diff options
Diffstat (limited to '3864/CH5')
| -rw-r--r-- | 3864/CH5/EX5.11/Ex5_11.sce | 55 | ||||
| -rw-r--r-- | 3864/CH5/EX5.12/Ex5_12.sce | 87 | ||||
| -rw-r--r-- | 3864/CH5/EX5.14/Ex5_14.sce | 62 | ||||
| -rw-r--r-- | 3864/CH5/EX5.16/Ex5_16.sce | 60 | ||||
| -rw-r--r-- | 3864/CH5/EX5.18/Ex5_18.sce | 72 | ||||
| -rw-r--r-- | 3864/CH5/EX5.2/Ex5_2.sce | 26 | ||||
| -rw-r--r-- | 3864/CH5/EX5.4/Ex5_4.sce | 27 |
7 files changed, 389 insertions, 0 deletions
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) |