ratio +P=12*10**3 //N Pull +Q=6*10**3 //N //Shear force + +//Calculations + +//Let d be the diameter of the shaft + +//Direct stress +//P_x=P*(%pi*4**-1*d**3)**-1 +//After substituting values and further simplifying we get +//P_x=48*10**3 + +//Now shear stress at the centre of bolt +//q=4*3**-1*q_av +//After substituting values and further simplifying we get +//q=32*10**3*(%pi*d**2)**-1 + +//Principal stresses are +//P1=P_x*2**-1+((P_x*2**-1)**2+q**2)**0.5 +//After substituting values and further simplifying we get +//p1=20371.833*(d**2)**-1 + +//P2=P_x*2**-1-((P_x*2**-1)**2+q**2)**0.5 +//After substituting values and further simplifying we get +//P2=-5092.984*(d**2)**-1 + +//q_max=((P_x*2**-1)**2+q**2)**0.5 + +//From Max Principal stress theory +//Permissible stress in Tension +P1=100 //N/mm**2 +d=(20371.833*P1**-1)**0.5 + +//Max strain theory +//e_max=P1*E**-1-mu*P2*E**-1 +//After substituting values and further simplifying we get +//e_max=21899.728*(d**2*E)**-1 + +//According to this theory,the design condition is +//e_max=P_e*(E*FOS)**-1 +//After substituting values and further simplifying we get +d2=(21899.728*3*300**-1)**0.5 //mm + +//Max shear stress theory +//e_max=shear stress at elastic*(FOS)**-1 +//After substituting values and further simplifying we get +d3=(12732.421*6*300**-1)**0.5 //mm + +//Result +printf("\n Diameter of Bolt by:Max Principal stress theory %0.2f mm",d) +printf("\n :Max strain theory %0.2f mm",d2) +printf("\n :Max shear stress theory %0.2f mm",d3) diff --git a/3864/CH10/EX10.2/Ex10_2.sce b/3864/CH10/EX10.2/Ex10_2.sce new file mode 100644 index 000000000..81c9db7dd --- /dev/null +++ b/3864/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,50 @@ +clear +// +// + +//Initilization of Variables + +M=40*10**6 //N-mm //Bending moment +T=10*10**6 //N-mm //TOrque +mu=0.25 //Poissons ratio +P_e=200 //N/mm**2 //Stress at Elastic Limit +FOS=2 + +//Calculations + +//Let d be the diameter of the shaft + +//Principal stresses are given by + +//P1=16*(%pi*d**3)**-1*(M+(M**2+T**2)**0.5) +//After substituting values and further simplifying we get +//P1=4.13706*10**8*(d**3)**-1 ............................(1) + +//P2=16*(%pi*d**3)**-1*(M-(M**2+T**2)**0.5) +//After substituting values and further simplifying we get +//P2=-6269718*(%pi*d**3)**-1 ..............................(2) + +//q_max=(P1-P2)*2**-1 +//After substituting values and further simplifying we get +//q_max=2.09988*10**8*(d**3)**-1 + +//Max Principal stress theory +//P1=P_e*(FOS)**-1 +//After substituting values and further simplifying we get +d=(4.13706*10**8*2*200**-1)**0.33333 //mm + +//Max shear stress theory +//q_max=shear stress at elastic limit*(FOS)**-1 +//After substituting values and further simplifying we get +d2=(2.09988*10**8*4*200**-1)**0.33333 + +//Max strain energy theory +//P_3=0 +//P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1 +//After substituting values and further simplifying we get +d3=(8.62444*10**12)**0.166666 + +//Result +printf("\n Diameter of shaft according to:MAx Principal stress theory %0.2f mm",d) +printf("\n :Max shear stress theory %0.2f mm",d2) +printf("\n :Max strain energy theory %0.2f mm",d3) diff --git a/3864/CH10/EX10.3/Ex10_3.sce b/3864/CH10/EX10.3/Ex10_3.sce new file mode 100644 index 000000000..19a6c0323 --- /dev/null +++ b/3864/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,67 @@ +clear +// +// + +//Initilization of Variables + +f_x=40 //N/mm**2 //Internal Fliud Pressure +d1=200 //mm //Internal Diameter +r1=d1*2**-1 //mm //Radius +q=300 //N/mm**2 //Tensile stress + +//Calculations + +//From Lame's Equation we have, + +//Hoop Stress +//f_x=b*(x**2)**-1+a ..........................(1) + +//Radial Pressure +//p_x=b*(x**2)**-1-a .........................(2) + +//the boundary conditions are +x=d1*2**-1 //mm +//After sub values in equation 1 and further simplifying we get +//40=b*100**-1-a ..........................(3) + +//Max Principal stress theory +//q*(FOS)**-1=b*100**2+a ..................(4) +//After sub values in above equation and further simplifying we get + +//From Equation 3 and 4 we get +a=80*2**-1 +//Sub value of a in equation 3 we get +b=(f_x+a)*100**2 + +//At outer edge where x=r_0 pressure is zero +r_0=(b*a**-1)**0.5 //mm + +//thickness +t=r_0-r1 //mm + +//Max shear stress theory +P1=b*(100**2)**-1+a //Max hoop stress +P2=-40 //pressure at int radius (since P2 is compressive) + +//Max shear stress +q_max=(P1-P2)*2**-1 + +//According max shear theory the design condition is +//q_max=P_e*2**-1*(FOS)**-1 +//After sub values in equation we get and further simplifying we get +//80=b*(100**2)**-1+a +//After sub values in equation 1 and 3 and further simplifying we get +b2=120*100**2*2**-1 + +//from equation(3) +a2=120*2**-1-a + +//At outer radius r_0,radial pressure=0 +r_02=(b2*a2**-1)**0.5 + +//thickness +t2=r_02-r1 + +//Result +printf("\n Thickness of metal by:Max Principal stress theory %0.2f mm",t) +printf("\n :Max shear stress thoery %0.2f mm",t2) diff --git a/3864/CH2/EX2.1/Ex2_1.sce b/3864/CH2/EX2.1/Ex2_1.sce new file mode 100644 index 000000000..573580d97 --- /dev/null +++ b/3864/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,21 @@ +clear +// +// + +//Initilization of Variables +P=45*10**3 //N //Load +E=200*10**3 //N/mm**2 //Modulus of elasticity of rod +L=500 //mm //Length of rod +d=20 //mm //Diameter of rod + +//Calculations + +A=%pi*d**2*4**-1 //mm**2 //Area of circular rod +p=P*A**-1 //N/mm**2 //stress +e=p*E**-1 //strain +dell_l=(P*L)*(A*E)**-1 + +//Result +printf("\n The stress in bar due to Load is %0.5f N/mm",p) +printf("\n The strain in bar due to Load is %0.5f N/mm",e) +printf("\n The Elongation in bar due to Load is %0.2f mm",dell_l) diff --git a/3864/CH2/EX2.11/Ex2_11.sce b/3864/CH2/EX2.11/Ex2_11.sce new file mode 100644 index 000000000..e3deff5c9 --- /dev/null +++ b/3864/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,26 @@ +clear +// +// + +//Initilization of Variables + +t=10 //mm //Thickness of steel +b1=60 //mm //width of plate1 +b2=40 //mm //width of plate2 +P=60*10**3 //Load +L=600 //mm //Length of plate +E=2*10**5 //N/mm**2 + +//Calculations + +//Extension of taperong bar of rectangular section +dell_l=P*L*(t*E*(b1-b2))**-1*log(b1*b2**-1) + +A_av=(b1*t+b2*t)*2**-1 //Average Area //mm**2 +dell_l2=P*L*(A_av*E)**-1 + +//PErcentage Error +e=(dell_l-dell_l2)*(dell_l)**-1*100 + +//Result +printf("\n The Percentage Error is %0.2f ",e) diff --git a/3864/CH2/EX2.12/Ex2_12.sce b/3864/CH2/EX2.12/Ex2_12.sce new file mode 100644 index 000000000..acb225448 --- /dev/null +++ b/3864/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,29 @@ +clear +// +// + +//Initilization of Variables + +L=1.5 //m //Length of steel bar +L1=1000 //m0 //Length of steel bar 1 +L2=500 //m //Length of steel bar 2 +d1=40 //Diameter of steel bar 1 +d2=20 //diameter of steel bar 2 +E=2*10**5 //N/mm**2 //Modulus of Elasticity +P=160*10**3 //N //Load + +//Calculations + +A1=%pi*4**-1*d1**2 //Area of Portion 1 + +//Extension of uniform Portion 1 +dell_l1=P*L1*(A1*E)**-1 //mm + +//Extension of uniform Portion 2 +dell_l2=4*P*L2*(%pi*d1*d2*E)**-1 //mm + +//Total Extension of Bar +dell_l=dell_l1+dell_l2 + +//Result +printf("\n The Elongation of the Bar is %0.2f mm",dell_l) diff --git a/3864/CH2/EX2.14/Ex2_14.sce b/3864/CH2/EX2.14/Ex2_14.sce new file mode 100644 index 000000000..b654e813d --- /dev/null +++ b/3864/CH2/EX2.14/Ex2_14.sce @@ -0,0 +1,51 @@ +clear +// +// + +//Initilization of Variables + +//Portion AB +L_AB=600 //mm //Length of AB +A_AB=40*40 //mm**2 //Cross-section Area of AB + +//Portion BC +L_BC=800 //mm //Length of BC +A_BC=30*30 //mm //Length of BC + +//Portion CD +L_CD=1000 //mm //Length of CD +A_CD=20*20 //mm //Area of CD + +P1=80*10**3 //N //Load1 +P2=60*10**3 //N //Load2 +P3=40*10**3 //N //Load3 + +E=2*10**5 //Modulus of Elasticity + +//Calculations + +P4=P1-P2+P3 //Load4 + +//Now Force in AB +F_AB=P1 + +//Force in BC +F_BC=P1-P2 + +//Force in CD +F_CD=P4 + +//Extension of AB +dell_l_AB=F_AB*L_AB*(A_AB*E)**-1 + +//Extension of BC +dell_l_BC=F_BC*L_BC*(A_BC*E)**-1 + +//Extension of CD +dell_l_CD=F_CD*L_CD*(A_CD*E)**-1 + +//Total Extension +dell_l=dell_l_AB+dell_l_BC+dell_l_CD + +//Result +printf("\n The Total Extension in Bar is %0.2f mm",dell_l) diff --git a/3864/CH2/EX2.15/Ex2_15.sce b/3864/CH2/EX2.15/Ex2_15.sce new file mode 100644 index 000000000..e93ab1cf5 --- /dev/null +++ b/3864/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,71 @@ +clear +// +// + +//Initilization of Variables + +L=800 //mm //Length of bar +F1=30*10**3 //N //Force acting on the bar +F2=60*10**3 //N //force acting on the bar +L=800 //mm //Length of bar +d=25 //mm //diameter of bar +L_AC=275 //mm //Length of AC +L_CD=150 //mm //Length of CD +L_DB=375 //mm //Length of DB +E=2*10**5 //Pa //Modulus of elasticity + +//Calculations + +//Let P be the Reaction on tne Bar from support at A + +//Shortening of Portion AC +//dell_l_AC1=P*L_AC*(A*E)**-1 + +//Shortening of Portion CD +//dell_l_CD1=(30+P)*L_CD*(A*E)**-1 + +//Extension of Portion DB +//dell_l_DB1=(30-P)*L_DB*(A*E)**-1 + +//Total Extensions=1*(A*E)**-1*(P*L_AC-(30+P)*L_CD+(30-P)*L_DB) +//As Supports are unyielding,Total Extensions=0 + +//After substituting values in above equation and Further simplifying we get +P=(30*375-150*30)*800**-1 + +//Reaction of support A +R_A=P + +//Reaction of support B +R_B=30-P + +//Cross-sectional Area +A=%pi*4**-1*d**2 + +//Stress in Portion AC +sigma1=P*10**3*A**-1 //N/mm**2 + +//Stress in Portion CD +sigma2=(30+P)*10**3*A**-1 //N/mm**2 + +//Stress in Portion DB +sigma3=(30-P)*10**3*A**-1 //N/mm**2 + +//Shortening of Portion AC +dell_l_AC2=P*10**3*L_AC*(A*E)**-1 //mm + +//Shortening of Portion CD +dell_l_CD2=(30+P)*10**3*L_CD*(A*E)**-1 //mm + +//Extension of Portion DB +dell_l_DB2=(30-P)*10**3*L_DB*(A*E)**-1 //mm + +//result +printf("\n The Reactios at two Ends are:R_A %0.2f KN",R_A) +printf("\n :R_B %0.2f KN",R_B) +printf("\n Stress in Portion AC %0.2f N/mm**2",sigma1) +printf("\n Stress in Portion CD %0.2f N/mm**2",sigma2) +printf("\n Stress in Portion DB %0.2f N/mm**2",sigma3) +printf("\n Shortening of Portion AC %0.3f mm",dell_l_AC2) +printf("\n Shortening of Portion CD %0.3f mm",dell_l_CD2) +printf("\n Shortening of Portion DB %0.3f mm",dell_l_DB2) diff --git a/3864/CH2/EX2.2/Ex2_2.sce b/3864/CH2/EX2.2/Ex2_2.sce new file mode 100644 index 000000000..b255a81fa --- /dev/null +++ b/3864/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,19 @@ +clear +// + +//Initilization of Variables + +A=15*0.75 //mm**2 //area of steel tape +P=100 //N //Force apllied +L=30*10**3 //mm //Length of tape +E=200*10**3 //N/m**2 //Modulus of Elasticity of steel tape +AB=150 //m //Measurement of Line AB + +//Calculations + +dell_l=P*L*(A*E)**-1 //mm //Elongation +l=L+dell_l*10**-3 //mm //Actual Length +AB1=AB*l*L**-1 //m Actual Length of AB + +//Result +printf("\n The Actual Length of Line AB is %0.2f m",AB1) diff --git a/3864/CH2/EX2.20/Ex2_20.sce b/3864/CH2/EX2.20/Ex2_20.sce new file mode 100644 index 000000000..9d5785310 --- /dev/null +++ b/3864/CH2/EX2.20/Ex2_20.sce @@ -0,0 +1,56 @@ +clear +// +// + +//Initilization of Variables + +sigma=150 //N/mm**2 //Stress +P=40*10**3 //N //Load + +//Calculations + +//LEt P_A.P_B,P_C,P_D be the forces developed in wires A,B,C,D respectively + +//Let sum of all Vertical Forces=0 +//P_A+P_B+P_C+P_D=40 ..........................(1) + +//Let x be the distance between each wires +//sum of all moments=0 +//P_B*x+P_C*2*x+P_D*3*x=40*2*x + +//After further simplifying we get +//P_B+2*P_C+3*P_D=80 ..........................(2) + +//As the equations of statics ae not enough to find unknowns,Consider compatibilit Equations + +//Let dell_l be the increse in elongation of wire + +//dell_l_B=dell_l_A+dell_l +//dell_l_C=dell_l_A+2*dell_l +//dell_l_D=dell_l_A+3*dell_l + +//Let P1 be the force required for the Elongation of wires,then +//P_B=P_A+P1 ] +//P_C=P_A+2*P1 ] +//P_D=P_A+3*P1 ] ................................(3) + +//from Equation (3) and (1) we get +//2*P_A+3*P1=20 ................................(4) + +//from Equation (3) and (2) we get +//6*P_A+14*P1=80 + +//subtracting 3 times equation (4) from (3) we get +P1=20*5**-1 + +//from Equation 4 we get +P_A=(80-14*P1)*6**-1 +P_B=P_A+P1 +P_C=P_A+2*P1 +P_D=P_A+3*P1 + +//Let d be the diameter required,then +d=(P_D*10**3*4*(%pi*150)**-1)**0.5 + +//result +printf("\n The Required Diameter is %0.2f mm",d) diff --git a/3864/CH2/EX2.21/Ex2_21.sce b/3864/CH2/EX2.21/Ex2_21.sce new file mode 100644 index 000000000..678fe0d54 --- /dev/null +++ b/3864/CH2/EX2.21/Ex2_21.sce @@ -0,0 +1,52 @@ +clear +// +// + +//Initilization of Variables + +P=20*10**3 //N //Load +d=6 //mm //diameter of wire +E=2*10**5 //N/mm**2 +L_BO=4000 //mm //Length of BO + +//Calculations + +//Let theta be the angle between OA and OB and also between OC and OB +theta=30 + +//Let P_OA,P_OB,P_OC be the Forces introduced in wires OA,OB,OC respectively +//Due to symmetry P_OA=P_OC (same angles) + +//Sum of all Vertical Forces=0 +//P_OA*cos(theta)+P_OB+P_OC*cos(theta)=P + +//After further simplifyinf we get +//2*P_OA*cos(theta)+P_OB=20 ...............(1) + +//Let oo1 be the extension of BO +//oo1=L_A1o1*(cos(theta))**-1 + +//From relation we get +//P_OB*L_BO=P_OA*L_AO*(cos(theta))**-1 + +//But L_AO=L_BO*(cos(theta))**-1 + +//After substituting value of L_AO in above equation we get +//P_OB=0.75*P_OA .......................(2) + +//substituting in Equation 1 we get +//2*P_OA*cos(theta)+0.75*P_OA=20 + +P_OA=20*(2*cos(theta*%pi*180**-1)+0.75)**-1 + +P_OB=0.75*P_OA + +A=%pi*4**-1*d**2 + +//Vertical displacement of Load +dell_l_BO=P_OB*10**3*L_BO*(A*E)**-1 + +//Result +printf("\n Forces in each wire is:P_OA %0.2f KN",P_OA) +printf("\n :P_OB %0.2f KN",P_OB) +printf("\n Vertical displacement of Loadis %0.2f mm",dell_l_BO) diff --git a/3864/CH2/EX2.22/Ex2_22.sce b/3864/CH2/EX2.22/Ex2_22.sce new file mode 100644 index 000000000..0949e768e --- /dev/null +++ b/3864/CH2/EX2.22/Ex2_22.sce @@ -0,0 +1,39 @@ +clear +// + +//Initilization of Variables + +A_a=50*20 //mm //Area of aluminium strip +A_s=50*15 //mm //Area of steel strip +P=50*10**3 //N //Load +E_a=1*10**5 //N/mm**2 //Modulus of aluminium +E_s=2*10**5 //N/mm**2 //Modulus of steel + +//Calculations + +//Let P_a and P_s br the Load shared by aluminium and steel strip +//P_a+P_s=P ..................(1) + +//For compatibility condition,dell_l_a=dell_l_s +//P_a*L_a*(A_a*E_a)**-1=P_s*L_s*(A_s*E_s)**-1 .....(2) + +//As L_a=L_s we get +//P_s=1.5*P_a .................(3) + +//From Equation 1 and 2 we get +P_a=P*2.5**-1 + +//Substituting in equation 1 we get +P_s=P-P_a + +//stress in aluminium strip +sigma_a=P_a*A_a**-1 + +//stress in steel strip +sigma_s=P_s*A_s**-1 + +//Now from the relation we get + +//result +printf("\n Stress in Aluminium strip is %0.2f N/mm**2",sigma_a) +printf("\n Stress in steel strip is %0.2f N/mm**2",sigma_s) diff --git a/3864/CH2/EX2.23/Ex2_23.sce b/3864/CH2/EX2.23/Ex2_23.sce new file mode 100644 index 000000000..b1ee4ddd9 --- /dev/null +++ b/3864/CH2/EX2.23/Ex2_23.sce @@ -0,0 +1,43 @@ +clear +// +// + +//Initilization of Variables + +D_s=20 //mm //Diameter of steel +D_Ci=20 //mm //Internal Diameter of Copper +t=5 //mm //THickness of copper bar +P=100*10**3 //N //Load +E_s=2*10**5 //N/mm**2 //modulus of elasticity of steel +E_c=1.2*10**5 //N/mm**2 //Modulus of Elasticity of Copper + +//Calculations + +A_s=%pi*4**-1*D_s**2 //mm**2 //Area of steel +D_Ce=D_s+2*t //mm //External Diameterof Copper Tube + +A_c=%pi*4**-1*(D_Ce**2-D_Ci**2) //mm**2 //Area of Copper + +//From static Equilibrium condition +//Let P_s and P_c be the Load shared by steel and copper in KN +//P_s+P_c=100 ....................................(1) + +//From compatibility Equation,dell_l_s=dell_l_c +//P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1 + +//Substituting values in above Equation we get +//P_s=1.3333*P_C + +//Now Substituting value of P_s in Equation (1),we get +P_c=100*2.3333**-1 //KN +P_s=100-P_c //KN + +//Stress in steel +sigma_s=P_s*10**3*A_s**-1 //N/mm**2 + +//Stress in copper +sigma_c=P_c*10**3*A_c**-1 //N/mm**2 + +//Result +printf("\n Stresses Developed in Two material are:sigma_s %0.2f N/mm**2",sigma_s) +printf("\n :sigma_c %0.2f N/mm**2",sigma_c) diff --git a/3864/CH2/EX2.24/Ex2_24.sce b/3864/CH2/EX2.24/Ex2_24.sce new file mode 100644 index 000000000..b75582839 --- /dev/null +++ b/3864/CH2/EX2.24/Ex2_24.sce @@ -0,0 +1,39 @@ +clear +// +// + +//Initilization of Variables + +A_C=230*400 //mm //Area of column +D_s=12 //mm //Diameter of steel Bar +P=600*10**3 //N //Axial compression +//E_s*E_c=18.67 +n=8 //number of steel Bars + +//Calculations + +A_s=%pi*4**-1*D_s**2*n //Area of steel //mm**2 +A_c=A_C-A_s //mm**2 //Area of concrete + +//From static Equilibrium condition +//P_s+P_c=600 .........(1) + +//Now from compatibility Equation dell_l_s=dell_l_c we get, +//P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1 + +//Substituting values in above Equation we get +//P_s=0.1854*P_c + +//Now Substituting value of P_s in Equation (1),we get +P_c=600*1.1854**-1 +P_s=600-P_c + +//Stress in steel +sigma_s=P_s*10**3*A_s**-1 //N/mm**2 + +//Stress in copper +sigma_c=P_c*10**3*A_c**-1 //N/mm**2 + +//Result +printf("\n Stresses Developed in Two material are:sigma_s %0.2f N/mm**2",sigma_s) +printf("\n :sigma_c %0.2f N/mm**2",sigma_c) diff --git a/3864/CH2/EX2.25/Ex2_25.sce b/3864/CH2/EX2.25/Ex2_25.sce new file mode 100644 index 000000000..8b3b0923d --- /dev/null +++ b/3864/CH2/EX2.25/Ex2_25.sce @@ -0,0 +1,56 @@ +clear +// + +//Initilization of Variables + +P=200*10**3 //N //Load +A_a=1000 //mm**2 //Area of Aluminium +A_s=800 //mm**2 //Area of steel +E_a=1*10**5 //N/mm**2 //Modulus of Elasticity of Aluminium +E_s=2*10**5 //N/mm**2 //Modulus of ELasticity of steel +sigma_a1=65 //N/mm**2 //stress in aluminium +sigma_s1=150 //N/mm**2 //Stress in steel + +//Calculations + +//Let P_a and P_s be the force in aluminium and steel pillar respectively + +//Now,sum of forces in Vertical direction we get +//2*P_a+P_s=200 .........................................(1) + +//By compatibility Equation dell_l_s=dell_l_a we get +//P_s=1.28*P_a ..........................................(2) + +//Now substituting value of P_s in Equation 1 we get +P_a=200*3.28**-1 //KN +P_s=200-2*P_a //KN + +//Stress developed in aluminium +sigma_a=P_a*10**3*A_a**-1 //N/mm**2 + +//Stress developed in steel +sigma_s=P_s*10**3*A_s**-1 //N/mm**2 + +//Part-2 + +//Let sigma_a1 and sigma_s1 be the stresses in Aluminium and steel due to Additional LOad + +P_a1=sigma_a1*A_a //Load carrying capacity of aluminium +P_s1=1.28*P_a1 + +//Total Load carrying capacity +P1=2*P_a1+P_s1 //N + +P_s2=sigma_s1*A_s //Load carrying capacity of steel +P_a2=P_s2*1.28**-1 + +//Total Load carrying capacity +P2=2*P_a2+P_s2 + +//Additional Load +P3=P1-P + +//Result +printf("\n Stresses Developed in Each Pillar is:sigma_a %0.2f N/mm**2",sigma_a) +printf("\n :sigma_s %0.2f N/mm**2",sigma_s) +printf("\n Additional Load taken by pillars is %0.2f N",P3) diff --git a/3864/CH2/EX2.26/Ex2_26.sce b/3864/CH2/EX2.26/Ex2_26.sce new file mode 100644 index 000000000..448652caf --- /dev/null +++ b/3864/CH2/EX2.26/Ex2_26.sce @@ -0,0 +1,38 @@ +clear +// +// + +//Initilization of Variables + +L=500 //mm //Length of assembly +D=16 //mm //Diameter of steel bolt +Di=20 //mm //internal Diameter of copper tube +Do=30 //mm //External Diameter of copper tube +E_s=2*10**5 //N/mm**2 //Modulus of Elasticity of steel +E_c=1.2*10**5 //N/mm**2 //Modulus of Elasticity of copper +p=2 //mm //Pitch of nut + +//Calculations + +//Let P_s be the Force in bolt and P_c be the FOrce in copper tube +//P_s=-P_s + +dell=1*4**-1*2 //Quarter turn of nut total movement + +//dell=dell_s+dell_c + +//Area of steel +A_s=%pi*4**-1*D**2 + +//Area of copper +A_c=%pi*4**-1*(Do**2-Di**2) + +//dell=P*L*(A_s*E_s)**-1+P*L*(A_c*E_c)**-1 +P=dell*(1*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1*L**-1 //LOad + +P_s=P*A_s**-1 +P_c=P*A_c**-1 + +//result +printf("\n stress introduced in bolt is %0.2f N/mm**2",P_s) +printf("\n stress introduced in tube is %0.2f N/mm**2",P_c) diff --git a/3864/CH2/EX2.27/Ex2_27.sce b/3864/CH2/EX2.27/Ex2_27.sce new file mode 100644 index 000000000..d02f8430d --- /dev/null +++ b/3864/CH2/EX2.27/Ex2_27.sce @@ -0,0 +1,42 @@ +clear +// +// + +//Initilization of Variables + +D=20 //mm //Diameter of Bolts +Di=25 //m //internal Diameter +t=10 //mm //Thickness of bolt +E_s=2*10**5 //N/mm**2 //Modulus of Elasticity +E_c=1.2*10**5 //N/mm**2 //Modulus of copper +p=3 //mm //Pitch +theta=30 //degree +L_c=500 //Lengh of copper +L_s=600 //Length of steel + +//Calculations + +//Let P_s be the Force in each bolt and P_c be the FOrce in copper tube +//From Static Equilibrium condition +//P_c=2*P_s + +//As nut moves by 60 degree.If nut moves by 360 degree its Longitudinal movement is by 3 mm +dell=theta*360**-1*p + +//From Compatibility Equaton we get +//dell=dell_c+dell_s + + +A_s=%pi*4**-1*Di**2 //mm**2 //Area of steel +A_c=%pi*4**-1*(45**2-Di**2) //mm**2 //Area of copper + +//Force introduced in steel +P_s=0.5*(2*L_c*(A_c*E_c)**-1+L_s*(A_s*E_s)**-1)**-1 //N +P_s2=P_s*A_s**-1 + +//Force introduced in copper +P_c=2*P_s*A_c**-1 //N + +//Result +printf("\n Stress introduced in bolt is %0.2f N/mm**2",P_s2) +printf("\n stress introduced in tube is %0.2f N/mm**2",P_c) diff --git a/3864/CH2/EX2.28/Ex2_28.sce b/3864/CH2/EX2.28/Ex2_28.sce new file mode 100644 index 000000000..a89771447 --- /dev/null +++ b/3864/CH2/EX2.28/Ex2_28.sce @@ -0,0 +1,29 @@ +clear +// + +//Initilization of Variables + +L=9 //m //Length of rigid bar +L_b=3000 //Length of bar +A_b=1000 //mm**2 //Area of bar +E_b=1*10**5 //N/mm**2 //Modulus of Elasticity of brasss bar +L_s=5000 //mm //Length of steel bar +A_s=445 //mm**2 //Area of steel bar +E_s=2*10**5 //N/mm**2 //Modulus of elasticity of steel bar +P=3000 //N //Load + +//Calculations + +//From static equilibrium Equation of the rod after appliying Load is +//P_b+P_s=P ......................(1) + +//P_b=1.8727*P_s ..................(2) + +//NOw substituting equation 2 in equation 1 we get +P_s=P*2.8727**-1 +P_b=P-P_s + +d=P_s*L*P**-1 + +//Result +printf("\n Distance at which Load applied even after which bar remains horizontal is %0.2f m",d) diff --git a/3864/CH2/EX2.3/Ex2_3.sce b/3864/CH2/EX2.3/Ex2_3.sce new file mode 100644 index 000000000..9ef330004 --- /dev/null +++ b/3864/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,25 @@ +clear +// +// + +//Initilization of Variables + +//Let y be the yield stress + +y=250 //N/mm**2 //yield stress +FOS=1.75 //Factor of safety +P=140*10**3 //N //compressive Load +D=101.6 //mm //External diameter + +//Calculations + +p=y*(FOS)**-1 //N/mm**2 //Permissible stress +A=P*p**-1 //mm**2 //Area of hollow tube + +//Let d be the internal diameter of tube +d=-((A*4*(%pi)**-1)-D**2) +X=d**0.5 +t=(D-X)*2**-1 //mm //Thickness of steel tube + +//result +printf("\n The thickness of steel tube is %0.2f mm",t) diff --git a/3864/CH2/EX2.30/Ex2_30.sce b/3864/CH2/EX2.30/Ex2_30.sce new file mode 100644 index 000000000..494f7c86a --- /dev/null +++ b/3864/CH2/EX2.30/Ex2_30.sce @@ -0,0 +1,35 @@ +clear +// + +//Initilization of Variables + +L=12.6 //m //Length of rail +t1=24 //Degree celsius +t2=44 //degree celsius +alpha=12*10**-6 //Per degree celsius +E=2*10**5 //N/mm**2 //Modulus of ELasticity +gamma=2 //mm //Gap provided for Expansion +sigma=20 //N/mm**2 //Stress + +//Calculations + +t=t2-t1 //Temperature Difference + +//Free Expansion of the rails +dell=alpha*t*L*1000 //mm + +//When no expansion joint is provided then +p=dell*E*(L*10**3)**-1 + +//When a gap of 2 mm is provided,then free expansion prevented is +dell_1=dell-gamma +p2=dell_1*E*(L*10**3)**-1 + +//When stress is developed,then gap left is +gamma2=-(sigma*L*10**3*E**-1-dell) + +//Result +printf("\n The minimum gap between the two rails is %0.2f mm",dell) +printf("\n Thermal Developed in the rials if:No expansionn joint is provided:p %0.2f N/mm**2",p) +printf("\n :If a gap of is provided then :p2 %0.2f N/mm**2",p2) +printf("\n When stress is developed gap left between the rails is %0.2f mm",gamma2) diff --git a/3864/CH2/EX2.31/Ex2_31.sce b/3864/CH2/EX2.31/Ex2_31.sce new file mode 100644 index 000000000..8a95fa359 --- /dev/null +++ b/3864/CH2/EX2.31/Ex2_31.sce @@ -0,0 +1,26 @@ +clear +// + +//Initilization of Variables + +t=20 //degree celsius +E_a=70*10**9 //N/mm**2 //Modulus of Elasticicty of aluminium +alpha_a=11*10**-6 //per degree celsius //Temperature coeff of aluminium +alpha_s=12*10**-6 //Per degree celsius //Temperature coeff of steel +L_a=1000 //mm //Length of aluminium +L_s=3000 //mm //Length of steel +E_a=7*10**4 //N/mm**2 //Modulus of Elasticity of aluminium +E_s=2*10**5 //N/mm*2 //Modulus of Elasticity of steel +A_a=600 //mm**2 //Area of aluminium +A_s=300 //mm**2 //Area of steel + +//Calculations + +//Free Expansion +dell=alpha_a*t*L_a+alpha_s*t*L_s + +//support Reaction +P=dell*(L_a*(A_a*E_a)**-1+L_s*(A_s*E_s)**-1)**-1 + +//Result +printf("\n Reaction at support is %0.2f N",P) diff --git a/3864/CH2/EX2.33/Ex2_33.sce b/3864/CH2/EX2.33/Ex2_33.sce new file mode 100644 index 000000000..4baa8d7d1 --- /dev/null +++ b/3864/CH2/EX2.33/Ex2_33.sce @@ -0,0 +1,50 @@ +clear +// +// + +//Initilization of Variables + +D=25 //mm //Diameter of Brass +De=50 //mm //External Diameter of steel tube +Di=25 //mm //Internal Diameter of steel tube +L=1.5 //m //Length of both bars +t1=30 //degree celsius //Initial Temperature +t2=100 //degree celsius //final Temperature +E_s=2*10**5 //N/mm**2 //Modulus of ELasticity of steel bar +E_b=1*10**5 //N/mm**2 //Modulus of Elasticity of brass bar +alpha_s=11.6*10**-6 //Temperature Coeff of steel +alpha_b=18.7*10**-6 //Temperature coeff of brass bar +d=20 //mm //diameter of pins + +//Calculations + +t=t2-t1 //Temperature Difference +A_s=%pi*4**-1*(De**2-Di**2) //mm**2 //Area of steel +A_b=%pi*4**-1*D**2 //mm**2 //Area of brass + +//Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar +//But from Equilibrium of Forces +//P_b=P_s=P + +//Let dell=dell_s+dell_b +dell=(alpha_b-alpha_s)*t*L*1000 + +P=dell*(1*(A_s*E_s)**-1+1*(A_b*E_b)**-1)**-1*(L*1000)**-1 +P_b=P +P_s=P +//Stress in steel +sigma_s=P*A_s**-1 + +//Stress in Brass +sigma_b=P_b*A_b**-1 + +//Area of Pins +A_p=%pi*4**-1*d**2 + +//Since,the force is resisted by two cross section of pins +tou=P*(2*A_p)**-1 + +//Result +printf("\n Stress in steel bar is %0.2f N/mm**2",sigma_s) +printf("\n Stress in Brass bar is %0.2f N/mm**2",sigma_b) +printf("\n Shear Stresss induced in pins is %0.2f N/mm**2",tou) diff --git a/3864/CH2/EX2.34/Ex2_34.sce b/3864/CH2/EX2.34/Ex2_34.sce new file mode 100644 index 000000000..9ce4dd115 --- /dev/null +++ b/3864/CH2/EX2.34/Ex2_34.sce @@ -0,0 +1,45 @@ +clear +// + +//Initilization of Variables + +b_s=60 //mm //width of steel Bar +t_s=10 //mm //thickness of steel Bar +b_c=40 //mm //width of copper bar +t_c=5 //mm //thickness of copper bar +E_s=2*10**5 //N/mm**2 //Modulus of Elasticity of steel bar +E_c=1*10**5 //N/mm**2 //Modulus of Elasticity of copper bar +alpha_s=12*10**-6 //Per degree celsius //Temperature coeff of steel bar +alpha_c=17*10**-6 //Per degree celsius //Temperature coeff of copper bar +L=1000 //mm //Length of bar +L_s=1000 //mm //Length of bar +t=80 //degree celsius + +//Calculations + +A_s=b_s*t_s //Area of steel bar +A_c=b_c*t_c //Area of copper bar + +//Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar +//The equilibrium of forces gives +//P_s=2*P_c + +//Let dell=dell_s+dell_b +dell=(alpha_c-alpha_s)*t + +P_c=dell*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1 +P_s=2*P_c + +//Stress in copper +sigma_c=P_c*A_c**-1 + +//Stress in steel +sigma_s=P_s*A_s**-1 + +//Change in Length of bar +dell_2=alpha_s*t*L+P_s*L_s*(A_s*E_s)**-1 + +//result +printf("\n Stress in copper is %0.2f N/mm**2",sigma_c) +printf("\n Stress in steel is %0.2f N/mm**2",sigma_s) +printf("\n the change in Length is %0.2f mm",dell_2) diff --git a/3864/CH2/EX2.35/Ex2_35.sce b/3864/CH2/EX2.35/Ex2_35.sce new file mode 100644 index 000000000..66abf1a79 --- /dev/null +++ b/3864/CH2/EX2.35/Ex2_35.sce @@ -0,0 +1,72 @@ +clear +// + +//Initilization of Variables + +A_c=500 //mm**2 //Area of each rod +A_s=500 +A=500 +P=2*10**5 //N //Weight +L=1 //m //Length of each rod +t=40 //degree celsius //temperature +E_s=2*10**5 //N/mm**2 //Modulus of Elasticity of steel rod +E_c=1*10**5 //N/mm**2 //modulus of Elastictiy of copper rod +alpha_s=1.2*10**-5 //Per degree Celsius //temp coeff of steel rod +alpha_c=1.8*10**-5 //Per degree Celsius //Temp coeff of copper rod + +//Calculations + +//Let P_s be the force in each one of the copper rods and P_s be the force in steel rod +//2*P_c+P_s=P .....................(1) + +//Extension of copper bar=Extension of steel bar +//P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1 +//after simplifying above equation we get +//P_s=2*P_c ........................(2) + +//Now substituting value of P_s in Equation 1 we get +P_c=P*4**-1 +P_s=2*P_c + +//Now EXtension due to copper Load +dell_1=P_c*L*1000*(A_c*E_c)**-1 + +//Part-2 + +//Due to rise of temperature of40 degree celsius + +//As bars are rigidly joined,let P_c1 be the compressive forccesdeveloped in copper bar and P_s1 be the tensile force in steel causing changes +//P_s1=2*P_c1 + +//dell_s+dell_c=(alpha_c-alpha_s)*t*L .......................................(3) +//P_s1*L*(A_s*E_s)**-1+P_c1*L*(A_c*E_c)**-1=(alpha_c-alpha_s)*t*L ................(4) +//After substituting values in above equation and further simplifying we get, +P_c1=(alpha_c-alpha_s)*t*L*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1 //.................(5) +P_s1=2*P_c1 + +//Extension of bar due to temperature rise +dell_2=alpha_s*t*L+P_s1*L*(A_s*E_s)**-1 + +//Amount by which bar will descend +dell_3=dell_1+dell_2 + +//Load carried by steel bar +P_S=P_s+P_s1 + +//Load carried by copper bar +P_C=P_c-P_c1 + +//Part-3 + +//Let P_c1_1=P_c //For convenience +//Rise in temperature if Load is to be carried out by steel rod alone +P_c1_1=P_c + +//From equation 5 +t=P_c1_1*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)*(alpha_c-alpha_s)**-1 + +//result +printf("\n Extension Due top copper Load %0.2f mm",dell_1) +printf("\n Load carried by each rod:P_s %0.2f N",P_s) +printf("\n :P_c %0.2f N",P_c) +printf("\n Rise in Temperature of steel rod should be %0.2f degree Celsius",t) diff --git a/3864/CH2/EX2.36/Ex2_36.sce b/3864/CH2/EX2.36/Ex2_36.sce new file mode 100644 index 000000000..6f16a2cc0 --- /dev/null +++ b/3864/CH2/EX2.36/Ex2_36.sce @@ -0,0 +1,36 @@ +clear +// + +//Initilization of Variables + +t=40 //degree celsius //temperature +A_s=400 //mm**2 //Area of steel bar +A_c=600 //mm**2 //Area of copper bar +E_s=2*10**5 //N/mm**2 //Modulus of Elasticity of steel bar +E_c=1*10**5 //N/mm**2 //Modulus of Elasticity of copper bar +alpha_s=12*10**-6 //degree celsius //Temperature coeff of steel bar +alpha_c=18*10**-6 //degree celsius //Temperature coeff of copper bar +L_c=800 //mm //Length of copper bar +L_s=600 //mm //Length of steel bar + +//Calculations + +//Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar +//Static Equilibrium obtained by taking moment about A +//P_c=2*P_s + +//From property of similar triangles we get +//(alpha_c*Lc-dell_c)*1**-1=(alpha_s*L_s-dell_s)*2**-1 +//After substituting values in above equations and further simplifying we get +P_s=(2*alpha_c*L_c-alpha_s*L_s)*t*(L_s*(A_s*E_s)**-1+4*L_c*(A_c*E_c)**-1)**-1 +P_c=2*P_s + +//Stress in steel rod +sigma_s=P_s*A_s**-1 //N/mm**2 + +//Stress in copper rod +sigma_c=P_c*A_c**-1 //N/mm**2 + +//Result +printf("\n Stress in steel rod is %0.2f N/mm**2",sigma_s) +printf("\n STress in copper rod is %0.2f N/mm**2",sigma_c) diff --git a/3864/CH2/EX2.37/Ex2_37.sce b/3864/CH2/EX2.37/Ex2_37.sce new file mode 100644 index 000000000..155d3e6c4 --- /dev/null +++ b/3864/CH2/EX2.37/Ex2_37.sce @@ -0,0 +1,36 @@ +clear +// +// + +//Initilization of Variables + +d=20 //mm //Diameter of bar +P=37.7*10**3 //N //Load +L=200 //mm //Guage Length +dell=0.12 //mm //Extension +dell_d=0.0036 //mm //contraction in diameter + +//Calculations + +//Area of bar +A=%pi*4**-1*d**2 + +//Let s and dell_s be the Linear strain and Lateral strain +s=dell*L**-1 +dell_s=dell_d*d**-1 +mu=dell_s*s**-1 //Poissons ratio + +//dell=P*L*(A*E)**-1 +E=P*L*(dell*A)**-1 //N/mm**2 //Modulus of Elasticity of bar + +//Modulus of Rigidity +G=E*(2*(1+mu))**-1 //N/mm**2 + +//Bulk Modulus +K=E*(3*(1-2*mu))**-1 //N/mm**2 + +//result +printf("\n Poissons ratio is %0.2f ",mu) +printf("\n The Elastic constant are:E %0.2f ",E) +printf("\n :G %0.2f ",G) +printf("\n :K %0.2f ",K) diff --git a/3864/CH2/EX2.38/Ex2_38.sce b/3864/CH2/EX2.38/Ex2_38.sce new file mode 100644 index 000000000..86842b02d --- /dev/null +++ b/3864/CH2/EX2.38/Ex2_38.sce @@ -0,0 +1,38 @@ +clear +// +// + +//Initilization of Variables + +d=100 //mm //Diameter of circular rod +P=1*10**6 //N //Tensile Force +mu=0.3 //Poissons ratio +E=2*10**5 //N/mm**2 //Youngs Modulus +L=500 //mm //Length of rod + +//Calculations + +//Modulus of Rigidity +G=E*(2*(1+mu))**-1 //N/mm**2 + +//Bulk Modulus +K=E*(3*(1-2*mu))**-1 //N/mm**2 + +A=%pi*4**-1*d**2 //mm**2 //Area of Circular rod +//Let sigma be the Longitudinal stress +sigma=P*A**-1 //N/mm**2 + +s=sigma*E**-1 //Linear strain +e_x=s + +//Volumetric strain +e_v=e_x*(1-2*mu) + +v=%pi*4**-1*d**2*L +//Change in VOlume +dell_v=e_v*v + +//Result +printf("\n Bulk Modulus is %0.2f N/mm**2",E) +printf("\n Modulus of Rigidity is %0.2f N/mm**2",G) +printf("\n The change in Volume is %0.2f mm**3",dell_v) diff --git a/3864/CH2/EX2.39/Ex2_39.sce b/3864/CH2/EX2.39/Ex2_39.sce new file mode 100644 index 000000000..1a55112ed --- /dev/null +++ b/3864/CH2/EX2.39/Ex2_39.sce @@ -0,0 +1,36 @@ +clear +// + +//Initilization of Variables + +L=500 //mm //Length of rectangular cross section bar +A=20*40 //mm**2 //Area of rectangular cross section bar +P1=4*10**4 //N //Tensile Force on 20mm*40mm Faces +P2=2*10**5 //N //compressive force on 20mm*500mm Faces +P3=3*10**5 //N //Tensile Force on 40mm*500mm Faces +E=2*10**5 //N/mm**2 //Youngs Modulus +mu=0.3 //Poissons Ratio + +//Calculations + +//Let P_x,P_y,P_z be the forces n x,y,z directions + +P_x=P1*A**-1 +P_y=P2*A**-1 +P_z=P3*A**-1 + +//Let e_x,e_y,e_z be the strains in x,y,z directions +e_x=1*E**-1*(50+mu*20-15*mu) +e_y=1*E**-1*(-mu*50-20-mu*15) +e_z=1*E**-1*(-mu*50+mu*20+15) + +//Volumetric strain +e_v=e_x+e_y+e_z + +//Volume +V=20*40*500 //mm**3 +//Change in Volume +dell_v=e_v*V //mm**3 + +//Result +printf("\n The change in Volume is %0.2f mm**3",dell_v) diff --git a/3864/CH2/EX2.4/Ex2_4.sce b/3864/CH2/EX2.4/Ex2_4.sce new file mode 100644 index 000000000..d8c5efd5a --- /dev/null +++ b/3864/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,39 @@ +clear +// +// + +//Initilization of Variables + +d=25 //mm //diameter of steel +d2=18 //mm //Diameter at neck +L=200 //mm //length of stee +P=80*10**3 //KN //Load +P1=160*10**3 //N //Load at Elastic Limit +P2=180*10**3 //N //Max Load +L1=56 //mm //Total Extension +dell_l=0.16 //mm //Extension + + +//Calculations + +A=%pi*d**2*4**-1 //Area of steel //mm**2 + +p=P1*A**-1 //Stress at Elastic Limit //N/mm**2 +Y=P*L*(A*dell_l)**-1 //Modulus of elasticity + +//Let % elongation be x +x=L1*L**-1*100 + +//Percentage reduction in area +//Let % A be a +a=((%pi*4**-1*d**2)-(%pi*4**-1*d2**2))*(%pi*4**-1*d**2)**-1*100 + +//Ultimate tensile stress +sigma=P2*A**-1 //N/mm**2 + +//result +printf("\n Stress at Elastic limit is %0.2f N/mm**2",p) +printf("\n Youngs Modulus is %0.2f N/mm**2",Y) +printf("\n Percentage Elongation is %0.2f ",a) +printf("\n Percentage reduction in area is %0.2f ",P2) +printf("\n Ultimate tensile stress %0.2f N/mm**2",sigma) diff --git a/3864/CH2/EX2.41/Ex2_41.sce b/3864/CH2/EX2.41/Ex2_41.sce new file mode 100644 index 000000000..c7360546a --- /dev/null +++ b/3864/CH2/EX2.41/Ex2_41.sce @@ -0,0 +1,21 @@ +clear +// + +//Initilization of Variables + +E=2.1*10**5 //N/mm**2 //Youngs Modulus +G=0.78*10**5 //N/mm**2 //Modulus of Rigidity + +//Calculations + +//Now using the relation +//E=2*G*(1+mu) +mu=E*(2*G)**-1-1 //Poissons ratio + +//Bulk Modulus +K=E*(3*(1-2*mu))**-1 //N/mm**2 + + +//Result +printf("\n The Poissons Ratio is %0.2f ",mu) +printf("\n The modulus of Rigidity %0.2f N/mm**2",K) diff --git a/3864/CH2/EX2.42/Ex2_42.sce b/3864/CH2/EX2.42/Ex2_42.sce new file mode 100644 index 000000000..f11a803cd --- /dev/null +++ b/3864/CH2/EX2.42/Ex2_42.sce @@ -0,0 +1,20 @@ +clear +// + +//Initilization of Variables + +G=0.4*10**5 //N/mm**2 //Modulus of rigidity +K=0.75*10**5 //N/mm**2 //Bulk Modulus + +//Calculations + +//Youngs Modulus +E=9*G*K*(3*K+G)**-1 + +//Now from the relation +//E=2*G(1+2*mu) +mu=E*(2*G)**-1-1 //Poissons ratio + +//result +printf("\n Youngs modulus is %0.2f N/mm**2",E) +printf("\n Poissons ratio is %0.2f ",mu) diff --git a/3864/CH2/EX2.43/Ex2_43.sce b/3864/CH2/EX2.43/Ex2_43.sce new file mode 100644 index 000000000..31a051d8c --- /dev/null +++ b/3864/CH2/EX2.43/Ex2_43.sce @@ -0,0 +1,82 @@ +clear +// + +//Initilization of Variables + +b=60 //mm //width of bar +d=30 //mm //depth of bar +L=200 //mm //Length of bar +A=30*60 //mm**2 //Area of bar +A2=30*200 //mm**2 //Area of bar along which expansion is restrained +P=180*10**3 //N //Compressive force +E=2*10**5 //N/mm**2 //Youngs Modulus +mu=0.3 //Poissons ratio + +//Calculations + +//The bar is restrained from expanding in Y direction +P_z=0 +P_x=P*A**-1 //stress developed in x direction + +//Now taking compressive strain as positive +//e_x=P_x*E**-1-mu*P_y*E**-1 .......................(1) +//e_y=-mu*P_x*E**-1+P_y*E**-1 ....................(2) +//e_z=-mu*P_x*E**-1-mu*P_y*E**-1 ......................(3) + +//Part-1 +//When it is fully restrained +e_y=0 +P_y=30 //N/mm**2 +e_x=P_x*E**-1-mu*P_y*E**-1 +e_z=-mu*P_x*E**-1-mu*P_y*E**-1 + +//Change in Length +dell_l=e_x*L //mm + +//Change in width +dell_b=b*e_y + +//change in Depth +dell_d=d*e_z + +//Volume of bar +V=b*d*L //mm**3 +//Change in Volume +e_v=(e_x+e_y+e_z)*V //mm**3 + +//Part-2 +//When 50% is restrained + +//Free strain in Y direction +e_y1=mu*P_x*E**-1 + +//As 50% is restrained,so +e_y2=-50*100**-1*e_y1 + +//But form Equation 2 we have e_y=-mu*P_x*E**-1+P_y*E**-1 +//After substituting values in above equation and furthe simplifying we get +P_y=e_y2*E+d + +e_x2=P_x*E**-1-mu*P_y*E**-1 +e_z2=-mu*P_x*E**-1-mu*P_y*E**-1 + +//Change in Length +dell_l2=e_x2*L //mm + +//Change in width +dell_b2=b*e_y2 + +//change in Depth +dell_d2=d*e_z2 + +//Change in Volume +e_v2=(e_x2+e_y2+e_z2)*V //mm**3 + +//REsult +printf("\n Change in Dimension of bar is:dell_l %0.2f mm",dell_l) +printf("\n :dell_b %0.4f mm",dell_b) +printf("\n :dell_d %0.2f mm",dell_d) +printf("\n Change in Volume is %0.2f mm**3",e_v) +printf("\n Changes in material when only 50% of expansion can be reatrained:dell_l2mm",dell_l2) +printf("\n :dell_b2 %0.4f mm",dell_b2) +printf("\n :dell_d2 %0.2f mm",dell_d2) diff --git a/3864/CH2/EX2.44/Ex2_44.sce b/3864/CH2/EX2.44/Ex2_44.sce new file mode 100644 index 000000000..8cd10f122 --- /dev/null +++ b/3864/CH2/EX2.44/Ex2_44.sce @@ -0,0 +1,32 @@ +clear +// +// + +//Initilization of Variables + +P=10*10**3 //N //Load +E=2*10**5 //N/mm**2 //Youngs Modulus +d2=12 //mm //Diameter of bar1 +d1=16 //mm //diameter of bar2 +L1=200 //mm //Length of bar1 +L2=500 //mm //Length of bar2 + +//Calculations + +//Let A1 and A2 be the cross Area of Bar1 & bar2 respectively +A1=%pi*4**-1*d1**2 //mm**2 +A2=%pi*4**-1*d2**2 //mm**2 + +//Let p1 and p2 be the stress in Bar1 nad bar2 respectively +p1=P*A1**-1 //N/mm**2 +p2=P*A2**-1 //N/mm**2 + +//Let V1 nad V2 be the Volume of of Bar1 and Bar2 +V1=A1*(L1+L1) +V2=A2*L2 + +//Let E be the strain Energy stored in the bar +E=p1**2*(2*E)**-1*V1+p2**2*V2*(2*E)**-1 + +//result +printf("\n The Strain Energy stored in Bar is %0.2f N-mm",E) diff --git a/3864/CH2/EX2.45/Ex2_45.sce b/3864/CH2/EX2.45/Ex2_45.sce new file mode 100644 index 000000000..bcb78b03e --- /dev/null +++ b/3864/CH2/EX2.45/Ex2_45.sce @@ -0,0 +1,66 @@ +clear +// +// + +//Initilization of Variables + +//Bar-A +d1=30 //mm //Diameter of bar1 +L=600 //mm //length of bar1 + +//Bar-B +d2=30 //mm //Diameter of bar2 +d3=20 //mm //Diameter of bar2 +L2=600 //mm //length of bar2 + +//Calculations + +//Area of bar-A +A1=%pi*4**-1*d1**2 + +//Area of bar-B +A2=%pi*4**-1*d2**2 +A3=%pi*4**-1*d3**2 + +//let SE be the Strain Energy +//Strain Energy stored in Bar-A +//SE=p**2*(2*E)**-1*V +//After substituting values and simolifying further we get +//SE=P**2*E**-1*0.4244 + +//Strain Energy stored in Bar-B +//SE2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1 +//After substituting values and simolifying further we get +//SE2=0.6897*P**2*E**-1 + +//Let X be the ratio of SE in Bar-B and SE in Bar-A +X=0.6897*0.4244**-1 + +//Part-2 + +//When Max stress is produced is same:Let p be the max stress produced + +//Stress in bar A is p throughout +//In bar B:stress in 20mm dia.portion=p2=p + +//Stress in 30 mm dia.portion +//p1=P*A2*A3**-1 +//After substituting values and simolifying further we get +//p1=4*9**-1*p + +//Strain Energy in bar A +//SE_1=p**2*(2*E)**-1*A1*L1 +//After substituting values and simolifying further we get +//SE_1=67500*p**2*%pi*E**-1 + +//Strain Energy in bar B +//SE_2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1 +//After substituting values and simolifying further we get +//SE_2=21666.67*%pi*p**2*E**-1 + +//Let Y be the Ratio of SE in bar B and SE in bar A +Y=21666.67*67500**-1 + +//result +printf("\n Gradually applied Load is %0.2f ",X) +printf("\n Gradually applied Load is %0.2f ",Y) diff --git a/3864/CH2/EX2.46/Ex2_46.sce b/3864/CH2/EX2.46/Ex2_46.sce new file mode 100644 index 000000000..11d592824 --- /dev/null +++ b/3864/CH2/EX2.46/Ex2_46.sce @@ -0,0 +1,34 @@ +clear +// +// + +//Initilization of Variables + +W=100 //N //Load +E=2*10**5 //N/mm**2 //Youngs Modulus +h=60 //mm //Height through Load falls down +L=400 //mm //Length of collar +d=30 //mm //diameter of bar + +//Calculations + +A=%pi*4**-1*d**2 //mm**2 //Area of bar + +//Instantaneous stress produced is +p=W*A**-1*(1+(1+(2*A*E*h*(W*L)**-1))**0.5) + +//Now the EXtension of the bar is neglected in calculating work doneby the Load,then +P=(2*E*h*W*(A*L)**-1)**0.5 + +//Let percentage error be denoted by E1 +//Percentage error in approximating is +E1=(p-P)*p**-1*100 + +//Instantaneous Extension produced is +dell_l=(P)*E**-1*L + + +//Result +printf("\n The Instantaneous stress is %0.2f N/mm",p) +printf("\n Percentage Error is %0.2f ",E1) +printf("\n The Instantaneous extension is %0.2f mm",dell_l) diff --git a/3864/CH2/EX2.47/Ex2_47.sce b/3864/CH2/EX2.47/Ex2_47.sce new file mode 100644 index 000000000..2ffb85808 --- /dev/null +++ b/3864/CH2/EX2.47/Ex2_47.sce @@ -0,0 +1,44 @@ +clear +// +// + +//Initilization of Variables + +d=20 //mm //Diameter of steel bar +L=1000 //mm //Length of bar +E=2*10**5 //N/mm**2 //Youngs Modulus +p=300 //N/mm**2 //max Permissible stress +h=50 //mm //Height through which weight will fall +w=600 //N //Load + +//Calculations + +//ARea of steel bar +A=%pi*4**-1*d**2 + +//Instantaneous extension is +dell_l=p*L*E**-1 //mm + +//Work done by Load +//W=W1*(h+dell_l) + +//Volume of bar +V=(A)*L + +//Let E1 be the strain Energy +E1=p**2*(2*E)**-1*V + +//Answer in Book for Strain Energy is Incorrect + +//Now Equating Workdone by Load to strain Energy +W1=E1*51.5**-1 + +//Now when w=600 N +//Let W2 be the Work done by the Load +//W2=w(h2*dell_l) + +h=E1*w**-1-dell_l + +//Result +printf("\n The Max Lodad which can Fall from a height of 50 mm on the collar is %0.2f N",W1) +printf("\n the Max Height from which a 600 N Load can fall on the collar is %0.2f mm",h) diff --git a/3864/CH2/EX2.48/Ex2_48.sce b/3864/CH2/EX2.48/Ex2_48.sce new file mode 100644 index 000000000..e5a667a1b --- /dev/null +++ b/3864/CH2/EX2.48/Ex2_48.sce @@ -0,0 +1,35 @@ +clear +// +// + +//Initilization of Variables + +D_s=30 //mm //Diameter of steel rod +d=30 //mm //Internal Diameter of copper tube +D=40//mm //External Diameter of copper tube +E_s=2*10**5 //N/mm**2 //Youngs Modulus of Steel rod +E_c=1*10**5//N/mm**2 //Youngs Modulus of copper tube +P=100 //N //Load +h=40 //mm //height from which Load falls +L=800 //mm //Length + +//Calculations + +//Area of steel rod +A_s=%pi*4**-1*D_s**2 + +//Area of copper tube +A_c=%pi*4**-1*(D**2-d**2) + +//But Dell_s=dell_c=dell +//p_s*E_s**-1*L=p_c*L*E_c +//After simplifying furthe we get +//p_s=2*p_c + +//Now Equating internal Energy to Workdone we get +p_c=(2*P*h*L**-1*(4*A_s*E_s**-1+A_c*E_c**-1))**0.5 +p_s=2*p_c + +//Result +printf("\n STress produced in steel is %0.2f N/mm**2",p_s) +printf("\n STress produced in copper is %0.2f N/mm**2",p_c) diff --git a/3864/CH2/EX2.49/Ex2_49.sce b/3864/CH2/EX2.49/Ex2_49.sce new file mode 100644 index 000000000..15102ddaf --- /dev/null +++ b/3864/CH2/EX2.49/Ex2_49.sce @@ -0,0 +1,63 @@ +clear +// +// + +dell=0.25 //mm //Instantaneous Extension + +//Bar-A +b1=25 //mm //width of bar +D1=500 //mm //Depth of bar + +//Bar-B +b2_1=25 //mm //width of upper bar +b2_2=15 //mm //Width of Lower Bar +L2=200 //mm //Length of upper bar +L1=300 //mm //Length of Lower bar + +E=2*10**5 //N/mm**2 //Youngs Modulus of bar + +//Calculations + +//Strain +e=dell*D1**-1 + +//Load +p=e*E + +//Area of bar-A +A=%pi*4**-1*25**2 + +//Volume of bar-A +V=A*D1 + +//Let E1 be the Energy of Blow +//Energy of Blow +E1=p**2*(E)**-1*V + +//Let p2 be the Max stress in bar B When this blow is applied. +//the max stress occurs in the 15mm dia. portion,Hence, the stress in 25 mm dia.portion is +//p2*%pi*4**-1*b2_2**2*(%pi*4**-1*b2_2**2=0.36*p + +//Strain Energy of bar B +//E2=p**2*(2*E)**-1*v1+1*(2*E)**-1*(0.36*p2)**2*v2 +//After substituting values and Further substituting values we get +//E2=0.1643445*p2**2 + +//Equating it to Energy of applied blow,we get +p2=(12271.846*0.1643445**-1)**0.5 + +//Stress in top portion +sigma=0.36*p2 + +//Extension in Bar-1 +dell_1=p2*E**-1*L1 + +//Extension in Bar-2 +dell_2=0.36*p2*E**-1*L2 + +//Extension of bar +dell_3=dell_1+dell_2 + +//Result +printf("\n Instantaneous Max stress is %0.2f N/mm**2",sigma) +printf("\n extension in Bar is %0.2f mm",dell_3) diff --git a/3864/CH2/EX2.6/Ex2_6.sce b/3864/CH2/EX2.6/Ex2_6.sce new file mode 100644 index 000000000..06fbd1663 --- /dev/null +++ b/3864/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,21 @@ +clear +// +// + +//Initilization of Variables + +P=40*10**3 //N //Load +L1=160 //mm //Length of Bar1 +L2=240 //mm //Length of bar2 +L3=160 //mm //Length of bar3 +d1=25 //mm //Diameter of Bar1 +d2=20 //mm //diameter of bar2 +d3=25 //mm //diameter of bar3 +dell_l=0.285 //mm //Total Extension of bar + +//Calculations + +E=P*4*(dell_l*%pi)**-1*(L1*(d1**2)**-1+L2*(d2**2)**-1+L3*(d3**2)**-1) + +//Result +printf("\n The Youngs Modulus of the material %0.2f N/mm**2",E) diff --git a/3864/CH2/EX2.7/Ex2_7.sce b/3864/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..89129ffa2 --- /dev/null +++ b/3864/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,27 @@ +clear +// + +//Initilization of Variables + +E1=2*10**5 //N/mm**2 //modulus of Elasticity of material1 +E2=1*10**5 //N/mm**2 //modulus of Elasticity of material2 +P=25*10**3 //N //Load +t=20 //mm //thickness of material +b1=40 //mm //width of material1 +b2=30 //mm //width of material2 +L1=500 //mm //Length of material1 +L2=750 //mm //Length of material2 + +//Calculations + +A1=b1*t //mm**2 //Area of materila1 +A2=b2*t //mm**2 //Area of material2 + +dell_l1=P*L1*(A1*E1)**-1 //Extension of Portion1 +dell_l2=P*L2*(A2*E2)**-1 //Extension of portion2 + +//Total Extension of Bar is +dell_l=dell_l1+dell_l2 + +//Result +printf("\n The Total Extension of the Bar is %0.2f mm",dell_l) diff --git a/3864/CH2/EX2.8/Ex2_8.sce b/3864/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..4d18eb885 --- /dev/null +++ b/3864/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,25 @@ +clear +// +// + +//Initilization of Variables + +L=1000 //mm //Length of Bar +l=400 //mm //Length upto which bire is drilled +D=30 //mm //diameter of bar +d1=10 //mm //diameter of bore +P=25*10**3 //N //Load +dell_l=0.185 //mm //Extension of bar + +//Calculations + +L1=L-l //Length of bar above the bore +L2=400 //mm //Length of bore + +A1=%pi*4**-1*D**2 //Area of bar +A2=%pi*4**-1*(D**2-d1**2) //Area of bore + +E=P*dell_l**-1*(L1*A1**-1+L2*A2**-1) + +//Result +printf("\n The Modulus of ELasticity is %0.2f N/mm**2",E) diff --git a/3864/CH4/EX4.1/Ex4_1.sce b/3864/CH4/EX4.1/Ex4_1.sce new file mode 100644 index 000000000..be8ad539f --- /dev/null +++ b/3864/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,47 @@ +clear +// + +//Initilization of Variables + +L=5000 //mm //Length of Beam +a=2000 //mm //Length of start of beam to Pt Load +b=3000 //mm //Length of Pt load to end of beam +A=150*250 //m**2 //Area of beam +b=150 //mm //Width of beam +d=250 //mm //Depth of beam +sigma=10//N/mm**2 //stress +l=2000 //m //Load applied from one end + +//Calculations + +//Moment of Inertia +I=1*12**-1*b*d**3 //m**4 + +//Distance from N.A to end +y_max=d*2**-1 //m + +//Section Modulus +Z=1*6**-1*b*d**2 //mm**3 + +//Moment Carrying Capacity +M=sigma*Z //N-mm + +//Let w be the Intensity of the Load in N/m,then Max moment +//M_max=w*L**2*8**-1 //N-mm +//After substituting values and further simplifying we get +//M_max=w*25*100*8**-1 + +//EQuating it to moment carrying capacity,we get max intensity load +w=M*(25*1000)**-1*8*10**-3 + +//Part-2 + +//Let P be the concentrated load,then max moment occurs under the load and its value +//M1=P*a*b*L**-1 //N-mm + +//Equting it to moment carrying capacity we get +P=M*1200**-1*10**-3 //N + +//Result +printf("\n Max Intensity of u.d.l it can carry %0.3f KN-m",w) +printf("\n MAx concentrated Load P apllied at 2 m from one end is %0.3f KN",P) diff --git a/3864/CH4/EX4.10/Ex4_10.sce b/3864/CH4/EX4.10/Ex4_10.sce new file mode 100644 index 000000000..d09c2efb0 --- /dev/null +++ b/3864/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,53 @@ +clear +// + +//Initilization of Variables +H=10 //mm //Height +A1=160*160 //mm**2 //area of square section at bottom +L1=160 //mm //Length of square section at bottom +b1=160 //mm //width of square section at bottom +A2=80*80 //mm**2 //area of square section at top +L2=80 //mm //Length of square section at top +b2=80 //mm //Width of square section at top +P=100 //N //Pull + +//Calculations + +//Consider a section at distance y from top. +//Let the side of square bar be 'a' +//a=L2+y*(H)**-1*(b1-b2) +//After further simplifying we get +//a=L2+8*y + +//Moment of Inertia +//I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3 +//After further simplifying we get +//I=a**4*12**-1 + +//Section Modulus +//Z=a**4*(12*a*(2)**0.5)**-1 +//After further simplifying we get +//Z=2**0.5*a**3*(12)**-1 //mm**3 + +//Bending moment at this section=100*y N-mm +//M=100*10**3*y //N-mm + +//But +//M=sigma*Z +//After sub values in above equation we get +//sigma=M*Z**-1 +//After further simplifying we get +//sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1) + +//For Max stress df*(dy)**-1=0 +//After taking Derivative of above equation we get +//df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8) +//After further simplifying we get +y=80*16**-1 //m + +//Max stress at this level is +sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1 + +//Result +printf("\n Max Bending stress is Developed at %0.3f m",y) +printf("\n Value of Max Bending stress is %0.3f N/mm**2",sigma) diff --git a/3864/CH4/EX4.12/Ex4_12.sce b/3864/CH4/EX4.12/Ex4_12.sce new file mode 100644 index 000000000..d1060e823 --- /dev/null +++ b/3864/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,43 @@ +clear +// + +//Initilization of Variables + +b=200 //mm //Width of timber +d=400 //mm //Depth of timber +t=6 //mm //Thickness +b2=200 //mm //width of steel plate +t2=20 //mm //Thickness of steel plate +M=40*10**6 //KN-mm //Moment +//Let E_s*E_t**-1=X +X=20 //Ratio of Modulus of steel to timber + +//Calculations + +//let y_bar be the Distance of centroidfrom bottom most fibre +y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 //mm + +//Moment of Inertia +I=1*12**-1*b*d**3+b*d*(b+t-(y_bar))**2+1*12**-1*t2*b2*t**3+b2*t2*t*((y_bar)-t*2**-1)**2 + + +//distance of the top fibre from N-A +y_1=d+t-y_bar //mm + +//Distance of the junction of timber and steel From N-A +y_2=y_bar-t //mm + +//Stress in Timber at the top +Y=M*I**-1*y_1 //N/mm**2 + +//Stress in the Timber at the junction point +Z=M*I**-1*y_2 + +//Coressponding stress in steel at the junction point +Z2=X*Z //N/mm**2 + +//The stress in Extreme steel fibre +Z3=X*M*I**-1*y_bar + +//Result +printf("\n Stress in Extreme steel Fibre %0.2f N/mm**2",Z3) diff --git a/3864/CH4/EX4.13/Ex4_13.sce b/3864/CH4/EX4.13/Ex4_13.sce new file mode 100644 index 000000000..39953ccc4 --- /dev/null +++ b/3864/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,41 @@ +clear +// + +//Initilization of Variables + +//Timber size +b=150 //mm //Width +d=300 //mm //Depth + +t=6 //mm //Thickness of steel plate +l=6 //m //Span + +//E_s*E_t**-1=20 +//m=E_s*E_t**-1 +m=20 +sigma_timber=8 //N/mm**2 //Stress in timber +sigma_steel=150 //N/mm**2 //Stress in steel plate + +//Let m*t=Y +Y=m*t //mm +L=(2*t+b)*m //mm //Width of flitched beam + +//Calculations + +//Due to synnetry cenroid,the neutral axis is half the depth +I=(1*12**-1*L*t**3+L*t*(b+t*2**-1)**2)*2+1*12**-1*(Y+b+Y)*d**3 //mm**4 + +y_max1=150 //mm //For timber +y_max2=156 //mm //For steel + +//stress in steel +f_t1=1*m**-1*sigma_steel //N/mm**2 + +//Moment of resistance +M=f_t1*(I*y_max2**-1) + +//load +w=8*M*(l**2)**-1*10**-6 //KN/m + +//Result +printf("\n Load beam can carry is %0.2f KN/m",w) diff --git a/3864/CH4/EX4.14/Ex4_14.sce b/3864/CH4/EX4.14/Ex4_14.sce new file mode 100644 index 000000000..dd7431bd4 --- /dev/null +++ b/3864/CH4/EX4.14/Ex4_14.sce @@ -0,0 +1,53 @@ +clear +// + +//Initilization of Variables + +L=6000 //mm //Span of beam +W=20*10**3 //N //Load +sigma=8 //N/mm**2 //Stress +b=200 //mm //Width of section +d=300 //mm //Depth of section + +//Calculations + +//let x be the distance from left side of beam + +//Bending moment +//M=W*2**-1*x //Nmm .......(1) + +//But M=sigma*Z ..........(2) + +//Equating equation 1 and 2 we get +//W*2**-1*x=sigma*Z ............(3) + +//Section Modulus +//Z=1*6*b*d**2 ...............(4) + +//Equating equation 3 and 4 we get +//b*d**2=3*W*x*sigma**-1 .............(5) + +//Beam of uniform strength of constant depth +//b=3*W*x*(sigma*d**2) + +//When x=0 +b=0 + +//When x=L*2**-1 +b2=3*W*L*(2*sigma*d**2)**-1 //mm + +//Beam with constant width of 200 mm + +//We have +//d=(3*W*x*(sigma*d)**-1)**0.5 +//thus depth varies as (x)**0.5 + +//when x=0 +d1=0 + +//when x=L*2**-1 +d2=(3*W*L*(2*sigma*200)**-1)**0.5 //mm + +//Result +printf("\n Cross section of rectangular beam is: %0.2f mm",b2) +printf("\n : %0.2f mm",d2) diff --git a/3864/CH4/EX4.15/Ex4_15.sce b/3864/CH4/EX4.15/Ex4_15.sce new file mode 100644 index 000000000..6991385fa --- /dev/null +++ b/3864/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,33 @@ +clear +// + +//Initilization of Variables + +L=800 //mm //Span +n=5 //number of leaves +b=60 //mm //Width +t=10 //mm //thickness +sigma=250 //N/mm**2 //Stress + +//Calculations + +//section Modulus +Z=n*6**-1*b*t**2 //mm**3 + +//from the relation +//sigma*Z=M ...................(1) +//M=P*L*4**-1 +//sub values of M in equation 1 we get +P=sigma*Z*4*L**-1*10**-3 //KN //Load + +//Length of Leaves +L1=0.2*L //mm +L2=0.4*L //mm +L3=0.6*L //mm +L4=0.8*L //mm +L5=L //mm + +//Result +printf("\n Max Load it can take is %0.2f KN",P) +printf("\n Length of leaves:L1 %0.2f mm",L1) +printf("\n :L2 %0.2f mm",L2) diff --git a/3864/CH4/EX4.17/Ex4_17.sce b/3864/CH4/EX4.17/Ex4_17.sce new file mode 100644 index 000000000..0477cf682 --- /dev/null +++ b/3864/CH4/EX4.17/Ex4_17.sce @@ -0,0 +1,54 @@ +clear +// +// + +//Initilization of Variables + +F=40*10**3 //N //shear Force + +//I-section + +//Flanges +b=80 //mm //Width of flange +t=20 //mm //Thickness + +//Web +d=200 //mm //Depth +t2=20 //mm //Thickness + +//Flange-2 +b2=160 //mm //Width +t3=20 //mm //Thickness + +D=240 //mm //Overall Depth + +//Calculations + +//Distance of N-A from Top Fibre +y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 //mm + +//Moment of Inertia +I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 //mm**4 + +//Shear stress bottom of flange +sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 //N/mm**2 + +//At same Level but in web +sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 //N/mm**2 + +//for shear stress at N.A +X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 //mm**3 +sigma3=F*X*(t2*I)**-1 //N/mm**2 + +//Shear stress at bottom of web + +X=b2*t3*((D-y)-t3*2**-1) //mm**3 + +//Stress at bottom of web +sigma4=F*X*(t2*I)**-1 //N/mm**2 + +//Stress at Lower flange +sigma5=F*X*(b2*I)**-1 //N/mm**2 + +//Result +printf("\n The Shear Force Diagram is the result") diff --git a/3864/CH4/EX4.18/Ex4_18.sce b/3864/CH4/EX4.18/Ex4_18.sce new file mode 100644 index 000000000..48f9ecb18 --- /dev/null +++ b/3864/CH4/EX4.18/Ex4_18.sce @@ -0,0 +1,36 @@ +clear +// + +//Initilization of Variables + +F=30*10**3 //N //Shear Force + +//Channel Section +d=400 //mm //Depth of web +t=10 //mm //THickness of web +t2=15 //mm //Thickness of flange +b=100 //mm //Width of flange + +//Rectangular Welded section +b2=80 //mm //Width +d2=60 //mm //Depth + +//Calculations + +//Distance of Centroid From Top Fibre +y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 //mm + +//Moment Of Inertia of the section about N-A +I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2 + +//Shear stress at level of weld +sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 //N/mm**2 + +//Max Shear Stress occurs at Neutral Axis +X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1 + +sigma_max=F*X*((b+t)*I)**-1 + +//Result +printf("\n Shear stress in the weld is %0.2f N/mm**2",sigma) +printf("\n Max shear stress is %0.2f N/mm**2",sigma_max) diff --git a/3864/CH4/EX4.19/Ex4_19.sce b/3864/CH4/EX4.19/Ex4_19.sce new file mode 100644 index 000000000..319ec29a0 --- /dev/null +++ b/3864/CH4/EX4.19/Ex4_19.sce @@ -0,0 +1,30 @@ +clear +// +// + +//Initilization of Variables + +//Wooden Section +b=300 //mm //Width +d=300 //mm //Depth + +D=100 //mm //Diameter of Bore +F=10*10**3 //N //Shear Force + +//Calculations + +//Moment Of Inertia Of Section +I=1*12**-1*b*d**3-%pi*64**-1*D**4 + +//Shear stress at crown of circle +sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1 + +//Let a*y_bar=X +X=b*d*2**-1*d*4**-1-%pi*8**-1*D**2*4*D*2**-1*(3*%pi)**-1 //mm**3 + +//Shear Stress at Neutral Axis +sigma2=F*X*((b-D)*I)**-1 //N/mm**2 + +//Result +printf("\n Shearing Stress at Crown of Bore %0.3f N/mm**2",sigma) +printf("\n Shear Stress at Neutral Axis %0.3f N/mm**2",sigma2) diff --git a/3864/CH4/EX4.2/Ex4_2.sce b/3864/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..029c9dacc --- /dev/null +++ b/3864/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,35 @@ +clear +// +// + +//Initilization of Variables + +D=70 //mm //External Diameter +t=8 //mm //Thickness of pipe +L=2500 //mm //span +sigma=150 //N/mm**2 //stress + +//Calculations + +//Internal Diameter +d=D-2*t //mm + +//M.I Of Pipe +I=%pi*64**-1*(D**4-d**4) //mm**4 + +y_max=D*2**-1 //mm +Z=I*(y_max)**-1 //mm**3 + +//Moment Carrying capacity +M=sigma*Z //N*mm + +//Max moment int the beam occurs at the mid-span and is equal to +//m=P*L*4**-1 + +//Equating Max moment to moment carrying capacity we get, +//M=P*2.5*L*4**-1 +//After substituting and simplifying we get +P=4*M*(L)**-1*10**-3 //N + +//Result +printf("\n Max concentrated load that can be applied at the centre of span is %0.3f KN",P) diff --git a/3864/CH4/EX4.20/Ex4_20.sce b/3864/CH4/EX4.20/Ex4_20.sce new file mode 100644 index 000000000..f4733fccf --- /dev/null +++ b/3864/CH4/EX4.20/Ex4_20.sce @@ -0,0 +1,48 @@ +clear +// + +//Initilization of Variables + +//flanges +b=200 //mm //width +t1=25 //mm //Thickness + +//web +d=450 //mm //Depth +t2=20 //mm //thickness + +D=500 //mm //Total Depth of section + +//Calculations + +//Moment Of Inertia of the section about N-A +I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 //mm**4 + +//Consider an element in the web at distance y from y from N-A +//Depth of web section=225-y + +//C.G From N-A +//y2=y+(((D*2**-1-t)-y)*2**-1) + +//ay_bar for section at y +//Let ay_bar be X +//X=X1 be of Flange + X2 be of web above y +//X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1 +//After Sub values and Further simplifying we get +//X=1187500+10*(225**2-y**2) + +//Shear stress at y +//sigma_y=F*(X)*(t2*I)**-1 + +//Shear Force resisted by the Element +//F1=F*X*t2*dy*(t2*I)**-1 + +//Shear stress resisted by web +//sigma=2*F*I**-1*(X)*dy + +//After Integrating above equation and further simplifying we get +//sigma=0.9578*F + +sigma=0.9578*100 + +//Result diff --git a/3864/CH4/EX4.21/Ex4_21.sce b/3864/CH4/EX4.21/Ex4_21.sce new file mode 100644 index 000000000..d8a38b8be --- /dev/null +++ b/3864/CH4/EX4.21/Ex4_21.sce @@ -0,0 +1,40 @@ +clear +// + +//Initilization of Variables + +//Wooden Beam + +b=150 //mm //width +d=250 //mm //Depth + +L=5000 //mm //span +m=11.2 //N/mm**2 //Max Bending stress +sigma=0.7 //N/mm**2 //Max shear stress + +//Calculations + +//Let 'a' be the distance from left support +//Max shear force +//F=R_A=W*(L-a)*L**-1 + +//Max Moment +//M=W*(L-a)*a*L**-1 + +//But M=sigma*Z +//W*(L-a)*a*L**-1=m*1*6**-1*b*d**2 .....................(1) + +//In Rectangular Section MAx stress is 1.5 times Avg shear stress +F=sigma*b*d*1.5**-1 + +//W*(L-a)*L**-1=F .....................(2) + +//Dividing Equation 1 nad 2 we get +a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1 + +//Sub above value in equation 2 we get +W=(L-a)**-1*L*F*10**-3 //KN + +//Result +printf("\n Load is %0.2f KN",W) +printf("\n Distance from Left support is %0.2f mm",a) diff --git a/3864/CH4/EX4.22/Ex4_22.sce b/3864/CH4/EX4.22/Ex4_22.sce new file mode 100644 index 000000000..d25a86e08 --- /dev/null +++ b/3864/CH4/EX4.22/Ex4_22.sce @@ -0,0 +1,38 @@ +clear +// + +//Initilization of Variables + +L=1000 //mm //span + +//Rectangular Section + +b=200 //mm //width +d=400 //mm //depth + +sigma=1.5 //N/mm**2 //Shear stress + +//Calculations + +//Let AB be the cantilever beam subjected to load W KN at free end + +//MAx shear Force +//F=W*10**3 //KN + +//Since Max shear stress in Rectangular section +//sigma_max=1.5*F*A**-1 +//After sub values and further simplifyng we get +W=1.5*b*d*(1.5*1000)**-1 //KN + +//Moment at fixwed end +M=W*1 //KN-m +y_max=d*2**-1 //mm + +//M.I +I=1*12**-1*b*d**3 //mm**3 + +//MAx Stress +sigma_max=M*10**6*I**-1*y_max + +//Result +printf("\n Concentrated Load is %0.2f N/mm**2",sigma_max) diff --git a/3864/CH4/EX4.24/Ex4_24.sce b/3864/CH4/EX4.24/Ex4_24.sce new file mode 100644 index 000000000..5b2e0cad5 --- /dev/null +++ b/3864/CH4/EX4.24/Ex4_24.sce @@ -0,0 +1,40 @@ +clear +// + +//Initilization of Variables + +L=4000 //mm //span + +//Rectangular Cross-section +b=100 //mm //Width +d=200 //mm //Thickness + +F_per=10 //N/mm**2 //Max Bending stress +q_max=0.6 //N/mm**2 //Shear stress + +//Calculations + +//If the Load W is in KN/m + +//Max shear Force +//F=w*l*2**-1 //KN +//After substituting values and further simplifying we get +//M=2*w //KN-m + +//Max Load from Consideration of moment +//M=1*6**-1*b*d**2*F_per +//After substituting values and further simplifying we get +w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 //KN/m + +//Max Load from Consideration of shear stress +//q_max=1.5*F*(b*d)**-1 //N +//After substituting values and further simplifying we get +F=q_max*(1.5)*b*d //N + +//If w is Max Load in KN/m,then +//2*w*1000=8000 +//After Rearranging and Further simplifying we get +w2=8000*(2*1000)**-1 //KN/m + +//Result +printf("\n Uniformly Distributed Load Beam can carry is %0.2f KN/m",w) diff --git a/3864/CH4/EX4.4/Ex4_4.sce b/3864/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..40556ec38 --- /dev/null +++ b/3864/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,52 @@ +clear +// + +//Initilization of Variables + +//Flange (Top) +b1=80 //mm //Width +t1=40 //mm //Thickness + +//Flange (Bottom) +b2=160 //mm //width +t2=40 //mm //Thickness + +//web +d=120 //mm //Depth +t3=20 //mm //Thickness + +D=200 //mm //Overall Depth +sigma1=30 //N/mm**2 //Tensile stress +sigma2=90 //N/mm**2 //Compressive stress +L=6000 //mm //Span + +//Calculations + +//Distance of centroid from bottom fibre +y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 //mm + +//Moment of Inertia +I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-(y_bar))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-(y_bar))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-(y_bar))**2 + + +//Extreme fibre distance of top and bottom fibres are y_t and y_c respectively + +y_t=y_bar //mm +y_c=D-y_bar //mm + +//Moment carrying capacity considering Tensile strength +M1=sigma1*I*y_t**-1*10**-6 //KN-m + +//Moment carrying capacity considering compressive strength +M2=sigma2*I*y_c**-1*10**-6 //KN-m + +//Max Bending moment in simply supported beam 6 m due to u.d.l +//M_max=w*L*10**-3*8**-1 +//After simplifying further we get +//M_max=4.5*w + +//Now Equating it to Moment carrying capacity, we get load carrying capacity +w=M1*4.5**-1 //KN/m + +//Result +printf("\n Max Uniformly Distributed Load is %0.3f KN/m",w) diff --git a/3864/CH4/EX4.5/Ex4_5.sce b/3864/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..0b6583da4 --- /dev/null +++ b/3864/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,49 @@ +clear +// +// + +//Initilization of Variables + +//Flanges +b=200 //mm //Width +t=25 //mm //Thickness + +D1=500 //mm //Overall Depth +t2=20 //mm //Thickness of web + +d=450 //mm //Depth of web + +//Calculations + +//Consider,Element of Thickness "y" at Distance "dy" from N.A +//Let Bending stress "sigma_max" + +//Stress on the element +//sigma=y*(D*2**-1)*sigma_max ..............(1) + +//Area of Element +//A=b*dy .................................(2) + +//Force on Element +//F=y*250**-1*sigma_max*b*dy + +//Let M be the Moment of resistance +//M=y*250**-1*sigma_max*b*dy*y + +//Moment of Resistance of top flange after simplification we gget +//M.R=2258333.3*f + +//M.I of I section +I=1*12**-1*(b*D1**3-180*d**3)*10**-8 + +//Moment acting on section +//After simplifying we get +//M=2865833.3*f + +//Percentage moment resistance +M1=2258333.3*2865833.3**-1*100 + +//Percentage moment resisted by web +M2=100-M1 + +//Result diff --git a/3864/CH4/EX4.6/Ex4_6.sce b/3864/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..d6659cef7 --- /dev/null +++ b/3864/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,63 @@ +clear +// +// + +//Initilization of Variables + +//Flanges +b1=200 //mm //Width +t1=10 //mm //Thickness + +//Web +d=380 //mm //Depth +t2=8 //mm //Thickness + +D=400 //mm //Overall Depth +sigma=150 //N/mm**2 + +//Calculations + +//Area +A=b1*t1+d*t2+b1*t1 //mm**2 + +//Moment of Inertia +I=1*12**-1*(b1*D**3-(b1-t2)*d**3) + +//Bending Moment +M=sigma*I*(D*2**-1)**-1 + +//Square Section + +//Let 'a' be the side +a=A**0.5 + +//Moment of Resistance of this section +M1=1*6**-1*a*a**2*sigma + +X=M*M1**-1 + +//Rectangular section +//Let 'a' be the side and depth be 2*a + +a=(A*2**-1)**0.5 + +//Moment of Rectangular secction +M2=1*6**-1*a*(2*a)**2*sigma + +X2=M*M2**-1 + +//Circular section +//A=%pi*d1**2*4**-1 + +d1=(A*4*%pi**-1)**0.5 + +//Moment of circular section +M3=%pi*32**-1*d1**3*sigma + +X3=M*M3**-1 + +//Result +printf("\n Moment of resistance of beam section %0.2f mm",M) +printf("\n Moment of resistance of square section %0.2f mm",X) +printf("\n Moment of resistance of rectangular section %0.2f mm",X2) +printf("\n Moment of resistance of circular section %0.2f mm",X3) diff --git a/3864/CH4/EX4.7/Ex4_7.sce b/3864/CH4/EX4.7/Ex4_7.sce new file mode 100644 index 000000000..e50fc587f --- /dev/null +++ b/3864/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,40 @@ +clear +// + +//Initilization of Variables + +F=12 //KN //Force at End of beam +L=2 //m //span + +//Square section +b=200 //mm //Width and depth of beam +d=200 + +//Rectangular section +b1=150 //mm //Width +d1=300 //mm //Depth + +//Calculations + +//Max bending Moment +M=F*L*10**6 //N-mm + +//M=sigma*b*d**2 +sigma=M*6*(b*d**2)**-1 //N/mm**2 + +//Let W be the central concentrated Load in simply supported beam of span L1=3 m +//MAx Moment +//M1=W*L1*4**-1 +//After Further simplifying we get +//M1=0.75*10**6 //N-mm + +//The section has a moment of resistance +M1=sigma*1*6**-1*b1*d1**2 + +//Equating it to moment of resistance we get max load W +//0.75*10**6*W=M1 +//After Further simplifying we get +W=M1*(0.75*10**6)**-1 + +//Result +printf("\n Minimum Concentrated Load required to brek the beam %0.2f KN",W) diff --git a/3864/CH4/EX4.8/Ex4_8.sce b/3864/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..73bf0efb6 --- /dev/null +++ b/3864/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,65 @@ +clear +// + +//Initilization of Variables + +L=3 //m //span +sigma_t=35 //N/mm**2 //Permissible stress in tension +sigma_c=90 //N/mm**2 //Permissible stress in compression + +//Flanges +t=30 //mm //Thickness +d=250 //mm //Depth + +//Web +t2=25 //mm //Thickness +b=600 //mm //Width + +//Calculations + +//Let y_bar be the Distance of N.A from Extreme Fibres +y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1 + +//Moment of Inertia +I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2 + +//Part-1 + +//If web is in Tension +y_t=y_bar //mm +y_c=d-y_bar //mm + +//Moment carrying caryying capacity From consideration of tensile stress +M=sigma_t*I*(y_bar)**-1 //N-mm + +//Moment carrying caryying capacity From consideration of compressive stress +M1=sigma_c*I*(y_c)**-1 //N-mm + +//If w KN/m is u.d.l in beam,Max bending moment +//M=wl**2*8**-1 +//After further simplifyng we get +//M=1.125*w*10**6 N-mm +w=M*(1.125*10**6)**-1 //KN + +//Part-2 + +//If web is in compression +y_t2=178.299 //mm +y_c2=71.71 //mm + +//Moment carrying caryying capacity From consideration of tensile stress +M2=sigma_t*I*(y_t2)**-1 //N-mm + +//Moment carrying caryying capacity From consideration of compressive stress +M3=sigma_c*I*(y_c2)**-1 //N-mm + +//Moment of resistance is M2 + +//Equating it to bending moment we get +//M2=1.125*10**6*w2 +//After further simplifyng we get +w2=M2*(1.125*10**6)**-1 + +//Result +printf("\n Uniformly Distributed Load carrying capacity if:web is in Tension %0.2f KN",w) +printf("\n :web is in compression %0.3f KN",w2) diff --git a/3864/CH4/EX4.9/Ex4_9.sce b/3864/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..93537e581 --- /dev/null +++ b/3864/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,48 @@ +clear +// +// + +//Initilization of Variables + +b1=200 //mm //Width at base +b2=100 //mm //Width at top + +L=8 //m Length +P=500 //N //Load + +//Calculations + +//Consider a section at y metres from top + +//At this section diameter d is +//d=b2+y*L**-1*(b1-b2) +//After Further simplifying we get +//d=b2+12.5*y //mm + +//Moment of Inertia +//I=%pi*64**-1*d**4 + +//Section Modulus +//Z=%pi*32**-1*(b1+12.5*y)**3 + +//Moment +//M=5*10**5*y //N-mm + +//Let sigma be the fibre stress at this section then +//M=sigma*Z +//After sub values in above equation and further simplifying we get +//sigma=5*10**5*32*%pi**-1*y*((b2+12.5*y)**3)**-1 + +//For sigma to be Max,d(sigma)*(dy)**-1=0 +//16*10**6*%pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5) +//After Further simplifying we get +//b2+12.5*y=37.5*y +//After Further simplifying we get +y=b2*25**-1 //m + +//Stress at this section +sigma=5*10**5*32*%pi**-1*y*((b2+12.5*y)**3)**-1 + +//Result +printf("\n Stress at Extreme Fibre is max %0.2f m",y) +printf("\n Max stress is %0.2f N/mm**2",sigma) diff --git a/3864/CH5/EX5.11/Ex5_11.sce b/3864/CH5/EX5.11/Ex5_11.sce new file mode 100644 index 000000000..ef6e75900 --- /dev/null +++ b/3864/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,55 @@ +clear +// +// + + +//Initilization of Variables + +L_CB=2 //m //Length of CB +L_AC=4 //m //Length of AB +M_C=15 //KN.m //Moment At Pt C +F_C=30 //KN +L=6 //m Span of Beam + +//Let X=E*I +X=10000 //KN-m**2 + +//Calculations + +//Let V_A and V_B be the reactions at A & B respectively +//V_A+V_B=30 + +//Taking Moment a A,we get +V_B=(F_C*L_AC+M_C)*L**-1 +V_A=30-V_B + +//Now Taking Moment at distacnce x from A +//M_x=7.5*x-30*(x-4)+15 + +//By using Macaulay's Method +//EI*(d**2*x/dx**2)=M_x=7.5*x-30*(x-4)+15 + +//Now Integrating above Equation we get +//EI*(dy/dx)=C1+7.5*x**2*2**-1-15*(x-4)**2+15*(x-4) ............(1) + +//Again Integrating above Equation we get +//EIy=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1..........(2) + +//Boundary Cinditions +x=0 +y=0 + +//Substituting above equations we get +C2=0 + +x=6 //m +y=0 + +C1=-(7.5*6**3*6**-1-5*2**3+15*2**2*2**-1)*6**-1 + +//EIy_c=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1 +//Sub values in Above equation we get +y_c=(93.3333*(X)**-1) + +//Result +printf("\n The Deflection at C %0.4f mm",y_c) diff --git a/3864/CH5/EX5.12/Ex5_12.sce b/3864/CH5/EX5.12/Ex5_12.sce new file mode 100644 index 000000000..755716ee5 --- /dev/null +++ b/3864/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,87 @@ +clear +// +// + + +//Initilization of Variables +L_AC=2 //m //Length of AC,CD,DB +L_DB=2 +L_CD=2 +F_C=40 //KN //Force at C +w=20 //KN/m //u.d.l +L=6 //m //span of beam + +//Let E*I=X +X=15000 //KN-m**2 + + +//Calculations + +//Let V_A & V_B be the reactions at A & B respectively +//V_A+V_B=80 + +//Taking Moment B,M_B +V_A=(F_C*(L_CD+L_DB)+w*L_DB*L_DB*2**-1)*L**-1 //KN +V_B=80-V_A //KN + +//Taking Moment at distance x from A +//M_x=33.333*x-40*(x-2)-20*(x-4)**2*2**-1 +//EI*(d**2/dx**2)=33.333*x-40*(x-2)-10*(x-4)**2 + +//Integrating above equation we get +//EI*(dy/dx)=C1+33.333*x**2*2**-1-20*(x-2)**2-10*3**-1*(x-4)**3 + +//Again Integrating above equation we get +//EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4 + +//At +x=0 +y=0 +C2=0 + +//At +x=6 +y=0 +C1=-760*6**-1 + +//Assuming Deflection to be max in portion CD and sustituting value of C1 in equation of slope we get +//EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4 +//0=-126.667+33.333*x**2**-1-20*(x-2)**2 + +//After rearranging and simplifying further we get + +//x**2-24*x+62=0 +//From above equations +a=1 +b=-24 +c=62 + +y=(b**2-4*a*c)**0.5 + +x1=(-b+y)*(2*a)**-1 +x2=(-b-y)*(2*a)**-1 + +//Taking x2 into account +x=2.945 //m +C1=-126.667 +C2=0 + +y_max=(C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3)*X**-1 //mm + +//Max slope occurs at the ends +//At A, +//EI*(dy/dx)_A=-126.667 +//At B +//EI*(dy/dx)_B=126.667+33.333*6**2*2**-1-20*4**2-10*2**3 +//After simplifying Further we get +//EI*(dy/dx)_B=73.3273 + +//Now Max slope is EI(dy/dx)_A=-126.667 +//15000*(dy/dx)_=-126.667 + +//Let Y=dy/dx +Y=-126.667*X**-1 //Radians + +//Result +printf("\n Maximum Deflection for Beam is %0.4f mm",y_max) +printf("\n Maximum Slope for beam is %0.4f radians",Y) diff --git a/3864/CH5/EX5.14/Ex5_14.sce b/3864/CH5/EX5.14/Ex5_14.sce new file mode 100644 index 000000000..598268c2f --- /dev/null +++ b/3864/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,62 @@ +clear +// +// + +//Initilization of Variables + +L_CD=2 //m //Length of CD +E=200 //KN/mm**2 +I=60*10**6 //mm**4 //M.I +F_C=20 //KN //Force at C +F_E=30 //KN //Force at E +w=10 //KN/m //u.d.l + +//Calculations + +X=E*I*10**-6 //KN-m**2 + +//Let V_A & V_B be the reactions at A & B respectively +//V_A+V_B=70 + +//Taking Moment at distance x from A +//M_x=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4) +//EI*(d**2y/dx**2)=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4) + +//Now Integrating Above equation,we get +//EI*(dy/dx)=C1+17*x**2-10*(x-1)**2-5*3**-1*(x-1)**3+5*3**-1*(x-3)**3-15*(x-4)**2 + +//Again Integrating Above equation,we get +//EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3 + +//At +x=0 +y=0 +C2=0 + +//At +x=5 //m +y=0 +C1=(-17*3**-1*x**3+10*3**-1*(x-1)**3+5*12**-1*(x-1)**4-5*12**-1*(x-3)**4+5*(x-4)**3)*5**-1 + +//EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3 +C2=0 +C1=-78 +x=1 +y_c=(-78*x+17*3**-1*x)*(X)**-1 + +//EI*y_D=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4 +x=3 +C1-78 +C2=0 +y_D=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4)*(X**-1) + +//EI*y_E=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4 +x=4 +C1-78 +C2=0 +y_E=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4)*X**-1 + +//Result +printf("\n Deflections at C %0.5f mm",y_c) +printf("\n Deflections at D %0.5f mm",y_D) +printf("\n Deflections at E %0.4f mm",y_E) diff --git a/3864/CH5/EX5.16/Ex5_16.sce b/3864/CH5/EX5.16/Ex5_16.sce new file mode 100644 index 000000000..244888a3d --- /dev/null +++ b/3864/CH5/EX5.16/Ex5_16.sce @@ -0,0 +1,60 @@ +clear +// +// + +//Initilization of Variables + +L_AC=2 //m //Length of BD,CB,AC +L_BD=2 +L_CB=2 +F_C=40 //KN //Force at C +F_D=10 //KN Force at D +L=6 //m spna of beam + +//EI is constant in this problem + +//Calculations + +//Let V_A & V_B be the reactions at A & B Respectively +//V_A+V_B=50 + +//Taking Moment at Pt A +V_B=(F_D*L+F_C*L_AC)*(L_AC+L_CB)**-1 +V_A=50-V_B + +//Now Taking Moment at distance x from A,M_x +//M_x=15*x-40*(x-2)+35*(x-4) +//EI*(d**2*y/dx**2)=15*x-40*(x-2)+35*(x-4) + +//Now Integrating above equation we get +//EI*(dy/dx)=C1+7.5*x**2-20*(x-2)**2+17.5(x-4)**2 + +//Again Integrating above equation we get +//EI*y=C2+C1*x+2.5*x**2-20*3**-1*(x-2)**3+17.5*(x-4)**3*3**-1 + +//At +x=0 +y=0 +//we get +C2=0 + +//At +x=4 +y=0 +//we get +C1=(2.5*4**3-20*3**-1*2**3)*4**-1 + +//Now Deflection at C +x=2 +C1=-26.667 +C2=0 +y_C=C2+C1*x+2.5*x**3 + +//Now Deflection at D +C1=-21.667 +C2=0 +y_D=-26.667*6+2.5*6**3-20*3**-1*4**3+17.5*2**3*3**-1 + +//Result +printf("\n Deflections Under Loads are:y_D %0.4f ",y_D) +printf("\n :y_C %0.2f ",y_C) diff --git a/3864/CH5/EX5.18/Ex5_18.sce b/3864/CH5/EX5.18/Ex5_18.sce new file mode 100644 index 000000000..26a9324a6 --- /dev/null +++ b/3864/CH5/EX5.18/Ex5_18.sce @@ -0,0 +1,72 @@ +clear +// +// + +//Initilization of Variables +L_ED=2 //m //Length of DB & AC +L_AC=2 +L_DB=2 +L_CD=4 //m //Length of CD +L_CE=2 //m //Length of CE +F_A=40 //KN //Force at C +F_B=20 //KN //Force at A +E=200*10**6 //KN/mm**2 //Modulus of Elasticity +I=50*10**-6 //m**4 //M.I + +//Calculations + +//LEt V_C & V_D be the reactions at C & D respectively +//V_C+V_D=60 + +//Taking Moment At D,M_D +V_C=-(-F_A*(L_AC+L_CE+L_ED)+F_B*L_DB)*L_CD**-1 +V_D=60-V_C + +//Now Taking Moment at Distance x from A, +//M_x=-40*x+50*(x-2)+10*(x-6) + +//EI*(d**2*y/dx**2)=-40*x+50*(x-2)+10*(x-6) + +//Now Integrating above Equation we get +//EI*(dy/dx)=C1+20*x**2-25*(x-2)+5*(x-6)**2 + +//Again Integrating above Equation we get +//EI*y=C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3 + +//At +x=0 +y=0 +//C2+2*C1=-53.33 ...............(1) + +//At +x=6 +y=0 +//C2+6*C1=906.667 ...............(2) + +//Subtracting Equation 1 from 2 we get +C1=853.333*4**-1 +C2=53.333-2*C1 +x=0 +y_A=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1 + +//Answer For y_A is incorrect in textbook + +//At Mid-span +C1=853.333*4**-1 +C2=53.333-2*C1 +x=4 +y_E=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1 + +//Answer For y_E is incorrect in textbook + +//At B +C1=853.333*4**-1 +C2=53.333-2*C1 +x=8 +y_B=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1 + + +//Result +printf("\n Deflection relative to the level of the supports:at End A %0.4f mm",y_A) +printf("\n :at End B %0.4f mm",y_B) +printf("\n :at Centre of CD %0.4f mm",y_E) diff --git a/3864/CH5/EX5.2/Ex5_2.sce b/3864/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..7d16b55d6 --- /dev/null +++ b/3864/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,26 @@ +clear +// +// + +//Initilization of Variables + +L=3000 //mm //span of beam +a=2000 //mm +W1=20*10**3 //N //Pt Load Acting on beam +W2=30*10**3 //N //Pt Load Acting on beam +E=2*10**5 //N/mm**2 //Youngs Modulus +I=2*10**8 //mm**4 //M.I + +//Calculations + +//Deflection at free End Due to W2 +dell1=W2*L**3*(3*E*I)**-1 //mm + +//Deflection at free end Due to W1 +dell2=W1*a**3*(3*E*I)**-1+(L-a)*W1*a**2*(2*E*I)**-1 //mm + +//Total Deflection at free end +dell=dell1+dell2 //mm + +//Result +printf("\n Deflection at Free End is %0.2f mm",dell) diff --git a/3864/CH5/EX5.4/Ex5_4.sce b/3864/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..46dcb2ca2 --- /dev/null +++ b/3864/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,27 @@ +clear +// +// + + +//Initilization of Variables + +E=2*10**5 //N/mm**2 //Youngs Modulus +I=180*10**6 //mm**4 //M.I +W1=20 //N/m //u.d.l +W2=20*10**3 //N //Pt load +L=3000 //m //Span of beam +a=2000 //m //Span of u.d.l + +//Calculations + +//Displacement of free End due to 20 KN Pt load at free end +dell1=W2*L**3*(3*E*I)**-1 //mm + +//Displacement of free end due to u.d.l +dell2=W1*a**4*(8*E*I)**-1+(L-a)*W1*a**3*(6*E*I)**-1 + +//Deflection at free end +dell=dell1+dell2 //mm + +//Result +printf("\n The Displacement of Free End of cantilever beam is %0.2f mm",dell) diff --git a/3864/CH6/EX6.1/Ex6_1.sce b/3864/CH6/EX6.1/Ex6_1.sce new file mode 100644 index 000000000..13d455f14 --- /dev/null +++ b/3864/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,25 @@ +clear +// +// + +//Initilization of Variables + +L=10000 //mm //Length of solid shaft +d=100 //mm //Diameter of shaft +n=150 //rpm +P=112.5*10**6 //N-mm/sec //Power Transmitted +G=82*10**3 //N/mm**2 //modulus of Rigidity + +//Calculations + +J=%pi*d**4*(32)**-1 //mm**3 //Polar Modulus +T=P*60*(2*%pi*n)**-1 //N-mm //Torsional moment + +r=50 //mm //Radius + +q_s=T*r*J**-1 //N/mm**2 //Max shear stress intensity +Theta=T*L*(G*J)**-1 //angle of twist + +//Result +printf("\n Max shear stress intensity %0.2f N/mm**2",q_s) +printf("\n Angle of Twist %0.3f radian",Theta) diff --git a/3864/CH6/EX6.11/Ex6_11.sce b/3864/CH6/EX6.11/Ex6_11.sce new file mode 100644 index 000000000..a3f927997 --- /dev/null +++ b/3864/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,45 @@ +clear +// +// + +//Initilization of Variables + +P=250*10**6 //N-mm/sec //Power transmitted +n=100 //rpm +q_s=75 //N/mm**2 //Shear stress + +//Calculations + +//From Equation of Power we have +T=P*60*(2*%pi*n)**-1 //N-mm //Torsional moment + +//Now from torsional moment equation we have +//T=j*q_s*(d/2**-1)**-1 +//After substituting values in above equation and further simplifying we get +//T=%pi*16**-1**d**3*q_s +d=(T*16*(%pi*q_s)**-1)**0.3333 //mm //Diameter of solid shaft + +//PArt-2 + +//Let d1 and d2 be the outer and inner diameter of hollow shaft +//d2=0.6*d1 + +//Again from torsional moment equation we have +//T=%pi*32**-1*(d1**4-d2**4)*q_s*(d1/2)**-1 +d1=(T*16*(%pi*(1-0.6**4)*q_s)**-1)**0.33333 +d2=0.6*d1 + +//Cross sectional area of solid shaft +A1=%pi*4**-1*d**2 //mm**2 + +//cross sectional area of hollow shaft +A2=%pi*4**-1*(d1**2-d2**2) + +//Now percentage saving in weight +//Let W be the percentage saving in weight +W=(A1-A2)*100*A1**-1 + +//Result +printf("\n Size of shaft is:solid shaft:d %0.3f mm",d) +printf("\n :Hollow shaft:d1 %0.3f mm",d1) +printf("\n : :d2 %0.3f mm",d2) diff --git a/3864/CH6/EX6.12/Ex6_12.sce b/3864/CH6/EX6.12/Ex6_12.sce new file mode 100644 index 000000000..a38b0a948 --- /dev/null +++ b/3864/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,43 @@ +clear +// +// + +//Initilization of Variables +d=100 //mm //Diameter of solid shaft +d1=100 //mm //Outer Diameter of hollow shaft +d2=50 //mm //Inner Diameter of hollow shaft + +//Calculations + +//Torsional moment of solid shaft +//T_s=J*q_s*(d*2**-1)**-1 +//After substituting values in above equation and further simplifying we get +//T_s=%pi*16*d**3*q_s ...............(1) + +//torsional moment for hollow shaft is +//T_h=J*q_s*(d1**4-d2**4)**-1*(d1*2**-1) +//After substituting values in above equation and further simplifying we get +//T_h=%pi*32**-1*2*d1**-1*(d1**4-d2**4)*q_s ...........(2) + +//Dividing Equation 2 by 1 we get +//Let the ratio of T_h*T_s**-1 Be X +X=1-0.5**4 + +//Loss in strength +//Let s be the loss in strength +//s=T_s*T_h*100*T_s**-1 +//After substituting values in above equation and further simplifying we get +s=(1-0.9375)*100 + +//Weight Ratio +//Let w be the Weight ratio +//w=W_h*W_s**-1 + +A_h=%pi*32**-1*(d1**2-d2**2) //mm**2 //Area of Hollow shaft +A_s=%pi*32**-1*d**2 //mm**2 //Area of solid shaft + +w=A_h*A_s**-1 + +//Result +printf("\n Loss in strength is %0.2f ",s) +printf("\n Weight ratio is %0.2f ",w) diff --git a/3864/CH6/EX6.13/Ex6_13.sce b/3864/CH6/EX6.13/Ex6_13.sce new file mode 100644 index 000000000..12f36afcd --- /dev/null +++ b/3864/CH6/EX6.13/Ex6_13.sce @@ -0,0 +1,36 @@ +clear +// +// + +//Initilization of Variables +T=8 //KN-m //Torque +d=100 //mm //Diameter of portion AB +d1=100 //mm //External Diameter of Portion BC +d2=75 //mm //Internal Diameter of Portion BC +G=80 //KN/mm**2 //Modulus of Rigidity +L1=1500 //mm //Radial Distance of Portion AB +L2=2500 //mm //Radial Distance ofPortion BC + +//Calculations + +R=d*2**-1 //mm //Radius of shaft + +//For Portion AB,Polar Modulus +J1=%pi*32**-1*d**4 //mm**4 + +//For Portion BC,Polar modulus +J2=%pi*32**-1*(d1**4-d2**4) //mm**4 + +//Now Max stress occurs in portion BC since max radial Distance is sme in both cases +q_max=T*J2**-1*R*10**6 //N/mm**2 + +//Let theta1 be the rotation in Portion AB and theta2 be the rotation in portion BC +theta1=T*L1*(G*J1)**-1 //Radians +theta2=T*L2*(G*J2)**-1 //Radians + +//Total Rotational at end C +theta=(theta1+theta2)*10**3 //Radians + +//Result +printf("\n Max stress induced is %0.2f N/mm**2",q_max) +printf("\n Angle of Twist is %0.3f radians",theta) diff --git a/3864/CH6/EX6.14/Ex6_14.sce b/3864/CH6/EX6.14/Ex6_14.sce new file mode 100644 index 000000000..62b0e8876 --- /dev/null +++ b/3864/CH6/EX6.14/Ex6_14.sce @@ -0,0 +1,40 @@ +clear +// +// + +//Initilization of Variables + +q_b=80 //N/mm**2 //Shear stress in Brass +q_s=100 //N/mm**2 //Shear stress in Steel +G_b=40*10**3 //N/mm**2 +G_s=80*10**3 +L_b=1000 //mm //Length of brass shaft +L_s=1200 //mm //Length of steel shaft +d1=80 //mm //Diameter of brass shaft +d2=60 //mm //Diameter of steel shaft + +//Calculations + +//Polar modulus of brass rod +J_b=%pi*32**-1*d1**4 //mm**4 + +//Polar modulus of steel rod +J_s=%pi*32**-1*d2**4 //mm**4 + +//Considering bras Rod:AB +T1=J_b*q_b*(d1*2**-1)**-1 //N-mm + +//Considering Steel Rod:BC +T2=J_s*q_s*(d2*2**-1)**-1 //N-mm + +//Max Torque that can be applied +T2 + +//Let theta_b and theta_s be the rotations in Brass and steel respectively +theta_b=T2*L_b*(G_b*J_b)**-1 //Radians +theta_s=T2*L_s*(G_s*J_s)**-1 //Radians + +theta=theta_b+theta_s //Radians //Rotation of free end + +//Result +printf("\n Total of free end is %0.3f Radians",theta) diff --git a/3864/CH6/EX6.15/Ex6_15.sce b/3864/CH6/EX6.15/Ex6_15.sce new file mode 100644 index 000000000..a96281897 --- /dev/null +++ b/3864/CH6/EX6.15/Ex6_15.sce @@ -0,0 +1,55 @@ +clear +// +// + +//Initilization of Variables + +G=80*10**3 //N/mm**2 //Modulus of Rigidity +d1=100 //mm //Outer diameter of hollow shft +d2=80 //mm //Inner diameter of hollow shaft +d=80 //mm //diameter of Solid shaft +d3=60 //mm //diameter of Solid shaft having L=0.5m +L1=300 //mm //Length of Hollow shaft +L2=400 //mm //Length of solid shaft +L3=500 //mm //LEngth of solid shaft of diameter 60mm +T1=2*10**6 //N-mm //Torsion in Shaft AB +T2=1*10**6 //N-mm //Torsion in shaft BC +T3=1*10**6 //N-mm //Torsion in shaft CD + +//Calculations + +//Now Polar modulus of section AB +J1=%pi*32**-1*(d1**4-d2**4) //mm**4 + +//Polar modulus of section BC +J2=%pi*32**-1*d**4 //mm**4 + +//Polar modulus of section CD +J3=%pi*32**-1*d3**4 //mm**4 + +//Now angle of twist of AB +theta1=T1*L1*(G*J1)**-1 //radians + +//Angle of twist of BC +theta2=T2*L2*(G*J2)**-1 //radians + +//Angle of twist of CD +theta3=T3*L3*(G*J3)**-1 //radians + +//Angle of twist +theta=theta1-theta2+theta3 //Radians + +//Shear stress in AB From Torsion Equation +q_s1=T1*(d1*2**-1)*J1**-1 //N/mm**2 + +//Shear stress in BC +q_s2=T2*(d*2**-1)*J2**-1 //N/mm**2 + +//Shear stress in CD +q_s3=T3*(d3*2**-1)*J3**-1 //N-mm**2 + +//As max shear stress occurs in portion CD,so consider CD + +//Result +printf("\n Angle of twist at free end is %0.5f Radian",theta) +printf("\n Max Shear stress %0.2f N/mm**2",q_s3) diff --git a/3864/CH6/EX6.16/Ex6_16.sce b/3864/CH6/EX6.16/Ex6_16.sce new file mode 100644 index 000000000..76268c16b --- /dev/null +++ b/3864/CH6/EX6.16/Ex6_16.sce @@ -0,0 +1,52 @@ +clear +// +// + +//Initilization of Variables + +L=1000 //mm //Length of bar +L1=600 //mm //Length of Bar AB +L2=400 //mm //Length of Bar BC +d1=60 //mm //Outer Diameter of bar BC +d2=30 //mm //Inner Diameter of bar BC +d=60 //mm //Diameter of bar AB +T=2*10**6 //N-mm //Total Torque + +//Calculations + +//Polar Modulus of Portion AB +J1=%pi*32**-1*d**4 //mm*4 + +//Polar Modulus of Portion BC +J2=%pi*32**-1*(d1**4-d2**4) //mm**4 + +//Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC +//Let theta1 and theta2 be the rotation of shaft in portion AB & BC + +//theta1=T1*L1*(G*J1)**-1 //radians +//After substituting values and further simplifying we get +//theta1=32*600*T1*(%pi*60**4*G)**-1 + +//theta2=T2*L*(J2*G)**-1 //Radians +//After substituting values and further simplifying we get +//theta2=32*400*T2*(%pi*60**4*(1-0.5**4)*G)**-1 + +//Now For consistency of Deformation,theta1=theta2 +//After substituting values and further simplifying we get +//T1=0.7111*T2 ..................................................(1) + +//But T1+T2=T=2*10**6 ...........................................(2) +//Substituting value of T1 in above equation + +T2=T*(0.7111+1)**-1 +T1=0.71111*T2 + +//Max stress in Portion AB +q_s1=T1*(d*2**-1)*(J1)**-1 //N/mm**2 + +//Max stress in Portion BC +q_s2=T2*(d1*2**-1)*J2**-1 //N/mm**2 + +//Result +printf("\n Stresses Developed in Portion:AB %0.2f N/mm**2",q_s1) +printf("\n :BC %0.2f N/mm**2",q_s2) diff --git a/3864/CH6/EX6.17/Ex6_17.sce b/3864/CH6/EX6.17/Ex6_17.sce new file mode 100644 index 000000000..7012e6601 --- /dev/null +++ b/3864/CH6/EX6.17/Ex6_17.sce @@ -0,0 +1,45 @@ +clear +// +// + +//Initilization of Variables + +d1=80 //mm //External Diameter of Brass tube +d2=50 //mm //Internal Diameter of Brass tube +d=50 //mm //Diameter of steel Tube +G_b=40*10**3 //N/mm**2 //Modulus of Rigidity of brass tube +G_s=80*10**3 //N/mm**2 //Modulus of rigidity of steel tube +T=6*10**6 //N-mm //Torque +L=2000 //mm //Length of Tube + +//Calculations + +//Polar Modulus of brass tube +J1=%pi*32**-1*(d1**4-d2**4) //mm**4 + +//Polar modulus of steel Tube +J2=%pi*32**-1*d**4 //mm**4 + +//Let T_s & T_b be the torque resisted by steel and brass respectively +//Then, T_b+T_s=T ............................................(1) + +//Since the angle of twist will be the same +//Theta1=Theta2 +//After substituting values and further simplifying we get +//Ts=0.360*Tb ...........................................(2) + +//After substituting value of Ts in eqn 1 and further simplifying we get +T_b=T*(0.36+1)**-1 //N-mm +T_s=0.360*T_b + +//Let q_s and q_b be the max stress in steel and brass respectively +q_b=T_b*(d1*2**-1)*J1**-1 //N/mm**2 +q_s=T_s*(d2*2**-1)*J2**-1 //N/mm**2 + +//Since angle of twist in brass=angle of twist in steel +theta_s=T_s*L*(J2*G_s)**-1 + +//Result +printf("\n Stresses Developed in Materials are:Brass %0.2f N/mm**2",q_b) +printf("\n :Steel %0.2f N/mm**2",q_s) +printf("\n Angle of Twist in 2m Length %0.3f Radians",theta_s) diff --git a/3864/CH6/EX6.18/Ex6_18.sce b/3864/CH6/EX6.18/Ex6_18.sce new file mode 100644 index 000000000..cf2139d10 --- /dev/null +++ b/3864/CH6/EX6.18/Ex6_18.sce @@ -0,0 +1,39 @@ +clear +// +// + +//Initilization of Variables + +d1=60 //mm //External Diameter of aluminium Tube +d2=40 //mm //Internal Diameter of aluminium Tube +d=40 //mm //Diameter of steel tube +q_a=60 //N/mm**2 //Permissible stress in aluminium +q_s=100 //N/mm**2 //Permissible stress in steel tube +G_a=27*10**3 //N/mm**2 +G_s=80*10**3 //N/mm**2 + +//Calculations + +//Polar modulus of aluminium Tube +J_a=%pi*32**-1*(d1**4-d2**4) //mm**4 + +//Polar Modulus of steel Tube +J_s=%pi*32**-1*d**4 //mm**4 + +//Now the angle of twist of steel tube = angle of twist of aluminium tube +//T_s*L_s*(J_s*theta_s)**-1=T_a*L_a*(J_a*theta_a)**-1 +//After substituting values in above Equation and Further simplifyin we get +//T_s=0.7293*T_a .....................(1) + +//If steel Governs the resisting capacity +T_s1=q_s*J_s*(d*2**-1)**-1 //N-mm +T_a1=T_s1*0.7293**-1 //N-mm +T1=(T_s1+T_a1)*10**-6 //KN-m //Total Torque in steel Tube + +//If aluminium Governs the resisting capacity +T_a2=q_a*J_a*(d1*2**-1) //N-mm +T_s2=T_a2*0.7293 //N-mm +T2=(T_s2+T_a2)*10**-6 //KN-m //Total Torque in aluminium tube + +//Result +printf("\n Steel Governs the torque carrying capacity %0.2f KN-m",T1) diff --git a/3864/CH6/EX6.2/Ex6_2.sce b/3864/CH6/EX6.2/Ex6_2.sce new file mode 100644 index 000000000..2c4184559 --- /dev/null +++ b/3864/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,26 @@ +clear +// +// + +//Initilization of Variables + +P=440*10**6 //N-m/sec //Power transmitted +n=280 //rpm +theta=%pi*180**-1 //radian //angle of twist +L=1000 //mm //Length of solid shaft +q_s=40 //N/mm**2 //Max torsional shear stress +G=84*10**3 //N/mm**2 //Modulus of rigidity + +//Calculations + +//P=2*%pi*n*T*(60)**-1 //Equation of Power transmitted +T=P*60*(2*%pi*n)**-1 //N-mm //torsional moment + +//From Consideration of shear stress +d1=(T*16*(%pi*40)**-1)**0.333333 + +//From Consideration of angle of twist +d2=(T*L*32*180*(%pi*84*10**3*%pi)**-1)**0.25 + +//result +printf("\n Diameter of solid shaft is %0.2f mm",d1) diff --git a/3864/CH6/EX6.20/Ex6_20.sce b/3864/CH6/EX6.20/Ex6_20.sce new file mode 100644 index 000000000..b8f895881 --- /dev/null +++ b/3864/CH6/EX6.20/Ex6_20.sce @@ -0,0 +1,31 @@ +clear +// +// + +//Initilization of Variables + +T=2*10**6 //N-mm //Torque transmitted +G=80*10**3 //N/mm**2 //Modulus of rigidity +d1=40 //mm +d2=80 //mm +r1=20 //mm +r2=40 //mm +L=2000 //mm //Length of shaft + +//Calculations + +//Angle of twist +theta=2*T*L*(r1**2+r1*r2+r2**2)*(3*%pi*G*r2**3*r1**3)**-1 //radians + +//If the shaft is treated as shaft of average Diameter +d_avg=(d1+d2)*2**-1 //mm + +theta1=T*L*(G*%pi*32**-1*d_avg**4)**-1 //Radians + +//Percentage Error +//Let Percentage Error be E +X=theta-theta1 +E=(X*theta**-1)*100 + +//Result +printf("Percentage Error is %0.3f",E) diff --git a/3864/CH6/EX6.21/Ex6_21.sce b/3864/CH6/EX6.21/Ex6_21.sce new file mode 100644 index 000000000..031493439 --- /dev/null +++ b/3864/CH6/EX6.21/Ex6_21.sce @@ -0,0 +1,29 @@ +clear +// +// + +//Initilization of Variables + +G=80*10**3 //N/mm**2 +P=1*10**9 //N-mm/sec //Power +n=300 +d1=150 //mm //Outer Diameter +d2=120 //mm //Inner Diameter +L=2000 //mm //Length of circular shaft + +//Calculations + +T=P*60*(2*%pi*n)**-1 //N-mm + +//Polar Modulus +J=%pi*32**-1*(d1**4-d2**4) //mm**4 + +q_s=T*J**-1*(d1*2**-1) //N/mm**2 + + +//Strain ENergy +U=q_s**2*(4*G)**-1*%pi*4**-1*(d1**2-d2**2)*L + +//Result +printf("\n Max shear stress is %0.2f N/mm**2",q_s) +printf("\n Strain Energy stored in the shaft is %0.2f N-mm",U) diff --git a/3864/CH6/EX6.22/Ex6_22.sce b/3864/CH6/EX6.22/Ex6_22.sce new file mode 100644 index 000000000..573d185e2 --- /dev/null +++ b/3864/CH6/EX6.22/Ex6_22.sce @@ -0,0 +1,32 @@ +clear +// +// + +//Initilization of Variables + +d=12 //mm //Diameter of helical spring +D=150 //mm //Mean Diameter +R=D*2**-1 //mm //Radius of helical spring +n=10 //no.of turns +G=80*10**3 //N/mm**2 +W=450 //N //Load + +//Calculations + +//Max shear stress +q_s=16*W*R*(%pi*d**3)**-1 //N/mm**2 + +//Strain Energy stored +U=32*W**2*R**3*n*(G*d**4)**-1 //N-mm + +//Deflection Produced +dell=64*W*R**3*n*(G*d**4)**-1 //mm + +//Stiffness Spring +k=W*dell**-1 //N/mm + +//Result +printf("\n Max shear stress is %0.2f N/mm**2",q_s) +printf("\n Strain Energy stored is %0.2f N-mm",U) +printf("\n Deflection Produced is %0.2f mm",dell) +printf("\n Stiffness spring is %0.2f N/mm",k) diff --git a/3864/CH6/EX6.23/Ex6_23.sce b/3864/CH6/EX6.23/Ex6_23.sce new file mode 100644 index 000000000..d93dc86f7 --- /dev/null +++ b/3864/CH6/EX6.23/Ex6_23.sce @@ -0,0 +1,35 @@ +clear +// +// + +//Initilization of Variables + +K=5 //N/mm //Stiffness +L=100 //mm //Solid Length +q_s=60 //N/mm**2 //Max shear stress +W=200 //N //Max Load +G=80*10**3 //N/mm**2 + +//Calculations + +//K=W*dell**-1 +//After substituting values and further simplifying we get +//d=0.004*R**3*n ........(1) //mm //Diameter of wire +//n=L*d**-1 ........(2) + +//From Shearing stress +//q_s=16*W*R*(%pi*d**3)**-1 +//After substituting values and further simplifying we get +//d**4=0.004*R**3*n .................(4) + +//From Equation 1,2,3 +//d**4=0.004*(0.0785*d**3)**3*100*d**-1 +//after further simplifying we get +d=5168.101**0.25 +n=100*d**-1 +R=(d**4*(0.004*n)**-1)**0.3333 + +//Result +printf("\n Diameter of Wire is %0.2f mm",d) +printf("\n No.of turns is %0.2f ",n) +printf("\n Mean Radius of spring is %0.2f mm",R) diff --git a/3864/CH6/EX6.24/Ex6_24.sce b/3864/CH6/EX6.24/Ex6_24.sce new file mode 100644 index 000000000..54c23cd26 --- /dev/null +++ b/3864/CH6/EX6.24/Ex6_24.sce @@ -0,0 +1,30 @@ +clear +// +// + +//Initilization of Variables + +m=5*10**5 //Wagon Weighing +v=18*1000*36000**-1 +d=300 //mm //Diameter of Beffer springs +n=18 //no.of turns +G=80*10**3 //N/mm**2 +dell=225 +R=100 //mm //Mean Radius + +//Calculations + +//Energy of Wagon +E=m*v**2*(9.81*2)**-1 //N-mm + +//Load applied +W=dell*G*d**4*(64*R**3*n)**-1 //N + +//Energy each spring can absorb is +E2=W*dell*2**-1 //N-mm + +//No.of springs required to absorb energy of Wagon +n2=E*E2**-1 *10**7 + +//Result +printf("\n No.of springs Required for Buffer is %0.2f ",n2) diff --git a/3864/CH6/EX6.25/Ex6_25.sce b/3864/CH6/EX6.25/Ex6_25.sce new file mode 100644 index 000000000..a1b4c1ee4 --- /dev/null +++ b/3864/CH6/EX6.25/Ex6_25.sce @@ -0,0 +1,27 @@ +clear +// +// + +//Initilization of Variables + +b=180 //mm //width of flange +d=10 //mm //Depth of flange +t=10 //mm //Thickness of flange +D=400 //mm //Overall Depth + +//Calculations + +I_xx=1*12**-1*(b*D**3-(b-t)*(D-2*d)**3) +I_yy=1*12**-1*((D-2*d)*t**3+2*t*b**3) + +//If warping is neglected +J=I_xx+I_yy //mm**4 + +//Since b/d>1.6,we get +J2=1*3**-1*d**3*b*(1-0.63*d*b**-1)*2+1*3**-1*t**3*(D-2*d)*(1-0.63*t*b**-1) + +//Over Estimation of torsional Rigidity would have been +T=J*J2**-1 + +//Result +printf("\n Error in assessing torsional Rigidity if the warping is neglected is %0.2f ",T) diff --git a/3864/CH6/EX6.26/Ex6_26.sce b/3864/CH6/EX6.26/Ex6_26.sce new file mode 100644 index 000000000..4f3185b4c --- /dev/null +++ b/3864/CH6/EX6.26/Ex6_26.sce @@ -0,0 +1,36 @@ +clear +// +// + +//Initilization of Variables + +d1=100 //mm //Outer Diameter +d2=95 //mm //Inner Diameter +T=2*10**6 //N-mm //Torque + +//Calculations + +J=%pi*32**-1*(d1**4-d2**4) //mm**4 //Polar Modulus + +//Shear stress +q_max=T*J**-1*d1*2**-1 //N/mm**2 + +//Now theta*L**-1=T*(G*J)**-1 +//After substituting values and further simplifying we get +//Let theta*L**-1=X +X=T*J**-1 + +//Now Treating it as very thin walled tube +d=(d1+d2)*2**-1 //mm + +r=d*2**-1 +t=(d1-d2)*2**-1 +q_max2=T*(2*%pi*r**2*t)**-1 //N/mm**2 + +X2=T*(2*%pi*r**3*t)**-1 + +//Result +printf("\n When it is treated as hollow shaft:Max shear stress %0.2f N/mm**2",q_max) +printf("\n :Angle of Twist per unit Length %0.3f ",X) +printf("\n When it is very thin Walled Tube :Max shear stress %0.2f N/mm**2",q_max2) +printf("\n :Angle of twist per Unit Length %0.3f ",X2) diff --git a/3864/CH6/EX6.3/Ex6_3.sce b/3864/CH6/EX6.3/Ex6_3.sce new file mode 100644 index 000000000..5d5ac6967 --- /dev/null +++ b/3864/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,27 @@ +clear +// +// + +//Initilization of Variables + +G=80*10**3 //N/mm**2 //Modulus of rigidity +q_s=80 //N/mm**2 //Max sheare stress +P=736*10**6 //N-mm/sec //Power transmitted +n=200 + +//Calculations + +T=P*60*(2*%pi*n)**-1 //N-mm //Torsional moment + +//Now From consideration of angle of twist +theta=%pi*180**-1 +//L=15*d + +d=(T*32*180*15*(%pi**2*G)**-1)**0.33333 + +//Now corresponding stress at the surface is +q_s2=T*32*d*(%pi*2*d**4)**-1 + +//Result +printf("\n Max diameter required is %0.2f mm",d) +printf("\n Corresponding shear stress is %0.2f N/mm**2",q_s2) diff --git a/3864/CH6/EX6.4/Ex6_4.sce b/3864/CH6/EX6.4/Ex6_4.sce new file mode 100644 index 000000000..0c51b1470 --- /dev/null +++ b/3864/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,28 @@ +clear +// +// + +//Initilization of Variables + +d=25 //mm //Diameter of steel bar +p=50*10**3 //N //Pull +dell_1=0.095 //mm //Extension of bar +l=200 //mm //Guage Length +T=200*10**3 //N-mm //Torsional moment +theta=0.9*%pi*180**-1 //angle of twist +L=250 //mm Length of steel bar + +//Calculations + +A=%pi*4**-1*d**2 //Area of steel bar //mm**2 +E=p*l*(dell_1*A)**-1 //N/mm**2 //Modulus of elasticity + +J=%pi*32**-1*d**4 //mm**4 //Polar modulus + +G=T*L*(theta*J)**-1 //Modulus of rigidity //N/mm**2 + +//Now from the relation of Elastic constants +mu=E*(2*G)**-1-1 + +//result +printf("\n The Poissons ratio is %0.3f ",mu) diff --git a/3864/CH6/EX6.5/Ex6_5.sce b/3864/CH6/EX6.5/Ex6_5.sce new file mode 100644 index 000000000..0ee4239e2 --- /dev/null +++ b/3864/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,26 @@ +clear +// +// + +//Initilization of Variables + +L=6000 //mm //Length of circular shaft +d1=100 //mm //Outer Diameter +d2=75 //mm //Inner Diameter +R=100*2**-1 //Radius of shaft +T=10*10**6 //N-mm //Torsional moment +G=80*10**3 //N/mm**2 //Modulus of Rigidity + +//Calculations + +J=%pi*32**-1*(d1**4-d2**4) //mm**4 //Polar Modulus + +//Max Shear stress produced +q_s=T*R*J**-1 //N/mm**2 + +//Angle of twist +theta=T*L*(G*J)**-1 //Radian + +//Result +printf("\n MAx shear stress produced is %0.2f N/mm**2",q_s) +printf("\n Angle of Twist is %0.2f Radian",theta) diff --git a/3864/CH6/EX6.6/Ex6_6.sce b/3864/CH6/EX6.6/Ex6_6.sce new file mode 100644 index 000000000..f16af5a6e --- /dev/null +++ b/3864/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,30 @@ +clear +// +// + +//Initilization of Variables + +d1=200 //mm //External Diameter of shaft +t=25 //mm //Thickness of shaft +n=200 //rpm +theta=0.5*%pi*180**-1 //Radian //angle of twist +L=2000 //mm //Length of shaft +G=84*10**3 //N/mm**2 +d2=d1-2*t //mm //Internal Diameter of shaft + +//Calculations + +J=%pi*32**-1*(d1**4-d2**4) //mm**4 //Polar Modulus + +//Torsional moment +T=G*J*theta*L**-1 //N/mm**2 + +//Power Transmitted +P=2*%pi*n*T*60**-1*10**-6 //N-mm + +//Max shear stress transmitted +q_s=G*theta*(d1*2**-1)*L**-1 //N/mm**2 + +//Result +printf("\n Power Transmitted is %0.2f N-mm",P) +printf("\n Max Shear stress produced is %0.2f N/mm**2",q_s) diff --git a/3864/CH6/EX6.7/Ex6_7.sce b/3864/CH6/EX6.7/Ex6_7.sce new file mode 100644 index 000000000..70abac79a --- /dev/null +++ b/3864/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,33 @@ +clear +// +// + +//Initilization of Variables + +P=3750*10**6 //N-mm/sec +n=240 //Rpm +q_s=160 //N/mm**2 //Max shear stress + +//Calculations + +//d2=0.8*d2 //mm //Internal Diameter of shaft + +//J=%pi*32**-1*(d1**4-d2**4) //mm**4 //Polar modulus +//After substituting value in above Equation we get +//J=0.05796*d1**4 + +T=P*60*(2*%pi*n)**-1 //N-mm //Torsional moment + +//Now from Torsion Formula +//T*J**-1=q_s*R**-1 ......................................(1) + +//But R=d1*2**-1 + +//Now substituting value of R and J in Equation (1) we get +d1=(T*(0.05796*q_s*2)**-1)**0.33333 + +d2=d1*0.8 + +//Result +printf("\n The size of the Shaft is:d1 %0.3f mm",d1) +printf("\n :d2 %0.3f mm",d2) diff --git a/3864/CH6/EX6.8/Ex6_8.sce b/3864/CH6/EX6.8/Ex6_8.sce new file mode 100644 index 000000000..51b52d047 --- /dev/null +++ b/3864/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,64 @@ +clear +// +// + +//Initilization of Variables + +P=245*10**6 //N-mm/sec //Power transmitted +n=240 //rpm +q_s=40 //N/mm**2 //Shear stress +theta=%pi*180**-1 //radian //Angle of twist +L=1000 //mm //Length of shaft +G=80*10**3 //N/mm**2 + +//Tmax=1.5*T + +//Calculations + +T=P*60*(2*%pi*n)**-1 //N-mm //Torsional Moment +Tmax=1.5*T + +//Now For Solid shaft +//J=%pi*32*d**4 + +//Now from the consideration of shear stress we get +//T*J**-1=q_s*(d*2**-1)**-1 +//After substituting value in above Equation we get +//T=%pi*16**-1*d**3*q_s + +//Designing For max Torque +d=(Tmax*16*(%pi*40)**-1)**0.33333 //mm //Diameter of shaft + +//For max Angle of Twist +//Tmax*J**-1=G*theta*L**-1 +//After substituting value in above Equation we get +d2=(Tmax*32*180*L*(%pi**2*G)**-1)**0.25 + +//For Hollow Shaft + +//d1_2=Outer Diameter +//d2_2=Inner Diameter + +//d2_2=0.5*d1_2 + +// Polar modulus +//J=%pi*32**-1*(d1_2**4-d2_2**4) +//After substituting values we get +//J=0.092038*d1_2**4 + +//Now from the consideration of stress +//Tmax*J**-1=q_s*(d1_2*2**-1)**-1 +//After substituting values and further simplifying we get +d1_2=(Tmax*(0.092038*2*q_s)**-1)**0.33333 + +//Now from the consideration of angle of twist +//Tmax*J**-1=G*theta*L**-1 +//After substituting values and further simplifying we get +d1_3=(Tmax*180*L*(0.092038*G*%pi)**-1)**0.25 + +d2_2=0.5*d1_2 + +//result +printf("\n Diameter of shaft is:For solid shaft:d %0.2f mm",d) +printf("\n :For Hollow shaft:d1_2 %0.3f mm",d1_2) +printf("\n : :d2_2 %0.3f mm",d2_2) diff --git a/3864/CH7/EX7.1/Ex7_1.sce b/3864/CH7/EX7.1/Ex7_1.sce new file mode 100644 index 000000000..98d6372db --- /dev/null +++ b/3864/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,35 @@ +clear +// +// + +//Initilization of Variables + +sigma1=30 //N/mm**2 //Stress in tension +d=20 //mm //Diameter +sigma2=90 //N/mm**2 //Max compressive stress +sigma3=25 //N/mm**2 + +//Calculations + +//In TEnsion + +//Corresponding stress in shear +P=sigma1*2**-1 //N/mm**2 + +//Tensile force +F=%pi*4**-1*d**2*sigma1 + +//In Compression + +//Correspong shear stress +P2=sigma2*2**-1 //N/mm**2 + +//Correspong compressive(axial) stress +p=2*sigma3 //N/mm**2 + +//Corresponding Compressive force +P3=p*%pi*4**-1*d**2 //N + +//Result +printf("\n Failure Loads are: %0.2f N",F) +printf("\n : %0.2f N",P3) diff --git a/3864/CH7/EX7.12/Ex7_12.sce b/3864/CH7/EX7.12/Ex7_12.sce new file mode 100644 index 000000000..40bfce331 --- /dev/null +++ b/3864/CH7/EX7.12/Ex7_12.sce @@ -0,0 +1,30 @@ +clear +// +// + +//Initilization of Variables + +//Direct stresses +p_x=120 //N/mm**2 //Tensile stress +p_y=-100 //N/mm**2 //Compressive stress +p1=160 //N/mm**2 //Major principal stress + +//Calculations + +//Let q be the shearing stress + +//p1=(p_x+p_y)*2**-1+((((p_x+p_y)*2**-1)**2)+q**2)**0.5 +//After further simplifying we get +q=(p1-((p_x+p_y)*2**-1))**2-((p_x-p_y)*2**-1)**2 //N/mm**2 +q2=(q)**0.5 //N/mm**2 + +//Minimum Principal stress +p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q2**2)**0.5 //N/mm**2 + +//Max shearing stress +q_max=(((p_x-p_y)*2**-1)**2+q2**2)**0.5 //N/mm**2 + +//Result +printf("\n Shearing stress of material %0.2f N/mm**2",q) +printf("\n Min Principal stress %0.2f N/mm**2",p2) +printf("\n Max shearing stress %0.2f N/mm**2",q_max) diff --git a/3864/CH7/EX7.14/Ex7_14.sce b/3864/CH7/EX7.14/Ex7_14.sce new file mode 100644 index 000000000..a000a0993 --- /dev/null +++ b/3864/CH7/EX7.14/Ex7_14.sce @@ -0,0 +1,40 @@ +clear +// +// + +//Initilization of Variables + +F=40*10**3 //N //Shear Force +M=20*10**6 //Bending Moment + +//Rectangular section +b=100 //mm //Width +d=200 //mm //Depth + +x=20 //mm //Distance from Top surface upto point +y=80 //mm //Distance from point to Bottom + +//Calculations + +I=1*12**-1*b*d**3 //mm**4 //M.I + +//At 20 mm Below top Fibre +f_x=M*I**-1*y //N/mm**2 //Stress + +//Assuming sagging moment ,f_x is compressive p_x=f_x=-24 //N/mm**2 +f_x=-24 //N/mm**2 +p_x=-24 + +//Shearing stress +q=F*(b*I)**-1*(b*x*(b-x*2**-1)) //N/mm**2 + +//Direct stresses + +p_y=0 //N/mm**2 + +p1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 +p2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//Result +printf("\n Directions of principal stresses at a point below 20mm is: %0.2f N/mm**2",p1) +printf("\n %0.2f N/mm**2",p2) diff --git a/3864/CH7/EX7.16/Ex7_16.sce b/3864/CH7/EX7.16/Ex7_16.sce new file mode 100644 index 000000000..d0f69cf48 --- /dev/null +++ b/3864/CH7/EX7.16/Ex7_16.sce @@ -0,0 +1,54 @@ +clear +// +// + +//Initilization of Variables + +L=8000 //mm //Span of beam +w=40*10**6 //N/mm //udl + +//I-section + +//Flanges +b=100 //mm //Width +t=10 //mm //Thickness + +D=400 //mm //Overall Depth +t2=10 //mm //thickness of web + +//Calculations + +//Let R_A and R_B be the Reactions at A & B respectively +R_A=w*2**-1*L*10**-9 //KN + +//Shear force at 2m for left support +F=R_A-2*w*10**-6 //KN + +//Bending Moment +M=R_A*2-2*w*10**-6 //KN-m + +//M.I +I=1*12**-1*b*D**3-1*12**-1*(b-t)*(D-2*t2)**3 //mm**4 + +//Bending stress at 100 mm above N_A +f=M*10**6*I**-1*b + +//Shear stress +q=F*10**3*(t*I)**-1*(b*t*(D-t)*2**-1 +t2*(b-t2)*145) //N/mm**2 + +p_x=-197.06 //N/mm**2 +p_y=0 //N/mm**2 +q=21.38 //N/mm**2 + +//Principal Stresses + +P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 +P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//Max shear stress +q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//Result +printf("\n Principal Stresses are: %0.2f N/mm**2",P1) +printf("\n %0.2f N/mm**2",P2) +printf("\n Max shear stress %0.2f N/mm**2",q_max) diff --git a/3864/CH7/EX7.18/Ex7_18.sce b/3864/CH7/EX7.18/Ex7_18.sce new file mode 100644 index 000000000..d2cefbbfc --- /dev/null +++ b/3864/CH7/EX7.18/Ex7_18.sce @@ -0,0 +1,26 @@ +clear +// +// + +//Initilization of Variables + +d=100 //mm //Diameter of shaft +M=3*10**6 //N-mm //B.M +T=6*10**6 //N-mm //Twisting Moment +mu=0.3 + +//Calculations + +//Max principal Stress + +P1=16*(%pi*d**3)**-1*(M+(M**2+T**2)**0.5) //N/mm**2 +P2=16*(%pi*d**3)**-1*(M-(M**2+T**2)**0.5) //N/mm**2 + +//Direct stress +P=(P1)-mu*(P2) //N/mm**2 + + +//Result +printf("\n Principal stresses are: %0.2f N/mm**2",P1) +printf("\n : %0.2f N/mm**2",P2) +printf("\n Stress Producing the same strain is %0.2f N/mm**2",P) diff --git a/3864/CH7/EX7.19/Ex7_19.sce b/3864/CH7/EX7.19/Ex7_19.sce new file mode 100644 index 000000000..63e01c0f3 --- /dev/null +++ b/3864/CH7/EX7.19/Ex7_19.sce @@ -0,0 +1,64 @@ +clear +// +// + +//Initilization of Variables + +d=75 //mm //diameter +P=30*10**6 //W //Power transmitted +W=6 //N-mm/sec //Load +L=1000 //mm +N=300 //r.p.m + +//Calculations + +//B.M +M=W*L*4**-1 //N-mm +T=P*60*(2*%pi*N)**-1 //Torque transmitted + +//M.I +I=%pi*64**-1*d**4 //mm**4 + +//Bending stress +f_A=M*I**-1*(d*2**-1) //N/mm**2 + +//At A +p_x=f_A +p_y=0 + +//Polar Modulus +J=%pi*32**-1*d**4 //mm**4 + +//Shearing stress +q=T*J**-1*(d*2**-1) //N/mm**2 + +//Principal Stresses +P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 +P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//Max shear stress +q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//Bending stress +p_x2=0 +p_y2=0 + +//Shearing stress +q2=T*J**-1*d*2**-1 //N/mm**2 + +//Principal stresses +P3=(p_x2+p_y2)*2**-1+(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 //N/mm**2 +P4=(p_x2+p_y2)*2**-1-(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 //N/mm**2 + +//Max shear stress +q_max2=(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 //N/mm**2 + +//Answer for Principal Stresses P1,P2 and Max stress i.e q_max is incorrect in Book + +//Result +printf("\n Principal Stresses at vertical Diameter:P1 %0.2f N/mm**2",P1) +printf("\n :P2 %0.2f N/mm**2",P2) +printf("\n Max stress at vertical Diameter : %0.2f N/mm**2",q_max) +printf("\n Principal Stresses at Horizontal Diameter:P3 %0.2f N/mm**2",P3) +printf("\n :P4 %0.2f N/mm**2",P4) +printf("\n Max stress at Horizontal Diameter : %0.2f N/mm**2",q_max2) diff --git a/3864/CH7/EX7.2/Ex7_2.sce b/3864/CH7/EX7.2/Ex7_2.sce new file mode 100644 index 000000000..8bb052951 --- /dev/null +++ b/3864/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,29 @@ +clear +// +// + +//Initilization of Variables + +d=25 //mm //Diameter of circular bar +F=20*10**3 //N //Axial Force +theta=30 //Degree //angle + +//Calculations + +//Axial stresses +p=F*(%pi*4**-1*d**2)**-1 //N/mm**2 + +//Normal Stress +p_n=p*(cos(30*%pi*180**-1))**2 + +//Tangential Stress +p_t=p*2**-1*sin(2*theta*%pi*180**-1) + +//Max shear stress occurs on plane where theta2=45 +theta2=45 +sigma_max=p*2**-1*sin(2*theta2*%pi*180**-1) + +//Result +printf("\n Stresses developed on a plane making 30 degree is: %0.2f N/mm**2",p_n) +printf("\n : %0.2f N/mm**2",p_t) +printf("\n stress on max shear stress is %0.2f N/mm**2",sigma_max) diff --git a/3864/CH7/EX7.20/Ex7_20.sce b/3864/CH7/EX7.20/Ex7_20.sce new file mode 100644 index 000000000..a8567c8a7 --- /dev/null +++ b/3864/CH7/EX7.20/Ex7_20.sce @@ -0,0 +1,49 @@ +clear +// +// + +//Initilization of Variables + +d1=100 //mm //External Diameter +d2=50 //mm //Internal Diameter +N=500 //mm //r.p.m +P=60*10**6 //N-mm/sec //Power +p=100 //N/mm**2 //principal stress + +//Calculations + +//M.I +I=%pi*(d1**4-d2**4)*64**-1 //mm**4 + +//Bending Stress +//f=M*I*d1*2**-1 //N/mm**2 + +//Principal Planes +//p_x=32*M*(%pi*(d1**4-d2**4))*d1 +//p_y=0 + +//Shear stress +//q=T*J**-1*(d1*2**-1) +//After sub values and further simplifying we get +//q=16*T*d1*(%pi*(d1**4-d2**4))*d1 + +//Principal stresses +//P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 +//After sub values and further simplifying we get +//P1=16*(%pi*(d1**4-d2**4))*d1*(M+(M**2+t**2)**0.5) ...............(1) + +//P=2*%pi*N*T*60**-1 +//After sub values and further simplifying we get +T=P*60*(2*%pi*N)**-1*10**-6 //N-mm + +//Again Sub values and further simplifying Equation 1 we get +M=(337.533)*(36.84)**-1 //KN-m + +//Min Principal stress +//P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 +//Sub values and further simplifying we get +P2=16*(%pi*(d1**4-d2**4))*d1*(M-(M**2+T**2)**0.5)*10**-11 + +//Result +printf("\n Bending Moment safely applied to shaft is %0.2f KN-m",M) +printf("\n Min Principal Stress is %0.3f N/mm**2",P2) diff --git a/3864/CH7/EX7.21/Ex7_21.sce b/3864/CH7/EX7.21/Ex7_21.sce new file mode 100644 index 000000000..2f8167ce7 --- /dev/null +++ b/3864/CH7/EX7.21/Ex7_21.sce @@ -0,0 +1,59 @@ +clear +// +// + +//Initilization of Variables + +d=150 //mm //Diameter +T=20*10**6 //N //Torque +M=12*10**6 //N-mm //B.M +F=200*10**3 //N //Axial Thrust + +//Calculations + +//M.I +I=(%pi*64**-1*d**4) + +//Bending stress +f_A=M*I**-1*(d*2**-1) //N/mm**2 +f_B=-f_A //N/mm**2 + +//Axial thrust due to thrust +sigma=F*(%pi*4**-1*d**2)**-1 + +//At A +p_x=f_A-sigma //N/mm**2 + +//At B +p_x2=f_B-sigma //N/mm**2 + +p_y=0 //At A and B + +//Polar Modulus +J=%pi*32**-1*d**4 //mm**4 + +//Shearing stress at A and B +q=T*J**-1*(d*2**-1) //N/mm**2 + + +//Principal Stresses +//At A +P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 +P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//Max shear stress +q_max1=(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//At B +P1_2=(p_x2+p_y)*2**-1+(((p_x2-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 +P2_2=(p_x2+p_y)*2**-1-(((p_x2-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//Max shear stress +q_max2=(((p_x2-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + + +//Result +printf("\n MAx Principal Stresses:P1 %0.2f N/mm**2",P1) +printf("\n :P2 %0.2f N/mm**2",P2) +printf("\n Min Principal Stresses:P1_2 %0.2f N/mm**2",P1_2) +printf("\n :P2_2 %0.2f N/mm**2",P2_2) diff --git a/3864/CH7/EX7.22/Ex7_22.sce b/3864/CH7/EX7.22/Ex7_22.sce new file mode 100644 index 000000000..99d992598 --- /dev/null +++ b/3864/CH7/EX7.22/Ex7_22.sce @@ -0,0 +1,39 @@ +clear +// +// + +//Initilization of Variables + +//strains +e_A=500 //microns +e_B=250 //microns +e_C=-150 //microns +E=2*10**5 //N/mm**2 //Modulus of Elasticity +mu=0.3 //Poissons ratio +theta=45 //Degrees + +//Calculations +e_x=500 +e_A=500 +e_45=250 +e_B=250 +e_y=-150 +e_C=-150 + +//e_45=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta) +//After sub values and further simplifying we get +rho_x_y=(e_45-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*%pi*180**-1))*(sin(2*theta*%pi*180**-1))**-1*2 + +//Principal strains are given by +e1=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 //microns +e2=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 //microns + +//Principal Stresses +sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 //N/mm**2 +sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 //N/mm**2 + +//Result +printf("\n Principal Strains are:e1 %0.2f N/mm**2",e1) +printf("\n :e2 %0.2f N/mm**2",e2) +printf("\n Principal Stresses are:sigma1 %0.2f N/mm**2",sigma1) +printf("\n :sigma2 %0.2f N/mm**2",sigma2) diff --git a/3864/CH7/EX7.23/Ex7_23.sce b/3864/CH7/EX7.23/Ex7_23.sce new file mode 100644 index 000000000..0520404bd --- /dev/null +++ b/3864/CH7/EX7.23/Ex7_23.sce @@ -0,0 +1,47 @@ +clear +// +// + +//Initilization of Variables + +//Strains +e_A=600 //microns +e_B=-450 //microns +e_C=100 //micron +E=2*10**5 //N/mm**2 //Modulus of Elasticity +mu=0.3 //Poissons ratio +theta=240 + +//Calculations + +e_x=600 +e_A=600 + +//e_A=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(theta)+rho_x_y*2**-1*sin(theta) +//After sub values and further simplifying we get +//-450=(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(1) + +//e_C=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta) +//After sub values and further simplifying we get +//100=(e_x+e_y)*2**-1-0.5*(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(2) + +//Adding Equation 1 and 2 we get equations as +//-350=e_x+e_y-(e_x-e_y)*2**-1 ...............(3) +//Further simplifying we get + +e_y=(-700-e_x)*3**-1 //micron + +rho_x_y=(e_C-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*%pi*180**-1))*(sin(2*theta*%pi*180**-1))**-1*2 //micron + +//Principal strains +e1=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 //microns +e2=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 //microns + +//Principal Stresses +sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 //N/mm**2 +sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 //N/mm**2 + + +//Result +printf("\n Principal Stresses are:sigma1 %0.2f N/mm**2",sigma1) +printf("\n :sigma2 %0.2f N/mm**2",sigma2) diff --git a/3864/CH7/EX7.4/Ex7_4.sce b/3864/CH7/EX7.4/Ex7_4.sce new file mode 100644 index 000000000..20c40bf75 --- /dev/null +++ b/3864/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,29 @@ +clear +// +// +// + +//Initilization of Variables + +//Direct Stresses +P1=60 //N/mm**2 +P2=100 //N/mm**2 + +Theta=25 //Degree //Angle + +//Calculations + +//Normal Stress +P_n=(P1-P2)*2**-1+(P1+P2)*2**-1*cos(2*Theta*%pi*180**-1) //N/mm**2 + +//Tangential Stress +P_t=(P1+P2)*2**-1*sin(Theta*2*%pi*180**-1) //N/mm**2 + +//Resultant stress +P=(P_n**2+P_t**2)**0.5 //N/mm**2 + +theta2=atan(P_n*P_t**-1)*(180*%pi**-1) + +//Result +printf("\n Stresses on the plane AC is: %0.2f N/mm**2",P_n) +printf("\n %0.2f N/mm**2",P_t) diff --git a/3864/CH7/EX7.7/Ex7_7.sce b/3864/CH7/EX7.7/Ex7_7.sce new file mode 100644 index 000000000..c07f45f71 --- /dev/null +++ b/3864/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,32 @@ +clear +// +// +// + +//Initilization of Variables + +//stresses +p_x=60 //N/mm**2 +p_y=-40 //N/mm**2 + +q=10 //N/mm**2 //shear stress + +//Calculations + +//Principal Stresses +p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 +p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//Max shear stress +q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2 + +//Inclination of principal stress to plane +theta=atan(2*q*(p_x-p_y)**-1)*(180*%pi**-1)//Degrees +theta2=(theta)*2**-1 //degrees + +theta3=(theta+180)*2**-1 //degrees + +//Result +printf("\n Principal Stresses are: %0.2f N/mm**2",p1) +printf("\n : %0.2f N/mm**2",p2) +printf("\n Max shear stresses %0.2f N/mm**2",q_max) diff --git a/3864/CH7/EX7.9/Ex7_9.sce b/3864/CH7/EX7.9/Ex7_9.sce new file mode 100644 index 000000000..d272a4ebd --- /dev/null +++ b/3864/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,42 @@ +clear +// +// +// + +//Initilization of Variables + +//stresses +p_x=-40 //N/mm**2 +p_y=80 //N/mm**2 + +q=48 //N/mm**2 //shear stress + +//Calculations + +//Max shear stress +q_max=((((p_x-p_y)*2**-1)**2)+q**2)**0.5 //N/mm**2 + +//Inclination of principal stress to plane +theta=atan(2*q*(p_x-p_y)**-1)*(180*%pi**-1)//Degrees +theta2=(theta)*2**-1 //degrees + +theta3=(theta+180)*2**-1 //degrees + +//Normal Corresponding stress +p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*(theta2+45)*%pi*180**-1)+q*sin(2*(theta2+45)*%pi*180**-1) //Degrees + +//Resultant stress +p=((p_n**2+q_max**2)**0.5) //N/mm**2 + +phi=atan(p_n*q_max**-1)*(180*%pi**-1) //Degrees + +//Inclination to the plane +alpha=((theta2+45))+(phi )//Degree + + +//Answer in book is incorrect of alpha ie41.25 + +//Result +printf("\n Planes of max shear stress: %0.2f N/mm**2",p_n) +printf("\n %0.2f N/mm*2",q_max) +printf("\n Resultant Stress is %0.2f N/mm**2",p) diff --git a/3864/CH8/EX8.1/Ex8_1.sce b/3864/CH8/EX8.1/Ex8_1.sce new file mode 100644 index 000000000..a58d18d29 --- /dev/null +++ b/3864/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,50 @@ +clear +// +// + +//Initilization of Variables + +L=3000 //mm //Length +d1=1000 //mm //Internal diameter +t=15 //mm //Thickness +P=1.5 //N/mm**2 //Fluid Pressure +E=2*10**5 //n/mm**2 //Modulus of elasticity +mu=0.3 //Poissons ratio + +//Calculations + +//Hoop stress +f1=P*d1*(2*t)**-1 //N/mm**2 + +//Longitudinal Stress +f2=P*d1*(4*t)**-1 //N/mm**2 + +//Max shear stress +q_max=(f1-f2)*2**-1 //N/mm**2 + +//Diametrical Strain +//Let e1=dell_d*d**-1 .....................(1) +e1=(f1-mu*f2)*E**-1 + +//Sub values in equation 1 and further simplifying we get +dell_d=e1*d1 //mm + +//Longitudinal strain +//e2=dell_L*L**-1 ......................(2) +e2=(f2-mu*f1)*E**-1 + +//Sub values in equation 2 and further simplifying we get +dell_L=e2*L //mm + +//Change in Volume +//Let Z=dell_V*V**-1 ................(3) +Z=2*e1+e2 + +//Sub values in equation 3 and further simplifying we get +dell_V=Z*%pi*4**-1*d1**2*L + +//Result +printf("\n Max Intensity of shear stress %0.2f N/mm**2",q_max) +printf("\n Change in the Dimensions of the shell is:dell_d %0.2f mm",dell_d) +printf("\n :dell_L %0.2f mm",dell_L) +printf("\n :dell_V %0.2f mm**3",dell_V) diff --git a/3864/CH8/EX8.11/Ex8_11.sce b/3864/CH8/EX8.11/Ex8_11.sce new file mode 100644 index 000000000..f02a0b052 --- /dev/null +++ b/3864/CH8/EX8.11/Ex8_11.sce @@ -0,0 +1,64 @@ +clear +// +// +// + +//Initilization of Variables + +d_o=300 //mm //Outside diameter +d2=200 //mm //Internal Diameter +p=14 //N/mm**2 //internal Fluid pressure +t=50 //mm //Thickness +r_o=150 //mm //Outside Diameter +r2=100 //mm //Internal Diameter + +//Calculations + +//From Lame's Equation +//p_x=b*(x**2)**-1-a //N/mm**2 ...................(1) +//F_x=b*(x**2)**-1+a //N/mm**2 ...................(2) + +//At +p_x=14 //N/mm**2 + +//Sub value of p_x in equation 1 we get +//14=(100)**-1*b-a ............................(3) + +//At +p_x2=0 //N/mm**2 + +//Sub value in equation 1 we get +//0=b*(150**2)**-1-a ......................(4) + +//From Equations 3 and 4 we get +//14=b*(100**2)**-1-b*(100**2)**-1 +//After sub values and further simplifying we get +b=14*100**2*150**2*(150**2-100**2)**-1 + +//From equation 4 we get +a=b*(150**2)**-1 + +//Hoop Stress +//F_x=b*(x**2)**-1+a //N/mm**2 + +//At +x=100 //mm +F_x=b*(x**2)**-1+a //N/mm**2 + +//At +x2=125 //mm +F_x2=b*(x2**2)**-1+a //N/mm**2 + +//At +x3=150 //mm +F_x3=b*(x3**2)**-1+a //N/mm**2 + +//If thin Cyclindrical shell theory is used,hoop stress is uniform and is given by +F=p*d2*(2*t)**-1 //N/mm**2 + +//Percentage error in estimating max hoop tension +E=(F_x-F)*F_x**-1*100 //% + +//Result +printf("\n Max Hoop Stress Developed in the cross-section is %0.2f N/mm**2",F) +printf("\n Plot of Variation of hoop stress") diff --git a/3864/CH8/EX8.12/Ex8_12.sce b/3864/CH8/EX8.12/Ex8_12.sce new file mode 100644 index 000000000..0a29dbff5 --- /dev/null +++ b/3864/CH8/EX8.12/Ex8_12.sce @@ -0,0 +1,43 @@ +clear +// +// + +//Initilization of Variables + +d_o=300 //mm //Outside diameter +d2=200 //mm //Internal Diameter +p=12 //N/mm**2 //internal Fluid pressure +F_max=16 //N/mm**2 //Tensile stress +r_o=150 //mm //Outside Diameter +r2=100 //mm //Internal Diameter + +//Calculations + +//Let p_o be the External Pressure applied. +//From LLame's theorem +//p_x=b*(x**2)**-1-a ..............(1) +//F_x=b*(x**2)**-1+a ...........................(2) + +//Now At +x=100 //mm +p_x=12 //N/mm**2 +//sub in equation 1 we get +//12=b*(100**2)**-1-a . ..................(3) + +//The Max Hoop stress occurs at least value of x where +//16=b*(100**2)**-1+a .......................(4) + +//From Equations 1 and 2 we get +//28=b*(100**2)**-1+b*(100**2)**-1 +//After furhter Simplifying we get +b=28*100**2*2**-1 + +//sub in equation 1 we get +a=-(12-(b*(100**2)**-1)) + +//Thus At +x2=150 //mm +p_o=b*(x2**2)**-1-a + +//Result +printf("\n Minimum External applied is %0.2f N/mm**2",p_o) diff --git a/3864/CH8/EX8.13/Ex8_13.sce b/3864/CH8/EX8.13/Ex8_13.sce new file mode 100644 index 000000000..320b02e37 --- /dev/null +++ b/3864/CH8/EX8.13/Ex8_13.sce @@ -0,0 +1,41 @@ +clear +// +// + +//Initilization of Variables + +d1=160 //mm //Internal Diameter +r1=80 //mm //External Diameter +p1=40 //N/mm**2 //Internal Diameter +P_max=120 //N/mm**2 //Allowable stress + +//Calculations + +//From Lame's Equation we have +//p_x=b*(x**2)**-1-a ..........................(1) +//F_x=b*(x**2)**-1+a ...........................(2) + +//At +//Sub in equation 1 we get +//120=b*(80**2)**-1+a ........................(3) + +//The hoop tension at inner edge is max stress +//Hence +//120=b*(80**2)**-1+a .............................(4) + +//From Equation 3 and 4 we get +b=160*80**2*2**-1 + +//Sub in equation 3 we get +a=-(40-(b*(80**2)**-1)) + +//Let External radius be r_o.Since at External Surface is Zero,we get +//0=b*(r_o)**-1-a +//After Further simplifying we get +r_o=(b*a**-1)**0.5 + +//Thickness of Cyclinder +t=r_o-r1 //mm + +//Result +printf("\n Thickness Required is %0.2f mm",t) diff --git a/3864/CH8/EX8.14/Ex8_14.sce b/3864/CH8/EX8.14/Ex8_14.sce new file mode 100644 index 000000000..b923e1abc --- /dev/null +++ b/3864/CH8/EX8.14/Ex8_14.sce @@ -0,0 +1,73 @@ +clear +// +// + +//Initilization of Variables + +d_o=300 //mm //Outside diameter +d1=180 //mm //Internal Diameter +p=12 //N/mm**2 //internal Fluid pressure +p_o=6 //N/mm**2 //External Pressure +r_o=150 //mm //Outside Diameter +r=90 //mm //Internal Diameter + + +//Calculations + +//From Lame's Equation we have +//p_x=b*(x**2)**-1-a ..........................(1) +//F_x=b*(x**2)**-1+a ...........................(2) + +//At +x=90 //N/mm**2 +r1=90 //N/mm**2 +p=42 //N/mm**2 +//Sub in equation 1 we get +//42=b*(90**2)**-1-a ..............................(3) + +//At +p2=6 //N/mm**2 +//sub in equation 1 we get +//6=b*(150**2)**-1-a ..............................(4) + +//From equations 3 and 4 weget +//36=b*(90**2)**-1-b2(150**2)**-1 +//After further simplifying we get +b=36*90**2*150**2*(150**2-90**2)**-1 + +//Sub value of b in equation 4 we get +a=b*(150**2)**-1-p_o + +//At +F_x=b*(x**2)**-1+a //N/mm**2 + +//At +x2=150 //mm +r=150 //mm + +F_x2=b*(x2**2)**-1+a //N/mm**2 + +//Now if External pressure is doubled i.e p_o2=12 //N/mm**2 We have +p_o2=12 //N/mm**2 +//sub in equation 4 we get +//12=b2*(150**2)**-1-a2 ..........................(5) + +//Max Hoop stress is to be 70.5 //N/mm**2,which occurs at x=r1=90 //mm +//Sub in equation 4 we get +//70.5=b*(90**2)**-1+a2 ................................(6) + +//Adding equation 5 and 6 +//82.5=b2*(150**2)**-1+b*(90**2)**-1 +//After furhter simplifying we get +b2=82.5*150**2*90**2*(150**2+90**2)**-1 + +//Sub in equation 5 we get +a2=b2*(150**2)**-1-12 + +//If p_i is the internal pressure required then from Lame's theorem +p_i=b2*(r1**2)**-1-a2 + +//Result +printf("\n Stresses int the material are:F_x %0.2f N/mm**2",F_x) +printf("\n :F_x2 %0.2f N/mm**2",F_x2) +printf("\n Internal Pressure that can be maintained is %0.2f N/mm**2",p_i) diff --git a/3864/CH8/EX8.16/Ex8_16.sce b/3864/CH8/EX8.16/Ex8_16.sce new file mode 100644 index 000000000..17ed164c5 --- /dev/null +++ b/3864/CH8/EX8.16/Ex8_16.sce @@ -0,0 +1,62 @@ +clear +// +// + +//Initilization of Variables + +do=200 //mm //Inner Diameter +r_o=100 //mm //Inner radius +d1=300 //mm //outer diameter +r1=150 //mm //Outer radius +d2=250 //mm //Junction Diameter +r2=125 //mm //Junction radius +E=2*10**5 //N/mm**2 //Modulus of Elasticity +p=30 //N/mm**2 //radial pressure + +//Calculations + +//from Lame's Equation we get +//p_x=b*(x**2)**-1-a ..........................(1) +//F_x=b*(x**2)**-1+a ...........................(2) + +//Then from Boundary condition +//p_x=0 at x=100 //mm +//0=b1*(100**2)**-1-a1 .....................(3) + +//p_x2=30 //N/mm**2 at x2=125 //mm +//30=b1*(125**2)**-1-a1 ................................(4) + +//From equation 3 and 4 we get +b1=30*125**2*100**2*(100**2-125**2)**-1 + +//From Equation 3 we get +a1=b1*(100**2)**-1 + +//therefore Hoop stress in inner cyclinder at junction +F_2_1=b1*(125**2)**-1+a1 //N/mm**2 + +//Outer Cyclinder +//p_x=b*(x**2)**-1-a ..........................(5) +//F_x=b*(x**2)**-1+a ...........................(6) + +//Now at x=125 //mm +//p_x3=30 //N/mm**2 +//30=b2*(125**2)**-1-a2 ..................................(7) + +//At x=150 //mm +//p_x4=0 +//0=b2*(150**2)**-1-a2 ...................................(8) + +//From equations 7 and 8 +b2=30*150**2*125**2*(150**2-125**2)**-1 + +//From eqauation 8 we get +a2=b2*(150**2)**-1 + +//Hoop stress at junction +F_2_0=b2*(125**2)**-1+a2 //N/mm**2 + +rho_r=(F_2_0-F_2_1)*E**-1*r2 + +//Result +printf("\n Shrinkage Allowance is %0.3f mm",rho_r) diff --git a/3864/CH8/EX8.17/Ex8_17.sce b/3864/CH8/EX8.17/Ex8_17.sce new file mode 100644 index 000000000..3d7860c79 --- /dev/null +++ b/3864/CH8/EX8.17/Ex8_17.sce @@ -0,0 +1,92 @@ +clear +// +// + +//Initilization of Variables + +d_o=500 //mm //Outer Diameter +r_o=250 //mm //Outer Radius +d1=300 //mm //Inner Diameter +r1=150 //mm //Inner Radius +d2=400 //mm //Junction Diameter +E=2*10**5 //N/mm**2 //Modulus ofElasticity +alpha=12*10**-6 //Per degree celsius +dell_d=0.2 //mm +dell_r=0.1 //mm + +//Calculations + +//Let p be the radial pressure developed at junction +//Let Lame's Equation for internal cyclinder be +//p_x=b*(x**2)**-1-a ................................(1) +//F_x=b*(x**2)**-1+a ...............................(2) + +//At +x=150 //mm +p_x=0 +//Sub in equation 1 we get +//0=b*(150**2)**-1-a .........................(3) + +//At +x2=200 //mm +//p_x2=p +//p=b*(200**2)**-1-a ......................(4) + +//From Equation 3 and 4 +//p=b*(200**2)**-1-b(150**2)**-1 +//after further simplifying we get +//b=-51428.571*p + +//sub in equation 3 we get +//a1=-2.2857*p + +//therefore hoop stress at junction is +//F_2_1=-21428.571*p*(200**2)**-1-2.2857*p +//after Further simplifying we geet +//F_2_1=3.5714*p + +//Let Lame's Equation for cyclinder be +//p_x=b*(x**2)**-1-a .........................5 +//F_x=b*(x**2)**-1+a .............................6 + +//At +x=200 //mm +//p_x=p2 +//p2=b2*(20**2)**-1-a2 ...................7 + +//At +x2=200 //mm +p_x2=0 +//0=b2*(250**2)**-1-a2 ....................8 + +//from equation 7 and 8 we get +//p2=b2*(200**2)**-1-b2*(250**2)**-1 +//After further simplifying we get +//p2=b2*(250**2-200**2)*(200**2*250**2)**-1 +//b2=111111.11*p + +//from equation 7 +//a2=b2*(250**2)**-1 +//further simplifying we get +//a2=1.778*p + +//At the junctionhoop stress in outer cyclinder +//F_2_0=b2*(200**2)**-1+a2 +//After further simplifying we get +//F_2_0=4.5556*p + +//Considering circumferential strain,the compatibility condition +//rho_r*r2**-1=1*E**-1*(F_2_1+F_2_0) +//where F_2_1 is compressive and F_2_0 is tensile +//furter simplifying we get +p=0.1*200**-1*2*10**5*(3.5714+4.5556)**-1 + +//Let T be the rise in temperature required +//dell_d=d*alpha*T +//After sub values and further simplifying we get +d=250 //mm +T=dell_d*(d*alpha)**-1 //Per degree celsius + +//Result +printf("\n Radial Pressure Developed at junction %0.2f N/mm**2",p) +printf("\n Min Temperatureto outer cyclinder %0.2f Per degree Celsius",T) diff --git a/3864/CH8/EX8.2/Ex8_2.sce b/3864/CH8/EX8.2/Ex8_2.sce new file mode 100644 index 000000000..ea64b7d60 --- /dev/null +++ b/3864/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,61 @@ +clear +// +// + +//Initilization of Variables + +L=2000 //mm //Length +d=200 //mm // diameter +t=10 //mm //Thickness +dell_V=25000 //mm**3 //Additional volume +E=2*10**5 //n/mm**2 //Modulus of elasticity +mu=0.3 //Poissons ratio + +//Calculations + +//Let p be the pressure developed + +//Circumferential Stress + +//f1=p*d*(2*t)**-1 //N/mm**2 +//After sub values and further simplifying +//f1=10*p + +//f1=p*d*(4*t)**-1 //N/mm**2 +//After sub values and further simplifying +//f1=5*p + +//Diameterical strain = Circumferential stress +//Let X=dell_d*d**-1 ................................(1) +//X=e1=(f1-mu*f2)*E**-1 +//After sub values and further simplifying +//e1=8.5*p*E**-1 + +//Longitudinal strain +//Let Y=dell_L*L**-1 ......................................(2) +//Y=e2=(f2-mu*f1)*E**-1 +//After sub values and further simplifying +//e2=2*p*E**-1 + +//Volumetric strain +//Let X=dell_V*V**-1 +//X=2*e1+e2 +//After sub values and further simplifying +//X=19*p*E**-1 +//After further simplifying we get +p=dell_V*(%pi*4**-1*d**2*L)**-1*E*19**-1 //N/mm**2 + +//Hoop Stress +f1=p*d*(2*t)**-1 + +//Sub value of X in equation 1 we get +dell_d=8.5*p*E**-1*d + +//Sub value of Y in equation 2 we get +dell_L=2*p*E**-1*L + +//Result +printf("\n Pressure Developed is %0.2f N/mm**2",p) +printf("\n Hoop stress Developed is %0.2f N/mm**2",f1) +printf("\n Change in diameter is %0.2f mm",dell_d) +printf("\n Change in Length is %0.2f mm",dell_L) diff --git a/3864/CH8/EX8.3/Ex8_3.sce b/3864/CH8/EX8.3/Ex8_3.sce new file mode 100644 index 000000000..cf692919b --- /dev/null +++ b/3864/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,23 @@ +clear +// +// + +//Initilization of Variables + +d=750 //mm //Diameter of water supply pipes +h=50*10**3 //mm //Water head +sigma=20 //N/mm**2 //Permissible stress +rho=9810*10**-9 //N/mm**3 + +//Calculations + +//Pressure of water +P=rho*h //N/mm**2 + +//Stress +//sigma=p*d*(2*t)**-1 +//After further simplifying +t=P*d*(2*sigma)**-1 //mm + +//Result +printf("\n Thickness of seamless pipe is %0.3f mm",t) diff --git a/3864/CH8/EX8.4/Ex8_4.sce b/3864/CH8/EX8.4/Ex8_4.sce new file mode 100644 index 000000000..536feaeea --- /dev/null +++ b/3864/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,26 @@ +clear +// +// + +//Initilization of Variables + +d=2500 //mm //Diameter of riveted boiler +P=1 //N/mm**2 //Pressure +rho1=0.7 //Percent efficiency +rho2=0.4 //Circumferential joints +sigma=150 //N/mm**2 //Permissible stress + +//Calculations + +//Equating Bursting force to longitudinal joint strength ,we get +//p*d*L=rho1*2*t*L*sigma +//After rearranging and further simplifying we get +t=P*d*(2*sigma*rho1)**-1 //mm + +//Considering Longitudinal force +//%pi*d**2*4**-1*P=rho2*%pi*d*t*sigma +//After rearranging and further simplifying we get +t2=P*d*(4*sigma*rho2)**-1 + +//Result +printf("\n Thickness of plate required is %0.2f mm",t) diff --git a/3864/CH8/EX8.5/Ex8_5.sce b/3864/CH8/EX8.5/Ex8_5.sce new file mode 100644 index 000000000..8a11dcc63 --- /dev/null +++ b/3864/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,23 @@ +clear +// +// + +//Initilization of Variables + +//Boiler Dimensions +t=16 //mm //Thickness +p=2 //N/mm**2 //internal pressure +f=150 //N/mm**2 //Permissible stress +rho1=0.75 //Longitudinal joints +rho2=0.45 //circumferential joints + +//Calculations + +//Equating Bursting force to longitudinal joint strength ,we get +d1=rho1*2*t*f*p**-1 //mm + +//Considering circumferential strength +d2=4*rho2*t*f*p**-1 //mm + +//Result +printf("\n Largest diameter of Boiler is %0.2f mm",d1) diff --git a/3864/CH8/EX8.6/Ex8_6.sce b/3864/CH8/EX8.6/Ex8_6.sce new file mode 100644 index 000000000..769bbd3c5 --- /dev/null +++ b/3864/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,53 @@ +clear +// +// + +//Initilization of Variables + +d=250 //mm //Diameter iron pipe +t=10 //mm //Thickness +d2=6 //mm //Diameter of steel +p=80 //N/mm**2 //stress +P=3 //N/mm**2 //Pressure +E_c=1*10**5 //N/mm**2 +mu=0.3 //Poissons ratio +E_s=2*10**5 //N/mm**2 +n=1 //No.of wires + +//Calculations + +L=6 //mm //Length of cyclinder + +//Force Exerted by steel wire at diameterical section +F=p*2*%pi*d2**2*1*4**-1 //N + +//Initial stress in cyclinder +f_c=F*(2*t*d2)**-1 //N/mm**2 + +//LEt due to fluid pressure alone stresses developed in steel wire be F_w and in cyclinder f1 and f2 +f2=P*d*(4*t)**-1 //N/mm**2 + +//Considering the equilibrium of half the cyclinder, 6mm long we get +//F_w*2*%pi*4**-1*d2**2*n+f1*2*t*d2=P*d*d2 +//After further simplifying we get +//F_w+2.122*f1=79.58 . ......................................(1) + +//Equating strain in wire to circumferential strain in cyclinder +//F_w=(f1-mu*f2)*E_s*E_c**-1 //N/mm**2 +//After further simplifying we get +//F_w=2*f1-11.25 ....................................(2) + +//Sub in equation in1 we get +f1=(79.58+11.25)*(4.122)**-1 //N/mm**2 +F_w=2*f1-11.25 //N/mm**2 + +//Final stresses +//1) In steel Wir +sigma=F_w+p //N/mm**2 + +//2) In Cyclinde +sigma2=f1-f_c + +//Result +printf("\n Final Stresses developed in:cyclinder is %0.2f N/mm**2",sigma) +printf("\n :Steel is %0.2f N/mm**2",sigma2) diff --git a/3864/CH8/EX8.7/Ex8_7.sce b/3864/CH8/EX8.7/Ex8_7.sce new file mode 100644 index 000000000..134c16ad3 --- /dev/null +++ b/3864/CH8/EX8.7/Ex8_7.sce @@ -0,0 +1,30 @@ +clear +// +// + +//Initilization of Variables + +d=750 //mm //Diameter of shell +t=8 //mm //THickness +p=2.5 //N/mm**2 +E=2*10**5 //N/mm**2 +mu=0.25 //Poissons ratio + +//Calculations + +//Hoop stress +f1=p*d*(4*t)**-1 //N/mm**2 +f2=p*d*(4*t)**-1 //N/mm**2 + +//Change in Diameter +dell_d=d*p*d*(1-mu)*(4*t*E)**-1 //mm + +//Change in Volume +dell_V=3*p*d*(1-mu)*(4*t*E)**-1*%pi*6**-1*d**3 + +//Answer for Change in diameter is incorrect in book + +//Result +printf("\n Stress introduced is %0.2f N/mm**2",f1) +printf("\n Change in Diameter is %0.2f N/mm**2",dell_d) +printf("\n Change in Volume is %0.2f mm**3",dell_V) diff --git a/3864/CH8/EX8.8/Ex8_8.sce b/3864/CH8/EX8.8/Ex8_8.sce new file mode 100644 index 000000000..4ae842bcf --- /dev/null +++ b/3864/CH8/EX8.8/Ex8_8.sce @@ -0,0 +1,18 @@ +clear +// +// + +//Initilization of Variables + +d=600 //mm //Diameter of sherical shell +t=10 //mm //Thickness +f=80 //N/mm**2 //Permissible stress +rho=0.75 //Efficiency joint + +//Calculations + +//Max Pressure +p=f*4*t*rho*d**-1 //N/mm**2 + +//Result +printf("\n Max Pressure is %0.2f N/mm**2",p) diff --git a/3864/CH8/EX8.9/Ex8_9.sce b/3864/CH8/EX8.9/Ex8_9.sce new file mode 100644 index 000000000..884b6515e --- /dev/null +++ b/3864/CH8/EX8.9/Ex8_9.sce @@ -0,0 +1,38 @@ +clear +// +// + +//Initilization of Variables + +L=1000 //mm //Length of shell +d=200 //mm //Diameter +t=6 //mm //Thickness +p=1.5 //N/mm**2 //Internal Pressure +E=2*10**5 //N/mm**2 +mu=0.25 //Poissons Ratio + +//Calculations + +//Change in Volume of sphere +dell_V_s=3*p*d*(1-mu)*(4*t*E)**-1*%pi*6**-1*d**3 + +//Hoop stress +f1=p*d*(2*t)**-1 //N/mm**2 + +//Longitudinal stress +f2=p*d*(4*t)**-1 //N/mm**2 + +//Principal strain +e1=(f1-mu*f2)*E**-1 +e2=(f2-mu*f1)*E**-1 + +V_c=1000 //mm**3 + +//Change in Volume of cyclinder +dell_V_c=(2*e1+e2)*%pi*4**-1*d**2*L + +//Total Change in Diameter +dell_V=dell_V_s+dell_V_c //mm**3 + +//Result +printf("\n Change in Volume is %0.2f mm**3",dell_V) diff --git a/3864/CH9/EX9.1/Ex9_1.sce b/3864/CH9/EX9.1/Ex9_1.sce new file mode 100644 index 000000000..aaad4d9c6 --- /dev/null +++ b/3864/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,23 @@ +clear +// +// + +//Initilization of Variables +L=5000 //mm //Length of strut +dell=10 //mm //Deflection +W=10 //N //Load + +//Calculations + +//Central Deflection of a simply supported beam with central concentrated load is +//dell=W*L**3*(48*E*I)**-1 + +//Let E*I=X +X=W*L**3*(48*dell)**-1 //mm + +//Euler's Load +//Let Euler's Load be P +P=%pi**2*X*(L**2)**-1 + +//Result +printf("\n Critical Load of Bar is %0.2f N",P) diff --git a/3864/CH9/EX9.10/Ex9_10.sce b/3864/CH9/EX9.10/Ex9_10.sce new file mode 100644 index 000000000..8bdea9929 --- /dev/null +++ b/3864/CH9/EX9.10/Ex9_10.sce @@ -0,0 +1,51 @@ +clear +// +// + +//Initilization of Variables + +sigma=326 //N/mm**2 //stress +E=2*10**5 //N/mm**2 //Modulus of Elasticity +FOS=2 //Factor of safety +a=1*7500**-1 //Rankine's constant +D=350 //mm //Overall Depth + +//Cover plates +b1=500 //mm //width +t1=10 //mm //Thickness + +d=220 //mm //Distance between two channels + +L=6000 //mm //Length of column + +A=5366 //mm**2 //Area of Column section +I_xx=100.08*10**6 //mm**4 //M.I of x-x axis +I_yy=4.306*10**6 //mm**4 //M.I of y-y axis +C_yy=23.6 //mm //Centroid at y-y axis + +//Calculations + +//Symmetric axes are the centroidal axes is + +//M.I of Channel at x-x axis +I_xx_1=2*I_xx+2*(1*12**-1*b1*t1**3+b1*t1*(D*2**-1+t1*2**-1)**2) + +//M.I of Channel at y-y axis +I_yy_1=2*(I_yy+A*(d*2**-1+C_yy)**2)+2*12**-1*t1*b1**3 + +//As I_yy<I_xx +//So +I=I_yy_1 //mm**4 + +A2=2*A+2*t1*b1 //Area of channel + +k=(I*A2**-1)**0.5 //mm + +//Critical Load +P=sigma*A2*(1+a*(L*k**-1)**2)**-1 + +//Safe Load +S=P*2**-1*10**-3 //KN + +//Result +printf("\n Safe Load carrying Capacity is %0.2f KN",S) diff --git a/3864/CH9/EX9.11/Ex9_11.sce b/3864/CH9/EX9.11/Ex9_11.sce new file mode 100644 index 000000000..48212a800 --- /dev/null +++ b/3864/CH9/EX9.11/Ex9_11.sce @@ -0,0 +1,31 @@ +clear +// +// + +//Initilization of Variables + +I=4.085*10**8 //mm**4 //M.I +A=20732.0 //mm**2 //area of column +f_y=250 //N/mm**2 +L=6000 //mm //Length of column + +//Calculations + +k=(I*A**-1)**0.5 //mm +lamda=L*k**-1 //Slenderness ratro + +//From Indian standard table +lamda_1=40 +sigma_a_c_1=139 //N/mm**2 +lamda_2=50 +sigma_a_c_2=132 //N/mm**2 + +//Linearly interpolating between these values for lambda=42.744 + +sigma_a_c_3=sigma_a_c_1-2.744*10**-1*(sigma_a_c_1-sigma_a_c_2) + +//Safe Load carrying capacity of column +P=sigma_a_c_3*A*10**-3 + +//Result +printf("\n Safe Load carrying capacity is %0.2f KN",P) diff --git a/3864/CH9/EX9.2/Ex9_2.sce b/3864/CH9/EX9.2/Ex9_2.sce new file mode 100644 index 000000000..2287a950a --- /dev/null +++ b/3864/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,54 @@ +clear +// +// + +//Initilization of Variables + +L=2000 //mm //Length of square column +E=12*10**3 //N/mm**2 //Modulus of Elasticity +sigma=12 //N/mm*2 //stress +W1=95*10**3 //N //Load1 +W2=200*10**3 //N //Load2 +FOS=3 + +//Calculations + +//From Euler's Formula +//P=%pi**2*E*I*(L**2)**-1 .........(1) + +//Working Load +//W=P*(FOS)**-1 + +//Part-1 + +//At W1=95*10**3 //N +//W1=P*(3*L**2)**-1 + +//Let 'a' be the side of the square +//I=1*12**-1*a**4 + +//sub value of I in Equation 1 and further rearranging we get +a=(W1*3*12*L**2*(%pi**2*E)**-1)**0.25 //mm + +//From Consideration of direct crushing +//sigma*a**2=W1 +//After Reaaranging the above equation we get +a2=(W1*(sigma)**-1)**0.5 //mm + +//required size is 103.67*103.67 i.e a*a + +//Part-2 + +//At W2=200*10**3 //N +//W2=P*(3*L**2)**-1 +//After substituting values and further Rearranging the above equation we get +a3=(W2*3*12*L**2*(%pi**2*E)**-1)**0.25 //mm + +//From consideration of direct compression,size required is +a4=(W2*sigma**-1)**0.5 + +//required size is 129.10*129.10 i.e a4*a4 + +//Result +printf("\n For W1 Load Required size is %0.2f mm**2",a*a) +printf("\n For W2 Load Required size is %0.2f mm**2",a4*a4) diff --git a/3864/CH9/EX9.3/Ex9_3.sce b/3864/CH9/EX9.3/Ex9_3.sce new file mode 100644 index 000000000..522658e0f --- /dev/null +++ b/3864/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,37 @@ +clear +// +// + +//Initilization of Variables + +//Flange +b=100 //mm //Width + +D=80 //mm //Overall Depth +t=10 //mm //Thickness of web and flanges +L=3000 //mm //Length of strut +E=200*10**3 //N/mm**2 //Modulus of Elasticity + +//Calculations + +//Let centroid be at depth y_bar from top fibre +y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1+t))*(b*t+(D-t)*t)**-1 //mm + +//M.I at x-x axis +I_x=1*12**-1*b*t**3+b*t*(y_bar-t*2**-1)**2+1*12**-1*t*((D-t))**3+t*((D-t))*((((D-t)*2**-1)+t)-y_bar)**2 + +//M.I at y-y axis +I_y=1*12**-1*t*b**3+1*12**-1*(D-t)*t**3 //mm**3 + +//Least M.I +I=I_y + +//Since both ends are hinged +//Feective Length=Actual Length +L=3000 //mm +l=3000 //mm +//Buckling Load +P=%pi**2*E*I*(l**2)**-1*10**-3 //KN + +//Result +printf("\n The Buckling Load for strut of tee section %0.2f KN",P) diff --git a/3864/CH9/EX9.4/Ex9_4.sce b/3864/CH9/EX9.4/Ex9_4.sce new file mode 100644 index 000000000..ea72d26dd --- /dev/null +++ b/3864/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,43 @@ +clear +// +// + +//Initilization of Variables + +D=400 //mm //Overall Depth + +//Flanges +b=300 //mm //Width +t=50 //mm //Thickness + +t2=30 //mm //Web Thickness + +dell=10 //mm //Deflection +w=40 //N/mm //Load +FOS=1.75 //Factor of safety +E=2*10**5 //N/mm**2 + +//Calculations + +//M.I at x-x axis +I_x=1*12**-1*(b*D**3-(b-t2)*b**3) //mm**4 + +//Central Deflection +//dell=5*w*L**4*(384*E*I)**-1 +//After sub values in above equation and further simplifying we get +L=(dell*384*E*I_x*(5*w)**-1)**0.25 + +//M.I aty-y axis +I=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 //mm**4 +I_y=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 //mm**4 + +//Both the Ends of column are hinged + +//Crippling Load +P=%pi**2*E*I*(L**2)**-1 //N + +//Safe Load +S=P*(FOS)**-1*10**-3 //N + +//Result +printf("\n Safe Load if I-section is used as column with both Ends hhinged %0.2f KN",S) diff --git a/3864/CH9/EX9.5/Ex9_5.sce b/3864/CH9/EX9.5/Ex9_5.sce new file mode 100644 index 000000000..b55ef7544 --- /dev/null +++ b/3864/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,42 @@ +clear +// +// + +//Initilization of Variables + +D=200 //mm //External Diameter +t=20 //mm //hickness +d=200-2*t //mm //Internal Diameter +E=1*10**5 //N/mm**2 +a=1*(1600)**-1 //Rankine's Constant +L=4.5 //m //Length +sigma=550 //N/mm**2 //Stress +FOS=2.5 + +//Calculations + +//Moment of Inertia +I=%pi*D**4*64**-1-%pi*d**4*64**-1 + +//Both Ends are fixed + +//Effective Length +l=1*2**-1*L*10**3 //mm + +//Euler's Critical Load +P_E=%pi**2*E*I*(l**2)**-1 + +A=%pi*4**-1*(D**2-d**2) //mm*2 + +k=(I*A**-1)**0.5 + +//Rankine's Critical Load +P_R=sigma*A*(1+a*(l*k**-1)**2)**-1 + +X=P_E*P_R**-1 + +//Safe Load using Rankine's Formula +S=P_R*(FOS)**-1*10**-3 //KN + +//Result +printf("\n Safe Load by Rankines Formula is %0.2f KN",S) diff --git a/3864/CH9/EX9.6/Ex9_6.sce b/3864/CH9/EX9.6/Ex9_6.sce new file mode 100644 index 000000000..cc66b9a26 --- /dev/null +++ b/3864/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,59 @@ +clear +// +// + +//Initilization of Variables + +L=3000 //mm //Length of column +W=800*10**3 //N //Load +a=1*1600**-1 //Rankine's constant +FOS=4 //Factor of safety +sigma=550 //N/mm**2 //stress + +//Calculations + +//Effective Length +l=L*2**-1 //mm + +//Let d1=outer diameter & d2=inner diameter +//d1=5*8**-1*d2 + +//M.I +//I=%pi*64**-1*(d1**4-d2**4) //mm**4 + +//Area of section +//A=pi4**-1*(d1**2-d2**2) //mm**2 + +//k=(I*A**-1) +//substituting values in above equation +//k=1*16**-1*(d1**2-d2**2) +//after simplifying further we get +//k=0.2948119.d1 + +//X=l*k**-1 +//substituting values in above equation and after simplifying further we get +//X=5087.9898*d1**-1 + +//Crtitcal Load +P=W*FOS //N + +//From Rankine's Load +//P2=sigma*A*(1+a*(X)**2)**-1 +//substituting values in above equation and after simplifying further we get +//d1**4-12156618*d1**4-1.96691*10**8=0 +//Solving Quadratic Equation we get +//d1**2-12156618*d1-196691000=0 +a=1 +b=-12156.618 +c=-196691000 + +Y=b**2-4*a*c + +d1_1=((-b+Y**0.5)*(2*a)**-1)**0.5 //mm +d1_2=((-b-Y**0.5)*(2*a)**-1) //mm + +d2=5*8**-1*d1_1 + +//Result +printf("\n Section of cast iron hollow cylindrical column is:d1_1 %0.2f mm",d1_1) +printf("\n :d2 %0.2f mm",d2) diff --git a/3864/CH9/EX9.7/Ex9_7.sce b/3864/CH9/EX9.7/Ex9_7.sce new file mode 100644 index 000000000..1793b5f17 --- /dev/null +++ b/3864/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,65 @@ +clear +// +// + +//Initilization of Variables + +//Let X=(P*A**-1) //Average Stress at Failure +Lamda_1=70 //Slenderness Ratio +Lamda_2=170 //Slenderness Ratio +X1=200 //N/mm**2 +X2=69 //N/mm**2 + +//Rectangular section +b=60 //mm //width +t=20 //mm //Thickness + +L=1250 //mm //Length of strut +FOS=4 //Factor of safety + +//Calculations + +//Slenderness ratio +//Lamda=L*k**-1 + +//The Rankine's Formula for strut +//P=sigma*A*(1+a*(L*k**-1)**-1 + +//From test result 1, +//After sub values in above equation we get and further simplifying we get +//sigma_1=200+980000*a ...................(1) + +//From test result 2, +//After sub values in above equation we get and further simplifying we get +//sigma_2=69+1994100*a ...................(2) + +//Substituting it in equation (1) we get +a=131*1014100**-1 + +//Substituting a in equation 1 +sigma_1=200+980000*a //N/mm**2 + +//Effective Length +l=1*2**-1*L //mm + +//Least of M.I +I=1*12**-1*b*t**3 //mm**4 + +//Area +A=b*t //mm**2 + +k=(I*A**-1)**0.5 + +//Slenderness ratio +Lamda=l*k**-1 + +//From Rankine's Ratio +P=sigma_1*A*(1+a*(Lamda)**2)**-1 + +//Safe Load +S=P*(FOS)**-1*10**-3 //N + +//Result +printf("\n Constant in the Formula is:a %0.6f ",a) +printf("\n :sigma_1 %0.2f ",sigma_1) +printf("\n Safe Load is %0.2f KN",S) diff --git a/3864/CH9/EX9.8/Ex9_8.sce b/3864/CH9/EX9.8/Ex9_8.sce new file mode 100644 index 000000000..cd2e3d419 --- /dev/null +++ b/3864/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,45 @@ +clear +// +// + +//Initilization of Variables + +D=200 //mm //Depth +b=140 //mm //width + +//Plate +b2=160 //mm //Width +t2=10 //mm //Thickness + +L=4000 //mm #Length +l=4000 //mm #Length +FOS=4 //Factor of safety +sigma=315 //N/mm**2 //stress +a2=1*7500**-1 +I_xx=26.245*10**6 //mm**4 //M.I at x-x +I_yy=3.288*10**6 //mm**4 //M.I at y-y +a=3671 //mm**2 //Area +k_x=84.6//mm +k_y=29.9 //mm + +//Calculations + +//Total Area +A=a+2*t2*b2 //mm**2 + +//M.I +I=I_yy+2*12**-1*t2*b2**3 //mm**4 + +k=(I*A**-1)**0.5 //mm + +//Let X=L*k**-1 +X=L*k**-1 + +//Appliying Rankine's Formula +P=sigma*A*(1+a2*(X)**2)**-1 //N + +//Safe Load +S=P*(FOS)**-1*10**-3 //KN + +//Result +printf("\n Safe axial Load is %0.2f KN",S) diff --git a/3864/CH9/EX9.9/Ex9_9.sce b/3864/CH9/EX9.9/Ex9_9.sce new file mode 100644 index 000000000..5d64122bd --- /dev/null +++ b/3864/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,30 @@ +clear +// +// + +//Initilization of Variables + +E=200*10**3 //N/mm**2 //Modulus of elasticity +sigma=330 //N/mm**2 //Stress +a=1*7500**-1 //Rankine's constant +A=5205 //mm**2 //area of column +I_xx=59.431*10**6 //mm**4 //M.I at x-x axis +I_yy=8.575*10**6 //mm**24//M.I at y-y axis + +//Calculations + +//Total M.I +I=I_xx+I_yy //mm**4 + +//Area of compound Section +A2=2*A //mm**2 + +k=(I*A2**-1)**0.5 //mm + +//Equating Euler's Load to Rankine's Load we get +//%pi**2*E*I*(L**2)**-1=sigma*A*(1+a*(L*k)**2)**-1 +//After Substitt=uting values and further simplifying we get +L=(39076198*(1-0.7975432)**-1)**0.5*10**-3 //m + +//Result +printf("\n Length of column for which Rankines formula and Eulers Formula give the same result is %0.2f m",L) |