diff options
Diffstat (limited to '3862')
145 files changed, 4127 insertions, 0 deletions
diff --git a/3862/CH1/EX1.1/Ex1_1.sce b/3862/CH1/EX1.1/Ex1_1.sce new file mode 100644 index 000000000..80d587abe --- /dev/null +++ b/3862/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,18 @@ +clear +//downstream direction as x +//direction across river as y + +// + +//variable declaration + +Vx= 8 //velocity of stream, km/hour +Vy=(20) //velocity of boat,km/hour + +V=sqrt((Vx**2)+(Vy**2)) //resultant velocity** km/hour +theta=Vy/Vx + +alpha= atan(theta)*180/%pi //angle, degrees + +printf("\n The resultant velocity : %0.2f km/hour",V) +printf("\n %0.2f °",alpha) diff --git a/3862/CH1/EX1.2/Ex1_2.sce b/3862/CH1/EX1.2/Ex1_2.sce new file mode 100644 index 000000000..4bbebabfb --- /dev/null +++ b/3862/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,13 @@ +clear +//components of force in horizontal and vertical components. +// +//variable declaration + +F= 20 //force in wire, KN + +//calculations +Fx= F*cos(60*%pi/180) +Fy= F*sin(60*%pi/180) + +printf("\n %0.3f KN totheleft ",Fx) +printf("\n %0.3f KN downward ",Fy) diff --git a/3862/CH1/EX1.4/Ex1_4.sce b/3862/CH1/EX1.4/Ex1_4.sce new file mode 100644 index 000000000..f5f561d0b --- /dev/null +++ b/3862/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,20 @@ +clear +//Let the magnitude of the smaller force be F. Hence the magnitude of the larger force is 2F + +// +//variable declaration +R1=260 //resultant of two forces,N +R2=(180) //resultant of two forces if larger force is reversed,N + + + +//calculations + +F=sqrt(((R1**2)+(R2**2))/10) +F1=F +F2=2*F +theta=acos(((R1**2)-(F1**2)-(F2**2))/(2*F1*F2))*180/%pi + +printf("\n F1= %0.3f N",F1) +printf("\n F2= %0.3f N",F2) +printf("\n theta= %0.1f °",theta) diff --git a/3862/CH1/EX1.5/Ex1_5.sce b/3862/CH1/EX1.5/Ex1_5.sce new file mode 100644 index 000000000..aa00cb0a1 --- /dev/null +++ b/3862/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,26 @@ +clear +//Let //ABC be the triangle of forces drawn to some scale +//Two forces F1 and F2 are acting at point A +//angle in degrees '°' + +// + +//variabble declaration +cnv=%pi/180 + +BAC = 20*cnv //Resultant R makes angle with F1 + +ABC = 130*cnv + +ACB = 30*cnv + +R = 500 //resultant force,N + +//calculations +//sinerule + +F1=R*sin(ACB)/sin(ABC) +F2=R*sin(BAC)/sin(ABC) + +printf("\n F1= %0.2f N",F1) +printf("\n F2= %0.2f N",F2) diff --git a/3862/CH1/EX1.6/Ex1_6.sce b/3862/CH1/EX1.6/Ex1_6.sce new file mode 100644 index 000000000..dcf86caf5 --- /dev/null +++ b/3862/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,19 @@ +clear +//Let ABC be the triangle of forces,'theta' be the angle between F1 and F2, and 'alpha' be the angle between resultant and F1 + +// + +//variable declaration +cnv= 180/%pi +F1=(400) //all forces are in newtons,'N' +F2=(260) +R=(520) + +//calculations + +theta=acos(((R**2)-(F1**2)-(F2**2))/(2*F1*F2))*cnv + +alpha=asin(F2*sin(theta*%pi/180)/R)*cnv + +printf("\n theta= %0.2f °",theta) +printf("\n alpha= %0.2f °",alpha) diff --git a/3862/CH1/EX1.7/Ex1_7.sce b/3862/CH1/EX1.7/Ex1_7.sce new file mode 100644 index 000000000..d7190ab81 --- /dev/null +++ b/3862/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,29 @@ +clear +//The force of 3000 N acts along line AB. Let AB make angle alpha with horizontal. + +// + +//variable declaration +F=3000 //force in newtons,'N' +BC=80 //length of crank BC, 'mm' +AB=200 //length of connecting rod AB ,'mm' +theta=60*%pi/180 //angle b/w BC & AC + +//calculations + +alpha=asin(BC*sin(theta)/200)*180/%pi + +HC=F*cos(alpha*%pi/180) //Horizontal component +VC= F*sin(alpha*%pi/180) //Vertical component + +//Components along and normal to crank +//The force makes angle alpha + 60 with crank. +alpha2=alpha+60 +CAC=F*cos(alpha2*%pi/180) // Component along crank +CNC= F*sin(alpha2*%pi/180) //Component normal to crank + + +printf("\n horizontal component= %0.1f N",HC) +printf("\n Vertical component = %0.1f N",VC) +printf("\n Component along crank = %0.1f N",CAC) +printf("\n Component normal to crank= %0.1f N",CNC) diff --git a/3862/CH10/EX10.1/Ex10_1.sce b/3862/CH10/EX10.1/Ex10_1.sce new file mode 100644 index 000000000..bbee7efbb --- /dev/null +++ b/3862/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,31 @@ +clear +//variable declaration + +//A simply supported beam of span 3.0 m has a cross-section 120 mm × 180 mm. If the permissible stress in the material of the beam is 10 N/mm^2 + +b=(120) +d=(180) + +//I=(b*d^3)/12,Ymax=d/2 + +Z=(b*(d**2))/6 +fper=(10) + +L=3 +Mmax=fper*Z + +//Let maximum udl beam can carry be w/metre length +//In this case, we know that maximum moment occurs at mid span and is equal to Mmax = (wL^2)/8 + +w=(Mmax*8)/((L**2)*1000000) + +printf("\n (i) w= %0.2f KN/m",w) + +// Concentrated load at distance 1 m from the support be P kN. + +a=(1) //distance of point at which load is applied from left,m +b=(2) //distance of point at which load is applied from right,m + +P=(L*Mmax)/(a*b*1000000) + +printf("\n (ii) P= %0.2f KN",P) diff --git a/3862/CH10/EX10.10/Ex10_10.sce b/3862/CH10/EX10.10/Ex10_10.sce new file mode 100644 index 000000000..9b286129c --- /dev/null +++ b/3862/CH10/EX10.10/Ex10_10.sce @@ -0,0 +1,21 @@ +clear +// +//A cantilever of 3 m span, carrying uniformly distributed load of 3 kN/m is to be designed using cast iron rectangular section. Permissible stresses in cast iron are f = 30 N/mm^2 in tension and fc = 90 N/mm^2 in compression + +L=(3) //Span of cantilever,m +w=(3) //uniformly distributed load,KN/m + +M=w*1000000*(L**2)/2 //Maximum moment**N-mm +//let b be the width and d the depth +//Z=b*(d**2)/6 + +//Since it is rectangular section, N-A lies at mid-depth, and stresses at top and bottom are same. Hence, permissible tensile stress value is reached earlier and it governs the design. +fper=30 //N/mm^2 +b=100 //mm +f=30 + +//f*Z=M + +d=sqrt((M*6)/(b*f)) + +printf("\n d= %0.1f mm",d) diff --git a/3862/CH10/EX10.11/Ex10_11.sce b/3862/CH10/EX10.11/Ex10_11.sce new file mode 100644 index 000000000..4b994d624 --- /dev/null +++ b/3862/CH10/EX10.11/Ex10_11.sce @@ -0,0 +1,15 @@ +clear +// + +//variable declaration + +// Let the diameter of the bar be ‘d’. Now, W = 800 N L = 1 m = 1000 mm +L=1000 +W=800 +M=W*L/4 //Maximum moment,N-mm +f=150 //permissible stress,N/mm^2 + +d=((((M*32)/(%pi*f)))**(0.33)) + +printf("\n d= %0.2f mm",d) +printf("\n select 25mm bar ") diff --git a/3862/CH10/EX10.2/Ex10_2.sce b/3862/CH10/EX10.2/Ex10_2.sce new file mode 100644 index 000000000..587addef3 --- /dev/null +++ b/3862/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,23 @@ +clear +// + +//variable declaration + +//A circular steel pipe of external diameter 60 mm and thickness 8 mm is used as a simply supported beam over an effective span of 2 m. If permissible stress in steel is 150 N/mm^2, + +D=(60) //external diameter,mm +d=(44) //Thickness,mm + +I=(%pi*((D**4)-(d**4)))/64 //Area moment of inertia**mm^4 +Ymax=(30) //extreme fibre distance,mm + +Z=I/Ymax +fper=(150) + +Mmax=fper*Z + +//Let maximum load it can carry be P kN. +L=(2) +P=(4*Mmax)/(L*1000000) + +printf("\n P= %0.2f KN",P) diff --git a/3862/CH10/EX10.3/Ex10_3.sce b/3862/CH10/EX10.3/Ex10_3.sce new file mode 100644 index 000000000..f1812eaac --- /dev/null +++ b/3862/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,24 @@ +clear +//the cross-section of a cantilever beam of 2.5 m span. Material used is steel for which maximum permissible stress is 150 N/mm^2 + +//variable declaration + +A=(180) //width of I-beam,mm +H=(400) //height of I-beam,mm +a=(170) //width of inter rectancle if I-beam consider as Rectangle with width 10,mm +h=(380) //Height of inter rectancle if I-beam consider as Rectangle with width 10,mm + +I=((A*(H**3))/12)-((a*(h**3))/12) +ymax=(200) //extreme fibre,mm + +Z=I/ymax +fper=(150) + +Mmax=fper*Z + +//If udl is w kN/m, maximum moment in cantilever + +L=2 //m + +w=Mmax/(L*1000000) +printf("\n w= %0.2f KN/m",w) diff --git a/3862/CH10/EX10.4/Ex10_4.sce b/3862/CH10/EX10.4/Ex10_4.sce new file mode 100644 index 000000000..d71fd3afc --- /dev/null +++ b/3862/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,57 @@ +clear +//Compare the moment carrying capacity of the section given in example 10.3 with equivalent section of the same area but (i) square section (ii) rectangular section with depth twice the width and (iii) a circular section. + +// +//variable declaration + +A=180.0*10.0+380.0*10.0+180.0*10.0 + +//If ‘a’ is the size of the equivalent square section, + +a=(sqrt(A)) //mm + +I=(a*(a**3))/12 //Moment of inertia of this section** mm^4 + +ymax=a/2 + +Z=I/ymax + +f=150.0 + +Mcc=f*Z //Moment carrying capacity + +MccI=136985000.0 + +Ratio=MccI/Mcc +printf("\n (i) Moment carryingcapacity of Isection/ Moment carryingcapacityof equivalent squaresection= %0.3f ",Ratio) + + +//Equivalent rectangular section of depth twice the width. Let b be the width,Depth d = 2b. Equating its area to area of I-section,we get +b=sqrt(7400/2) + +ymax=b + +I=b*(((2*b)**3))/12 + +M=f*I/ymax + + +MccI=136985000 + +Ratio=MccI/M +printf("\n (ii) Moment carryingcapacity of I-section/ Moment carryingcapacityof equivalent squaresection= %0.3f ",Ratio) + +//Equivalent circular section. Let diameter be d. + +d=sqrt(7400*4/%pi) + +I=(%pi*(d**4))/64 +ymax=d/2 +Z=I/ymax +fper=(150) +M=fper*Z + +MccI=136985000 + +Ratio=MccI/M +printf("\n (i) Moment carryingcapacity of Isection/ Moment carryingcapacityof equivalent squaresection= %0.3f ",Ratio) diff --git a/3862/CH10/EX10.5/Ex10_5.sce b/3862/CH10/EX10.5/Ex10_5.sce new file mode 100644 index 000000000..288ee40dc --- /dev/null +++ b/3862/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,30 @@ +clear +//variable declaration + +//A symmetric I-section of size 180 mm × 40 mm, 8 mm thick is strengthened with 240 mm × 10 mm rectangular plate on top flange. If permissible stress in the material is 150 N/mm^2, determine how much concentrated load the beam of this section can carry at centre of 4 m span. + +b1=(240) +b=(180) +t=(10) +h=(400) +w=(8) + +A=(240*10+180*8+384*8+180*8) //Area of section,A + +Y=(240*10*405+180*8*(400-4)+384*8*200+180*8*4)/A + +I=(b1*(t**3)/12)+(b1*t*((((h+5)-Y)**2)))+(b*(w**3)/12)+(b*w*((((h-4)-Y)**2)))+(w*((h-16)**3)/12)+((h-16)*w*((((h/2)-Y)**2)))+(b*(w**3)/12)+(b*w*(((4-Y)**2))) + +ytop=(h+t/2)-Y +ybottom=Y +ymax=Y + +Z=I/ymax +fper=150 +M=fper*Z/1000000 //Momnent carrying capacity of the section + +//Let P kN be the central concentrated load the simply supported beam can carry. Then max bending movement in the beam + +P=M*4/(w/2) + +printf("\n P= %0.3f KN",P) diff --git a/3862/CH10/EX10.6/Ex10_6.sce b/3862/CH10/EX10.6/Ex10_6.sce new file mode 100644 index 000000000..77cc47027 --- /dev/null +++ b/3862/CH10/EX10.6/Ex10_6.sce @@ -0,0 +1,47 @@ +clear +//The cross-section of a cast iron beam. The top flange is in compression and bottom flange is in tension. Permissible stress in tension is 30 N/mm^2 and its value in compression is 90 N/mm^2 +//variable declaration +// +b1=(75) +h1=50 +h2=50 +b2=(150) +t=(25) +h=(200) + + +A=(75*50+25*100+150*50) //Area of section,A + +Y=(75*50*175+25*100*100+150*50*25)/A + +I=(b1*(h1**3)/12)+(b1*h1*((((h-(h1/2))-Y)**2)))+(t*((h-h1-h2)**3)/12)+(t*(h-h1-h2)*((((h/2)-Y)**2)))+(b2*(h2**3)/12)+(b2*h2*((((h2/2)-Y)**2))) + + + +ytop=(h-Y) +ybottom=Y + + +Z1=I/ytop +fperc=90 +//Top fibres are in compression. Hence from consideration of compression strength, moment carrying capacity of the beam is given by + +M1=fperc*Z1/1000000 //Momnent carrying capacity of the section,KN-m. + +//Bottom fibres are in tension. Hence from consideration of tension, moment carrying capacity of the section is given by + +Z2=I/ybottom + +fpert=30 + +M2=fpert*Z2/1000000 //Momnent carrying capacity of the section,KN-m. + + +//Actual moment carrying capacity is the lower value of the above two values. Hence moment carrying capacity of the section is +Mmax=min(M1,M2) + +L=(5) +w=sqrt(Mmax*8/(L**2)) + +printf("\n w= %0.3f KN/m",w) +printf("\n calculation mistake in book") diff --git a/3862/CH10/EX10.7/Ex10_7.sce b/3862/CH10/EX10.7/Ex10_7.sce new file mode 100644 index 000000000..169f0f1f1 --- /dev/null +++ b/3862/CH10/EX10.7/Ex10_7.sce @@ -0,0 +1,23 @@ +clear +//The diameter of a concrete flag post varies from 240 mm at base to 120 mm at top. The height of the post is 10 m. If the post is subjected to a horizontal force of 600 N at top +//Consider a section y metres from top. Diameter at this section is d. +//d=120+12*y +//I=%pi*(d**4)/64 +//Z=I*2/d=%pi*(d**3)/32 +//variable declaration +//M=600*1000*y //moment,N-mm +//f*Z=M,f is extreme fibre stress. +// +y=(5) +printf("\n y= %0.2f m",y) + +//Stress at this section f is given by +P=600 +M=P*y*1000 +d=120+12*y +I=%pi*(d**4)/64 +Z=I*2/d + +f=M/Z + +printf("\n f= %0.3f N/mm^2",f) diff --git a/3862/CH10/EX10.9/Ex10_9.sce b/3862/CH10/EX10.9/Ex10_9.sce new file mode 100644 index 000000000..b78881941 --- /dev/null +++ b/3862/CH10/EX10.9/Ex10_9.sce @@ -0,0 +1,18 @@ +clear +//Design a timber beam is to carry a load of 5 kN/m over a simply supported span of 6 m. Permissible stress in timber is 10 N/mm2. Keep depth twice the width. + +//variable declaration +w=(5) //KN/m +L=(6) //m + +M=w*1000000*(L**2)/8 //Maximum bending moment**N-mm + +//Let b be the width and d the depth. Then in this problem d = 2b. +//Z=b*(d**2)/6=2*(b**3)/3 +f=10 //N/mm^2 +//f*Z=M +b=(((M*3)/(2*f))**(0.3333)) +printf("\n b= %0.0f mm",b) + +d=2*b +printf("\n d= %0.0f mm",d) diff --git a/3862/CH11/EX11.10/Ex11_10.sce b/3862/CH11/EX11.10/Ex11_10.sce new file mode 100644 index 000000000..e7b003d81 --- /dev/null +++ b/3862/CH11/EX11.10/Ex11_10.sce @@ -0,0 +1,80 @@ +clear +// + +P1=(20) //vertical loading from A at distance of 1m,KN. +P2=(20) //vertical loading from A at distance of 2m,KN. +P3=(20) //vertical loading from A at distance of 3m,KN. +Ra=(P1+P2+P3)/2 //Due to symmetry + +Rb=Ra +//At section 1.5 m from A +F=(Ra-P1)*1000 +M=((Ra*1.5-P1*0.5)*1000000) +b=(100) +h=(180) + +I=((b*(h**3))/12) + +// Bending stress +//f=M*y/I +y11=0 +f1=(-1)*M*y11/I +y22=45 +f2=(-1)*M*y22/I +y33=90 +f3=(-1)*M*y33/I +//Shearing stress at a fibre ‘y’ above N–A is +//q=(F/(b*I))*(A*y1) +//at y=0, +y1=45 +A1=b*90 +q1=(F/(b*I))*(A1*y1) +//at y=45 +y2=(90-45/2) +A2=b*45 +q2=(F/(b*I))*(A2*y2) +//at y=90 +q3=0 + +//(a) At neutral axis (y = 0) : The element is under pure shear + +py=0 + +p1=(f1+py)/2+sqrt((((f1-py)/2)**2)+(q1**2)) + +p2=(f1+py)/2-sqrt((((f1-py)/2)**2)+(q1**2)) +printf("\n (i) p1= %0.4f N/mm^2",p1) +printf("\n p2= %0.4f N/mm^2",p2) + +theta1=45 +theta2=theta1+90 +printf("\n theta= %0.0f ° and %0.0f °",theta1,theta2) + +//(b) At (y = 45) +py=0 + +p1=(f2+py)/2+sqrt((((f2-py)/2)**2)+(q2**2)) + +p2=(f2+py)/2-sqrt((((f2-py)/2)**2)+(q2**2)) +printf("\n (ii) p1= %0.4f N/mm^2",p1) +printf("\n p2= %0.4f N/mm^2",p2) + +thetab1=(atan((2*q2)/(f2-py))*180)/(%pi*2) +thetab2=thetab1+90 +printf("\n theta= %0.0f ° and %0.0f °",thetab1,thetab2) +//mistake in book +printf("\n mistake in book") + +//(c) At Y=90 + +py=0 + +p1=(f3+py)/2+sqrt((((f3-py)/2)**2)+(q3**2)) + +p2=(f3+py)/2-sqrt((((f3-py)/2)**2)+(q3**2)) +printf("\n (iii) p1= %e N/mm^2",p1) +printf("\n p2= %0.4f N/mm^2",p2) + +thetac1=(atan((2*q3)/(f3-py))*180)/(%pi*2) +thetac2=thetac1+90 +printf("\n theta= %0.0f ° and %0.0f °",thetac1,thetac2) diff --git a/3862/CH11/EX11.11/Ex11_11.sce b/3862/CH11/EX11.11/Ex11_11.sce new file mode 100644 index 000000000..d7780da3c --- /dev/null +++ b/3862/CH11/EX11.11/Ex11_11.sce @@ -0,0 +1,39 @@ +clear +// + +//variable declaration +L=(6) //m +w=(60) //uniformly distributed load,KN/m +Rs=L*w/2 //Reaction at support,KN + +//Moment at 1.5 m from support +M =( Rs*1.5-(w*(1.5**2)/2)) +//Shear force at 1.5 m from support +F=Rs-1.5*w + +B=(200) //width of I-beam,mm +H=(400) //height or I-beam,mm +b=(190) +h=(380) +I= (B*(H**3)/12)-(b*(h**3)/12) + +//Bending stress at 100 mm above N–A +y=100 + +f=M*1000000*y/I + +//Thus the state of stress on an element at y = 100 mm, as px = f,py=0 +px=-f +py=0 +A=200*10*195+10*90*145 +q=(F*1000*(A))/(10*I) //shearing stress,N/mm^2 + +p1=(px+py)/2+sqrt((((px-py)/2)**2)+(q**2)) +p2=(px+py)/2-sqrt((((px-py)/2)**2)+(q**2)) +printf("\n p1= %0.2f N/mm^2",p1) +printf("\n p2= %0.2f N/mm^2",p2) + + +qmax=sqrt((((px-py)/2)**2)+(q**2)) + +printf("\n qmax= %0.2f N/mm^2",qmax) diff --git a/3862/CH11/EX11.2/Ex11_2.sce b/3862/CH11/EX11.2/Ex11_2.sce new file mode 100644 index 000000000..93cae8d59 --- /dev/null +++ b/3862/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,52 @@ +clear +// + +//A material has strength in tension, compression and shear as 30N/mm2, 90 N/mm2 and 25 N/mm2, respectively. If a specimen of diameter 25 mm is tested in tension and compression identity the failure surfaces and loads. + +//variable declaration + +//In tension: Let axial direction be x direction. Since it is uniaxial loading, py = 0, q = 0 and only px exists.when the material is subjected to full tensile stress, px = 30 N/mm^2. + +pt=(30) +pc=(90) +ps=(25) + +d=(25) +px=(30) //N/mm^2 +py=0 +q=0 +p1=(px+py)/2+sqrt((((px-py)/2)**2)+(q**2)) + +p2=(px+py)/2-sqrt((((px-py)/2)**2)+(q**2)) + +qmax=(px-py)/2 + +//Hence failure criteria is normal stress p1 + +A=%pi*(d**2)/4 + +//Corresponding load P is obtained by +p=p1 +P=p1*A + +printf("\n (a) P= %0.2f N",P) + +//In case of compression test, + +px=-pc + +P=-px*A + +printf("\n (b) P= %0.2f N compressive",(-P)) + +//at this stage + +qmax=sqrt((((px-py)/2**2))+(q**2)) + +printf("\n Material fails because of maximum shear and not by axial compression.") +qmax=25 +px=2*qmax + +P=px*A +printf("\n P= %0.0f N",P) +printf("\n The plane of qmax is at 45° to the plane of px. ") diff --git a/3862/CH11/EX11.3/Ex11_3.sce b/3862/CH11/EX11.3/Ex11_3.sce new file mode 100644 index 000000000..9ebd4586b --- /dev/null +++ b/3862/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,26 @@ +clear +//The direct stresses at a point in the strained material are 120 N/mm2 compressive and 80 N/mm2 tensile. There is no shear stress. + +// +//variable declaration + +//The plane AC makes 30° (anticlockwise) to the plane of px (y-axis). Hence theta= 30°. px = 80 N/mm^2 py = – 120 N/mm^2 ,q = 0 + +px=(80) +py=(-120) +q=(0) +theta=30 +pn=((px+py)/2)+((px-py)/2)*cos(2*theta*%pi/180)+q*sin(2*theta*%pi/180) + +printf("\n pn= %0.0f N/mm^2",pn) + +pt=((px-py)/2)*sin(2*theta*%pi/180)-q*cos(2*theta*%pi/180) + +printf("\n pt= %0.1f N/mm^2",pt) +p=sqrt((pn**2)+(pt**2)) + +printf("\n p= %0.2f N/mm^2",p) + +alpha=atan(pn/pt)*180/%pi + +printf("\n alpha= %0.1f °",alpha) diff --git a/3862/CH11/EX11.4/Ex11_4.sce b/3862/CH11/EX11.4/Ex11_4.sce new file mode 100644 index 000000000..fdc03672c --- /dev/null +++ b/3862/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,24 @@ +clear +// +//variable declaration +//Let the principal plane make anticlockwise angle theta with the plane of px with y-axis. Then + +px=(200) //N/mm^2 +py=(150) //N/mm^2 +q=(100) //N/mm^2 + +theta1=(atan((2*q)/(px-py))*180)/(%pi*2) +theta2=90+theta1 +printf("\n theta= %0.2f ° and %0.2f °",theta1,theta2) + +p1=(px+py)/2+sqrt((((px-py)/2)**2)+(q**2)) + +printf("\n p1= %0.2f N/mm^2",p1) + +p2=(px+py)/2-sqrt((((px-py)/2)**2)+(q**2)) + +printf("\n p2= %0.2f N/mm^2",p2) + +qmax=sqrt((((px-py)/2**2))+(q**2)) + +printf("\n qmax= %0.2f N/mm^2",qmax) diff --git a/3862/CH11/EX11.6/Ex11_6.sce b/3862/CH11/EX11.6/Ex11_6.sce new file mode 100644 index 000000000..04a7891f6 --- /dev/null +++ b/3862/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,28 @@ +clear +// +//variable declaration +//Let the principal plane make anticlockwise angle theta with the plane of px with y-axis. Then + +px=(-100) //N/mm^2 +py=(-75) //N/mm^2 +q=(-50) //N/mm^2 + + +p1=(px+py)/2+sqrt((((px-py)/2)**2)+(q**2)) + +printf("\n p1= %0.2f N/mm^2",p1) + +p2=(px+py)/2-sqrt((((px-py)/2)**2)+(q**2)) + +printf("\n p2= %0.2f N/mm^2",p2) + +qmax=sqrt((((px-py)/2**2))+(q**2)) + +printf("\n qmax= %0.2f N/mm^2",qmax) + +//let theta be the inclination of principal stress to the plane of px. + + +theta1=(atan((2*q)/(px-py))*180)/(%pi*2) +theta2=90+theta1 +printf("\n theta= %0.2f ° and %0.2f °",theta1,theta2) diff --git a/3862/CH11/EX11.7/Ex11_7.sce b/3862/CH11/EX11.7/Ex11_7.sce new file mode 100644 index 000000000..7faeff669 --- /dev/null +++ b/3862/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,53 @@ +clear +// +//variable declaration +//Let the principal plane make anticlockwise angle theta with the plane of px with y-axis. Then + +px=(-50) //N/mm^2 +py=(100) //N/mm^2 +q=(75) //N/mm^2 + + +p1=(px+py)/2+sqrt((((px-py)/2)**2)+(q**2)) + +printf("\n (i) p1= %0.2f N/mm^2",p1) + +p2=(px+py)/2-sqrt((((px-py)/2)**2)+(q**2)) + +printf("\n p2= %0.2f N/mm^2",p2) + +qmax=sqrt((((px-py)/2**2))+(q**2)) + +printf("\n (ii) qmax= %0.2f N/mm^2",qmax) + +//let theta be the inclination of principal stress to the plane of px. + + +theta1=(atan((2*q)/(px-py))*180)/(%pi*2) + +printf("\n theta= %0.2f ° clockwise",theta1) + +//Plane of maximum shear makes 45° to it + +theta2=theta1+45 +printf("\n theta2= %0.2f °",theta2) + +//Normal stress on this plane is given by + +pn=((px+py)/2)+((px-py)/2)*cos(2*theta2*%pi/180)+q*sin(2*theta2*%pi/180) + +pt=qmax + +//Resultant stress +p=sqrt((pn**2)+(pt**2)) + +printf("\n p= %0.2f N/mm^2",p) + +//Let ‘p’ make angle phi to tangential stress (maximum shear stress plane). + +phi=atan(pn/pt)*180/%pi + +printf("\n phi= %0.1f °",phi) + +//there is mistake in book +printf("\n mitake in book answer is wrong") diff --git a/3862/CH11/EX11.9/Ex11_9.sce b/3862/CH11/EX11.9/Ex11_9.sce new file mode 100644 index 000000000..e783ab20c --- /dev/null +++ b/3862/CH11/EX11.9/Ex11_9.sce @@ -0,0 +1,29 @@ +clear +// + +//variable declaration + +w=(100) //wide of rectangular beam,mm +h=(200) //height or rectangular beam dude,mm + +I=w*(h**3)/12 + +//At point A, which is at 30 mm below top fibre +y=100-30 +M=(80*1000000) //sagging moment,KN-m + +fx=M*y/I + +px=-fx +F=(100*1000 ) //shear force,N +b=(100) +A=b*30 +y1=100-15 + +q=(F*(A*y1))/(b*I) //shearing stress,N/mm^2 + +py=0 +p1=(px+py)/2+sqrt((((px-py)/2)**2)+(q**2)) +p2=(px+py)/2-sqrt((((px-py)/2)**2)+(q**2)) +printf("\n p1= %0.2f N/mm^2",p1) +printf("\n p2= %0.2f N/mm^2",p2) diff --git a/3862/CH2/EX2.10/Ex2_10.sce b/3862/CH2/EX2.10/Ex2_10.sce new file mode 100644 index 000000000..75bb34d51 --- /dev/null +++ b/3862/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,33 @@ +clear +// + +//variable declaration + +PA=800.0 //Vertical down loading at A,N +PC=400.0 //vertical up loading at B,N +HD=600.0 //Horizontal left loading at A,N +HB=200.0 //Horizontal right loading at B,N +a=1.0 //length of side,m + +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Fx=HB-HD +Fy=PC-PA + + +R=sqrt((Fx**2)+(Fy**2)) +printf("\n R= %0.2f N",R) + +theta=atan(Fy/Fx)*180/%pi +printf("\n theta= %0.0f °",theta) + +//moment at A + +MA=PC*a+HD*a + +//Let x be the distance from A along x axis, where resultant cuts AB. + +x=MA/Fy + +printf("\n x= %0.1f m",(-x)) diff --git a/3862/CH2/EX2.11/Ex2_11.sce b/3862/CH2/EX2.11/Ex2_11.sce new file mode 100644 index 000000000..b8fca588f --- /dev/null +++ b/3862/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,31 @@ +clear +// + +//variable declaration + +PB=2.0 //loading at B,KN +PC=sqrt(3.0) //loading at C,KN +PD=5.0 //loading at D,KN +PE=PC //loading at E,KN +PF=PB //loading at F,KN + +//Let O be the centre of the encircling circle A, B, C, D, E and F. In regular hexagon each side is equal to the radius AO. Hence OAB is equilateral triangle. + +angleoab=60.0*%pi/180 +anglecab=angleoab/2.0 +theta1=anglecab +theta2=(angleoab-theta1) +theta3=theta1 +theta4=theta1 + +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Fx=PB*cos(theta1+theta2)+PC*cos(theta2)+PD+PE*cos(theta3)+PF*cos(theta3+theta4) + +Fy=-PB*sin(theta1+theta2)-PC*sin(theta2)+0+PE*sin(theta3)+PF*sin(theta3+theta4) + +R=sqrt((Fx**2)+(Fy**2)) +printf("\n R= %0.2f KN",R) + +theta=atan(Fy/Fx)*180/%pi diff --git a/3862/CH2/EX2.12/Ex2_12.sce b/3862/CH2/EX2.12/Ex2_12.sce new file mode 100644 index 000000000..b590fb8c7 --- /dev/null +++ b/3862/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,36 @@ +clear +// + +//variable declaration + +P1=2.0 //loading at 1,KN +P2=1.5 //loading at 2,KN +P3=5.0 //loading at 3,KN +a=10.0 //side length,mm + +// If theta1, theta2 and theta3 are the slopes of the forces 2 kN, 5 kN and 1.5 kN forces with respect to x axis, then + + +theta1=atan(a/a) +theta2=atan((3*a)/(4*a)) +theta3=atan((a)/(2*a)) + + +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Fx=P1*cos(theta1)+P3*cos(theta2)-P2*cos(theta3) + +Fy=P1*sin(theta1)-P3*sin(theta2)-P2*sin(theta3) + +R=sqrt((Fx**2)+(Fy**2)) +printf("\n R= %0.2f N",R) + +alpha=atan(Fy/Fx)*180/%pi +printf("\n alpha= %0.2f °",(-alpha)) + +//Distance d of the resultant from O is given by +//Rd=sum of moment at A + +d=((a*3)*P1*cos(theta1)+(5*a)*P3*sin(theta2)+P2*(a)*sin(theta3))/(4.66) +printf("\n d= %0.2f mm",d) diff --git a/3862/CH2/EX2.13/Ex2_13.sce b/3862/CH2/EX2.13/Ex2_13.sce new file mode 100644 index 000000000..1768197a8 --- /dev/null +++ b/3862/CH2/EX2.13/Ex2_13.sce @@ -0,0 +1,35 @@ +clear +// + +//variable declaration + +PB=20.0 //loading at B,KN +PC=30.0 //loading at C,KN +PD=40.0 //loading at D,KN +PA=60.0 //loading at E,KN +AB=1.0 +BC=2.0 +CD=1.0 +//length are in m + +// Let x and y axes be selected +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Rx=0 +Ry=PA+PB+PC+PD + +R=sqrt((Rx**2)+(Ry**2)) +printf("\n R= %0.2f KN",R) + + +//Taking clockwise moment as positive, +//sum of moment at A + +MA=(0)*PA+(AB)*PB+PC*(AB+BC)+PD*(AB+BC+CD) +printf("\n MA= %0.2f KN-m",MA) + +// The distance of resultant from A is, + +x=MA/R +printf("\n x= %0.1f m",x) diff --git a/3862/CH2/EX2.14/Ex2_14.sce b/3862/CH2/EX2.14/Ex2_14.sce new file mode 100644 index 000000000..acbfb4b48 --- /dev/null +++ b/3862/CH2/EX2.14/Ex2_14.sce @@ -0,0 +1,38 @@ +clear +// + +//variable declaration + +PB=30.0 //up loading at B,KN +PC=40.0 //down loading at C,KN +PD=50.0 //up loading at D,KN +PA=80.0 //down loading at A,KN +PE=60.0 //down loading at E,KN +AB=2.0 +BC=2.0 +CD=4.0 +DE=2.0 +//length are in m + +// Let x and y axes be selected +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Rx=0 +Ry=PA-PB+PC-PD+PE + +R=sqrt((Rx**2)+(Ry**2)) +printf("\n R= %0.2f KN in y-direction",R) + + +//Taking clockwise moment as positive, +//sum of moment at A + +MA=(0)*PA-(AB)*PB+PC*(AB+BC)-PD*(AB+BC+CD)+PE*(AB+BC+CD+DE) + +printf("\n MA= %0.2f KN-m",MA) + +// The distance of resultant from A is, + +x=MA/R +printf("\n x= %0.0f m",x) diff --git a/3862/CH2/EX2.15/Ex2_15.sce b/3862/CH2/EX2.15/Ex2_15.sce new file mode 100644 index 000000000..638f5be2a --- /dev/null +++ b/3862/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,27 @@ +clear +// + +//variable declaration +P1=500.0 //Loading at inclined to 60.0°,N +P2=1000.0 //vertical loading at 150 distance from O,N +P3=1200.0 //vertical loading at 150 distance from O,N +H=700.0 //Horizontal loading at 300 ditance from O,N +a=150.0 +theta=60.0*%pi/180 +//assume Resulat R at distance x from O, +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Rx=P1*cos(theta)-H +Ry=-P3-P2-P1*sin(theta) + +R=sqrt((Rx**2)+(Ry**2)) +printf("\n R= %0.2f KN in y-direction",R) + +alpha=atan(Ry/Rx)*180/%pi +printf("\n alpha %0.2f °",alpha) + +//Let the point of application of the resultant be at a distance x from the point O along the horizontal arm. Then, + +x=(P1*sin(theta)*(2*a)+P2*a-P3*a*cos(theta)+H*a*2*sin(theta))/(-Ry) +printf("\n x= %0.3f mm",x) diff --git a/3862/CH2/EX2.16/Ex2_16.sce b/3862/CH2/EX2.16/Ex2_16.sce new file mode 100644 index 000000000..390cdc522 --- /dev/null +++ b/3862/CH2/EX2.16/Ex2_16.sce @@ -0,0 +1,30 @@ +clear +// + +//variable declaration +P1=1120.0 //vertical down Loading at 2m distance from O,KN +P2=120.0 //vertical up loading at 4m distance from O,KN +P3=420.0 //vertical downloading at 5m distance from O,KN +H=500.0 //Horizontal loading at 4m ditance from O,KN +ah=4.0 +a1=2.0 +a2=4.0 +a3=5.0 +a=7.0 +//assume Resulat R at distance x from O, +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Rx=H +Ry=P1-P2+P3 + +printf("\n Ry= %0.2f KN downward",Ry) + +//Let x be the distance from O where the resultant cuts the base. +//moment at O +x=(H*ah+P1*a1-P2*a2+P3*a3)/(Ry) + +printf("\n x= %0.3f m",x) + +printf("\n The resultant passes through the middle third of the base i.e., between 7/3m, and 2*7/3 m.Hence, the dam is safe.") + diff --git a/3862/CH2/EX2.17/Ex2_17.sce b/3862/CH2/EX2.17/Ex2_17.sce new file mode 100644 index 000000000..72db5cc58 --- /dev/null +++ b/3862/CH2/EX2.17/Ex2_17.sce @@ -0,0 +1,31 @@ +clear +// + +//variable declaration +P1=5.0 //Inclined at 45° down Loading at 3m distance from A,KN +P2=10.0 //Inclined at 45° down Loading at 2m distance from A,KN +P3=10.0 //Inclined at 45° down Loading at 1m distance from A,KN +P4=5.0 //Inclined at 45° down Loading A,KN +P8=5.0 //Inclined at 45° UP Loading at 3m distance from A,KN +P7=10.0 //Inclined at 45° UP Loading at 2m distance from A,KN +P6=10.0 //Inclined at 45° UP Loading at 1m distance from A,KN +P5=5.0 //Inclined at 45° UP Loading A,KN +a=1.0 + +theta=45.0*%pi/180.0 +//The roof is inclined at 45° to horizontal and loads are at 90° to the roof. Hence, the loads are also inclined at 45° to vertical/horizontal. + +//assume Resulat R at distance d from A, +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Rx=(P1+P2+P3+P4+P5+P6+P7+P8)*cos(theta) +Ry=-(P1+P2+P3+P4)*sin(theta)+(P5+P6+P7+P8)*sin(theta) + +printf("\n R= %0.3f KN",Rx) +//and its direction is horizontal +//Let R be at a distance d from the ridge A +//moment at A +d=((P1*3*cos(theta)*a+P2*cos(theta)*2*a+P3*cos(theta)*a)*2)/(Rx) + +printf("\n d= %0.1f m \n Resultant is a horizontal force of magnitude %0.3f at %0.1f m below A.",d,Rx,d) diff --git a/3862/CH2/EX2.18/Ex2_18.sce b/3862/CH2/EX2.18/Ex2_18.sce new file mode 100644 index 000000000..9f2b7260d --- /dev/null +++ b/3862/CH2/EX2.18/Ex2_18.sce @@ -0,0 +1,48 @@ +clear +// + +//variable declaration +//The two 40 kN forces acting on the smooth pulley may be replaced by a pair of 40 kN forces acting at centre of pulley C and parallel to the given forces, since the sum of moments of the two given forces about C is zero + +PA=20.0 //inclined at 45° loading at A,KN +PB=30.0 //inclined at 60° loading at B,KN + +PC1=40.0 //inclined at 30° loading at C,KN +PC2=40.0 //inclined at 20° loading at C,KN +PD=50.0 //inclined at 30.0 at distance 2m form A,KN +PE=20.0 //inclined at alpha at distance xm form A,KN +P=20.0 //vertical loading at distance 4m,KN + + + +thetaA=45.0*%pi/180.0 +thetaB=60.0*%pi/180.0 +thetaC1=30.0*%pi/180.0 +thetaC2=20.0*%pi/180.0 +thetaD=30.0*%pi/180.0 +AD=2.0 +AC=3.0 +AB=6.0 + +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Fx=PA*cos(thetaA)-PB*cos(thetaB)-PD*cos(thetaD)-PC1*sin(thetaC1)+PC2*cos(thetaC2) + +Fy=-PA*sin(thetaA)-P+P-PB*sin(thetaB)-PD*sin(thetaD)-PC2*sin(thetaC2)-PC1*cos(thetaC1) + + +R=sqrt((Fx**2)+(Fy**2)) +printf("\n R= %0.2f KN",R) + +alpha=atan(Fy/Fx)*180/%pi +printf("\n alpha= %0.2f °",alpha) + +//Let the resultant intersect AB at a distance x from A. Then, + + +X=(-P*4+P*4+PB*sin(thetaB)*AB+PD*sin(thetaD)*AD-PD*cos(thetaD)*AD+PC2*AC*cos(thetaC2)-PC1*AC*sin(thetaC1))/R + +printf("\n x= %0.2f m",X) + +printf("\n The equilibriant is equal and opposite to the resultant in which E = 116.515 kN, alpha= 76.82° and x= %0.2f m",X) diff --git a/3862/CH2/EX2.19/Ex2_19.sce b/3862/CH2/EX2.19/Ex2_19.sce new file mode 100644 index 000000000..919d71221 --- /dev/null +++ b/3862/CH2/EX2.19/Ex2_19.sce @@ -0,0 +1,13 @@ +clear +// + +//Free body diagram of the sphere shows all the forces moving away from the centre of the ball. Applying Lami’s theorem to the system of forces. + +//variable declaration +W=100.0 //weight of sphere,N +theta=15.0*%pi/180 //angle of inclination of string with wall + +T=(W*sin((%pi/2)))/sin((%pi/2)+theta) +R=(W*sin((%pi-theta)))/sin((%pi/2)+theta) +printf("\n T= %0.2f N",T) +printf("\n R= %0.2f N",R) diff --git a/3862/CH2/EX2.2/Ex2_2.sce b/3862/CH2/EX2.2/Ex2_2.sce new file mode 100644 index 000000000..521430bff --- /dev/null +++ b/3862/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,14 @@ +clear +// +//F force +//hd horizontal distance +//vd vertical distance +//O angle +//M moment of force +//Taking clockwise moment as positive +//calculations +F=5000.0 +o=37 +M=8000.0 +hd=M/(F*cos(o*3.14/180)) +printf("\n %s %.2f %s" ,"\n \n Distance = %0.3f ",hd,"m") diff --git a/3862/CH2/EX2.20/Ex2_20.sce b/3862/CH2/EX2.20/Ex2_20.sce new file mode 100644 index 000000000..a6a5d8eb3 --- /dev/null +++ b/3862/CH2/EX2.20/Ex2_20.sce @@ -0,0 +1,18 @@ +clear +// + +//The body is in equilibrium under the action of applied force P, self-weight 1500 N and normal reaction R from the plane. Since R, which is normal to the plane, makes 30° with the vertical (or 60° with the horizontal), + +//variable declaration +W=1500.0 //weight of block,N +theta=30.0*%pi/180 //angle of inclination + +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +R=W/cos(theta) +printf("\n R= %0.2f N",R) + +P=R*sin(theta) +printf("\n P= %0.2f N",P) diff --git a/3862/CH2/EX2.21/Ex2_21.sce b/3862/CH2/EX2.21/Ex2_21.sce new file mode 100644 index 000000000..e220de538 --- /dev/null +++ b/3862/CH2/EX2.21/Ex2_21.sce @@ -0,0 +1,25 @@ +clear +// + +//A bar can develop a tensile force or a compressive force. Let the force developed be a compressive force S (push on the cylinder). + +//variable declaration +W=10.0 //weight of Roller,KN +IL=7.0 //inclined loading at angle of 45°,KN +H=5.0 //Horizontal loading ,KN + +theta=45.0*%pi/180 //angle of loading of IL +thetaS=30.0*%pi/180.0 + +//Since there are more than three forces in the system, Lami’s equations cannot be applied. Consider the components in horizontal and vertical directions. +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +S=(-H+IL*cos(theta))/cos(thetaS) +printf("\n S= %0.3f N",S) + +printf("\n Since the value of S is negative the force exerted by the bar is not a push, but it is pull (tensile force in bar) of magnitude %0.3f kN.",-S) + +R=W+IL*sin(theta)-S*sin(thetaS) +printf("\n R= %0.3f kN",R) diff --git a/3862/CH2/EX2.22/Ex2_22.sce b/3862/CH2/EX2.22/Ex2_22.sce new file mode 100644 index 000000000..722f6b559 --- /dev/null +++ b/3862/CH2/EX2.22/Ex2_22.sce @@ -0,0 +1,25 @@ +clear +// + +//The pulley C is in equilibrium under the action of tensile forces in CA and CB and vertical downward load 200 N. The tensile forces in segment CA and CB are the same since the pulley is frictionless. Now consider the equilibrium of pulley C +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +//variable declaration +L=200.0 //suspended load at C,N +AB=3.0 +BI=1.0 +ACB=5.0 //Length of cord,m +DE=3.0 +BE=4.0 +theta=asin(4.0/5.0) +//assume T is tension in string making angle theta1 & theta2,solving horizontal we find theta1=theta2,lets called them theta ,as triangleCFD=triangle=CFA.so, CD=AC + +HI=BI*DE/BE +AH=DE-HI +x=AH/2 +printf("\n x= %0.3f m",x) + +T=L/(2*sin(theta)) +printf("\n T= %0.0f N",T) diff --git a/3862/CH2/EX2.23/Ex2_23.sce b/3862/CH2/EX2.23/Ex2_23.sce new file mode 100644 index 000000000..2615e9871 --- /dev/null +++ b/3862/CH2/EX2.23/Ex2_23.sce @@ -0,0 +1,31 @@ +clear +// + +//When the roller is about to turn over the curb, the contact with the floor is lost and hence there is no reaction from the floor. The reaction R from the curb must pass through the intersection of P and the line of action of self weight, since the body is in equilibrium under the action of only three forces (all the three forces must be concurrent). + +//variable declaration +W=2000.0 //weight of roller,N +r=300.0 //radius of roller,mm +h=150.0 // height of curb,mm +OC=r-h +AO=r + +alpha=acos(OC/AO) + +//angleOAB=angleOBA,Since OA=OB, +angleOBA=(alpha)/2 + +//the reaction makes 30° with the vertical +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +R=W/cos(angleOBA) +P=R*sin(angleOBA) + +printf("\n P= %0.2f N",P) + +//Least force through the centre of wheel: Now the reaction from the curb must pass through the centre of the wheel since the other two forces pass through that point. Its inclination to vertical is theta = 60°. If the triangle of forces ABC representing selfweight by AB, reaction R by BC and pull P by AC, it may be observed that AC to be least, it should be perpendicular to BC. In other words, P makes 90° with the line of action of R. +//From triangle of forces ABC, we get +P=W*sin(alpha) +printf("\n P= %0.2f N",P) diff --git a/3862/CH2/EX2.24/Ex2_24.sce b/3862/CH2/EX2.24/Ex2_24.sce new file mode 100644 index 000000000..1d8eca541 --- /dev/null +++ b/3862/CH2/EX2.24/Ex2_24.sce @@ -0,0 +1,26 @@ +clear +// + +//variable declaration +PB=200.0 //Vertical loading at B,N +PD=250.0 //Vertical loading at D,N +thetabc=30.0*%pi/180.0 +thetabd=60.0*%pi/180.0 +thetaed=45.0*%pi/180.0 +//Free body diagrams of points B and D . Let the forces in the members be as shown in the figure. Applying Lami’s theorem to the system of forces at point D, + +T1=PD*sin(%pi-thetabd)/sin(thetaed+(%pi/2)-thetabd) +T2=PD*sin(%pi-thetaed)/sin(thetaed+(%pi/2)-thetabd) + +printf("\n T1= %0.2f N",T1) +printf("\n T2= %0.2f N",T2) + +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +T3=(PB+T2*cos(thetabd))/cos(thetabc) +printf("\n T3= %0.2f N",T3) + +T4=(T2*sin(thetabd))+T3*sin(thetabc) +printf("\n T4= %0.2f N",T4) diff --git a/3862/CH2/EX2.25/Ex2_25.sce b/3862/CH2/EX2.25/Ex2_25.sce new file mode 100644 index 000000000..48941d59c --- /dev/null +++ b/3862/CH2/EX2.25/Ex2_25.sce @@ -0,0 +1,26 @@ +clear +// + +//variable declaration + +PC=1500.0 //Vertical loading at C,N +CD=2.0 +AC=1.5 +BD=1.0 +AB=4.0 + +x=(((AC**2)-(BD**2))/4)+1 +y=sqrt((AC**2)-(x**2)) + +alpha=acos(x/AC) +beta1=acos((CD-x)/BD) + +//Applying Lami’s theorem to the system of forces acting at point C + +T1=PC*sin(%pi/2)/sin(%pi-alpha) +T2=PC*sin((%pi/2)+alpha)/sin(%pi-alpha) +T3=T2*sin(%pi/2)/sin((%pi/2)+beta1) +W=T2*sin(%pi-beta1)/sin((%pi/2)+beta1) + + +printf("\n W= %0.2f N",W) diff --git a/3862/CH2/EX2.26/Ex2_26.sce b/3862/CH2/EX2.26/Ex2_26.sce new file mode 100644 index 000000000..b43586981 --- /dev/null +++ b/3862/CH2/EX2.26/Ex2_26.sce @@ -0,0 +1,28 @@ +clear +// + +//variable declaration + +PB=20.0 //vertical loadng at point B,KN +PC=30.0 //vertical loadng at point C,KN + +thetaab=30.0 *%pi/180.0 +thetabc=50.0*%pi/180.0 + +//applying lami's thereom + +T1=PB*sin(thetabc)/sin(%pi-thetabc+thetaab) +T2=PB*sin(%pi-thetaab)/sin(%pi-thetabc+thetaab) +theta=atan((T2*sin(thetabc))/(PC-T2*cos(thetabc)))*180/%pi + + +printf("\n T1= %0.2f KN",T1) + +printf("\n T2= %0.2f KN",T2) + +//Writing equations of equilibrium for the system of forces at C + +printf("\n theta= %0.2f °",theta) + +T3=(PC-T2*cos(thetabc))/cos(theta*%pi/180) +printf("\n T3= %0.2f KN",T3) diff --git a/3862/CH2/EX2.27/Ex2_27.sce b/3862/CH2/EX2.27/Ex2_27.sce new file mode 100644 index 000000000..727450cfb --- /dev/null +++ b/3862/CH2/EX2.27/Ex2_27.sce @@ -0,0 +1,26 @@ +clear +// + +//variable declaration + +PB=20.0 //vertical loadng at point B,KN + +PC=25.0 //vertical loadng at point C,KN + +thetaab=30.0*%pi/180.0 +thetadc=60.0*%pi/180.0 + +//Writing equations of equilibrium for the system of forces at joints B and C +//T1*sin(thetaab)=T3*sin(thetadc) + +T3=(PB+PC)/((sin(thetadc)*cos(thetaab)/sin(thetaab))+cos(thetadc)) +printf("\n T3= %0.2f KN",T3) + +T1=T3*sin(thetadc)/sin(thetaab) +printf("\n T1= %0.2f KN",T1) + +theta=(atan((T3*sin(thetadc))/(PC-T3*cos(thetadc))))*180/%pi +printf("\n theta= %0.2f °",theta) + +T2=T3*sin(thetadc)/(sin(theta*%pi/180)) +printf("\n T2= %0.2f KN",T2) diff --git a/3862/CH2/EX2.28/Ex2_28.sce b/3862/CH2/EX2.28/Ex2_28.sce new file mode 100644 index 000000000..dda74db1d --- /dev/null +++ b/3862/CH2/EX2.28/Ex2_28.sce @@ -0,0 +1,28 @@ +clear +// + +//variable declaration +W=600.0 //weight of cyclinder,N +r=150.0 //radius of cylinder,mm +a=600.0 //mm +b=300.0 //mm + +//Free body diagram of sphere and frame + +////sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +RB=600.0 +//As the frame is in equilibrium under the action of three forces only, they must be concurrent forces. In other words, reaction at D has line of action alone OD. Hence, its inclination to horizontal is given by: +printf("\n RB= %0.2f N",RB) +alpha=atan((a-r)/r) +printf("\n alpha= %0.4f °",alpha) + +RD=W/sin(alpha) +printf("\n RD= %0.3f N",RD) + +RC=RD*cos(alpha) +RA=RC +printf("\n RC= %0.0f N",RC) +printf("\n RA= %0.0f N",RA) diff --git a/3862/CH2/EX2.29/Ex2_29.sce b/3862/CH2/EX2.29/Ex2_29.sce new file mode 100644 index 000000000..21a9df515 --- /dev/null +++ b/3862/CH2/EX2.29/Ex2_29.sce @@ -0,0 +1,39 @@ +clear +// + + +// Let O1 and O2 be the centres of the first and second spheres. Drop perpendicular O1P to the horizontal line through O2. show free body diagram of the sphere 1 and 2, respectively. Since the surface of contact are smooth, reaction of B is in the radial direction, i.e., in the direction O1O2. Let it make angle a with the horizontal. Then, + +//Variable declaration + +W=100.0 //weight of spheres,N + +r=100.0 //radius of spheres,mm + +d=360.0 // horizontal channel having vertical walls, the distance b/w,mm + +O1A=100.0 +O2D=100.0 +O1B=100.0 +BO2=100.0 + +O2P=360.0-O1A-O2D +O1O2=O1B+BO2 + +alpha=acos(O2P/O1O2) + +//////sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +RB=W/sin(alpha) +RA=RB*cos(alpha) +printf("\n RB= %0.2f N",RB) +printf("\n RA= %0.2f N",RA) + +RC=100+RB*sin(alpha) + +RD=RB*cos(alpha) + +printf("\n RC= %0.0f N",RC) + +printf("\n RD= %0.2f N",RD) diff --git a/3862/CH2/EX2.30/Ex2_30.sce b/3862/CH2/EX2.30/Ex2_30.sce new file mode 100644 index 000000000..9711396bc --- /dev/null +++ b/3862/CH2/EX2.30/Ex2_30.sce @@ -0,0 +1,22 @@ +clear +// + +// Two cylinders, A of weight 4000 N and B of weight 2000 N rest on smooth inclines. They are connected by a bar of negligible weight hinged to each cylinder at its geometric centre by smooth pins + +//variable declaration + +WA=4000.0 //weight of cylinder A,N +WB=2000.0 //weight of cylinder B,N + +thetaWA=60.0*%pi/180.0 //inclination of wall with cylinderA,° +thetaWB=45.0*%pi/180.0 //inclination of wall with cylinderB,° +thetaAb=15.0*%pi/180.0 //angle inclination bar with cylinder A ,N +thetaBb=15.0*%pi/180.0 //angle inclination bar with cylinder B ,N + +//he free body diagram of the two cylinders. Applying Lami’s theorem to the system of forces on cylinder A, we get + +C=WA*sin(thetaWA)/sin(thetaWA+(%pi/2)-thetaAb) + +//Consider cylinder B. Summation of the forces parallel to the inclined plane +P=(-WB*cos(thetaWB)+C*cos(thetaWA))/cos(thetaBb) +printf("\n P= %0.1f N",P) diff --git a/3862/CH2/EX2.32/Ex2_32.sce b/3862/CH2/EX2.32/Ex2_32.sce new file mode 100644 index 000000000..8abd82138 --- /dev/null +++ b/3862/CH2/EX2.32/Ex2_32.sce @@ -0,0 +1,22 @@ +clear +// + +//variable declaration + +//A cable car used for carrying materials in a hydroelectric project is at rest on a track formed at an angle of 30° with the vertical. The gross weight of the car and its load is 60 kN and its centroid is at a point 800 mm from the track half way between the axles. The car is held by a cable . The axles of the car are at a distance 1.2 m. Find the tension in the cables and reaction at each of the axles neglecting friction of the track. + +W=60.0 //gross weight of car,KN +theta=60.0*%pi/180.0 + + +T=W*sin(theta) +printf("\n T= %0.4f KN",T) + +//Taking moment equilibrium condition about upper axle point on track, we get + +R1=(-T*600.0+W*sin(theta)*800.0+W*cos(theta)*600.0)/1200.0 +printf("\n R1= %0.4f KN",R1) + +//Sum of forces normal to the plane = 0, gives +R2=W*cos(theta)-R1 +printf("\n R2= %0.4f KN",R2) diff --git a/3862/CH2/EX2.33/Ex2_33.sce b/3862/CH2/EX2.33/Ex2_33.sce new file mode 100644 index 000000000..b2a9df1a9 --- /dev/null +++ b/3862/CH2/EX2.33/Ex2_33.sce @@ -0,0 +1,34 @@ +clear +// + +// A hollow right circular cylinder of radius 800 mm is open at both ends and rests on a smooth horizontal plane. Inside the cylinder there are two spheres having weights 1 kN and 3 kN and radii 400 mm and 600 mm, respectively. The lower sphere also rests on the horizontal plane. +// Join the centres of spheres, O1 and O2 and drop O1D perpendicular to horizontal through O2. + +//variable declaration +R=800.0 +W1=1.0 +r1=400.0 +W2=3.0 +r2=600.0 +O1O2=1000 //mm +O2D=600 //mm + +//If alpha is the inclination of O2O1 to horizontal +alpha=acos(O2D/O1O2) + +//Free body diagrams of cylinder and spheres are shown. Considering the equilibrium of the spheres. +//Sum of Moment at O2 + +R1=W1*O2D/(O1O2*sin(alpha)) +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +R2=R1 +R3=W1+W2 + +//Sum of Moment at A + +W=R1*O1O2*sin(alpha)/R + +printf("\n W= %0.2f KN",W) diff --git a/3862/CH2/EX2.4/Ex2_4.sce b/3862/CH2/EX2.4/Ex2_4.sce new file mode 100644 index 000000000..24953b2e7 --- /dev/null +++ b/3862/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,24 @@ +clear +// +//R resultant force +//Rx resultant horizontal component +//Ry resultant vertical component +//f1 force +//f2 force +//f3 force +//o1 angle with the line +//o2 angle with the line +//o3 angle with the line +//O angle of resultant force with line +f1=70.0 +f2=80.0 +f3=50.0 +o1=50.0 +o2=25.0 +o3=-45.0 +Rx=(f1*cos(o1/180*3.14)+f2*cos(o2/180*3.14)+f3*cos(o3/180*3.14)) +Ry=(f1*sin(o1/180*3.14)+f2*sin(o2/180*3.14)+f3*sin(o3/180*3.14)) +R=sqrt(Rx**2+Ry**2) +O=atand(Ry/Rx) +printf("\n %s %.2f %s" ,"\n \n Resultant Force = %0.3f ",R,"N") +printf("\n %s %.2f %s" ,"\n \n Resultant angle = %0.3f ",O,"degrees") diff --git a/3862/CH2/EX2.6/Ex2_6.sce b/3862/CH2/EX2.6/Ex2_6.sce new file mode 100644 index 000000000..45a634e53 --- /dev/null +++ b/3862/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,14 @@ +clear +// +R=1000.0 //Resultant force +F1=500.0 //Force +F2=1000.0 //force +o=45.0*3.14/180.0 //angle resultant makes with x axis +o1=30.0*3.14/180.0 //angle F1 makes with x axis +o2=60.0*3.14/180.0 //angle F2 makes with x axis +//F3coso3=Rcoso-F1coso1-F2sino2 +//F3sino=Rsino-F1sino1-F2coso2 +F3=((R*cos(o)-F1*cos(o1)-F2*cos(o2))**2+(R*sin(o)-F1*sin(o1)-F2*sin(o2))**2)**0.5 +printf("\n Force %0.3f N",F3) +o3=180/3.14*atan((R*sin(o)-F1*sin(o1)-F2*sin(o2))/(R*cos(o)-F1*cos(o1)-F2*cos(o2))) +printf("\n At an angle %0.3f degrees",o3) diff --git a/3862/CH2/EX2.7/Ex2_7.sce b/3862/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..7983a3544 --- /dev/null +++ b/3862/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,20 @@ +clear +// + +//variable declaration + +P1=300.0 +P2=500.0 +thetaI=30.0*%pi/180.0 +thetaP2=30.0*%pi/180 +thetaP1=40.0*%pi/180 +// Let the x and y axes be If the resultant is directed along the x axis, its component in y direction is zero. +//Taking horizontal direction towards left as x axis and the vertical downward direction as y axis. +////sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +F=(P2*sin(thetaP2))/(P1) +theta=(asin((F/(cos(20*%pi/180)*2)))*180/%pi)-20 + +printf("\n theta= %0.2f °",theta) diff --git a/3862/CH2/EX2.8/Ex2_8.sce b/3862/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..6f0d373be --- /dev/null +++ b/3862/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,34 @@ +clear +// + +//variable declaration + +P1=20.0 +P2=30.0 +P3=20.0 +theta3=60.0*%pi/180.0 + +//Taking horizontal direction towards left as x axis and the vertical downward direction as y axis. +////sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +Fx=20.0*cos(theta3) +Fy=P1+P2+P3*sin(theta3) + + +R=sqrt((Fx**2)+(Fy**2)) +printf("\n R= %0.4f KN",R) + +alpha=atan(Fy/Fx)*180/%pi +printf("\n alpha= %0.2f °",alpha) + +//moment at A + +MA=P1*1.5+P2*3.0+P3*sin(theta3)*6.0 + +//The distance of the resultant from point O is given by: + +d=MA/R +x=d/sin(alpha*%pi/180) +printf("\n x= %0.3f m",x) diff --git a/3862/CH2/EX2.9/Ex2_9.sce b/3862/CH2/EX2.9/Ex2_9.sce new file mode 100644 index 000000000..4c53b8a90 --- /dev/null +++ b/3862/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,37 @@ +clear +// + +//variable declaration + +PA=100.0 //inclined up loading at 60° at A, N +PB1=80.0 //Vertical down loading at B,N +PB2=80.0 //Horizontal right loading at at B,N +PC=120.0 //inclined down loading at 30° at C,N + +thetaA=60.0*%pi/180.0 +thetaB=30.0*%pi/180.0 + + + +//Taking horizontal direction towards left as x axis and the vertical downward direction as y axis. +////sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up + +Fx=PB2-PA*cos(thetaA)-PC*cos(thetaB) +Rx=-Fx + +Fy=PB1+PC*sin(thetaB)-PA*sin(thetaA) +Ry=Fy + + +R=sqrt((Rx**2)+(Ry**2)) +printf("\n R= %0.2f KN",R) + +alpha=atan(Fy/Fx)*180/%pi +printf("\n alpha= %0.2f °",(-alpha)) + +//Let x be the distance from A at which the resultant cuts AC. Then taking A as moment centre, + +x=(PB1*100*sin(thetaA)+PB2*50+PC*sin(thetaB)*100)/Ry +printf("\n x= %0.3f mm",x) diff --git a/3862/CH3/EX3.1/Ex3_1.sce b/3862/CH3/EX3.1/Ex3_1.sce new file mode 100644 index 000000000..9dcac0f34 --- /dev/null +++ b/3862/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,44 @@ +clear +// + +//variable declaration + +//Determine the inclinations of all inclined members + +theta=atan(1)*180/%pi + +printf("\n theta= %0.2f °",theta) + +//Now at joints C, there are only two unknowns,forces in members CB and CD, say FCB and FCD. +//Now there are two equations of equilibrium for the forces meeting at the joint and two unknown forces. Hence, the unknown forces can be determined. At joint C sum V= 0 condition shows that the force FCB should act away from the joint C so that its vertical component balances the vertical downward load at C. + +P=40.0 +FCB=P/sin(theta*%pi/180) + +printf("\n FCB= %0.2f KN",FCB) + +//Now sum H=0 indicates that FCD should act towards C. + +FCD=FCB*cos(theta*%pi/180) + +printf("\n FCD= %0.2f KN",FCD) + +//In the present case, near the joint C, the arrows are marked on the members CB and CD to indicate forces FCB and FCD directions as found in the analysis of joint C. Then reversed directions are marked in the members CB and CD near joints B and D, respectively. + +FDB=40.0 +FDE=40.0 + +printf("\n FDB= %0.2f KN",FDB) + +printf("\n FDE= %0.2f KN",FDE) + +//In the present case, after marking the forces in the members DB and DE, we find that analysis of joint B can be taken up. + +FBE=(FCB*sin(theta*%pi/180)+P)/(sin(theta*%pi/180)) + +FBA=FCB*cos(theta*%pi/180)+FBE*cos(theta*%pi/180) + +printf("\n FBE= %0.2f KN",FBE) +printf("\n FBA= %0.2f KN",FBA) +//Determine the nature of forces in each member and tabulate the results. Note that if the arrow marks on a member are towards each other, then the member is in tension and if the arrow marks are away from each other, the member is in compression. + diff --git a/3862/CH3/EX3.10/Ex3_10.sce b/3862/CH3/EX3.10/Ex3_10.sce new file mode 100644 index 000000000..da5656eff --- /dev/null +++ b/3862/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,36 @@ +clear +// + +//Each load is 20 kN. + +//variable declaration + +P=20.0 +AB=18.0 +A=3.0 + +RA=P*7/2 +RB=RA + +theta1=30.0*%pi/180 +a=(3*A)/(4*cos(theta1)) +//Take Section (A)–(A) and consider the equilibrium of left hand side part of the French Truss +//Drop perpendicular CE on AB. + +CE=3*A*tan(theta1) +DE=A + +theta=atan(CE/DE)*180/%pi +printf("\n theta= %0.0f °",theta) + +//moment at point A + +F2=(P*a*cos(theta1)*6)/(A*2*sin(theta*%pi/180)) +printf("\n F2= %0.4f KN (Tension)", F2) + +//sum of all vertical forces & sum of all horizotal forces is zero +F1=(F2*sin(theta*%pi/180)+RA-P*3)/(sin(theta1)) +printf("\n F1= %0.4f KN (Comp)",F1) + +F3=F1*cos(theta1)-F2*cos(theta*%pi/180) +printf("\n F3= %0.4f KN (Tension)",F3) diff --git a/3862/CH3/EX3.11/Ex3_11.sce b/3862/CH3/EX3.11/Ex3_11.sce new file mode 100644 index 000000000..31e1fca85 --- /dev/null +++ b/3862/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,60 @@ +clear +// + +//variable declaration + +PA=15.0 //vertical loading at point A,KN +PB=30.0 //vertical loading at point B,KN +PC=30.0 //vertical loading at point C,KN +PD=30.0 //vertical loading at point D,KN +PE=15.0 //vertical loading at point E,KN + +//Due to symmetry, the reactions are equal +RA=(PA+PB+PC+PD+PE)/2 +RB=RA +//Drop perpendicular CH on AF. +//in traingle ACH + +angleACH=45.0*%pi/180 //angleACH,° +angleFCV=30.0*%pi/180 // FC is inclined at 30° to vertical i.e., 60° to horizontal and CH = 5 m +CH=5.0 +angleFCH=60.0*%pi/180 + +//It is not possible to find a joint where there are only two unknowns. Hence, consider section (1)–(1). +//For left hand side part of the frame +//moment at C + +FAE=(RA*CH-PA*CH-PB*CH/2)/(CH) +printf("\n FAE= %0.0f KN (Tension)",FAE) + +//Assuming the directions for FFC and FBC +//sum of vertical & sum of horizontal forces is zero + +//FFC=FBC*sqrt(2)-RA + +FBC=(RA*sin(angleFCH)-PA)/(sqrt(2)*sin(angleFCH)-(1/sqrt(2))) +printf("\n FBC= %0.2f KN (Comp.)",FBC) + +FFC=FBC*sqrt(2)-RA +printf("\n FFC= %0.2f KN (Tension)",FFC) + +//Assumed directions of FBC and FFC are correct. Therefore, FBC is in compression and FFC is in tension. Now we can proceed with method of joints to find the forces in other members. Since it is a symmetric truss, analysis of half the truss is sufficient. Other values may be written down by making use of symmetrry. + +//Joint B: sum of forces normal to AC = 0, gives + +FBF=PC*cos(angleACH) + +//sum of forces parallel to AC = 0, gives + +FAB=FBC+PC*sin(angleACH) + +printf("\n FAB= %0.2f KN (Comp.)",FAB) + + + +//JOINT A +//sum of vertical & sum of horizontal forces is zero + +FAF=(FAB*sin(angleACH)+PA-RA)/sin(angleFCV) + +printf("\n FAF= %0.2f KN (Tension)",FAF) diff --git a/3862/CH3/EX3.2/Ex3_2.sce b/3862/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..91f36f6f2 --- /dev/null +++ b/3862/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,42 @@ +clear +// + +//Now, we cannot find a joint with only two unknown forces without finding reactions. +//Consider the equilibrium of the entire frame,Sum of moments about A is zero,Horizontal forces & Vertical forces is zero. + +//variable declaration + +PB=40.0 +PC=50.0 +PE=60.0 + +theta=60.0 + +RD=(PC*3+PE*2+PB*1)/(4.0) + +RA=PB+PC+PE-RD + +FAB=RA/sin(theta*%pi/180) + +printf("\n FAB= %0.4f KN (Comp.)",FAB) + +FAE=FAB*cos(theta*%pi/180) + +printf("\n FAE= %0.4f KN (Tension)",FAE) + +FDC=RD/sin(theta*%pi/180) + +printf("\n FDC= %0.4f KN (Comp.) ",FDC) + +FDE=FDC*cos(theta*%pi/180) + +printf("\n FDE= %0.4f KN (Tension) ",FDE) + +FBE=(FAB*sin(theta*%pi/180)-PB)/sin(theta*%pi/180) + +FBC=(FAB+FBE)*(0.5) +printf("\n FBC= %0.4f KN (Comp.)",FBC) + + +FCE=(FDC*sin(theta*%pi/180)-PC)/(sin(theta*%pi/180)) +printf("\n FCE= %0.4f KN (Tension)",FCE) diff --git a/3862/CH3/EX3.3/Ex3_3.sce b/3862/CH3/EX3.3/Ex3_3.sce new file mode 100644 index 000000000..9c0f84347 --- /dev/null +++ b/3862/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,53 @@ +clear +// + +//variable declaration + +PB=20.0 //Load at point B,KN +PC=10.0 //Load at point C,KN +thetaA=60.0 //angleBAC +thetaD=30.0 //angleBDC + +AC=3.0 //length,m +CD=3.0 //length,m + +AB=(AC+CD)*cos(thetaA*%pi/180) +BD=(AC+CD)*cos(thetaD*%pi/180) +//mistake in book +//angleBCA=angleABC=theta + +theta=(180.0-thetaA)/(2.0) + +//Taking moment about A, we get +RD=(PC*AC+PB*AC*cos(thetaA*%pi/180))/(AC+CD) + +RA=PC+PB-RD +//Joint A +//vertical & horizontal forces sum to zero + +FAB=RA/sin(thetaA*%pi/180) + +printf("\n FAB= %0.2f KN [Comp.]",FAB) +FAC=FAB*cos(thetaA*%pi/180) +printf("\n FAC= %0.2f KN [Tensile]",FAC) + +//Joint D +//vertical & horizontal forces sum to zero + +FDB=RD/sin(thetaD*%pi/180) + +printf("\n FDB= %0.2f KN [Comp.]",FDB) +FDC=FDB*cos(thetaD*%pi/180) +printf("\n FDC= %0.2f KN [Tensile]",FDC) + +//Joint C +//vertical & horizontal forces sum to zero + +FCB=PC/sin(theta*%pi/180) + +printf("\n FCB= %0.2f KN ",FCB) + +//CHECK + +FCB=(FDC-FAC)/cos(theta*%pi/180) +printf("\n FCB= %0.2f KN Checked",FCB) diff --git a/3862/CH3/EX3.4/Ex3_4.sce b/3862/CH3/EX3.4/Ex3_4.sce new file mode 100644 index 000000000..6f799932b --- /dev/null +++ b/3862/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,68 @@ +clear +// + +//Now, we cannot find a joint with only two unknown forces without finding reactions. +//Consider the equilibrium of the entire frame,Sum of moments about A is zero,Horizontal forces & Vertical forces is zero. + +//variable declaration + +PB=30.0 //vertical load at point B,KN +PC=50.0 //vertical load at point C,KN +PDv=40.0 //vertical load at point D,KN +PDh=20.0 //Horizontal load at point D,KN +PF=30.0 //vertical load at point F,KN +HA=PDh + +RE=(PC*4+PDv*8+PDh*4+PF*4)/(8.0) + +VA=PB+PC+PDv+PF-RE + +//joint A +//sum of vertical & sum of horizontal forces is zero. + +FAB=VA +FAF=HA + +//joint E +//sum of vertical & sum of horizontal forces is zero. + +FED=RE +FEF=0 + +//Joint B: Noting that inclined member is at 45° +//sum of vertical & sum of horizontal forces is zero. + +theta=45.0 +FBF=(VA-PB)/sin(theta*%pi/180) + +printf("\n FBF= %0.4f KN (Tension) ",FBF) + +FBC=FBF*cos(theta*%pi/180) + +printf("\n FBC= %0.4f KN (Comp.) ",FBC) + +//Joint C: +//sum of vertical & sum of horizontal forces is zero. + + +FCF=PC + +printf("\n FCF= %0.4f KN (Comp.) ",FCF) + +FCD=FBC + +printf("\n FCD= %0.4f KN (Comp.) ",FCD) + +//Joint D: Noting that inclined member is at 45° +//sum of vertical & sum of horizontal forces is zero. + +theta=45.0 +FDF=(RE-PDv)/cos(theta*%pi/180) + +printf("\n FDF= %0.4f KN (Tensile) ",FDF) + +//check + +FDF=(FCD+PDh)/cos(theta*%pi/180) + +printf("\n FDF= %0.4f KN Checked ",FDF) diff --git a/3862/CH3/EX3.5/Ex3_5.sce b/3862/CH3/EX3.5/Ex3_5.sce new file mode 100644 index 000000000..7868fe5bf --- /dev/null +++ b/3862/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,74 @@ +clear +// + +//All inclined members have the same inclination to horizontal. Now, length of an inclined member is BF + +//variable declaration + +PE=20.0 +AF=3.0 +FE=3.0 +AB=4.0 +FD=4.0 +BD=3.0 +CD=4.0 + +BF=sqrt((AF**2)+(AB**2)) +DE=BF +BC=DE + +//sin(theta)=AB/BF +//cos(theta)=AF/BF + +theta=asin(AB/BF) +//As soon as a joint is analysed the forces on the joint are marked on members + +//Joint E +//Consider the equilibrium of the entire frame,Sum of moments about A is zero,Horizontal forces & Vertical forces is zero. + +FED=PE/sin(theta) +printf("\n FED= %0.0f KN (Tension)",FED) + +FEF=FED*cos(theta) +printf("\n FEF= %0.0f KN (Comp.)",FEF) + +//At this stage as no other joint is having only two unknowns, no further progress is possible. Let us find the reactions at the supports considering the whole structure. Let the reaction be RC HORIZONTAL at point C,VA,HA at point A Vertically & Horizontally respectively. +//Taking moment at point A, + +RC=PE*6/8 +//sum of vertical & sun of horizontal forces is zero. + +VA=PE +HA=RC + +//Joint A +//sum of vertical & sun of horizontal forces is zero. +FAB=VA +printf("\n FAB= %0.0f KN (Comp.)",FAB) + +FAF=HA +printf("\n FAF= %0.0f KN (Comp.)",FAF) + +//Joint C +//sum of vertical & sun of horizontal forces is zero. +FCB=RC/cos(theta) +printf("\n FCB= %0.0f KN (Comp.)",FCB) + +FCD=FCB*sin(theta) +printf("\n FCD= %0.0f KN (Tension)",FCD) + +//Joint B +//sum of vertical & sun of horizontal forces is zero. + +FBF=(FCB*sin(theta)-FAB)/sin(theta) + +printf("\n FBF= %e ",FBF) + +FBD=FCB*cos(theta) +printf("\n FBD= %0.0f KN (Tension)",FBD) + +//joint F +//sum of vertical & sun of horizontal forces is zero. + +FFD=FBF +printf("\n FFD= %e ",FFD) diff --git a/3862/CH3/EX3.6/Ex3_6.sce b/3862/CH3/EX3.6/Ex3_6.sce new file mode 100644 index 000000000..7f1930dc4 --- /dev/null +++ b/3862/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,86 @@ +clear +// + +//variable declaration + +AB=2.0 //length of beam AB,m +BD=2.0 //length of beam BD,m +DF=2.0 //length of beam DF,m +FH=3.0 //length of beam FH,m +FG=4.0 //length of beam FG,m +PF=12.0 //Vertical Load at point F,KN +PH=20.0 //Vertical Load at point H,KN + +//mistake in book FG=4.0 , given FG=2.0 + +theta1=atan(FG/(AB+BD+DF)) +theta3=atan(FG/FH) +theta2=theta3 + +//sum of all vertical forces & sum of all horizotal forces is zero + +//joint H + +FHG=PH/sin(theta3) +printf("\n FHG= %0.0f KN (Comp.)",FHG) + +FHF=FHG*cos(theta2) +printf("\n FHF= %0.0f KN (Tension)",FHF) + +//taking moment at G + +RA=PH*FH/(AB+BD+DF) + +RG=RA+PF+PH + +//joint A +//sum of all vertical forces & sum of all horizotal forces is zero + +FAC=RA/sin(theta1) +printf("\n FAC= %0.4f KN (Comp.)",FAC) + +FAB=FAC*cos(theta1) +printf("\n FAB= %0.0f KN (Tension)",FAB) + +//joint B +//sum of all vertical forces & sum of all horizotal forces is zero + +FBC=0 +printf("\n FBC= %e ",FBC) +FBA=FAB +FBD=FBA +printf("\n FBD=FBA %0.0f KN (Tension)",FBD) + +//Joint C: Sum of Forces normal to AC = 0, gives FCD =0 since FBC = 0 ,sum of Forces parallel to CE =0 + +FCA=FAC +FCE=FCA +printf("\n FCE=FCA %0.4f KN (Comp.)",FCE) + + +//joint D +//sum of all vertical forces & sum of all horizotal forces is zero + +FDE=0 +printf("\n FDE= %e ",FDE) + +FDB=FBD +FDF=FDB + +printf("\n FDF=FDB %0.0f KN (Tension)",FDF) + +//Joint E: sum of Forces normal to CG = 0, gives FEF = 0 and sum of Forces in the direction of CG = 0, gives + +FEF=0 + + +FEG=FCE + +printf("\n FEG=FCE= %0.4f KN (Comp.)",FEG) + +//Joint F: +//sum of all vertical forces & sum of all horizotal forces is zero + +FFG=PF + +printf("\n FFG= %0.0f KN (Tension)",FFG) diff --git a/3862/CH3/EX3.7/Ex3_7.sce b/3862/CH3/EX3.7/Ex3_7.sce new file mode 100644 index 000000000..f3534f490 --- /dev/null +++ b/3862/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,99 @@ +clear +// + +// Since all members are 3 m long, all triangles are equilateral and hence all inclined members are at 60° to horizontal. Joint-by-joint analysis is carried out . Then nature of the force is determined. + +//variable declaration + +AB=3.0 +BC=AB +AC=AB +BD=BC +CD=BD +CE=CD +DE=CE +EF=DE +DF=DE +EG=DE +FG=DF + +theta=60.0*%pi/180 //angles BAC,BCA,DCE,DEC,FEG,FGE,° + +PB=40.0 //Vertical Loading at point B,KN +PD=30.0 //Vertical Loading at point D,KN +HF=10.0 //Horizontal Loading at point F,KN +PG=20.0 //Vertical Loading at point G,KN + +//joint G +//sum of all vertical forces & sum of all horizotal forces is zero + +FGF=PG/sin(theta) + +printf("\n FGF= %0.4f KN (Tension)",FGF) + +FGE=FGF*cos(theta) + +printf("\n FGE= %0.4f KN (Comp.)",FGE) + +//joint F + +//sum of all vertical forces & sum of all horizotal forces is zero + +FFG=FGF + +printf("\n FFG= %0.4f KN (Comp.)",FFG) + +FFE=FGF +FFD=FGF*cos(theta)+FFE*cos(theta)-HF +printf("\n FFD= %0.4f KN (Tension)",FFD) + +//Now, without finding reaction we cannot proceed. Hence, consider equilibrium of the entire truss +//moment about point A + +RE=((PB*AC/2)-(HF*EF*sin(theta))+(PD*(AC+CE/2))+(PG*(AC+CE+EG)))/(AC+CE) + +VA=PB+PD+PG-RE + +HA=HF + +//joint A +//sum of all vertical forces & sum of all horizotal forces is zero + +FAB=VA/sin(theta) + +printf("\n FAB= %0.4f KN (Comp.)",FAB) + +FAC=FAB*cos(theta)-HF + +printf("\n FAC= %0.4f KN (Tension)",FAC) + + +//joint B +//sum of all vertical forces & sum of all horizotal forces is zero + +FBC=(PB-FAB*sin(theta))/sin(theta) + +printf("\n FBC= %0.4f KN (Comp.)",FBC) + +FBA=FAB +FBD=-FBC*cos(theta)+FBA*cos(theta) + +printf("\n FBD= %0.4f KN (Comp.)",FBD) + +//joint C +//sum of all vertical forces & sum of all horizotal forces is zero + +FCD=FBC*sin(theta)/sin(theta) + +printf("\n FCD= %0.4f KN (Tension)",FCD) + +FCE=FCD*cos(theta)+FBC*cos(theta)-FAC + +printf("\n FCE= %0.4f KN (Comp.)",FCE) + +//joint D +//sum of all vertical forces & sum of all horizotal forces is zero + +FDE=(PD+FCD*sin(theta))/sin(theta) + +printf("\n FDE= %0.4f KN (Comp.)",FDE) diff --git a/3862/CH3/EX3.8/Ex3_8.sce b/3862/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..acf2e8f75 --- /dev/null +++ b/3862/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,40 @@ +clear +// + +//Each load is 10 kN and all triangles are equilateral with sides 4 m. + +//variable declaration + +PB=10.0 +PD=PB +PF=PD +AB=4.0 +BC=AB +AC=BC +BD=BC +CD=BC +DE=CD +CE=CD +DF=DE +EF=DE +EG=DE +FG=EF +//Take section (A)–(A), which cuts the members FH, GH and GI and separates the truss into two parts. +AG=AC+CE+EG +BG=CE+EG+AC/2 +DG=EG+CE/2 +FG1=EG/2 +RA=PB*7/2 +RO=RA +theta=60.0*%pi/180 +//moment at point G +FFH=(RA*AG-PB*BG-PD*DG-PF*FG1)/(FG*sin(theta)) +printf("\n FFH= %0.4f KN (Comp.)",FFH) + +//sum of all vertical forces & sum of all horizotal forces is zero + +FGH=(RA-PB-PD-PF)/(sin(theta)) +printf("\n FGH= %0.4f KN (Comp.)",FGH) + +FGI=FFH+FGH*cos(theta) +printf("\n FGI= %0.4f KN (Tensile)",FGI) diff --git a/3862/CH3/EX3.9/Ex3_9.sce b/3862/CH3/EX3.9/Ex3_9.sce new file mode 100644 index 000000000..f097d8ab8 --- /dev/null +++ b/3862/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,55 @@ +clear +// + +//To determine reactions, consider equilibrium equations + +//variable declaration +//all Vertical loading are in KN +PL1=200.0 +PL2=200.0 +PL3=150.0 +PL4=100.0 +PL5=100.0 + +//length in m +UL1=6.0 +UL2=8.0 +UL3=9.0 +UL4=UL2 +UL5=UL1 + +L1=6.0 +L2=6.0 +L3=6.0 +L4=6.0 +L5=6.0 +L6=6.0 + +//moment at point LO + +R2=(PL1*L1+PL2*(L1+L2)+PL3*(L1+L2+L3)+PL4*(L1+L2+L3+L4)+PL5*(L1+L2+L3+L4+L5))/(L1+L2+L3+L4+L5+L6) + +R1=PL1+PL2+PL3+PL4+PL5-R2 + +//Take the section (1)–(1) and consider the right hand side part. + +U3U4=sqrt((1**2)+(UL1**2)) +theta1=asin(1/U3U4) + +L3U4=sqrt((UL1**2)+(UL2**2)) +theta2=asin(6/L3U4) + +//moment at U4 + +FL3L4=(R2*(L5+L6)-PL4*L4)/UL4 + +printf("\n FL3L4= %0.1f KN (Tension)",FL3L4) + +//moment at L3 +FU4U3=(-PL4*L4-PL5*(L4+L5)+R2*(L4+L5+L6))/(cos(theta1)*UL3) +printf("\n FU4U3= %0.1f KN (Comp.)",FU4U3) + +//sum of horizontal forces +FL4L3=FL3L4 +FU4L3=(-FL4L3+FU4U3*cos(theta1))/sin(theta2) +printf("\n FU4L3= %0.1f KN (Tension)",FU4L3) diff --git a/3862/CH4/EX4.1/Ex4_1.sce b/3862/CH4/EX4.1/Ex4_1.sce new file mode 100644 index 000000000..265a46521 --- /dev/null +++ b/3862/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,27 @@ +clear +// + +//variable declaration + +L1=600.0 //length of wire AB,mm +L2=200.0 //length of wire BC,mm +L3=300.0 //length of wire CD,mm +theta=45*%pi/180 +//The wire is divided into three segments AB, BC and CD. Taking A as origin the coordinates of the centroids of AB, BC and CD are (X1,Y1),(X2,Y2),(X3,Y3) + +X1=300.0 +X2=600.0 +X3=600.0-150*cos(theta) +Y1=0 +Y2=100 +Y3=200+150*sin(theta) +L=L1+L2+L3 //Total length,mm + +xc=(L1*X1+L2*X2+L3*X3)/L + +printf("\n xc= %0.2f mm",xc) + + +yc=(L1*Y1+L2*Y2+L3*Y3)/L + +printf("\n yc= %0.2f mm",yc) diff --git a/3862/CH4/EX4.10/Ex4_10.sce b/3862/CH4/EX4.10/Ex4_10.sce new file mode 100644 index 000000000..bff9ed699 --- /dev/null +++ b/3862/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,30 @@ +clear +// If xc and yc are the coordinates of the centre of the circle, centroid also must have the coordinates xc and yc as per the condition laid down in the problem. The shaded area may be considered as a rectangle of size 200 mm × 150 mm minus a triangle of sides 100 mm × 75 mm and a circle of diameter 100 mm. + +// +//variable declaration + +Ap=200.0*150.0 //Area of plate,mm^2 +At=100.0*75.0/2.0 //Area of triangle,mm^2 +Ah=%pi*(100**2)/4.0 //Area of hole **mm^2 + +A=Ap-At-Ah + + +X1=100.0 +X2=200.0-100.0/3.0 +//X3=Xc + +Y1=75.0 +Y2=150.0-25.0 +//Y3=Yc + +A=Ap-At-Ah + +xc=(Ap*X1-At*X2)/(Ah+A) + +printf("\n xc= %0.2f mm",xc) + +yc=(Ap*Y1-At*Y2)/(Ah+A) + +printf("\n yc= %0.2f mm",yc) diff --git a/3862/CH4/EX4.11/Ex4_11.sce b/3862/CH4/EX4.11/Ex4_11.sce new file mode 100644 index 000000000..c39288a06 --- /dev/null +++ b/3862/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,33 @@ +clear +// + +//variable declaration +X=40.0 +A1=14.0*12.0*(X**2) //Area of rectangle**mm^2 +A2=6.0*4.0*(X**2)/2.0 //Area of triangle**mm^2 +A3=-4*4*(X**2) //Area of removed subtracted**mm^2 +A4=-%pi*(4*X**2)/2.0 //Area of semicircle to be subtracted**mm^2 +A5=-%pi*(4*X**2)/4.0 //Area of quarter of circle to be subtracted**mm^2 + +X1=7.0*X +X2=14*X+2*X +X3=2*X +X4=6.0*X +X5=14.0*X-(16*X/(3*%pi)) + +Y1=6.0*X +Y2=4.0*X/3.0 +Y3=8.0*X+2.0*X +Y4=(16.0*X)/(3*%pi) +Y5=12*X-4*(4*X/(3*%pi)) + +A=A1+A2+A3+A4+A5 + +xc=(A1*X1+A2*X2+A3*X3+A4*X4+A5*X5)/A + +printf("\n xc= %0.2f m",xc) + + +yc=(A1*Y1+A2*Y2+A3*Y3+A4*Y4+A5*Y5)/A + +printf("\n yc= %0.2f m",yc) diff --git a/3862/CH4/EX4.12/Ex4_12.sce b/3862/CH4/EX4.12/Ex4_12.sce new file mode 100644 index 000000000..571e8eddb --- /dev/null +++ b/3862/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,38 @@ +clear +//The given composite section can be divided into two rectangles + +// +//variable declaration + + +A1=150.0*10.0 //Area of 1,mm^2 +A2=140.0*10.0 //Area of 2,mm^2 +A=A1+A2 //Total area,mm^2 +//Due to symmetry, centroid lies on the symmetric axis y-y. The distance of the centroid from the top most fibre is given by: + +Y1=5.0 +Y2=10.0+70.0 + +yc=(A1*Y1+A2*Y2)/A + +//Referring to the centroidal axis x-x and y-y, the centroid of A1 is g1 (0.0, yc-5) and that of A2 is g2 (0.0, 80-yc) + +//Moment of inertia of the section about x-x axis Ixx = moment of inertia of A1 about x-x axis + moment of inertia of A2 about x-x axis. + + +Ixx=(150*(10**3)/12)+(A1*((yc-5)**2))+(10*(140**3)/12)+(A2*((80-yc)**2)) + +printf("\n Ixx= %0.1f mm^4",Ixx) + +Iyy=(10*(150**3)/12)+(140*(10**3)/12) + +printf("\n Iyy= %0.1f mm^4",Iyy) + +//Hence, the moment of inertia of the section about an axis passing through the centroid and parallel to the top most fibre is Ixxmm^4 and moment of inertia of the section about the axis of symmetry is Iyy mm^4. +//The radius of gyration is given by + +kxx=sqrt(Ixx/A) +printf("\n kxx= %0.2f mm",kxx) + +kyy=sqrt(Iyy/A) +printf("\n kyy= %0.2f mm",kyy) diff --git a/3862/CH4/EX4.13/Ex4_13.sce b/3862/CH4/EX4.13/Ex4_13.sce new file mode 100644 index 000000000..10766b63c --- /dev/null +++ b/3862/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,41 @@ +clear +//The given composite section can be divided into two rectangles + + +//variable declaration + + +A1=125.0*10.0 //Area of 1,mm^2 +A2=75.0*10.0 //Area of 2,mm^2 +A=A1+A2 //Total area,mm^2 + +//First, the centroid of the given section is to be located. Two reference axis (1)–(1) and (2)–(2) + +//The distance of centroid from the axis (1)–(1) + +X1=5.0 +X2=10.0+75.0/2 + +xc=(A1*X1+A2*X2)/A + +//Similarly, the distance of the centroid from the axis (2)–(2) + +Y1=125.0/2 +Y2=5.0 + +yc=(A1*Y1+A2*Y2)/A + +//With respect to the centroidal axis x-x and y-y, the centroid of A1 is g1 (xc-5, (85/2)-xc) and that of A2 is g2 ((135/2)-yc, yc-5). +Ixx=(10*(125**3)/12)+(A1*(21.56**2))+(75.0*(10.0**3.0)/12)+(A2*((39.94)**2)) + +printf("\n Ixx= %0.1f mm^4",Ixx) + +Iyy=(125*(10**3)/12)+(A1*(15.94**2))+(10*(75**3)/12)+(A2*(26.56**2)) + +printf("\n Iyy= %0.1f mm^4",Iyy) + +//Izz=Polar moment of inertia + +Izz=Ixx+Iyy + +printf("\n Izz= %0.1f mm^4",Izz)
\ No newline at end of file diff --git a/3862/CH4/EX4.15/Ex4_15.sce b/3862/CH4/EX4.15/Ex4_15.sce new file mode 100644 index 000000000..405e99812 --- /dev/null +++ b/3862/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,34 @@ +clear +// + +//variable declaration + + +A1=100.0*13.5 //Area of 1,mm^2 +A2=(400.0-27.0)*8.1 //Area of 2,mm^2 +A3=100.0*13.5 //Area of 3,mm^2 + +A=A1+A2+A3 //Total area,mm^2 + +//The given section is symmetric about horizontal axis passing through the centroid g2 of the rectangle A2. + +X1=50.0 +X2=8.1/2.0 +X3=50.0 + +xc=(A1*X1+A2*X2+A3*X3)/A + +Y1=386.5+13.5/2.0 +Y2=200.0 +Y3=13.5/2 + +yc=(A1*Y1+A2*Y2+A3*Y3)/A + +//With reference to the centroidal axis x-x and y-y + +Ixx=(100.0*(13.5**3)/12.0)+(A1*((200-(13.5/2))**2))+(8.1*(373**3.0)/12.0)+0+(100*(13.5**3)/12.0)+(A3*((200-(13.5/2))**2)) +printf("\n Ixx= %0.1f mm^4",Ixx) + +Iyy=(13.5*(100.0**3)/12.0)+(A1*((50-xc)**2))+(373.0*(8.1**3.0)/12.0)+A2*(21.68**2)+(13.5*(100**3)/12.0)+(A3*((50-xc)**2)) + +printf("\n Iyy= %0.1f mm^4",Iyy) diff --git a/3862/CH4/EX4.16/Ex4_16.sce b/3862/CH4/EX4.16/Ex4_16.sce new file mode 100644 index 000000000..ce6ff8492 --- /dev/null +++ b/3862/CH4/EX4.16/Ex4_16.sce @@ -0,0 +1,47 @@ +clear +// The section is divided into three rectangles A1, A2 and A3 + +// + +//variable declaration + + +A1=80.0*12.0 //Area of 1,mm^2 +A2=(150.0-22.0)*12.0 //Area of 2,mm^2 +A3=120.0*10.0 //Area of 3,mm^2 + +A=A1+A2+A3 //Total area,mm^2 + +//Due to symmetry, centroid lies on axis y-y. The bottom fibre (1)–(1) is chosen as reference axis to locate the centroid + +Y1=150-6 +Y2=(128/2) +10 +Y3=5 +X1=60.0 +X2=60.0 +X3=60.0 + +yc=(A1*X1+A2*X2+A3*X3)/A + + + +xc=(A1*Y1+A2*Y2+A3*Y3)/A + +//With reference to the centroidal axis x-x and y-y, the centroid of the rectangles A1 is g1 (0.0, 150-6-yc), that of A2 is g2 (0.0, 75-yc) and that of A3 is g3 (0.0, yc-5 ). + +Iyy=(12*((80**3))/12)+(128*((12**3))/12)+(10*((120**3))/12) + +Ixx=(80.0*(12.0**3)/12.0)+(A1*((150-6-yc)**2))+(12*(128**3.0)/12.0)+(A2*((75-yc)**2))+(120*(10**3)/12.0)+(A3*((150-10-6-yc)**2)) + + + +PolarmomentofInertia=Ixx+Iyy + +printf("\n Polar moment of Inertia= %0.0f mm^4",PolarmomentofInertia) + +kxx=sqrt(Ixx/A) +printf("\n kxx= %0.2f mm",kxx) + + +kyy=sqrt(Iyy/A) +printf("\n kyy= %0.2f mm",kyy) diff --git a/3862/CH4/EX4.17/Ex4_17.sce b/3862/CH4/EX4.17/Ex4_17.sce new file mode 100644 index 000000000..ac0c13a12 --- /dev/null +++ b/3862/CH4/EX4.17/Ex4_17.sce @@ -0,0 +1,43 @@ +clear +// + +//The given composite section may be divided into simple rectangles and triangle + +//variable declaration + + +A1=100.0*30.0 //Area of 1,mm^2 +A2=100.0*25.0 //Area of 2,mm^2 +A3=200.0*20.0 //Area of 3,mm^2 +A4=87.5*20.0/2.0 //Area of 4,mm^2 +A5=87.5*20.0/2.0 //Area of 5,mm^2 + +A=A1+A2+A3+A4+A5 //Total area,mm^2 + +//Due to symmetry, centroid lies on the axis y-y. A reference axis (1)–(1) is choosen as shown in the figure. The distance of the centroidal axis from (1)–(1) + +X1=100.0 +X2=100.0 +X3=100.0 +X4=2.0*87.5/3.0 +X5=200-X4 +xc=(A1*X1+A2*X2+A3*X3+A4*X4+A5*X5)/A + +Y1=135.0 +Y2=70.0 +Y3=10.0 +Y4=(20.0/3.0)+20.0 +Y5=Y4 + +yc=(A1*Y1+A2*Y2+A3*Y3+A4*Y4+A5*Y5)/A + +//With reference to the centroidal axis x-x and y-y, the centroid of the rectangle A1 is g1 (0.0,135.0-yc), that of A2 is g2(0.0,70.00-yc), that of A3 is g3 (0.0, yc-10.0), the centroid of triangle A4 is g4 (41.66,yc-20.0-(20.0/3.0) ) and that of A5 is g5 (41.66,yc-20.0-(20.0/3.0)). + + +Ixx=(100.0*(30**3)/12.0)+(A1*((135.0-yc)**2))+(25.0*(100**3.0)/12.0)+(A2*((70.0-yc)**2))+(200*(20**3)/12.0)+(A3*((yc-10.0)**2))+((87.5*(20**3)/36.0)+(A4*((yc-20.0-(20.0/3.0))**2)))*2 + +printf("\n Ixx= %0.1f mm^4",Ixx) + +Iyy=(30.0*(100**3)/12.0)+(100.0*(25**3.0)/12.0)+(20*(200**3)/12.0)+((20.0*(87.5**3)/36.0)+(A4*((41.66)**2)))*2 + +printf("\n Iyy= %0.1f mm^4",Iyy) diff --git a/3862/CH4/EX4.18/Ex4_18.sce b/3862/CH4/EX4.18/Ex4_18.sce new file mode 100644 index 000000000..a9d56d4cd --- /dev/null +++ b/3862/CH4/EX4.18/Ex4_18.sce @@ -0,0 +1,15 @@ +clear +//In this problem, it is required to find out the moment of inertia of the section about an axis AB. So there is no need to find out the position of the centroid. +//The given section is split up into simple rectangles +//Moment of inertia about AB = Sum of moments of inertia of the rectangle about AB + +//variable declaration + +A1=400*20.0 +A2=100*10 +A3=10*380.0 +A4=100*10.0 + +IAB=(400.0*(20**3)/12)+(A1*(10**2))+((100*(10**3)/12)+(A2*(25**2)))*2+((10*(380**3)/12)+(A3*(220**2)))*2+((100*(10**3)/12)+(A4*(415**2)))*2 + +printf("\n IAB= %0.0f mm^4",IAB) diff --git a/3862/CH4/EX4.19/Ex4_19.sce b/3862/CH4/EX4.19/Ex4_19.sce new file mode 100644 index 000000000..e55f95842 --- /dev/null +++ b/3862/CH4/EX4.19/Ex4_19.sce @@ -0,0 +1,25 @@ +clear +// The built-up section is divided into six simple rectangles + +//variable declaration + + +A1=250.0*10.0 //Area of 1,mm^2 +A2=40.0*10.0 //Area of 2,mm^2 + +A=A1*2+A2*4 //Total area,mm^2 + + +Y1=5.0 +Y2=30.0 +Y3=15.0 +Y4=255.0 +Y5=135.0 + +yc=(A1*Y1+2*A2*Y2+A2*Y3+A2*Y4+A1*Y5)/A + +//Now, Moment of inertia about the centroidalaxis=Sum of the moment of inertia of the individual rectangles + +Ixx=(250.0*(10**3)/12.0)+(A1*((yc-5)**2))+((10.0*(40**3.0)/12.0)+(A2*((yc-30.0)**2)))*2+(40*(10**3)/12.0)+(A2*((yc-15.0)**2))+(10.0*(250.0**3.0)/12.0)+(A1*((yc-135.0)**2))+(40.0*(10.0**3)/12)+(A2*((yc-255)**2)) + +printf("\n Ixx= %0.1f mm^4",Ixx) diff --git a/3862/CH4/EX4.2/Ex4_2.sce b/3862/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..fa6b88563 --- /dev/null +++ b/3862/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,30 @@ +clear +// + +//The composite figure is divided into three simple figures and taking A as origin coordinates of their centroids + +//variable declaration + +L1=400.0 //length of wire AB,mm +L2=150.0*%pi //length of wire BC,mm +L3=250.0 //length of wire CD,mm +theta=30*%pi/180 + +//The wire is divided into three segments AB, BC and CD. Taking A as origin the coordinates of the centroids of AB, BC and CD are (X1,Y1),(X2,Y2),(X3,Y3) +X1=200.0 +X2=475.0 +X3=400+300.0+250*cos(theta)/2 + +Y1=0 +Y2=2*150/%pi +Y3=125*sin(theta) +L=L1+L2+L3 //Total length,mm + +xc=(L1*X1+L2*X2+L3*X3)/L + +printf("\n xc= %0.2f mm",xc) + + +yc=(L1*Y1+L2*Y2+L3*Y3)/L + +printf("\n yc= %0.2f mm",yc) diff --git a/3862/CH4/EX4.20/Ex4_20.sce b/3862/CH4/EX4.20/Ex4_20.sce new file mode 100644 index 000000000..6d9ef22f4 --- /dev/null +++ b/3862/CH4/EX4.20/Ex4_20.sce @@ -0,0 +1,26 @@ +clear +//Each angle is divided into two rectangles + +//variable declaration + +A1=600.0*15.0 //Area of 1,mm^2 +A2=140.0*10.0 //Area of 2,mm^2 +A3=150.0*10.0 +A4=400.0*20.0 +A=A1+A2*2+A3*2+A4 //Total area,mm^2 + +//The distance of the centroidal axis from the bottom fibres of section + +Y1=320.0 +Y2=100.0 +Y3=25.0 +Y4=10.0 + +yc=(A1*Y1+2*A2*Y2+A3*Y3*2+A4*Y4)/A +printf("\n yc") +//Now, Moment of inertia about the centroidalaxis=Sum of the moment of inertia of the individual rectangles + +Ixx=(15.0*(600**3)/12.0)+(A1*((yc-320)**2))+((10.0*(140**3.0)/12.0)+(A2*((yc-100.0)**2)))*2+((150*(10**3)/12.0)+(A3*((yc-15.0)**2)))*2+(400.0*(20.0**3.0)/12.0)+(A4*((yc-10.0)**2)) + + +printf("\n Ixx= %0.1f mm^4",Ixx) diff --git a/3862/CH4/EX4.21/Ex4_21.sce b/3862/CH4/EX4.21/Ex4_21.sce new file mode 100644 index 000000000..202ef8b1e --- /dev/null +++ b/3862/CH4/EX4.21/Ex4_21.sce @@ -0,0 +1,24 @@ +clear +// + +//The rectangle is divided into four triangles +//The lines AE and FC are parallel to x-axis + +//variable declaration + +theta=asin(4.0/5.0) + +AB=100.0 +BK=AB*sin((90*%pi/180)-theta) +ND=BK +FD=60.0/sin(theta) +AF=150.0-FD +AE=AB/cos((90*%pi/180)-theta) +FC=AE +A=125.0*60.0/2.0 + +//Moment of inertia of the section about axis x-x=Sum of the momentsof inertia of individual triangular areasabout axis + +Ixx=(125*(60**3)/36)+(A*((ND*4.0/3.0)**2))+(125*(60**3)/36)+(A*((ND*2.0/3.0)**2))+(125*(60**3)/36)+(A*((ND*1.0/3.0)**2))+(125*(60**3)/36)+(A*((ND*1.0/3.0)**2)) + +printf("\n Ixx= %0.0f mm^4",Ixx) diff --git a/3862/CH4/EX4.22/Ex4_22.sce b/3862/CH4/EX4.22/Ex4_22.sce new file mode 100644 index 000000000..7a011265b --- /dev/null +++ b/3862/CH4/EX4.22/Ex4_22.sce @@ -0,0 +1,10 @@ +clear +// + +//The section is divided into a triangle PQR, a semicircle PSQ having base on axis AB and a circle having its centre on axis AB + +//variable declaration +//Now,Moment of inertia of the section about axis AB +IAB=(80*(80**3)/12)+(%pi*(80**4)/128)-(%pi*(40**4)/64) + +printf("\n IAB= %0.0f mm^4",IAB) diff --git a/3862/CH4/EX4.23/Ex4_23.sce b/3862/CH4/EX4.23/Ex4_23.sce new file mode 100644 index 000000000..4ffbca00e --- /dev/null +++ b/3862/CH4/EX4.23/Ex4_23.sce @@ -0,0 +1,41 @@ +clear +// + +//The section is divided into three simple figures viz., a triangle ABC, a rectangle ACDE and a semicircle. + +//variable declaration + +r=20.0 //radius of semicircle +A1=80.0*20.0/2 //Area of triangle ABC +A3=40.0*80.0 //Area of rectangle ACDE +A4=%pi*(r**2)/2 //Area of semicircle +At1=30.0*20.0/2.0 +At2=50.0*20.0/2.0 +A=A1+A3-A4 //Total area + +X1=2.0*30.0/3.0 +X2=50.0*30.0/3.0 +X3=40.0 +X4=40.0 + +xc=(At1*X1+At2*X2+A3*X3-A4*X4)/A +//mistake in book + +Y1=(20.0/3.0)+40.0 +Y3=20.0 +Y4=(4.0*20.0)/(3.0*%pi) + +yc=(A1*Y1+A3*Y3-A4*Y4)/A +printf("\n %0.3f %0.3f ",xc,yc) + +// +//Moment of inertia of the section about axis x-x=Sum of the momentsof inertia of individual triangular areasabout axis + +Ixx=(80.0*(20.0**3)/36) +A1*((60.0-(2*20.0/3.0)-yc)**2)+(80*(40**3)/12)+(A3*((yc-20.0)**2))-((0.0068598*(20**4))+(A4*((yc-Y4)**2))) + +printf("\n Ixx= %0.0f mm^4",Ixx) + + +Iyy=(20.0*(30.0**3)/36) +At1*((xc-(2*30.0/3.0))**2)+(20*(50**3)/36)+(At2*((xc-(30.0+(50/3)))**2))+((40*(80**3)/12)+(A3*((xc-40)**2)))-((%pi*(40**4))/(2*64))-(A4*((40-xc)**2)) + +printf("\n Iyy= %0.0f mm^4",Iyy) diff --git a/3862/CH4/EX4.27/Ex4_27.sce b/3862/CH4/EX4.27/Ex4_27.sce new file mode 100644 index 000000000..615d9a78d --- /dev/null +++ b/3862/CH4/EX4.27/Ex4_27.sce @@ -0,0 +1,39 @@ +clear +//A concrete block of size 0.60 m × 0.75 m × 0.5 m is cast with a hole of diameter 0.2 m and depth 0.3 m +//The hole is completely filled with steel balls weighing 2500 N. Locate the centre of gravity of the body. + +// + +//variable declaration + +W=25000.0 // weight of concrete=25000, N/m^3 +W1=0.6*0.75*0.5*W //Weight of solid concrete block +W2=%pi*(0.2**2)*0.3*W/4 //Weight of concrete (W2) removed for making hole: +W3=2500 + +//Taking origin as shown in the figure, the centre of gravity of solid block is (0.375, 0.3, 0.25) and that of hollow portion is (0.5, 0.4, 0.15) + +X1=0.375 +X2=0.5 +X3=0.5 + +Y1=0.3 +Y2=0.4 +Y3=0.4 + +Z1=0.25 +Z2=0.15 +Z3=0.15 + +Wt=W3+W1-W2 +printf("\n %0.3f %0.3f %0.3f %0.3f ",W,W1,W2,Wt) + +xc=(W1*X1-W2*X2+W3*X3)/Wt + +yc=(W1*Y1-W2*Y2+W3*Y3)/Wt + +zc=(W1*Z1-W2*Z2+W3*Z3)/Wt + +printf("\n xc= %0.3f m",xc) +printf("\n yc= %0.3f m",yc) +printf("\n zc= %0.3f m",zc) diff --git a/3862/CH4/EX4.3/Ex4_3.sce b/3862/CH4/EX4.3/Ex4_3.sce new file mode 100644 index 000000000..1bac15a13 --- /dev/null +++ b/3862/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,38 @@ +clear +// +// The length and the centroid of portions AB, BC and CD +// portion AB is in x-z plane, BC in y-z plane and CD in x-y plane. AB and BC are semi circular in shape + +//variable declaration + +L1=100.0*%pi //length of wire AB,mm +L2=140.0*%pi //length of wire BC,mm +L3=300.0 //length of wire CD,mm +theta=45*%pi/180 + +//The wire is divided into three segments AB, BC and CD. Taking A as origin the coordinates of the centroids of AB, BC and CD are (X1,Y1),(X2,Y2),(X3,Y3) +X1=100.0 +X2=0 +X3=300*sin(theta) + +Y1=0 +Y2=140 +Y3=280+300*cos(theta) +Z1=2*100/%pi +Z2=2*140/%pi +Z3=0 + +L=L1+L2+L3 //Total length,mm + +xc=(L1*X1+L2*X2+L3*X3)/L + +printf("\n xc= %0.2f mm",xc) + + +yc=(L1*Y1+L2*Y2+L3*Y3)/L + +printf("\n yc= %0.2f mm",yc) + +zc=(L1*Z1+L2*Z2+L3*Z3)/L + +printf("\n zc= %0.2f mm",zc) diff --git a/3862/CH4/EX4.5/Ex4_5.sce b/3862/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..60a719ee0 --- /dev/null +++ b/3862/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,24 @@ +clear +//variable declaration + +A1=150.0*12.0 //Area of 1 ,mm^2 +A2=(200.0-12.0)*12.0 //Area of 2,mm^2 + +X1=75 +X2=6 + +Y1=6 +Y2=12+(200-12)/2 + +A=A1+A2 + +xc=(A1*X1+A2*X2)/A + +printf("\n xc= %0.2f ",xc) + +yc=(A1*Y1+A2*Y2)/A + +printf("\n yc= %0.2f mm",yc) + +printf("\nThus, the centroid is at x = 36.62 mm and y = 61.62 mm ") + diff --git a/3862/CH4/EX4.6/Ex4_6.sce b/3862/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..c52f5dddc --- /dev/null +++ b/3862/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,25 @@ +clear +//variable declaration + +A1=100.0*20 //Area of 1 ,mm^2 +A2=100.0*20.0 //Area of 2,mm^2 +A3=150.0*30.0 //Area of 3,mm^2 + +//Selecting the coordinate system, due to symmetry centroid must lie on y axis, + +X1=0 +X2=0 + +Y1=30+100+20/2 +Y2=30+100/2 +Y3=30/2 + +A=A1+A2+A3 + + +yc=(A1*Y1+A2*Y2+A3*Y3)/A + +printf("\n yc= %0.2f mm",yc) + +printf("\n Thus, the centroid is on the symmetric axis at a distance 59.71 mm from the bottom") + diff --git a/3862/CH4/EX4.7/Ex4_7.sce b/3862/CH4/EX4.7/Ex4_7.sce new file mode 100644 index 000000000..bdd81f46d --- /dev/null +++ b/3862/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,32 @@ +clear +// Note that it is convenient to take axis in such a way that the centroids of all simple figures are having positive coordinates. If coordinate of any simple figure comes out to be negative, one should be careful in assigning the sign of moment of area + +//variable declaration + +A1=2.0*6.0*1.0/2.0 //Area of 1,m^2 +A2=2.0*7.5 //Area of 2,m^2 +A3=3.0*5.0*1.0/2 //Area of 3,m^2 +A4=1.0*4.0 //Area of 4,m^2 + +//The composite figure can be conveniently divided into two triangles and two rectangle + +X1=2.0*2.0/3.0 +X2=2.0+1.0 +X3=2.0+2.0+(1.0*3.0/3.0) +X4=4.0+4.0/2.0 + +Y1=6.0/3.0 +Y2=7.5/2.0 +Y3=1.0+5.0/3.0 +Y4=1/2.0 + +A=A1+A2+A3+A4 + +xc=(A1*X1+A2*X2+A3*X3+A4*X4)/A + +printf("\n xc= %0.3f m",xc) + + +yc=(A1*Y1+A2*Y2+A3*Y3+A4*Y4)/A + +printf("\n yc= %0.3f m",yc) diff --git a/3862/CH4/EX4.8/Ex4_8.sce b/3862/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..2d226987d --- /dev/null +++ b/3862/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,30 @@ +clear +// The composite section is divided into three simple figures, a triangle, a rectangle and a semicircle + +// +//variable declaration + +A1=1.0*3.0*4.0/2.0 //Area of 1,m^2 +A2=6.0*4.0 //Area of 2,m^2 +A3=1.0*%pi*(2**2)/2 //Area of 3**m^2 + +//The coordinates of centroids of these three simple figures are: + +X1=6.0+3.0/3.0 +X2=3.0 +X3=-(4*2)/(3.0*%pi) + +Y1=4.0/3.0 +Y2=2.0 +Y3=2.0 + +A=A1+A2+A3 + +xc=(A1*X1+A2*X2+A3*X3)/A + +printf("\n xc= %0.4f m",xc) + + +yc=(A1*Y1+A2*Y2+A3*Y3)/A + +printf("\n yc= %0.3f m",yc) diff --git a/3862/CH4/EX4.9/Ex4_9.sce b/3862/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..0177d4636 --- /dev/null +++ b/3862/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,41 @@ +clear +//The composite area is equal to a rectangle of size 160 × 280 mm plus a triangle of size 280 mm base width and 40 mm height and minus areas of six holes. In this case also the can be used for locating centroid by treating area of holes as negative. The area of simple figures and their centroids are + +// + +//variable declaration + +Ar=160.0*280.0 //Area of rectangle,mm^2 +At=280.0*40.0/2.0 //Area of triangle,mm^2 +d=21.5 //diameter of hole,mm +Ah=-%pi*(d**2)/4 //Area of hole**mm^2 + +A=Ar+At+Ah*6 + + +Xr=140.0 +Xt=560/3.0 +Xh1=70.0 +Xh2=140.0 +Xh3=210.0 +Xh4=70.0 +Xh5=140.0 +Xh6=210.0 + +Yr=80.0 +Yt=160.0+40.0/3.0 +Yh1=50.0 +Yh2=50.0 +Yh3=50.0 +Yh4=120.0 +Yh5=130.0 +Yh6=140.0 + +xc=(Ar*Xr+At*Xt+Ah*(Xh1+Xh2+Xh3+Xh4+Xh5+Xh6))/A + +printf("\n xc= %0.2f m",xc) + + +yc=(Ar*Yr+At*Yt+Ah*(Yh1+Yh2+Yh3+Yh4+Yh5+Yh6))/A + +printf("\n yc= %0.2f m",yc) diff --git a/3862/CH5/EX5.1/Ex5_1.sce b/3862/CH5/EX5.1/Ex5_1.sce new file mode 100644 index 000000000..2bd85cb19 --- /dev/null +++ b/3862/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,33 @@ +clear +// +Wa=1000.0 //weight of block a +Wb=2000.0 //weight of block b +uab=1.0/4.0 //coefficient of friction between A and B +ubg=1.0/3.0 //coefficient of friction between g and B + +//When P is horizontal +//considering equilibrium of block A +N1=Wa //Normal Reaction on block A from block B +F1=uab*N1 //limiting Friction between A and B +T=F1 //tension +//considering equilibrium of block B +N2=N1+ Wb //Normal Reaction on block B from G + +F2=ubg*N2 //limiting Friction between A and g + +P=F1+F2 +printf("\n P= %0.3f N",P) +//When P is inclined at angle o +o=30.0*3.14/180.0 +//considering equilibrium of block A +N1=Wa //Normal Reaction on block A from block B +F1=uab*N1 //limiting Friction between A and B +T=F1 //tension +//considering equilibrium of block B +//from +//N2+Psin30=N1+Wb +//Pcos30=F1+F2 +//F1=ubg*N2 +N2=(N1+Wb-F1*tan(o))/(1+ubg*tan(o)) +P=(N1+Wb-N2)/sin(o) +printf("\n P= %0.3f N",P) diff --git a/3862/CH5/EX5.11/Ex5_11.sce b/3862/CH5/EX5.11/Ex5_11.sce new file mode 100644 index 000000000..6261d9690 --- /dev/null +++ b/3862/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,23 @@ +clear +// +l=6.0 //length of ladder +u1=0.4 //coefficient of friction between the wall and the ladder +w=200.0 //weight of ladder +u2=0.25 //coefficient of friction between floor and the ladder +wl=900.0 //weight of load +ll=5.0 //distance of load +//force balancing +//Na Nb normal reaction at A and B +//Fa Fb friction at A and B +//Fa=u2*Na +//Fb=u1*Nb +//Na+Fb=w+wl +//Fa=Nb +Nb=(wl+w)*u2/(1+u2*u1) +Na=Nb/u2 +Fa=u2*Na +Fb=u1*Nb +//sum of all moments about a is =0 +temp=((w*l*0.5)+(wl*ll)-(Fb*l))/(Nb*l) +o=atan(temp)*180/3.14 +printf("\n Angle of inclination is %0.3f degrees",o) diff --git a/3862/CH5/EX5.12/Ex5_12.sce b/3862/CH5/EX5.12/Ex5_12.sce new file mode 100644 index 000000000..259ca17f7 --- /dev/null +++ b/3862/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,14 @@ +clear +// +o=45.0*3.14/180.0 //angle of inclination +u=0.5 //coefficient of friction +r=1.5 //ratio of mans weight to ladders weight +o1=45.0*%pi/180.0 //angle of inclination +//from law of friction +//Fa = μNa +//Fb = μNb +//Fa – Nb = 0 +//Na + Fb = W + r W +//ΣMA = 0 +o=(((u*u+u)*(1+r)/((1+u)))-1.0/2.0)/r +printf("\n length will %0.3f times",o) diff --git a/3862/CH5/EX5.13/Ex5_13.sce b/3862/CH5/EX5.13/Ex5_13.sce new file mode 100644 index 000000000..bee2a03f5 --- /dev/null +++ b/3862/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,15 @@ +clear +// +n=1.25 //number of turns +o=2*3.14*n //angle of contact +u=0.3 //coefficient of friction +t=600.0 //force at the other end of the rope +//if the impending motion of the weight be downward. +T2=t*%e**(u*o) +W=T2 +printf("Maximum weight is %f",W) +printf("\n answer in textbook is wrong") +//if the impending motion of weight be upwards +T1=t*%e**(-1*u*o) +W=T1 +printf("\n Minimum weight is %f",W) diff --git a/3862/CH5/EX5.14/Ex5_14.sce b/3862/CH5/EX5.14/Ex5_14.sce new file mode 100644 index 000000000..882e71c85 --- /dev/null +++ b/3862/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,23 @@ +clear +// +ur=0.20 //The coefficient of friction between the rope and the fixed drum +uo=0.30 //The coefficient of friction between other surfaces +cosa=4.0/5.0 //cos of angle of inclination +sina=3.0/5.0 //sin of angle of inclination +Ww=1000.0 //weight +o=3.14 //angle of contact of rope with pulley +//for unknown weight +//force balance perpendicular to the plane +//N1 = W cos α +//fr=uoN1 +//force balance along the plane +//T1 = F1 + W sin α +//for 1000 N body +//force balance perpendicular to the plane +//N2=N1+Wwcosa +//fr2=uoN2 +//force balance along the plane +//T2= Wwsina -F1 -F2 +//T2=T1*e^(ur*o) +W=(Ww*sina-uo*Ww*cosa)/(((uo*cosa+sina)*(2.71**(uo*o)))+(uo*cosa+uo*cosa)) +printf("\n Weight is %0.3f N",W) diff --git a/3862/CH5/EX5.15/Ex5_15.sce b/3862/CH5/EX5.15/Ex5_15.sce new file mode 100644 index 000000000..4eae92a1f --- /dev/null +++ b/3862/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,16 @@ +clear +// +u=0.3 //coefficient of friction +r=250 //radius of brake drum +l=300 //length of lever arm +M=300000.0 //torque +o=r*3.14/180.0 +l2=50.0 +//using +//T2 = T1e^(μθ) T1 and T2 are tension +//(T2-T1)r=M +T1=M/(r*(2.71**(u*o)-1)) +T2=(2.71**(u*o))*T1 +//Consider the lever arm. Taking moment about the hinge +p=T2*l2/l //force P applied at the end of the brake lever +printf("force P applied at the end of the brake lever %0.3f N",p) diff --git a/3862/CH5/EX5.16/Ex5_16.sce b/3862/CH5/EX5.16/Ex5_16.sce new file mode 100644 index 000000000..f7a55c7a0 --- /dev/null +++ b/3862/CH5/EX5.16/Ex5_16.sce @@ -0,0 +1,16 @@ +clear +// +d1=500.0 //diameter of a shaft +d2=100.0 //diameter of a shaft +D=3000.0 //distance between shafts in mm +T=1000.0 //Maximum permissible tension in the belt +U=0.25 //coefficient of friction between the belt and the pulley +R=220.0 //revlution per minute of larger shaft +//Length of belt = Arc length DC + Arc length FE + 2BG +O1=3.14+2*asin((d1+d2)/(2*D)) +L=(d1/2+d2/2)*O1+2*D*cos(asin((d1+d2)/(2*D))) +printf("\n Length of belt is %0.3f mm",L) +T1=T/(2.71**(U*O1)) +Velocity_of_the_belt =d1/2*(R*2*3.14/60.0) +Power_transmitted=(T-T1)*Velocity_of_the_belt +printf("\n Power Transmitted %0.3f Watt",Power_transmitted) diff --git a/3862/CH5/EX5.17/Ex5_17.sce b/3862/CH5/EX5.17/Ex5_17.sce new file mode 100644 index 000000000..381270730 --- /dev/null +++ b/3862/CH5/EX5.17/Ex5_17.sce @@ -0,0 +1,17 @@ +clear +// +d1=500.0 //diameter of a shaft +d2=100.0 //diameter of a shaft +D=3000.0 //distance between shafts in mm +T=1000.0 //Maximum permissible tension in the belt +U=0.25 //coefficient of friction between the belt and the pulley +R=220.0 //revlution per minute of larger shaft +O1=3.14+2*asin((d1-d2)/(2*D)) +O2=3.14-2*asin((d1-d2)/(2*D)) +//Length of belt = Arc length DC + Arc length FE + 2BG +L=(d1/2*O1+d2/2*O2)+2*D*cos(asin((d1-d2)/(2*D))) +printf("\n Length of belt is %0.3f mm",L) +T1=T/(2.71**(U*O2)) +Velocity_of_the_belt =d1/2*(R*2*3.14/60.0) +Power_transmitted=(T-T1)*Velocity_of_the_belt +printf("\n Power Transmitted %0.3f Watt",Power_transmitted) diff --git a/3862/CH5/EX5.2/Ex5_2.sce b/3862/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..a2aedc35b --- /dev/null +++ b/3862/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,14 @@ +clear +// +Wa=300.0 //weight of upper block +Wb=900.0 //weight of lower block +u1=1.0/3.0 //coefficient of friction between upper block and lower block +u2=1.0/3.0 //coefficient of friction between g and lower block + +//using +//N1=Wacoso Normal Reaction +//F1=u1*N1 Friction +//N2=Wbcoso+N1 +//F2=u2*N2 +o=atan((u1*Wa+u2*Wb+u2*Wa)/Wb)*180/3.14 +printf("\n %0.3f °",o) diff --git a/3862/CH5/EX5.3/Ex5_3.sce b/3862/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..284963ee9 --- /dev/null +++ b/3862/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,22 @@ +clear +// +W=500.0 //weight of block +F1=200.0 //force up the inclined plane when block is moving down +F2=300.0 //force up the inclined plane when block is at rest +//When block starts moving down the plane +//sum of all forces perpendicular to the plane = 0 +//N =Wcoso +//sum of all forces parallel to the plane = 0 +//Fr+F1=Wsino +//sino-ucoso=F1/w 1 +//When block starts moving up the plane +//sum of all forces perpendicular to the plane = 0 +//N =Wcoso +//sum of all forces parallel to the plane = 0 +//Wsino+Wucoso=F2 +//using these equations +o=asin((F1*0.5/W)+(F2*0.5/W)) //angle of inclination +printf("\n Angle of inclination is %0.3f ",o*180/3.14) +//using 1 +u=sin(o)-F1/W +printf("\n coefficient of friction is %0.3f ",u) diff --git a/3862/CH5/EX5.4/Ex5_4.sce b/3862/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..98f73d9b8 --- /dev/null +++ b/3862/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,17 @@ +clear +// +uag=0.5 //coefficient of friction between block A and the plane +ubg=0.2 //coefficient of friction between block B and the plane +Wb=500.0 //weight of block B +Wa=1000.0 //weight of block A +//Considering equilibrium of block A, +//sum of all forces along the plane is 0 +//N1=Wacoso ,Fr=uagN1 +//sum of all forces perpendicaular to the plane is 0 +//T=uagWacoso-Wasino +//Considering equilibrium of block A, +//sum of all forces along the plane is 0 +//N2=Wbcoso ,Fr=uagN2 +//sum of all forces perpendicaular to the plane is 0 +//T=Wbsino-ubgwbsino +o=atan((uag*Wa+ubg*Wb)/(Wa+Wb))*180.0/3.14 diff --git a/3862/CH5/EX5.5/Ex5_5.sce b/3862/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..cff5d1ab8 --- /dev/null +++ b/3862/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,16 @@ +clear +// +Wl=750.0 //weight of lower block +Wu=500.0 //weight of upper block +o1=60.0*3.14/180.0 //angle of inclined plane +o2=30.0 *3.14/180.0 // anlge at which pull is applied +u=0.2 //coefficient of friction +//for 750 N block +//Σ Forces normal to the plane = 0 +N1=Wl*cos(o1) +F1=u*N1 +//Σ Forces parallel to the plane = 0 +T=F1+Wl*sin(o1) +//Σ Forces horizontal to the plane = 0 +P=(T+u*Wu)/(cos(o2)+u*sin(o2)) +printf("\n %0.3f N",P) diff --git a/3862/CH5/EX5.6/Ex5_6.sce b/3862/CH5/EX5.6/Ex5_6.sce new file mode 100644 index 000000000..56751a264 --- /dev/null +++ b/3862/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,27 @@ +clear +// +o1=60.0*3.14/180.0 //angle of inclination of plane AC +o2=30.0*3.14/180.0 //angle of inclination of plane BC +Wbc=1000.0 //weight of block on plane BC +ubc=0.28 //coefficient of friction between the load and the plane BC +uac=0.20 //coefficient of friction between the load and the plane AC +//for least weight +N1=Wbc*cos(o2) //Normal Reaction +F1=ubc*N1 //frictional Force +T=Wbc*sin(o2)-F1 //Tension +//for block on plane AC +//N2=Wcoso1 +//F2=uac*N2 +//T=F2+W sino2 +W=T/(uac*cos(o1)+sin(o1)) +printf("\n Least Weight is %0.3f N",W) +//for greatest weight +N1=Wbc*cos(o2) //Normal Reaction +F1=ubc*N1 //frictional Force +T=Wbc*sin(o2)+F1 //Tension +//for block on plane AC +//N2=Wcoso1 +//F2=uac*N2 +//T=F2+W sino2 +W=T/(-1*uac*cos(o1)+sin(o1)) +printf("\n Greatest Weight is %0.3f N",W) diff --git a/3862/CH5/EX5.7/Ex5_7.sce b/3862/CH5/EX5.7/Ex5_7.sce new file mode 100644 index 000000000..9781c9674 --- /dev/null +++ b/3862/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,22 @@ +clear +// +u=0.4 //The coefficient of friction on the horizontal plane +oi=30 //angle of inclined plane +o=20.0 //The limiting angle of friction for block B on the inclined plane +wb=5000.0 //weight of block b +ub=tan(o*3.14/180.0) //coefficcient of friction on plane +//for block B +//N1 N2 N3 are normal reaction +//F1 F2 are frictional forces +//F1=ub*N1 +//N1 sinoi + F1 cos oi=wb +N1=wb/(sin(oi*3.14/180.0)+ub*cos(oi*3.14/180.0)) +F1=ub*N1 +C=N1*cos(oi*3.14/180.0)-F1*sin(oi*3.14/180.0) + +//force balance on A in horizontal balance +F2=C +N2=F2/u +//force balance on A in vertical balance +W=N2 +printf("\n Weight %0.3f N",W) diff --git a/3862/CH5/EX5.8/Ex5_8.sce b/3862/CH5/EX5.8/Ex5_8.sce new file mode 100644 index 000000000..a20ff3285 --- /dev/null +++ b/3862/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,14 @@ +clear +// +w=20000.0 //weight of upper block +o=15.0 //The angle of friction for all surfaces of contact +u=tan(o) //coefficient of friction +//R1 R2 are forces +Or1=15.0 //angle force R1 makes with x axis +Or2=35.0 //angle force R2 makes with Y axis +R2=w*sin((90-Or1)*3.14/180.0)/sin((90+Or1+Or2)*3.14/180.0) +//applyig lamis theorem on block B +Or1=15.0 //angle force R3 makes with Y axis +Or2=35.0 //angle force R2 makes with Y axis +P=R2*sin((180-Or1-Or2)*3.14/180.0)/sin((90+Or1)*3.14/180.0) +printf("\n Force = %0.3f N",P) diff --git a/3862/CH6/EX6.1/Ex6_1.sce b/3862/CH6/EX6.1/Ex6_1.sce new file mode 100644 index 000000000..72f566c04 --- /dev/null +++ b/3862/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,20 @@ +clear +// +W = 10000.0 //Load +P = 500.0 //Effort +D = 20.0 //Distance moved by the effort +d = 0.8 //Distance moved by the load +MA=W/P //Mechanical advantage +VR=D/d //Velocity Ratio +Efficiency=MA/VR +Pi =W/VR //Ideal effort +Wi = P*VR //ideal load +efl=P-Pi //Effort lost in friction +Fr=Wi-W //frictional resistance +printf(" Mechanical advantage-- %0.3f",MA) +printf("\n Velocity Ratio %0.3f",VR) +printf("\n Efficiency %0.3f",Efficiency) +printf("\n Ideal Load %0.3f",Wi) +printf("\n Ideal Effort %0.3f",Pi) +printf("\n Effort lost in friction %0.3f",efl) +printf("\n frictional resistance %0.3f",Fr) diff --git a/3862/CH6/EX6.10/Ex6_10.sce b/3862/CH6/EX6.10/Ex6_10.sce new file mode 100644 index 000000000..204e86659 --- /dev/null +++ b/3862/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,10 @@ +clear +// +W = 2500.0 //Load +N1=2.0 //number of movable pulleys in system 1 in figure B +N2=2.0 //number of movable puleys in system 2 in figure C +VR=2**N1-1+2**N2-1 //Velocity Ratio +Efficiency=0.70 +MA=Efficiency*VR //Mechanical advantage +P = W/MA //Effort +printf("\n Effort is %0.3f N",P) diff --git a/3862/CH6/EX6.11/Ex6_11.sce b/3862/CH6/EX6.11/Ex6_11.sce new file mode 100644 index 000000000..ea7601285 --- /dev/null +++ b/3862/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,13 @@ +clear +D=500.0 //diameter of the wheel +d=200.0 //diameter of axle +tcw=6.0 //thickness of the cord on the wheel +tca=20.0 //thickness of the cord on the axle +W=1200 //effort +ED=D+tcw //Effective diameter of the wheel +Ed=d+tca //Effectivediameter of axle +VR=ED/Ed //Velocity Ratio +Efficiency=0.7 +MA=Efficiency*VR //Mechanical advantage +P = W/MA //Effort +printf("\n Effort is %0.3f N",P) diff --git a/3862/CH6/EX6.12/Ex6_12.sce b/3862/CH6/EX6.12/Ex6_12.sce new file mode 100644 index 000000000..38e254ee3 --- /dev/null +++ b/3862/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,12 @@ +clear +D=800.0 //diameter of the wheel +d1=250.0 //diameter of axle 1 +d2=300.0 //diameter of axle 2 + +W=20000.0 //effort + +VR=(2*D)/(d2-d1) //Velocity Ratio +Efficiency=0.55 +MA=Efficiency*VR //Mechanical advantage +P = W/MA //Effort +printf("\n Effort is %0.3f N",P) diff --git a/3862/CH6/EX6.13/Ex6_13.sce b/3862/CH6/EX6.13/Ex6_13.sce new file mode 100644 index 000000000..0cee86d44 --- /dev/null +++ b/3862/CH6/EX6.13/Ex6_13.sce @@ -0,0 +1,11 @@ +clear +D=500.0 //diameter of the wheel +d=200.0 //diameter of axle + +W=5000.0 //effort + +VR=(2*D)/(D-d) //Velocity Ratio +Efficiency=0.6 +MA=Efficiency*VR //Mechanical advantage +P = W/MA //Effort +printf("\n Effort is %0.3f N",P) diff --git a/3862/CH6/EX6.14/Ex6_14.sce b/3862/CH6/EX6.14/Ex6_14.sce new file mode 100644 index 000000000..6d9d7fdae --- /dev/null +++ b/3862/CH6/EX6.14/Ex6_14.sce @@ -0,0 +1,9 @@ +clear +D=40.0 //Screw diameter +l=20.0 //Screw lwngth +p=l/3.0 //Lead of the screw +W=40000.0 //effort +R = 400 //Lever length +u = 0.12 //coefficient of friction between screw and nut +P = (D/(2*R))*W*((u+(p/(3.14*D)))/(1-u*(p/(3.14*D)))) //Effort +printf("\n Effort is %0.3f N",P) diff --git a/3862/CH6/EX6.16/Ex6_16.sce b/3862/CH6/EX6.16/Ex6_16.sce new file mode 100644 index 000000000..7badb22ec --- /dev/null +++ b/3862/CH6/EX6.16/Ex6_16.sce @@ -0,0 +1,12 @@ +clear +// +p1=5.0 //Pitch of smaller screw +p2=10.0 //Pitch of larger screw +R=500.0 //Lever arm length from centre of screw +W=15000.0 //Load +P=185.0 //Effort +VR=2*3.14*R/(p2-p1) //Velocity Ratio +MA=W/P //Mechanical advantage +Efficiency=MA/VR*100.0 + +printf("\n Efficiency %0.3f percentage",Efficiency) diff --git a/3862/CH6/EX6.17/Ex6_17.sce b/3862/CH6/EX6.17/Ex6_17.sce new file mode 100644 index 000000000..b82fd0e3d --- /dev/null +++ b/3862/CH6/EX6.17/Ex6_17.sce @@ -0,0 +1,24 @@ +clear +d=200.0 //Diameter of the load drum +R = 1200.0 // Length of lever arm +T1 = 10.0 //Number of teeth on pinion, +T2 = 100.0 //Number of teeth on spur wheel +VR=R*T2/(d*T1)*2.0 //Velocity Ratio +printf("\n Velocity Ratio is %0.3f",VR) +W1 = 3000.0 //Load 1 +P1= 100.0 //Effort1 + +W2 = 9000.0 //Load 2 +P2= 160.0 //Effort2 + +//law of machine is given by P=mW+C +m=(P2-P1)/(W2-W1) +C=P2-m*W2 +printf("\n Law of machine is P= %0.3f W + %0.3f ",m,C) +MA=W1/P1 //Mechanical advantage +Efficiency=MA/VR*100.0 +printf("\n Efficiency for first case %0.3f percentage",Efficiency) +MA=W2/P2 //Mechanical advantage +Efficiency=MA/VR*100.0 + +printf("\n Efficiency for second case %0.3f percentage",Efficiency) diff --git a/3862/CH6/EX6.18/Ex6_18.sce b/3862/CH6/EX6.18/Ex6_18.sce new file mode 100644 index 000000000..21c3fe117 --- /dev/null +++ b/3862/CH6/EX6.18/Ex6_18.sce @@ -0,0 +1,14 @@ +clear +d=150.0 //Diameter of the load drum +R = 400.0 // Length of lever arm +T1 = 15.0 //Number of teeth on pinion, +T3 = 20.0 //Number of teeth on pinion, +T2 = 45.0 //Number of teeth on spur wheel +T4 = 40.0 //Number of teeth on spur wheel +P= 250.0 //Effort +Efficiency=0.4 +VR=R*T2/(d*T1)*2.0*T4/T3 //Velocity Ratio + +W=VR*Efficiency*P //Load + +printf("\n LOad %0.3f N",W) diff --git a/3862/CH6/EX6.2/Ex6_2.sce b/3862/CH6/EX6.2/Ex6_2.sce new file mode 100644 index 000000000..809d5f09d --- /dev/null +++ b/3862/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,25 @@ +clear +// +W1 = 2400.0 //Load 1 +P1= 150.0 //Effort1 + +W2 = 3000.0 //Load 2 +P2= 180.0 //Effort2 +P3= 200.0 //Effort3 +//law of machine is given by P=mW+C +m=(P2-P1)/(W2-W1) +C=P2-m*W2 +printf("\n Law of machine is P= %0.3f W + %0.3f ",m,C) +W3=(P3-C)/m //Load 2 +printf("\n Load is %0.3f N",W3) +MA=W3/P3 //Mechanical advantage +VR=30.0 //Velocity Ratio +Efficiency=MA/VR*100 +Pi =W3/VR //Ideal effort +printf("\n Ideal effort is %0.3f N",Pi) + +efl=P3-Pi //Effort lost in friction + +printf("\n Effort lost in friction %0.3f",efl) +printf("\n Efficiency %0.3f",Efficiency) +printf("\n Mechanical advantage-- %0.3f",MA) diff --git a/3862/CH6/EX6.3/Ex6_3.sce b/3862/CH6/EX6.3/Ex6_3.sce new file mode 100644 index 000000000..e6f89678f --- /dev/null +++ b/3862/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,25 @@ +clear +// +W1 = 7700.0 //Load 1 +P1= 150.0 //Effort1 +MA=W1/P1 //Mechanical advantage +printf("\n Mechanical advantage-- %0.3f",MA) + +Efficiency=0.6 +VR=MA/Efficiency //Velocity Ratio +W2 = 13200.0 //Load 2 +P2= 250.0 //Effort2 +MA=W2/P2 +Efficiency=MA/VR*100 +//law of machine is given by P=mW+C +m=(P2-P1)/(W2-W1) + + +MMA=1/m //Maximum Mechanical advantage + +MaxEfficiency=MMA/VR*100 + +printf("\n Velocity Ratio %0.3f",VR) +printf("\n Efficiency %0.3f",Efficiency) +printf("\n Maximum Mechanical advantage-- %0.3f",MMA) +printf("\n Maximum Efficiency %0.3f",MaxEfficiency) diff --git a/3862/CH6/EX6.7/Ex6_7.sce b/3862/CH6/EX6.7/Ex6_7.sce new file mode 100644 index 000000000..09efd610f --- /dev/null +++ b/3862/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,10 @@ +clear +// +W = 12000.0 //Load +N=3.0 //number of movable pulleys +VR=2*N //Velocity Ratio +L=0.05 //Efficiency loss in each pulley +Efficiency=0.85 +MA=Efficiency*VR //Mechanical advantage +P = W/MA //Effort +printf("\n Effort is %0.3f N",P) diff --git a/3862/CH6/EX6.8/Ex6_8.sce b/3862/CH6/EX6.8/Ex6_8.sce new file mode 100644 index 000000000..b60127654 --- /dev/null +++ b/3862/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,11 @@ +clear +// +W = 12000.0 //Load +N1=2.0 //number of movable pulleys in system 1 +N2=2.0 //number of movable puleys in system 2 +VR=2*N1+2*N2 //Velocity Ratio +L=0.05 //Efficiency loss in each pulley +Efficiency=0.78 +MA=Efficiency*VR //Mechanical advantage +P = W/MA //Effort +printf("\n Effort is %0.3f N",P) diff --git a/3862/CH6/EX6.9/Ex6_9.sce b/3862/CH6/EX6.9/Ex6_9.sce new file mode 100644 index 000000000..e1722bfd2 --- /dev/null +++ b/3862/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,13 @@ +clear +// +W = 1000.0 //Load +N=3.0 //number of pulleys +VR=2**N-1 //Velocity Ratio +P = 180.0 //Effort +MA=W/P //Mechanical advantage +Efficiency=MA/VR*100 +Pi =W/VR //Ideal effort + +efl=P-Pi //Effort lost in friction +printf("\n Efficiency %0.3f",Efficiency) +printf("\n Effort lost in friction %0.3f",efl) diff --git a/3862/CH8/EX8.1/Ex8_1.sce b/3862/CH8/EX8.1/Ex8_1.sce new file mode 100644 index 000000000..b5268249f --- /dev/null +++ b/3862/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,19 @@ +clear +// + +//variable declaration + +P=(40000) //Load,N +E=(200000) //Modulus of elasticity for steel,N/mm^2 +L=500 //length of circular rod,mm +d=(16) //diameter of rod,mm + +A=(%pi*((d**2)))/4 //sectional area** mm^2 +p=P/A //stress, N/mm^2 +e=p/E //strain +delta=(P*L)/(A*E) //Elongation,mm + +printf("\n sectional area= %0.2f mm^2",A) +printf("\n stress= %0.2f N/mm^2",p) +printf("\n strain= %0.10f ",e) +printf("\n Elongation= %0.3f mm",delta) diff --git a/3862/CH8/EX8.11/Ex8_11.sce b/3862/CH8/EX8.11/Ex8_11.sce new file mode 100644 index 000000000..f97272901 --- /dev/null +++ b/3862/CH8/EX8.11/Ex8_11.sce @@ -0,0 +1,26 @@ +clear +// + +//variable declaration + +P=(200) //loading,KN +E=200*1000 +d1=40 //Young's modulus,N/mm^2 +A= %pi*(d1**2)/4 //Area of uniform portion**mm^2 +L1=1500 //length of uniform portion,mm +d2=60 //diameter of tapered section,mm +L2=500 //length of tapered section,mm +//Extensions of uniform portion and tapering portion are worked out separately and then added to get extension of the given bar. + +//Extension of uniform portion + +delta1=(P*1000*L1)/(A*E) + +printf("\n delta1= %0.3f mm",delta1) + +delta2=(P*1000*4*L2)/(E*%pi*d1*d2) + +printf("\n delta2= %0.3f mm",delta2) + +T=delta1 + delta2 +printf("\n Total extension %0.3f mm",T) diff --git a/3862/CH8/EX8.13/Ex8_13.sce b/3862/CH8/EX8.13/Ex8_13.sce new file mode 100644 index 000000000..64318da59 --- /dev/null +++ b/3862/CH8/EX8.13/Ex8_13.sce @@ -0,0 +1,30 @@ +clear +// + +//variable declaration + +P=(60) //load,KN +d=(25) //diameter,mm +A=%pi*(d**2)/4 //Area**mm^2 +L=(200) //gauge length,mm + +delta=0.12 //extension,mm +deltad=0.0045 //contraction in diameter,mm +Linearstrain=delta/L +Lateralstrain=deltad/d + +Pr=Lateralstrain/Linearstrain + +printf("\n Poissons ratio= %0.1f ",Pr) + +E=(P*1000*L)/(A*delta) + +printf("\n E= %0.2f N/mm^2",E) + +G=E/(2*(1+Pr)) //Rigidity modulus + +printf("\n G= %0.1f N/mm^2",G) + +K=E/(3*(1-(2*Pr))) //bulk modulus + +printf("\n K= %0.2f N/mm^2",K) diff --git a/3862/CH8/EX8.14/Ex8_14.sce b/3862/CH8/EX8.14/Ex8_14.sce new file mode 100644 index 000000000..2cd35001f --- /dev/null +++ b/3862/CH8/EX8.14/Ex8_14.sce @@ -0,0 +1,37 @@ +clear +// + +//variable declaration + +E=(2*100000) //Young's modulus,N/mm^2 +Pr=(0.3) //poisson's ratio + +G=E/(2*(1+Pr)) //Rigidity modulus + +K=E/(3*(1-2*(Pr))) //Bulk modulus + +printf("\n G= %0.1f N/mm^2",G) + +printf("\n K= %0.2f N/mm^2 ", K) + +P=60 //Load,kN +A=%pi*(25**2)/4 //Area**mm^2 + +Stress=P*1000/A //N/mm^2 +//Linear strain,ex + +ex=Stress/E + +//Lateralstrain,ey,ez + +ey=-1*Pr*ex +ez=-1*Pr*ex + +//volumetric strain,ev=ex+ey+ez + +ev=ex+ey+ez + +v=%pi*(25**2)*500/4 +Changeinvolume=ev*v + +printf("\n change in volume %0.2f mm^3",Changeinvolume) diff --git a/3862/CH8/EX8.15/Ex8_15.sce b/3862/CH8/EX8.15/Ex8_15.sce new file mode 100644 index 000000000..4f49987ab --- /dev/null +++ b/3862/CH8/EX8.15/Ex8_15.sce @@ -0,0 +1,35 @@ +clear +//variable declaration +// Let the x, y, z be the mutually perpendicular directions + +pr=(0.3) +PX=(15) //Loading in x-direction,KN +PY=(80) //Loading in Y-direction(compressive),KN +PZ=(180) //Loading in Z-direction,KN + +//Area in X-,Y-,Z-Direction is AX,AY,AZ respectively,mm^2 + +AX=(10*30) +AY=(10*400) +AZ=(30*400) + +//stress devoloped in X-,Y-,Z- direction as px,py,pz respectively,N/mm^2 + +px=PX*1000/AX +py=PY*1000/AY +pz=PZ*1000/AZ + +//Noting that a stress produces a strain of p/E in its own direction, the nature being same as that of stress and µ p E in lateral direction of opposite nature, and taking tensile stress as +ve, we can write expression for strains ex, ey, ez. +E=2*100000 //young's modulus,N/mm^2 + +ex=(px/E)+(pr*py/E)-(pr*pz/E) +ey=(-pr*px/E)-(py/E)-(pr*pz/E) +ez=(-pr*px/E)+(pr*py/E)+(pz/E) + +ev=ex+ey+ez //Volumetric strain + +volume=10*30*400 + +Changeinvolume=ev*volume + +printf("\n Change in volume= %0.2f mm^3",Changeinvolume) diff --git a/3862/CH8/EX8.17/Ex8_17.sce b/3862/CH8/EX8.17/Ex8_17.sce new file mode 100644 index 000000000..a444c31bc --- /dev/null +++ b/3862/CH8/EX8.17/Ex8_17.sce @@ -0,0 +1,13 @@ +clear +//variable declaration + +E=(2.1*100000) //Young’s modulus of the material,N/mm^2 +G=(0.78*100000) //modulus of rigidity,N/mm^2 + +pr=(E/(2*G))-1 + +printf("\n poissons Ratio= %0.3f ",pr) + +K=E/(3*(1-2*pr)) + +printf("\n Bulk modulus= %0.3f N/mm^2",K) diff --git a/3862/CH8/EX8.18/Ex8_18.sce b/3862/CH8/EX8.18/Ex8_18.sce new file mode 100644 index 000000000..2be3a616a --- /dev/null +++ b/3862/CH8/EX8.18/Ex8_18.sce @@ -0,0 +1,14 @@ +clear +//variable declaration + +G=(0.4*100000) //modulus of rigidity of material,N/mm^2 +K=(0.8*100000) //bulk modulus,N/mm^2 + +E=(9*G*K)/(3*K+G) + + +printf("\n Youngs modulus= %0.3f N",E) + +pr=(E/(2*G))-1 + +printf("\n Poissons Ratio %0.4f ",pr) diff --git a/3862/CH8/EX8.19/Ex8_19.sce b/3862/CH8/EX8.19/Ex8_19.sce new file mode 100644 index 000000000..506660ea3 --- /dev/null +++ b/3862/CH8/EX8.19/Ex8_19.sce @@ -0,0 +1,26 @@ +clear +//variable declaration + +L=(600) //compound bar of length,mm +P=(60) //compound bar when axial tensile force ,KN + +Aa=(40*20) //area of aluminium strip,mm^2 +As=(60*15) //area of steel strip,mm^2 + +Ea=1*100000 // elastic modulus of aluminium,N/mm^2 +Es=2*100000 // elastic modulus of steel,N/mm^2 + +//load shared by aluminium strip be Pa and that shared by steel be Ps. Then from equilibrium condition Pa+Ps=P +//From compatibility condition, deltaAL=deltaS +Pa=(P*1000)/(1+((As*Es)/(Aa*Ea))) +Ps=Pa*((As*Es)/(Aa*Ea)) + +Sias=Pa/Aa +printf("\n Stress in aluminium strip= %0.2f N/mm^2",Sias) +Siss=Ps/As +printf("\n Stress in steel strip= %0.2f N/mm^2",Siss) + +L=600 +//Extension of the compound bar +deltal=(Pa*L)/(Aa*Ea) +printf("\n Extension of the compound bar= %0.3f mm",deltal) diff --git a/3862/CH8/EX8.2/Ex8_2.sce b/3862/CH8/EX8.2/Ex8_2.sce new file mode 100644 index 000000000..5f8d1fb8a --- /dev/null +++ b/3862/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,18 @@ +clear +//variable declaration + +P=(120) // force applied during measurement,N +E=(200000) //Modulus of elasticity for steel,N/mm^2 +L=(30) //length of Surveyor’s steel tape,mm + + +A=15*0.75 //area, mm^2 +delta=((P*L*1000)/(A*E)) //Elongation,mm + +printf("\n area= %0.2f mm^2",A) +printf("\n Elongation= %0.3f mm",delta) + +printf("\n Hence, if measured length is %0.3f m.", L) +printf("\n Actual length is %0.6f m",(L+(delta/1000))) + +printf("\n Actual length of line AB= %0.3f m.",(150*(L+(delta/1000))/30)) diff --git a/3862/CH8/EX8.20/Ex8_20.sce b/3862/CH8/EX8.20/Ex8_20.sce new file mode 100644 index 000000000..8800b01b4 --- /dev/null +++ b/3862/CH8/EX8.20/Ex8_20.sce @@ -0,0 +1,25 @@ +clear +// + +//variable declaration + +Es=(2*100000) //Young's modulus of steel rod ,N/mm^2 +Ec=(1.2*100000) //Young's modulus of copper tube,N/mm^2 + +di=(25) //internal diameter,mm +de=(40) //external diameter,mm + +As=%pi*(di**2)/4 //Area of steel rod**mm^2 +Ac=%pi*((de**2)-(di**2))/4 //Area of copper tube**mm^2 +P=120 //load, KN +//From equation of equilibrium, Ps+Pc=P,where Ps is the load shared by steel rod and Pc is the load shared by the copper tube. +//From compatibility condition,deltaS=deltaC + +Pc=(P*1000)/(1+((As*Es)/(Ac*Ec))) +Ps=Pc*((As*Es)/(Ac*Ec)) + +SIC=Pc/Ac //stress in copper, N/mm^2 +SIS=Ps/As //stress in steel,N/mm^2 + +printf("\n stress in Copper= %0.2f N/mm^2",SIC) +printf("\n stress in Steel= %0.2f N/mm^2",SIS) diff --git a/3862/CH8/EX8.21/Ex8_21.sce b/3862/CH8/EX8.21/Ex8_21.sce new file mode 100644 index 000000000..5c7a1f805 --- /dev/null +++ b/3862/CH8/EX8.21/Ex8_21.sce @@ -0,0 +1,23 @@ +clear +// + +//variable declaration +//Es/Ec=18(given) +Er=(18) //young modulus ratio Er=Es/Ec +d=(16) //steel bar diameter,mm +//8 steel bars +As=8*%pi*(d**2)/4 //Area of steel bar**mm^2 +Ac=(300*500)-As //Area of concrete,mm^2 + +P=800 //Compressive force, KN +//From equation of equilibrium, Ps+Pc=P,where Ps is the load shared by steel bar and Pc is the load shared by the Concrete +//From compatibility condition,deltaS=deltaC + +Pc=(P*1000)/(1+((As*Er)/(Ac))) +Ps=Pc*((As*Er)/(Ac)) + +SIC=Pc/Ac //stress in Concrete, N/mm^2 +SIS=Ps/As //stress in steel,N/mm^2 + +printf("\n stress in Concrete= %0.2f N/mm^2",SIC) +printf("\n stress in Steel= %0.2f N/mm^2",SIS) diff --git a/3862/CH8/EX8.22/Ex8_22.sce b/3862/CH8/EX8.22/Ex8_22.sce new file mode 100644 index 000000000..9c4014752 --- /dev/null +++ b/3862/CH8/EX8.22/Ex8_22.sce @@ -0,0 +1,21 @@ +clear +//variable declaration + +Es=(2*100000) //Young's modulus of steel ,N/mm^2 +Ea=(1*100000) //Young's modulus of aluminium,N/mm^2 +Ls=240 //length of steel,mm +La=160 //length of aluminium,mm +Aa=1200 //Area of aluminium,mm^2 +As=1000 //Area of steel,mm^2 +P=250 //load, KN +//From equation of equilibrium, Ps+2Pa=P,et force shared by each aluminium pillar be Pa and that shared by steel pillar be Ps. +//From compatibility condition,deltaS=deltaC + +Pa=(P*1000)/(2+((As*Es*La)/(Aa*Ea*Ls))) +Ps=Pa*((As*Es*La)/(Aa*Ea*Ls)) + +SIA=Pa/Aa //stress in aluminium, N/mm^2 +SIS=Ps/As //stress in steel,N/mm^2 + +printf("\n stress in Aluminium= %0.2f N/mm^2",SIA) +printf("\n stress in Steel= %0.2f N/mm^2",SIS) diff --git a/3862/CH8/EX8.23/Ex8_23.sce b/3862/CH8/EX8.23/Ex8_23.sce new file mode 100644 index 000000000..2911ca612 --- /dev/null +++ b/3862/CH8/EX8.23/Ex8_23.sce @@ -0,0 +1,28 @@ +clear +// + +//variable declaration + +// Let the force shared by bolt be Ps and that by tube be Pc. Since there is no external force, static equilibrium condition gives Ps + Pc = 0 or Ps = – Pc i.e., the two forces are equal in magnitude but opposite in nature. Obviously bolt is in tension and tube is in compression. +//Let the magnitude of force be P. Due to quarter turn of the nut + +//[Note. Pitch means advancement of nut in one full turn] + +Ls=(600) //length of whole assembly,mm +Lc=(600) //length of whole assembly,mm +delta=(0.5) +ds=(20) //diameter,mm +di=(28) //internal diameter,mm +de=(40) //external diameter,mm +Es=(2*100000) //Young's modulus, N/mm^2 +Ec=(1.2*100000) +As=%pi*(ds**2)/4 //area of steel bolt**mm^2 +Ac=%pi*((de**2)-(di**2))/4 //area of copper tube**mm^2 + +P= (delta*(1/Ls))/((1/(As*Es))+(1/(Ac*Ec))) //Load,N + +ps=P/As //stress,N/mm^2 +pc=P/Ac //copper,N/mm^2 + +printf("\n ps= %0.2f N/mm^2",ps) +printf("\n pc= %0.2f N/mm^2",pc) diff --git a/3862/CH8/EX8.24/Ex8_24.sce b/3862/CH8/EX8.24/Ex8_24.sce new file mode 100644 index 000000000..16fa6bc55 --- /dev/null +++ b/3862/CH8/EX8.24/Ex8_24.sce @@ -0,0 +1,32 @@ +clear +//variable declaration +E=(2*100000) //Young's modulus,N/mm^2 +alpha=(0.000012) //expansion coeffecient,/°c +L=(12) //length,m +t=(40-18) //temperature difference,°c + +delta=alpha*t*L*1000 //free expansion of the rails,mm +// Provide a minimum gap of 3.168 mm between the rails, so that temperature stresses do not develop + +// a) If no expansion joint is provided, free expansion prevented is equal to 3.168 mm + +//delta=(P*L)/(A*E) & p=P/A where p is stress, P,A is load,area + +p1=(delta*E)/(L*1000) //stress developed , N/mm^2 + +printf("\n (a) p= %0.1f N/mm^2",p1) + +//(b) If a gap of 1.5 mm is provided, free expansion prevented delta2 = 3.168 – 1.5 = 1.668 mm. + +delta2=1.668 //mm +//delta2=(P*L)/(A*E) & p=P/A where p is stress, P,A is load,area + +p2=(delta2*E)/(L*1000) //stress developed , N/mm^2 + +printf("\n (b) p= %0.1f N/mm^2",p2) + +// If the stress developed is 20 N/mm2, then p = P/ A +p3=20 //stress developed,N/mm^2 +delta3=delta-(p3*L*1000/E) + +printf("\n (iii) delta= %0.3f mm",delta3) diff --git a/3862/CH8/EX8.26/Ex8_26.sce b/3862/CH8/EX8.26/Ex8_26.sce new file mode 100644 index 000000000..56ebcca94 --- /dev/null +++ b/3862/CH8/EX8.26/Ex8_26.sce @@ -0,0 +1,20 @@ +clear +//variable declaration + +Ea=70*1000 //Young's modulus of aluminium,N/mm^2 +Es=200*1000 //Young's modulus of steel,N/mm^2 + +alphaa=(0.000011) //expansion coefficient,/°C +alphas=(0.000012) //expansion coefficient,/°C + +Aa=600 //Area of aluminium portion,mm^2 +As=400 //Area of steel, mm^2 +La=(1.5) //length of aluminium portion,m +Ls=(3.0) //length of steel portion,m +t=18 //temperature,°C + +delta=(alphaa*t*La*1000)+(alphas*t*Ls*1000) //mm + +P=(delta)/(((La*1000)/(Aa*Ea))+((Ls*1000)/(As*Es))) + +printf("\n P= %0.1f N",P) diff --git a/3862/CH8/EX8.27/Ex8_27.sce b/3862/CH8/EX8.27/Ex8_27.sce new file mode 100644 index 000000000..a119b6d77 --- /dev/null +++ b/3862/CH8/EX8.27/Ex8_27.sce @@ -0,0 +1,21 @@ +clear +// + +//variable declaration + +d1=(25) // variation linearly in diameter from 25 mm to 50 mm +d2=(50) +L=(500) //length,mm +alpha=(0.000012) //expansion coeffecient,/°C +t=25 //rise in temperture,°C +E=2*100000 //Young's modulus,N/mm^2 + +delta=alpha*t*L + +//If P is the force developed by supports, then it can cause a contraction of 4*P*L/(%pi*d1*d2*E) + +P=(delta*%pi*d1*d2*E)/(4*L) +Am=%pi*(d1**2)/4 +Ms=P/Am + +printf("\n Corresponding maximum stress = %0.1f N/mm^2",Ms) diff --git a/3862/CH8/EX8.28/Ex8_28.sce b/3862/CH8/EX8.28/Ex8_28.sce new file mode 100644 index 000000000..c9511fb9d --- /dev/null +++ b/3862/CH8/EX8.28/Ex8_28.sce @@ -0,0 +1,32 @@ +clear +// + +//variable declaration + +Db=(20) //diameter of brass rod,mm +Dse=(40) //external diameter of steel tube,mm +Dsi=(20) //internal diameter of steel tube,mm +Es=(2*100000 ) //Young's modulus steel, N/mm^2 +Eb=(1*100000 ) //Young's modulus brass, N/mm^2 +alphas=(0.0000116) //coeffcient of expansion of steel,/°C +alphab=(0.0000187) //coeffcient of expansion of brass,/°C +t=60 //raise in temperature, °C +As=%pi*((Dse**2)-(Dsi**2))/4 //Area of steel tube** mm^2 +Ab=%pi*((Db**2))/4 //Area of brass rod**mm^2 +L=1200 //length,mm +//Since free expansion of brass is more than free expansion of steel , compressive force Pb develops in brass and tensile force Ps develops in steel to keep the final position at CC + +//Horizontal equilibrium condition gives Pb = Ps, say P. + +P=((alphab-alphas)*t*L)/((L/(As*Es))+(L/(Ab*Eb))) + +ps=P/As +pb=P/Ab + +printf("\n stress in steel= %0.2f N/mm^2",ps) +printf("\n Stress in brass= %0.2f N/mm^2",pb) + +//the pin resist the force P at the two cross- sections at junction of two bars. + +Shearstress=P/(2*Ab) +printf("\n Shear stress in pin %0.2f N/mm^2",Shearstress) diff --git a/3862/CH8/EX8.29/Ex8_29.sce b/3862/CH8/EX8.29/Ex8_29.sce new file mode 100644 index 000000000..5e832dde3 --- /dev/null +++ b/3862/CH8/EX8.29/Ex8_29.sce @@ -0,0 +1,22 @@ +clear +//variable declaration + +L=(1000) //length of the bar at normal temperature,mm +As=(50*10) //Area of steel,mm^2 +Ac=(40*5) //Area of copper,mm^2 +//Ac = Free expansion of copper is greater than free expansion of steel . To bring them to the same position, tensile force Ps acts on steel plate and compressive force Pc acts on each copper plate. +alphas=(0.000012) //Expansion of coeffcient of steel,/°C +alphac=(0.000017 ) //Expansion of coeffcient of copper,/°C +t=80 //raise by temperature, °C +Es=2*100000 //Young's modulus of steel,N/mm^2 +Ec=1*100000 //Young's modulus of copper,N/mm^2 +Pc=((alphac-alphas)*t*L)/((2*L/(As*Es)) +(L/(Ac*Ec))) +Ps=2*Pc + +pc=Pc/Ac //Stress in copper,N/mm^2 +ps=Ps/As //Stress in steel, N/mm^2 + +Changeinlength=alphas*t*L+(Ps*L/(As*Es)) + + +printf("\n Change in length= %0.2f mm",Changeinlength) diff --git a/3862/CH8/EX8.3/Ex8_3.sce b/3862/CH8/EX8.3/Ex8_3.sce new file mode 100644 index 000000000..a33b10123 --- /dev/null +++ b/3862/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,34 @@ +clear +// + +//variable declaration + +Y=(250) //Yield stress, N/mm^2 +FOS=(1.75) //Factor of safety +P=(160) //Load,KN + +p=Y/FOS + +printf("\n Therefore, permissible stress") + +printf("\n p= %0.3f N/mm^2 ",p) +printf("\n Load P= %0.3f N",P*1000) + +//p=P/A + +A=P*1000/p //area,mm^2 + +printf("\n A= %0.0f mm^2",A) + +//For hollow section of outer diameter ‘D’ and inner diameter ‘d’ A=%pi*(D^2-d^2)/4 +D=(101.6) //outer diameter,mm + +d=sqrt((D**2)-(4*A/%pi)) + +printf("\n d= %0.2f mm",d) + +t=(D-d)/2 +printf("\n t= %0.2f mm",t) + +printf("\n Hence, use of light section is recommended.") + diff --git a/3862/CH8/EX8.30/Ex8_30.sce b/3862/CH8/EX8.30/Ex8_30.sce new file mode 100644 index 000000000..23541ca6b --- /dev/null +++ b/3862/CH8/EX8.30/Ex8_30.sce @@ -0,0 +1,10 @@ +clear +//variable declaration + +p=(2) //internal pressure, N/mm^2 +t=12 //thickness of thin cylinder,mm +D=(1000) //internal diameter,mm + +f=(p*D)/(2*t) //Hoop stress,N/mm^2 + +printf("\n Hoop stress f= %0.2f N/mm^2",f) diff --git a/3862/CH8/EX8.4/Ex8_4.sce b/3862/CH8/EX8.4/Ex8_4.sce new file mode 100644 index 000000000..7b85cefc2 --- /dev/null +++ b/3862/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,36 @@ +clear +// + +//variable declaration + +d=(20) //Diameter ,mm +Loadatelasticlimit=(102) //Load at elastic limit,KN +P=80 //Load for extension of o.25mm , KN +delta=(0.25) //extension in specimen of steel,mm +L=200 //gauge length of specimen of steel,mm +Finalextension=(56) //total extension at fracture,mm + + +A=(%pi*(d**2))/4 //Area**mm^2 +printf("\n Area= %0.2f mm^2",A) + +Stressatelasticlimit=Loadatelasticlimit*1000/A //Stress at elastic limit,N/mm^2 +printf("\n Stress at elastic limit= %0.2f N/mm^2",Stressatelasticlimit) + +E=(P*1000/A)/(delta/L) //Young’s modulus ,N/mm^2 +printf("\n Youngs modulus E= %0.2f N/mm^22",E) + +Percentageelongation=Finalextension*100/L //percentage elongation,% +printf("\n Percentage elongation= %0.3f percentage",Percentageelongation ) + +Initialarea=(%pi*(d**2))/4 + +Finalarea=(%pi*(15**2))/4 // total extension at fracture is 56 mm and diameter at neck is 15 mm. +Percentagereductionina=(Initialarea-Finalarea)*100/Initialarea + +printf("\n Percentage reduction in area= %0.3f percentage",Percentagereductionina ) + +UltimateLoad=130 //Maximum Load=130,kN +UltimateTensileStress=UltimateLoad*1000/A + +printf("\n Ultimate Tensile Stress= %0.2f N/mm^2",UltimateTensileStress) diff --git a/3862/CH8/EX8.5/Ex8_5.sce b/3862/CH8/EX8.5/Ex8_5.sce new file mode 100644 index 000000000..cd677a9cc --- /dev/null +++ b/3862/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,22 @@ +clear +// + +//variable declaration + +P=(40) //Load,KN +L1=150 //length of 1st portion,mm +A1=%pi*(25**2)/4 //Area of 1st portion**mm^2 +L2=250 //length of 2nd portion,mm +A2=%pi*(20**2)/4 //Area of 2nd portion**mm^2 +L3=150 //length of 3rd portion,mm +A3=%pi*(25**2)/4 //Area of 3rd portion**mm^2 + +//E,Young's modulus ,N/mm^2 + +//Total extension= Extension of portion 1+Extension of portion 2+Extension of portion 3 + +//Extension=(P*1000*L)/(A*E) + +E=((P*1000*L1/A1)+(P*1000*L2/A2)+(P*1000*L3/A3))/0.28 + +printf("\n E= %0.2f N/mm^2",E) diff --git a/3862/CH8/EX8.6/Ex8_6.sce b/3862/CH8/EX8.6/Ex8_6.sce new file mode 100644 index 000000000..12c961f0b --- /dev/null +++ b/3862/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,23 @@ +clear +// + +//variable declaration + +P=(30) //Load,KN +L1=600 //length of 1st portion,mm +A1=40*20 //Area of 1st portion,mm^2 + +E1=200000 // material 1 Young’s modulus,N/mm^2 + +E2=100000 // material 2 Young’s modulus,N/mm^2 + + +L2=800 //length of 2nd portion,mm +A2=30*20 //Area of 2nd portion,mm^2 + +Extensionofportion1=(P*1000*L1)/(A1*E1) //mm +Extensionofportion2=(P*1000*L2)/(A2*E2) //mm + +Totalextensionofthebar= Extensionofportion1 + Extensionofportion2 + +printf("\n Total extension of the bar= %0.4f mm",Totalextensionofthebar) diff --git a/3862/CH9/EX9.1/Ex9_1.sce b/3862/CH9/EX9.1/Ex9_1.sce new file mode 100644 index 000000000..1c3b111e7 --- /dev/null +++ b/3862/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,28 @@ +clear +// + +//variable declaration + +//summation of all horizontal forces is zero & vertical forces is zero. +P1=(10) //Vertical down Load at 4m from A,KN +P2=(15) //Inclined down Load at angle 30° at 6m from A,KN +P3=(20) //Inclined down Load at angle 45° at 10m from A,KN +theta2=30 +theta3=45 +//horizontal,vertical component at A is Ha,Va respectively. + +Ha=P2*cos(theta2*%pi/180)+P3*cos(theta3*%pi/180) +Rb=(P1*4+P2*6*sin(theta2*%pi/180)+P3*10*sin(theta3*%pi/180))/12 //reaction at B point,KN + +printf("\n RB= %0.4f KN",Rb) + +//now vertical component +Va=P2*sin(theta2*%pi/180)+P3*sin(theta3*%pi/180)+P1-Rb + +Ra=sqrt((Ha**2)+(Va**2)) + +printf("\n RA= %0.4f KN",Ra) + +alpha=(atan(Va/Ha))*180/%pi + +printf("\n alpha= %0.2f °",alpha) diff --git a/3862/CH9/EX9.2/Ex9_2.sce b/3862/CH9/EX9.2/Ex9_2.sce new file mode 100644 index 000000000..0162a26c7 --- /dev/null +++ b/3862/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,29 @@ +clear +// + +//variable declaration + +//summation of all horizontal forces is zero & vertical forces is zero. +P1=(60) //inclined down to right Load at angle 60 at 1m from A,KN +P2=(80) //Inclined down to left Load at angle 75° at 3m from A,KN +P3=(50) //Inclined down to left Load at angle 60° at 5.5m from A,KN +theta1=60 +theta2=75 +theta3=60 +thetaRb=60 +//horizontal,vertical component at A is Ha,Va respectively. + +Rb=(P1*1*sin(theta1*%pi/180)+P2*3*sin(theta2*%pi/180)+P3*5.5*sin(theta3*%pi/180))/(6*sin(thetaRb*%pi/180)) //reaction at B point,KN +Ha=-P1*cos(theta1*%pi/180)+P2*cos(theta2*%pi/180)-P3*cos(theta3*%pi/180)+Rb*cos(thetaRb*%pi/180) +printf("\n RB= %0.4f KN",Rb) + +//now vertical component +Va=P1*sin(theta1*%pi/180)+P2*sin(theta2*%pi/180)+P3*sin(theta3*%pi/180)-Rb*sin(thetaRb*%pi/180) + +Ra=sqrt((Ha**2)+(Va**2)) + +printf("\n RA= %0.4f KN",Ra) + +alpha=(atan(Va/Ha))*180/%pi + +printf("\n alpha= %0.2f °",alpha) diff --git a/3862/CH9/EX9.3/Ex9_3.sce b/3862/CH9/EX9.3/Ex9_3.sce new file mode 100644 index 000000000..1eb307d46 --- /dev/null +++ b/3862/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,30 @@ +clear +// + +//variable declaration + +//summation of all horizontal forces is zero & vertical forces is zero. +P1=(20) //vertical down Load at 2m from A,KN +P2=(30) //uniform distributed load from 2m to 6m from A,KN/m(in 4m of span) +P3=(60) //Inclined down to right Load at angle 45° at 7m from A,KN + +theta3=45 +//horizontal,vertical component at B is Hb,Vb respectively. + +Ra=(P1*7+P2*4*5+P3*2*sin(theta3*%pi/180))/(9) //reaction at B point,KN + +printf("\n RA= %0.4f KN",Ra) + +Hb=P3*cos(theta3*%pi/180) +printf("\n HB= %0.4f KN",Hb) +//now vertical component +Vb=P1+P2*4+P3*sin(theta3*%pi/180)-Ra +printf("\n VB= %0.4f KN",Vb) + +Rb=sqrt((Hb**2)+(Vb**2)) + +printf("\n RB= %0.4f KN",Rb) + +alpha=(atan(Vb/Hb))*180/%pi + +printf("\n alpha= %0.2f °",alpha) diff --git a/3862/CH9/EX9.4/Ex9_4.sce b/3862/CH9/EX9.4/Ex9_4.sce new file mode 100644 index 000000000..9563245b2 --- /dev/null +++ b/3862/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,15 @@ +clear +//variable declaration +//Let the reactions at A be Ha, Va and Ma +//summation of all horizontal forces is zero & vertical forces is zero. + +P1=(20) //vertical down Load at 2m from A,KN +P2=(12) //vertical down Load at 3m from A,KN +P3=(10) //vertical down Load at 4m from A,KN +Pu=(16) //uniform distributed load from A to 2m from A,KN/m(in 2m of span) +////horizontal,vertical component at A is Ha,Va respectively. +printf("\n no horizontal force HA=0") +Va=Pu*2+P1+P2+P3 +printf("\n VA= %0.2f KN",Va) +Ma=Pu*2*1+P1*2+P2*3+P3*4 +printf("\n MA= %0.2f KN-m",Ma) diff --git a/3862/CH9/EX9.5/Ex9_5.sce b/3862/CH9/EX9.5/Ex9_5.sce new file mode 100644 index 000000000..250cf900d --- /dev/null +++ b/3862/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,15 @@ +clear +//variable declaration +//Let the reactions at A be Va and Ma +//summation of all horizontal forces is zero & vertical forces is zero. + +P1=(15) //vertical down Load at 3m from A,KN +P2=(10) //vertical down Load at 5m from A,KN +M=(30) //CW moment at 4m distance from A, KN-m +Pu=(20) //uniform distributed load from A to 2m from A,KN/m(in 2m of span) +////horizontal,vertical component at A is Ha,Va respectively. +printf("\n no horizontal force HA=0") +Va=Pu*2+P1+P2 +printf("\n VA= %0.2f KN",Va) +Ma=Pu*2*1+P1*3+P2*5+M +printf("\n MA= %0.2f KN-m",Ma) diff --git a/3862/CH9/EX9.6/Ex9_6.sce b/3862/CH9/EX9.6/Ex9_6.sce new file mode 100644 index 000000000..6d9c4a4b1 --- /dev/null +++ b/3862/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,15 @@ +clear +//variable declaration + +//As supports A and B are simple supports and loading is only in vertical direction, the reactions RA and RB are in vertical directions only. + +//summation of all horizontal forces is zero & vertical forces is zero. + +P1=(30) //vertical down Load at 1m from A,KN +P2=(40) //vertical down Load at 6.5m from A,KN +Pu=(20) //uniform distributed load from 2m to 5m from A,KN/m(in 3m of span). + +Rb=(Pu*3*3.5+P1*1+P2*6.5)/5 +printf("\n RB= %0.2f KN",Rb) +Ra=Pu*3+P1+P2-Rb +printf("\n RA= %0.2f KN",Ra) diff --git a/3862/CH9/EX9.9/Ex9_9.sce b/3862/CH9/EX9.9/Ex9_9.sce new file mode 100644 index 000000000..3a650583f --- /dev/null +++ b/3862/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,20 @@ +clear +//variable declaration + +//summation of all horizontal forces is zero & vertical forces is zero. + +//Let the left support C be at a distance x metres from A. + +P1=(30) //vertical down load at A,KN +Pu=(6) //uniform distributed load over whole span,KN/m,(20m of span) +P2=(50) //vertical down load at B, KN + +//Rc=Rd(given) reaction at C & D is equal. + +Rc=(P1+P2+Pu*20)/2 +Rd=Rc + +//taking moment at A +x=(((Pu*20*10+P2*20)/100)-12)/2 + +printf("\n X= %0.2f m",x) |