# Chapter 6 : Shearing Stresses in Beams and Thin-Walled Members¶

## Example 6.01, Page number 386¶

In [1]:
import math
from sympy import symbols
#Variable declaration
l=0.020                                                                           # Length(m)
V=500                                                                             # Vertical shear(N)
y=0.060                                                                           # Distance(m)

#Calculation
A=l*b                                                                             # Area(m**2)
Q=A*y                                                                             # First moment of an area with respect to a given axis
I=(1/12.0)*(0.020)*(0.1**3) + 2*((1/12.0)*(0.1)*(0.02**3) + (0.020*0.1)*(0.06**2))    # Moment of inertia(m**4)
q=(V*Q)/(I)
F=(0.025)*q                                                                       # Shearing force in each nail(N)

# Result
print ('Shearing force in each nail is = %lf N' %F)

Shearing force in each nail is = 92.592593 N


## Example 6.02, Page number 389¶

In [2]:
import math
from sympy import symbols
#Variable declaration
Vmax=4.50                                                                           # Force maximum(kips)
b=3.5                                                                               # Actual width of beam(in)
h=14.55                                                                             # Depth(in)

#Calculation
Tmax=(3/2.0)*(Vmax/(b*h))                                                             # Maximum shearing stress(ksi)

# Result
print ('Maximum shearing stress in a narrow rectangular beam = %lf ksi' %Tmax)

Maximum shearing stress in a narrow rectangular beam = 0.132548 ksi


## Example 6.03, Page number 389¶

In [3]:
import math
from sympy import symbols
#Variable declaration
tw=5.8                                                     # Distance(mm)
d=349                                                      # Distance(mm)
Vmax=58                                                    # Force(kN)

#Calculation
Aweb=d*tw                                                  # Area(mm*2)
Tmax=(Vmax/Aweb)*(1000)                                    # Maximum shearing stress(ksi)

# Result
print ('Maximum allowable shearing stress for steel beam = %lf MPa' %Tmax)

Maximum allowable shearing stress for steel beam = 28.653295 MPa


## SAMPLE PROBLEM 6.1, Page number 392¶

In [3]:
import math

#Variable declaration
A=B=1.5                                                                       # Force(kN)
V=1.5                                                                         # Force(kN)
y1=0.0417                                                                     # Distance(m)
y2=0.0583                                                                     # Distance(m)
AreaA=0.100*0.020                                                             # Area(m**2)
AreaB=0.060*0.020                                                             # Area(m**2)
I=8.63*(pow(10,-6))                                                           # Moment of inertia(m**2)
t=0.020                                                                       # Distance(m)

#Calculation
#Shearing Stress in Joint a
Qa=AreaA*y1
taweA=round((V*Qa)/(I*t),0)                                                   # Shearing stress in joint a(kPa)

#Shearing Stress in Joint b
Qb=AreaB*y2
taweB=round((V*Qb)/(I*t),0)                                                   # Shearing stress in joint b(kPa)

# Result
print ('Shearing stress in joint a = %lf kPa' %taweA)
print ('Shearing stress in joint b = %lf kPa' %taweB)

Shearing stress in joint a = 725.000000 kPa
Shearing stress in joint b = 608.000000 kPa


## SAMPLE PROBLEM 6.2, Page number 393¶

In [10]:
import math
from sympy import symbols

#Variable declaration
#Maximum Shear and Bending Moment
I=symbols('I')                                                             # Variable declaration
b=symbols('b')                                                             # Variable declaration
d=symbols('d')                                                             # Variable declaration
c=symbols('c')                                                             # Variable declaration
S=symbols('S')                                                             # Variable declaration
Mmax=90                                                                    # Maximum bending moment(kip.in)
Vmax=3                                                                     # Force(kips)
sall=1800                                                                  # Stress all(psi)
tall=120                                                                   # Shearing stress(psi)
b=3.5                                                                      # Width(in)

#Calculation
#Design Based on Allowable Normal Stress
I=(1/12.0)*(b)*(d**3)                                                        # Moment of inertia(m**4)
S=round((1/6.0)*(b),4)*(d**2)                                                # Section modulus
d=round(sqrt(((90*(pow(10,3)))/(1800))*(1/0.5833)),2)                      # Distance(in)

#Check Shearing Stress
Area=(b)*(d)                                                               # Area(in**2)
tm=round((3/2.0)*((Vmax*1000)/Area),1)                                       # Allowable shearing stress(psi)

#Design Based on Allowable Shearing Stress
d=round((3*3000)/(2*3.5*120),2)                                            # Distance(in)

# Result
print ('Minimum required depth d of beam = %lf in' %d)

Minimum required depth d of beam = 10.710000 in


## Example 6.04, Page number 401¶

In [11]:
import math
from sympy import symbols

#Variable declaration
l=0.75                                                     # Distance(in)
V=600                                                      # Vertical shear(lb)
y=1.875                                                    # Distance(in)

#Calculation
Q=l*b*y
I=(1/12.0)*((4.5**4)-(3**4))                                 # Moment of inertia(in**4)
q=(V*Q)/(I)
F=(1.75)*(46.15)                                           # Shearing force(lb)

# Result
print ('Shearing force in each nail = %lf lb' %F)

Shearing force in each nail = 80.762500 lb


## SAMPLE PROBLEM 6.3, Page number 406¶

In [4]:
import math
from sympy import symbols
#Variable declaration

#Calculation
Q=round((4.31)*(0.770)*(4.815),2)
t=round(((50)*(15.98))/(394*0.770),2)                                             # Normal shearing stress(ksi)

# Result
print ('Horizontal shearing stress = %lf ksi' %t)

Horizontal shearing stress = 2.630000 ksi


## SAMPLE PROBLEM 6.4, Page number 406¶

In [13]:
import math

#Variable declaration
tf=0.770                                                                      # Distance(in)

#Calculation
I=round(394 + 2*((1/12)*(12)*pow(0.75,3) + (12)*(0.75)*(pow(5.575,2))),0)     # Centroidal moment of inertia(in**4)
t=2*tf                                                                        # Distance(in)
Q=round(2*(4.31*0.770*4.815) + (12)*(0.75)*(5.575),1)
t=round(((50)*(82.1))/((954)*(1.54)),2)                                       # Shearing stress(ksi)

# Result
print ('Horizontal shearing stress = %lf ksi' %t)

Horizontal shearing stress = 2.790000 ksi


## SAMPLE PROBLEM 6.5, Page number 407¶

In [5]:
import math

#Variable declaration
cosB=12/13.0

#Calculation
#Centroid
Y=round((2*65*3*30)/((2*65*3)+((50)*(3))),2)                                         # Distance y(mm)
#Centroidal Moment of Inertia
b=(3)/(cosB)                                                                         # Distance(mm)
I=2*((1/12.0)*(3.25)*(pow(60,3)) + (3.25)*(60)*(pow(8.33,2))) +  ((1/12.0)*(50)*(pow(3,3)) + (50)*(3)*(pow(21.67,2)))# Moment of inertia(mm**4)
I=(I/(pow(10,12)))                                                                   # Moment of inertia(m**4)
#Shearing Stress at A
ta=0
#Maximum Shearing Stress
Q=round(3.25*38.33*(38.33/2.0),0)
tE=((5)*(2.387*(pow(10,-6))))/((0.2146*(pow(10,-6)))*(0.003))
tE=round(tE/1000.0,2)                                                                  # Largest shearing stress

# Result
print ('Case(a) Shearing stress at A = %lf ksi' %ta)
print ('Case(a) Maximum shearing stress = %lf MPa' %tE)

Case(a) Shearing stress at A = 0.000000 ksi
Case(a) Maximum shearing stress = 18.540000 MPa


## Example 6.05, Page number 417¶

In [14]:
import math
from sympy import symbols
#Variable declaration
b=4                                                                                      # Distance(in)
h=6                                                                                      # Distance(in)
t=0.15                                                                                   # Thickness(in)
s=symbols('s')                                                                           # Variable declaration
V=symbols('V')                                                                           # Variable declaration
I=symbols('I')                                                                           # Variable declaration
b=symbols('b')                                                                           # Variable declaration

#Calculation
q=(V*s*t*h)/(2.0*I)                                                                        # First moment
F=integrate(q,(s,0,b))                                                                   # Magnitude of shearing force
e=(F*h)/(V)                                                                              # Distance e from the center line of the web BD to the shear center O
I=((1/12.0)*t*(h**3)) + 2*(((1/12.0)*b*(t**3)) + (b*t*((h/2)**2)))                           # Moment of inertia
e=(4)/(2+0.5)                                                                            # Distance(in)

# Result
print ('Shear center O of a channel section of uniform thickness = %lf in' %e)

Shear center O of a channel section of uniform thickness = 1.600000 in


## Example 6.06, Page number 418¶

In [6]:
import math
from sympy import symbols

#Variable declaration
V=2.5                                                                             # Force(kips)
b=4                                                                               # Distance(in)
t=0.15                                                                            # Thickness(in)
h=6                                                                               # Height(in)

#Calculation
tB=(6*V*b)/((t*h)*(6.0*b+h))                                                        # Horizontal shearing stress(ksi)
tMAX=(3*(V)*(4*b+h))/(2.0*t*h*(6.0*b+h))                                              # Shearing stress in web(ksi)

# Result
print ('Shearing stress in flanges = %lf ksi' %tB)
print ('Shearing stress in web = %lf ksi' %tMAX)

Shearing stress in flanges = 2.222222 ksi
Shearing stress in web = 3.055556 ksi


## Example 6.07, Page number 419¶

In [7]:
import math
from sympy import Symbol
#Variable declaration
V=2.5                                                              # Force(kip)
OC=1.6+1.143                                                       # Distance(in)
a=4+6+4.0                                                            # Distance(in)
b=0.15                                                             # Distance(in)

#Calculation
T=V*OC                                                             # Torque(kip.in)
TmaxBend=3.06                                                      # Maximum value of stress(ksi)
c1=(1/3.0)*(1-(0.630*(b/a)))
tmaxTwist=(T/(c1*a*(b**2)))                                        # Stress due to twisting(ksi)
tmax=3.06 + 65.8                                                   # Combined stress(ksi)

# Result
print ('Maximum shearing stress = %lf ksi' %tmax)

Maximum shearing stress = 68.860000 ksi


## SAMPLE PROBLEM 6.6, Page number 422¶

In [10]:
import math
from sympy import symbols,cos,factor

#Variable declaration
b=symbols('b')                                                               # Variable declaration
h=symbols('h')                                                               # Variable declaration
t=symbols('t')                                                               # Variable declaration
a=symbols('a')                                                               # Variable declaration
y=symbols('y')                                                               # Variable declaration
P=symbols('P')                                                               # Variable declaration
t1=symbols('t1')                                                             # Variable declaration

#Calculation
#Principal Axes
Inn=(1/12.0)*(b)*(pow(h,3))                                                    # Moment of inertia
Imm=(1/3.0)*(b)*(pow(h,3))                                                     # Moment of inertia
Iy=(2)*(1/3.0)*(t/cos((math.pi)/4.0))*(pow((a*cos((math.pi)/4.0)),3))              # Centroidal moment of inertia at y
Iz=(2)*(1/12.0)*(t/cos((math.pi)/4.0))*(pow((a*cos((math.pi)/4.0)),3))             # Centroidal moment of inertia at z

#Shearing Stresses Due to Vy
Y=(1/2.0)*(a+y)*(cos((math.pi)/4.0))-(1/2.0)*(a)*(cos((math.pi)/4.0))                # Distance
Q=(1/2.0)*(t)*(a-y)*(y)*cos((math.pi)/4.0)
Vy=P*(cos((math.pi)/4.0))
t1=(Vy*Q)/(Iz*t)                                                             # Stress

#Shearing Stresses Due to Vz
Z=(1/2.0)*(a+y)*cos((math.pi)/4.0)
Q=(1/2.0)*(a**2-y**2)*(t)*(cos((math.pi)/4.0))
Vz=P*(cos((math.pi)/4.0))
t2=(Vz*Q)/(Iy*t)                                                             # Stress

#Combined Stresses.
#Along the Vertical Leg
te=factor(t1+t2)                                                             # Stress

#Along the Horizontal Leg.
tf=factor(t2-t1)                                                             # Stress

# Result
print ('Combined stress along the vertical leg :-' )
print(te)
print ('Combined stress along the horizontal leg :-')
print(tf)

Combined stress along the vertical leg :-
-3.0*P*(-0.25*a + 0.25*y)*(0.25*a + 1.25*y)/(a**3*t)
Combined stress along the horizontal leg :-
3.0*P*(-0.25*a + 0.25*y)*(-0.25*a + 0.75*y)/(a**3*t)