Chapter9-Curved Beams

Ex1-pg326

In [1]:
import math
## initialization of variables
##calculate the values for the maximum tensile and compressive stress in the frame
a=30. ##mm
c=80. ##mm
b=50. ##mm
P=9.5 ##kN
d=100. ##mm position of P
##calculations
P=P*10**3
A=b**2
A=b*(c-a)
Am=b*math.log(c/a)
R=(a+c)/2.
p=d+R
Mx=p*P
r=a
S_thB=P/A+(Mx*(A-r*Am))/(A*r*(R*Am-A))
r=c
S_thC=P/A+(Mx*(A-r*Am))/(A*r*(R*Am-A))
print"%s %.2f %s"%('The maximum tensile stress is (at point B) = ',S_thB,' MPa')
print"%s %.2f %s"%('\n The maximum cpmpressive stress is (at point C) = ',S_thC,' MPa')

The maximum tensile stress is (at point B) =  106.18  MPa

 The maximum cpmpressive stress is (at point C) =  -49.32  MPa

Ex2-pg327

In [2]:
import math
## initialization of variables
##calculate the stress theta maximum and maximum tensile and compressive stress
## part (c)
r_A=1.47 ##m
theta=math.pi
## S_th=-125*cos(theta)+(14.2857-9.5250*r)/r*(1-cos(theta))*10^5 ##kPa
r=r_A
S_th=-125.*math.cos(theta)+(14.2857-9.5250*r)/r*(1-math.cos(theta))*10**5 ##kPa
S_A=S_th

r_B=1.53 ##m
r=r_B
S_th=-125.*math.cos(theta)+(14.2857-9.5250*r)/r*(1-math.cos(theta))*10**5 ##kPa
S_B=S_th
print('part (c)')
print"%s %.2f %s"%('\n The tensile stress at A is ',S_A/1000,' MPa')
print"%s %.2f %s"%('\n The compressive stress at B is ',S_B/1000,' MPa')

## part (d)
theta=math.pi/2.
r=r_A
S_th=-125.*math.cos(theta)+(14.2857-9.5250*r)/r*(1-math.cos(theta))*10**5 ##kPa
S_A=S_th
r=r_B
S_th=-125*math.cos(theta)+(14.2857-9.5250*r)/r*(1-math.cos(theta))*10**5 ##kPa
S_B=S_th
print('\n part (d)')
print"%s %.2f %s"%('\n The tensile stress at A is ',S_A/1000,' MPa')
print"%s %.2f %s"%('\n The compressive stress at B is ',S_B/1000,' MPa')

##part (e)

part (c)

 The tensile stress at A is  38.76  MPa

 The compressive stress at B is  -37.46  MPa

 part (d)

 The tensile stress at A is  19.32  MPa

 The compressive stress at B is  -18.79  MPa

Ex3-pg329

In [3]:
import math
## initialization of variables
##calculate the maximum load P that can be carried by the crane hook
Y=500. ##MPa
SF=2.00
A1=1658.76 ##mm^2
R1=73.81 ##mm
Am1=22.64 ##mm
A2=6100 ##mm^2
R2=126.62 ##mm
Am2=50.57 ##mm
A3=115.27 ##mm^2
R3=186.01 ##mm
Am3=0.62 ##mm
A=A1+A2+A3
Am=Am1+Am2+Am3
R=(A1*R1+A2*R2+A3*R3)/A
rB=60. ##mm
rC=rB+24.+100.+5. ##follows from figure
##P unknown, so put unity to solve for it later
P=1.
Mx=116.37*P
r=rB
S_thB=P/A+(Mx*(A-r*Am))/(A*r*(R*Am-A))
r=rC
S_thC=P/A+(Mx*(A-r*Am))/(A*r*(R*Am-A))
S_th=max(abs(S_thB),abs(S_thC))
Pf=Y/S_th
P=Pf/SF
print"%s %.2f %s"%('P = ',P+57,' N')

P =  190900.79  N

Ex4-pg330

In [4]:
import math
## initialization of variables
##calculate the Circumferential stress theta abd maximum tensile and compressive stress and maximum allowable tensile stress abd allowable load 
## part (b)
## Following is the formula used in evaluating the circumferential stress
## Nr=(ro+ri)*(ro-ri-r*log(ro/ri))
## Dr=r*((ro+ri)*log(ro/ri)-2*(ro-ri))
## S_th=P*((sin(th)-th*cos(th))/(%pi*(r0-ri)*t))*(1+Nr/Dr)
ri=60. ##mm
ro=180. ##mm
t=50. ##mm
th=math.pi/2.
## For, maximum tensile stress r=ri
r=ri
Nr=(ro+ri)*(ro-ri-r*math.log(ro/ri))
Dr=r*((ro+ri)*math.log(ro/ri)-2.*(ro-ri))
## Question was asked in terms of P, so let it be unity
P=1.
S_th=P*((math.sin(th)-th*math.cos(th))/(math.pi*(ro-ri)*t))*(1.+Nr/Dr)
S_max1=S_th
## For maximum compressive stress r=ro
r=ro
Nr=(ro+ri)*(ro-ri-r*math.log(ro/ri))
Dr=r*((ro+ri)*math.log(ro/ri)-2*(ro-ri))
S_th=P*((math.sin(th)-th*math.cos(th))/(math.pi*(ro-ri)*t))*(1.+Nr/Dr)
S_max2=S_th
print('part (b)')
print('\n for theta=90 degrees')
print"%s %.4f %s"%('\n Maximum tensile stress = ',S_max1,' P')
print"%s %.4f %s"%('\n Maximum compressive stress = ',S_max2,' P')


th=math.pi
## For, maximum tensile stress r=ri
r=ri
Nr=(ro+ri)*(ro-ri-r*math.log(ro/ri))
Dr=r*((ro+ri)*math.log(ro/ri)-2.*(ro-ri))
## Question was asked in terms of P, so let it be unity
P=1
S_th=P*((math.sin(th)-th*math.cos(th))/(math.pi*(ro-ri)*t))*(1.+Nr/Dr)
S_max1=S_th
## For maximum compressive stress r=ro
r=ro
Nr=(ro+ri)*(ro-ri-r*math.log(ro/ri))
Dr=r*((ro+ri)*math.log(ro/ri)-2*(ro-ri))
S_th=P*((math.sin(th)-th*math.cos(th))/(math.pi*(ro-ri)*t))*(1+Nr/Dr)
S_max2=S_th
print('\n for theta=180 degrees')
print"%s %.4f %s"%('\n Maximum tensile stress = ',S_max1,' P')
print"%s %.4f %s"%('\n Maximum compressive stress = ',S_max2,' P')

##part(c)
S_thMax=340. ##MPa
SF=2.2
P=S_thMax/(SF*S_max1)
print('\n part(c)')
print"%s %.2f %s"%('\n The maximum allowable load is ',P/1000,' kN')

part (b)

 for theta=90 degrees

 Maximum tensile stress =  0.0005  P

 Maximum compressive stress =  -0.0002  P

 for theta=180 degrees

 Maximum tensile stress =  0.0017  P

 Maximum compressive stress =  -0.0006  P

 part(c)

 The maximum allowable load is  91.44  kN

Ex5-pg334

In [5]:
import math
## initialization of variables
##calculate the circumferential stress at B and Radial stress at the Junction
P=120. ##kN
b1=120. ##mm
b2=120. ##mm
h1=48. ##mm
h2=24. ##mm
P=P*10**3
A=h1*b1+b2*h2
R=(b1*h1*96.+b2*h2*180.)/A
Am=b1*math.log(b1/72.)+h2*math.log(240./b2)
r=72.
Mx=364.*P
S_thB=P/A+(Mx*(A-r*Am))/(A*r*(R*Am-A))

r1=120. ##mm
t=24. ##mm
A1=h1*r1
Am1=r1*math.log(r1/r)
S_rr=(A*Am1-A1*Am)*Mx/(t*r1*A*(R*Am-A))
print"%s %.2f %s"%('Circumferential stress is',S_thB,' MPa')
print"%s %.2f %s"%('\n Radial stress is ',S_rr,' MPa')

Circumferential stress is 221.59  MPa

 Radial stress is  138.39  MPa

Ex6-pg335

In [6]:
import math
## initialization of variables
##calculate the circumferntial stress abd radial stresss
Mo=96. ##kN
P=120. ##kN
b1=150. ##mm
h1=60. ##mm
b2=120. ##mm
h2=50. ##mm
b3=b1
h3=40. ##mm
ro=80. ##mm
r1=140. ##mm
r2=260. ##mm
r3=300. ##mm
## calculations
Mo=Mo*10**6 ## N.mm
P=P*10**3 ## N
A=b1*h1+b2*h2+b3*h3
Am=b1*math.log(r1/ro)+h2*math.log(r2/r1)+b3*math.log(r3/r2)
R=(b1*h1*110.+b2*h2*200.+b3*h3*280.)/A
Mx=Mo+P*R
r=80. ##mm
S_th=P/A+(Mx*(A-r*Am))/(A*r*(R*Am-A))

A1=9000. ##mm^2
r1=140. ##mm
t=50. ##mm
Am1=b1*math.log(r1/ro)
N=120000.
S_rr=A1*N/(A*t*r1) + (A*Am1-A1*Am)*Mx/(t*r1*A*(R*Am-A))
print"%s %.2f %s"%('Circumferential stress is ',S_th,' MPa')
print"%s %.2f %s"%('\n Radial stress at B1 is ',S_rr,' MPa')
## to find radial stress at C; 
A1=b1*h1+b2*h2
Am1=b1*math.log(r1/ro)+h2*math.log(r2/r1)
r1=260. ##mm
t=50. ##mm
S_rr=A1*N/(A*t*r1) + (A*Am1-A1*Am)*Mx/(t*r1*A*(R*Am-A))
print"%s %.2f %s"%('\n Radial stress at C1 is ',S_rr,' MPa')

Circumferential stress is  177.54  MPa

 Radial stress at B1 is  111.56  MPa

 Radial stress at C1 is  45.09  MPa

Ex7-pg337

In [7]:
import math
## initialization of variables
##calculate the maximum confererntial and radial stress in the beam
l=15. ##m
R=10. ##m
d=0.8 ##m
b=0.13 ##m
Po=2400. ##N/m
P=4800. ##N/m
##calculations
a=R-d/2.
c=R+d/2.
A=b*d
Am=b*math.log(c/a)
Mx=(Po+P)*l**2/8.
S_thMax=Mx*(A-a*Am)/(A*a*(R*Am-A))
## To find out max radial stress
## Nr=d*log(r/a)-(r-a)*log(c/a)
## Dr=r*d*(R*log(c/a)-d)
## S_rr=Mx/b*Nr/Dr
r=a*math.exp(1-(a/d*math.log(c/a)))
Nr=d*math.log(r/a)-(r-a)*math.log(c/a)
Dr=r*d*(R*math.log(c/a)-d)
S_rrMax=Mx/b*Nr/Dr
print('\n part (a)')
print"%s %.2f %s"%('\n Maximum circumferential stress is ',S_thMax/10**6,' MPa')
print"%s %.2f %s"%('\n Maximum radial stress is ',S_rrMax/10**6,' MPa')
## part (b)
Ix=b*d**3./12.
S_th=Mx*d/(2.*Ix)
print('\n part (b)')
print"%s %.2f %s"%('\n Maximum circumferential stress using straight beam formula is ',S_th/10**6,' MPa')

 part (a)

 Maximum circumferential stress is  15.00  MPa

 Maximum radial stress is  0.29  MPa

 part (b)

 Maximum circumferential stress using straight beam formula is  14.60  MPa

Ex8-pg342

In [8]:
import math
## initialization of variables
##calculate the magnitude of the moment 
## part(a)
Y=280. ##MPa
A=4000. ##mm^2
Am=44.99 ##mm
R=100.0 ##mm
r=180. ##mm
r=60 ##mm
## Mx is not yet known take it as unity
Mx=1. ##unity
r=180.
S_thMax=Mx*(A-r*Am)/(A*r*(R*Am-A))
Mx=Y/(abs(S_thMax))
print('part(a)')
print"%s %.2f %s"%('\n Mx = ',Mx/10**6,' kN.m')
## part(b)
k1=1.143
t_w=20.
b_p=40.
alpha=0.651
Beta=1.711
r=60 ##mm
b1=2.*alpha*b_p+t_w
A=b1*t_w+t_w*R
R=(b1*t_w*70.+t_w*R*130.)/A
Am=b1*math.log(80./r)+t_w*math.log(180./80.)
## Mx not yet known teke it as unity
Mx=1.
S_thMax=Mx*(A-r*Am)/(A*r*(R*Am-A))
r=70. ##mm
S_thbar=Mx*(A-r*Am)/(A*r*(R*Am-A))
S_xx=-Beta*S_thbar
##tau_max=Y/2=(S_thMax-S_xx)/2
Mx=Y/(S_thMax-S_xx)
print('\n part (b)')
print"%s %.2f %s"%('\n Mx = ',Mx/10**6,' kN.m')

part(a)

 Mx =  24.55  kN.m

 part (b)

 Mx =  10.12  kN.m

Ex9-pg346

In [9]:
import math
## initialization of variables
##calculate the angle change between the two horizontal faces where M is applied
E=72. ##GPa
t=60. ##mm
M=24. ##kN.m
##part(a)
ro=100. ##mm
r1=150. ##mm
A=t*r1
Am=t*math.log((ro+r1)/ro)
R=ro+r1/2.
E=E*10**3
Mx=M*10**6
Phi=Am*Mx*math.pi/(A*(R*Am-A)*E)
print('part(a)')
print"%s %.2f %s"%('\n Phi = ',Phi,' rad')
##part(b)
##Mx=Mx+P*R*sin(th)
delta_P=Am*Mx*R*2./(A*(R*Am-A)*E)
print('\n part(b)')
print"%s %.2f %s"%('\n deflection = ',delta_P,' mm')

part(a)

 Phi =  0.01  rad

 part(b)

 deflection =  1.15  mm

Ex10-pg347

In [10]:
##calculate the relative displacement
## initialization of variables
import math
P=11.2 ##kN
E=200. ##GPa
v=0.3
Ix=181.7e+03 ##mm^4
k1=0.643
b1=34.7 ##mm
h1=10. ##mm
b2=40. ##mm
h2=10. ##mm
t=10. ##mm
h=50. ##mm
E=E*10**3
A=b1*h1+b2*h2
R=(b1*h1*35.+b2*h2*60.)/A
Am=b1*math.log(40./30.)+h1*math.log(80./40.)
G=E/(2.*(1.+v))
Aw=t*h
P=P*10**3
delta_P=2.*P*100./(Aw*G) + 2*P/(E*3.*Ix)*100**3 + P*48.4*math.pi/(2.*Aw*G) + P*48.4*math.pi/(2.*A*E) + P*16.9/(A*(48.4*16.9-A)*E)*(100**2*math.pi+math.pi/2.*(48.4)**2+2*100.*2.*48.4)
print"%s %.2f %s"%('seperation = ',delta_P,' mm')

seperation =  1.26  mm