Chapter 7:Transverse Shear¶

Example 7.1 Page No 369¶

In [13]:
#Given
co = 0.05		#mm, outside radius
ci = 0.02		 #mm, inside radius
t1 = 0.1 	#mm, thickness
t2=0.06

#Part (a)
#Section Properties
import math
Isolid=1/4.0*(math.pi)*co**4
Itube = 1/4.0*(math.pi)*(co**4-ci**4)
Qsolid=4*co/(3*math.pi)*(math.pi*co**2/2.0)
Qtube=4*co/(3*math.pi)*(math.pi*co**2/2.0)-4*ci/(3*math.pi)*(math.pi*ci**2/2.0)
Tsolid=V*10**3*Qsolid/(Isolid*t1)
Ttube=V*10**3*Qsolid/(Itube*t2)

#Display
print"The shear stress in solid    = ",round(Tsolid/1000,1),"KPa"
print'The shear stress in tube  = ',round(Ttube/1000000,2),"MPa"

The shear stress in solid    =  679.1 KPa
The shear stress in tube  =  1.16 MPa


Example 7.3 Page No 370¶

In [17]:
#Given
thick_1 = 20/1000.0 	#m, thickness
thick_2 = 15/1000.0 	#m
l = 300/1000.0 		#m, length
y = 100/1000.0 		#m
h = 2*y
y_dash = y +thick_1/2.0

#Part(a)
I1 = (thick_2*(h**3))/12.0
I2 = (l*(thick_1**3))/12.0
I3 = (l*thick_1*(y_dash)**2)
I = I1+2*(I2+I3) #Moment of inertia
Q_b = y_dash*l*thick_1
#At B'
tou_b_dash = (V*Q_b)/(I*l*1000)
#At B
tou_b = (V*Q_b)/(I*thick_2*1000)

#At C
Q_c = (y_dash*l*thick_1)+(y*thick_2*y/2.0)
tou_c = (V*Q_c)/(I*thick_2*1000)

#Display
print"The shear stress at B dash     = ",round(tou_b_dash,1),"MPa"
print"The shear stress at B         = ",round(tou_b,1),"MPa"
print"The shear stress at C     = ",round(tou_c,1),"MPa"

The shear stress at B dash     =  1.1 MPa
The shear stress at B         =  22.6 MPa
The shear stress at C     =  25.2 MPa


Example 7.4 Page No 372¶

In [20]:
#Given
udl = 6.5		 #kN, force
l_bc = 8 		#m, length
l = 150/1000.0		#m
t = 30/1000.0		#m, thickness

#Calculation
#Internal Shear
w = udl*l_bc/2.0
l_wc = l_bc/4.0
l_bw = l_bc - l_wc
V = (w*l_bw)/l_bc
R_b = w - V

#Section Properties
y1= l/2.0
A = (l*t)
y2= l+(t/2.0)
y_dash = (y1*A + y2*A)/(2*A)
I1 = (t*l**3)/12.0
I2 = (A*(y_dash-y1)**2)
I3 = (l*t**3)/12.0
I4 = (A*(y2 - y_dash)**2)
I = I1+I2+I3+I4
Q = ((l+t)-(t/2.0)-y_dash)*A
#Shear Stress
tou_max = (V*Q)/(I*t*1000)

#Display
print"The maximum shear stress in the glue necessary to hold the boards together",round(tou_max,2),"MPa"

The maximum shear stress in the glue necessary to hold the boards together 4.88 MPa


Example 7.5 Page No 380¶

In [8]:
#Given
V = 850            #kN, force
#The given dimension are
l1 =250/1000.0     #m.
l2 = 300/1000.0     #m
l3 = 125/1000.0    #m
t = 10/1000.0      #m
h = 200/1000.0     #m

#Calculation
A1 = l1*t
A2 = l2*t
A3 = l3*t
y1 = l2+(t/2.0)
y2 = l2/2.0
y3 = h+(t/2.0)
y_dash = (2*y2*A2 + A1*y1 + A3*y3)/(2*A2 + A1 + A3)
I1 = ((l1*t**3)/12.0) +(A1 * (l2+(t/2.0)-y_dash)**2)
I2 = ((t*l2**3)/12.0) +(A2 * (y_dash - (l2/2.0))**2)
I3 = ((l3*t**3)/12.0) +(A1 * (h+(t/2.0)-y_dash)**2)
I = 2*I2 + I1 + I3
Q_b = (l2+(t/2.0) - y_dash)*A1 #Q = y'A'
Q_c = (h+(t/2.0) - y_dash)*A3 #Q = y'A'

#Shear Flow
q_b = (V*Q_b)/I
q_c = (V*Q_c)/I
q_b = q_b/(2*1000)
q_c = q_c/(2*1000)

#Display
print"The shear flow at B, resisted by the glue is  ",round(q_b,2),"MN/m"
print"The shear flow at C, resisted by the glue is  ",round(q_c,4),"MN/m"

The shear flow at B, resisted by the glue is   1.31 MN/m
The shear flow at C, resisted by the glue is   0.0498 MN/m


Example 7.6 Page No 381¶

In [10]:
#Given
#The given dimension are
t = 1.5    #inch
a = 7.5    #inch
b = a-2*t  #inch
F_nail= 30 #lb

#Calculation
#Section Properties
I = (a*a**3 - b*b**3 )/12.0
Q_b = (((a-2*t)/2.0)+(t/2.0))*a*t #Q = y'A'
Q_c = (((a-2*t)/2.0)+(t/2.0))*(a-2*t)*t #Q = y'A'

#Shear Flow
q_b = (V*Q_b)/I
q_c = (V*Q_c)/I
s_b = F_nail/(q_b/2.0)
s_c = F_nail/(q_c/2.0)

#Display
print"The maximum spacing of nails required at B is    =",s_b,"inch"
print"The maximum spacing of nails required at C is    =",s_c,"inch"

The maximum spacing of nails required at B is    = 5.1 inch
The maximum spacing of nails required at C is    = 8.5 inch


Example 7.7 Page No 382¶

In [14]:
#Given
F = 40        #lb, force
#The other dimension are
s = 9.0         #inch
h = 5         #inch
t = 0.5       #inch
w = 3         #inch
w_3 = w/3.0   #inch

#Calculations
I = (w*h**3)/12.0 - (2*w_3*(h - 2*t)**3)/12.0
#Case 1
Q1 = ((h-t)/2.0)*(w*t)
V1 =((F/s)*I)/Q1   #q = VQ/I

#Case2
Q2 = ((h-t)/2.0)*(w_3*t)
V2 =((F/s)*I)/Q2   #q = VQ/I

#Display
print"The largest vertical shear that can be supported in Case 1    = ",round(V1,1),"lb"
print"The largest vertical shear that can be supported in Case 2    = ",round(V2,1),"lb"

The largest vertical shear that can be supported in Case 1    =  27.1 lb
The largest vertical shear that can be supported in Case 2    =  81.3 lb


Example 7.8 Page No 381¶

In [30]:
#Given
b1 = 6        #inch
h1 = 8        #inch
t = 1         #inch
b2 = b1-2*t
h2 = h1-2*t   #inch
b3 = 4         #inch

#Calculations
I = ((b2/2.0*(b1+t)**3))/12.0 +2*((b3+t)*t*((h1-t)/2.0)**2)
q_b = 0
Q_c = (h1-t)/2.0*(b3+t)*t
q_c = 1/2.0*(V*Q_c)/I
Q_d = 2*((h1-t)/4.0)*(h1-t)/2.0*t+(h1-t)/2.0*(b3+t)*t
q_d = 1/2.0*(V*Q_d)/I #Q = VQ/I

#Display
print"Variation of shear flow at B  = ",q_b,"kip/inch"
print"Variation of shear flow at C  = ",round(q_c,3),"kip/inch"
print'Variation of shear flow at D  = ',round(q_d,3),"kip/inch"

Variation of shear flow at B  =  0 kip/inch
Variation of shear flow at C  =  0.487 kip/inch
Variation of shear flow at D  =  0.828 kip/inch