from math import pi
#variable declaration
P=float(40000) #Load,N
E=float(200000) #Modulus of elasticity for steel,N/mm^2
L=500 #length of circular rod,mm
d=float(16) #diameter of rod,mm
A=(pi*(pow(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
print "sectional area=",round(A,2),"mm^2"
print "stress=",round(p,2),"N/mm^2"
print "strain=",e,"N/mm^2"
print "Elongation=",round(delta,3),"mm"
#variable declaration
P=float(120) # force applied during measurement,N
E=float(200000) #Modulus of elasticity for steel,N/mm^2
L=float(30) #length of Surveyor’s steel tape,mm
A=15*0.75 #area, mm^2
delta=((P*L*1000)/(A*E)) #Elongation,mm
print "area=",round(A,2),"mm^2"
print "Elongation=",round(delta,3),"mm"
print "Hence, if measured length is", L,"m."
print "Actual length is" ,round((L+(delta/1000)),6),"m"
print "Actual length of line AB=",round((150*(L+(delta/1000))/30),3),"m."
from math import pi,sqrt
#variable declaration
Y=float(250) #Yield stress, N/mm^2
FOS=float(1.75) #Factor of safety
P=float(160) #Load,KN
p=Y/FOS
print "Therefore, permissible stress"
print "p=",round(p,3), "N/mm^2"
print "Load P=",P*1000,"N"
#p=P/A
A=P*1000/p #area,mm^2
print "A=",round(A),"mm^2"
#For hollow section of outer diameter ‘D’ and inner diameter ‘d’ A=pi*(D^2-d^2)/4
D=float(101.6) #outer diameter,mm
d=sqrt(pow(D,2)-(4*A/pi))
print "d=",round(d,2),"mm"
t=(D-d)/2
print "t=",round(t,2),"mm"
print "Hence, use of light section is recommended."
from math import pi
#variable declaration
d=float(20) #Diameter ,mm
Loadatelasticlimit=float(102) #Load at elastic limit,KN
P=80 #Load for extension of o.25mm , KN
delta=float(0.25) #extension in specimen of steel,mm
L=200 #gauge length of specimen of steel,mm
Finalextension=float(56) #total extension at fracture,mm
A=(pi*pow(d,2))/4 #Area,mm^2
print "Area=", round(A,2),"mm^2"
Stressatelasticlimit=Loadatelasticlimit*1000/A #Stress at elastic limit,N/mm^2
print "Stress at elastic limit=",round(Stressatelasticlimit,2),"N/mm^2"
E=(P*1000/A)*(delta*L) #Young’s modulus ,N/mm^2
print "Young's modulus E=", round(E,2),"N/mm^22"
Percentageelongation=Finalextension*100/L #percentage elongation,%
print "Percentage elongation=", round(Percentageelongation,2),"%"
Initialarea=(pi*pow(d,2))/4
Finalarea=(pi*pow(15,2))/4 # total extension at fracture is 56 mm and diameter at neck is 15 mm.
Percentagereductioninarea=(Initialarea-Finalarea)*100/Initialarea
print "Percentage reduction in area=",round(Percentagereductioninarea,2),"%"
UltimateLoad=130 #Maximum Load=130,kN
UltimateTensileStress=UltimateLoad/A
print"Ultimate Tensile Stress=",round(UltimateTensileStress,2),"N/mm^2"
from math import pi
#variable declaration
P=float(40) #Load,KN
L1=150 #length of 1st portion,mm
A1=pi*pow(25,2)/4 #Area of 1st portion,mm^2
L2=250 #length of 2nd portion,mm
A2=pi*pow(20,2)/4 #Area of 2nd portion,mm^2
L3=150 #length of 3rd portion,mm
A3=pi*pow(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)
print "E=",round(E,2),"N/mm^2"
from math import pi
#variable declaration
P=float(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
print"Total extension of the bar=",round(Totalextensionofthebar,4),"mm"
from math import pi
#variable declaration
P=float(30) #Load,KN
L1=600 #length of 1st portion,mm
A1=pi*pow(30,2)/4 #Area of 1st portion,mm^2
L2=400 #length of 2nd portion,mm
A2=pi*(pow(30,2)-pow(10,2))/4 #Area of 2nd portion,mm^2
#E,Young's modulus ,N/mm^2
#Total extension= Extension of portion 1+Extension of portion 2
#Extension=(P*1000*L)/(A*E)
T=float(0.222) #Total extension of the bar,mm
E=((P*1000*L1/A1)+(P*1000*L2/A2))/T
print "E=",round(E,2),"N/mm^2"
import math
#variable declaration
t=10 #steel flat thickness,mm
b1=float(60) #tapering from b1 to b2
b2=40
L=600 #steel flat length
P=float(80) #Load,KN
E=2*100000 #Young's Modulus,N/mm^2
#Extension of the tapering bar of rectangular section
delta1=(P*1000*L*math.log((b1/b2),10))/(t*E*(b1-b2))
print "delta1=",round(delta1,4),"mm"
print "there is calculation mistake in book"
#If averages cross-section is considered instead of tapering cross-section, extension is given by
Aav=(b1+b2)*t/2 #mm^2
delta2=(P*1000*L)/(Aav*E) #mm
print"delta2=",round(delta2,3),"mm^2"
P= (delta2-delta1)*100/delta2
print"Percentage error=",round(P,3),"%"
print "there is calculation mistake in book"
from math import pi
#variable declaration
P=float(200) #loading,KN
E=200*1000
d1=40 #Young's modulus,N/mm^2
A= pi*pow(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)
print "delta1=",round(delta1,3),"mm"
delta2=(P*1000*4*L2)/(E*pi*d1*d2)
print "delta2=",round(delta2,3),"mm"
T=delta1 + delta2
print "Total extension",round(T,3),"mm"
from math import pi
#variable declaration
P=float(60) #load,KN
d=float(25) #diameter,mm
A=pi*pow(d,2)/4 #Area,mm^2
L=float(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
print "Poisson's ratio=",round(Pr,1)
E=(P*1000*L)/(A*delta)
print "E=",round(E,2),"N/mm^2"
G=E/(2*(1+Pr)) #Rigidity modulus
print "G=",round(G,1),"N/mm^2"
K=E/(3*(1-(2*Pr))) #bulk modulus
print "K=",round(K,2),"N/mm^2"
from math import pi
#variable declaration
E=float(2*100000) #Young's modulus,N/mm^2
Pr=float(0.3) #poisson's ratio
G=E/(2*(1+Pr)) #Rigidity modulus
K=E/(3*(1-2*(Pr))) #Bulk modulus
print "G=", round(G,1),"N/mm^2"
print "K=", round(K,2), "N/mm^2"
P=60 #Load,kN
A=pi*pow(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*pow(25,2)*500/4
Changeinvolume=ev*v
print"change in volume",round(Changeinvolume,2),"mm^3"
#variable declaration
# Let the x, y, z be the mutually perpendicular directions
pr=float(0.3)
PX=float(15) #Loading in x-direction,KN
PY=float(80) #Loading in Y-direction(compressive),KN
PZ=float(180) #Loading in Z-direction,KN
#Area in X-,Y-,Z-Direction is AX,AY,AZ respectively,mm^2
AX=float(10*30)
AY=float(10*400)
AZ=float(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
print "Change in volume=",round(Changeinvolume,2),"mm^3"
#variable declaration
E=float(2.1*100000) #Young’s modulus of the material,N/mm^2
G=float(0.78*100000) #modulus of rigidity,N/mm^2
pr=(E/(2*G))-1
print "poisson's Ratio=",round(pr,3)
K=E/(3*(1-2*pr))
print "Bulk modulus=",round(K,3),"N/mm^2"
#variable declaration
G=float(0.4*100000) #modulus of rigidity of material,N/mm^2
K=float(0.8*100000) #bulk modulus,N/mm^2
E=(9*G*K)/(3*K+G)
print "Young's modulus=",round(E,3),"N"
pr=(E/(2*G))-1
print "Poisson's Ratio",round(pr,4)
#variable declaration
L=float(600) #compound bar of length,mm
P=float(60) #compound bar when axial tensile force ,KN
Aa=float(40*20) #area of aluminium strip,mm^2
As=float(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
print "Stress in aluminium strip=",round(Sias,2),"N/mm^2"
Siss=Ps/As
print "Stress in steel strip=",round(Siss,2),"N/mm^2"
L=600
#Extension of the compound bar
deltal=(Pa*L)/(Aa*Ea)
print"Extension of the compound bar=",round(deltal,3),"mm"
from math import pi
#variable declaration
Es=float(2*100000) #Young's modulus of steel rod ,N/mm^2
Ec=float(1.2*100000) #Young's modulus of copper tube,N/mm^2
di=float(25) #internal diameter,mm
de=float(40) #external diameter,mm
As=pi*pow(di,2)/4 #Area of steel rod,mm^2
Ac=pi*(pow(de,2)-pow(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
print "stress in Copper=",round(SIC,2),"N/mm^2"
print "stress in Steel=",round(SIS,2),"N/mm^2"
from math import pi
#variable declaration
#Es/Ec=18(given)
Er=float(18) #young modulus ratio Er=Es/Ec
d=float(16) #steel bar diameter,mm
#8 steel bars
As=8*pi*pow(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
print "stress in Concrete=",round(SIC,2),"N/mm^2"
print "stress in Steel=",round(SIS,2),"N/mm^2"
#variable declaration
Es=float(2*100000) #Young's modulus of steel ,N/mm^2
Ea=float(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
print "stress in Aluminium=",round(SIA,2),"N/mm^2"
print "stress in Steel=",round(SIS,2),"N/mm^2"
from math import pi
#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=float(600) #length of whole assembly,mm
Lc=float(600) #length of whole assembly,mm
delta=float(0.5)
ds=float(20) #diameter,mm
di=float(28) #internal diameter,mm
de=float(40) #external diameter,mm
Es=float(2*100000) #Young's modulus, N/mm^2
Ec=float(1.2*100000)
As=pi*pow(ds,2)/4 #area of steel bolt,mm^2
Ac=pi*(pow(de,2)-pow(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
print "ps=",round(ps,2),"N/mm^2"
print "pc=",round(pc,2),"N/mm^2"
#variable declaration
E=float(2*100000) #Young's modulus,N/mm^2
alpha=float(0.000012) #expansion coeffecient,/°c
L=float(12) #length,m
t=float(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
print "(a) p=", round(p1,1)," N/mm^2"
#(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
print "(b) p=", round(p2,1)," N/mm^2"
# 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)
print " (iii) delta=",round(delta3,3),"mm"
from math import pi
#variable declaration
# Let D be the diameter of ring after heating and ‘d’ be its diameter before heating
D=float(1.2*1000) #mm
#Circumference of ring after heating Ca= pi*D & Circumference of ring before heating Cb= pi*d
Ca=pi*D
Cb=pi*d
alphas=float(0.000012) #coefficient of expansion,/°C
t=150 #temperature change,°C
Es=2*100000 #young's modulus,N/mm^2
d=(Ca-Cb)/(alphas*t*pi)
#when it cools expansion prevented
#delta=pi*(D-d)
delta=alphas*t*pi*d
p=(delta*Es)/(pi*d) #stress,N/mm^2
print "stress p=",round(p,2),"N/mm^2"
#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=float(0.000011) #expansion coefficient,/°C
alphas=float(0.000012) #expansion coefficient,/°C
Aa=600 #Area of aluminium portion,mm^2
As=400 #Area of steel, mm^2
La=float(1.5) #length of aluminium portion,m
Ls=float(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)))
print "P=" ,round(P,1),"N"
from math import pi
#variable declaration
d1=float(25) # variation linearly in diameter from 25 mm to 50 mm
d2=float(50)
L=float(500) #length,mm
alpha=float(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*pow(d1,2)/4
Ms=P/Am
print "Corresponding maximum stress = ",round(Ms,1),"N/mm^2"
from math import pi
#variable declaration
Db=float(20) #diameter of brass rod,mm
Dse=float(40) #external diameter of steel tube,mm
Dsi=float(20) #internal diameter of steel tube,mm
Es=float(2*100000 ) #Young's modulus steel, N/mm^2
Eb=float(1*100000 ) #Young's modulus brass, N/mm^2
alphas=float(0.0000116) #coeffcient of expansion of steel,/°C
alphab=float(0.0000187) #coeffcient of expansion of brass,/°C
t=60 #raise in temperature, °C
As=pi*(pow(Dse,2)-pow(Dsi,2))/4 #Area of steel tube, mm^2
Ab=pi*(pow(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
print "stress in steel=",round(ps,2),"N/mm^2"
print "Stress in brass=",round(pb,2),"N/mm^2"
#the pin resist the force P at the two cross- sections at junction of two bars.
Shearstress=P/(2*Ab)
print "Shear stress in pin",round(Shearstress,2),"N/mm^2"
#variable declaration
L=float(1000) #length of the bar at normal temperature,mm
As=float(50*10) #Area of steel,mm^2
Ac=float(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=float(0.000012) #Expansion of coeffcient of steel,/°C
alphac=float(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))
print"Change in length=",round(Changeinlength,2),"mm"
#variable declaration
p=float(2) #internal pressure, N/mm^2
t=12 #thickness of thin cylinder,mm
D=float(1000) #internal diameter,mm
f=(p*D)/(2*t) #Hoop stress,N/mm^2
print "Hoop stress f=",round(f,2),"N/mm^2"