# chapter 7:Shear stress in Beams and Related Problems¶

## Example 7.1 page number 365¶

In [1]:
#Given
shear_v = 3000      #N - Transmitted vetical shear
shear_al = 700      #N - The maximum allowable
#We will divide this into two parts
l_1 = 50.0   #mm
l_2 = 200.0  #mm
b_1 = 200.0  #mm
b_2 = 50.0   #mm
A_1 = l_1* b_1  #mm2 - area of part_1
y_1 = 25.0      #mm com distance
A_2 =l_2*b_2    #mm2 - area of part_1
y_2 = 150.0     #in com distance
y_net = (A_1*y_1  +A_2*y_2)/(A_1+A_2)               #mm - The com of the whole system
c_max = (4-y_net)                                   #mm - The maximum distace from com to end
c_min  = y_net                                      #mm - the minimum distance from com to end
I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2)        #Parallel axis theorm
I_2 =  b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)
I_net = I_1 + I_2      #mm4 - the total moment of inertia
Q = A_1*(-y_1+y_net)   #mm3
q = shear_v*Q/I_net    #N/mm - Shear flow
d = shear_al/q         # The space between the nails
print "The minimal space between the nails ",round(d,0) ,"mm"

The minimal space between the nails  42.0 mm


## Example 7.2 pagenumber 365¶

In [2]:
#Given
l = 6        #m -length of the beam
p = 3        #KN-m _ the load applied
R_a = l*p/2  #KN -The reaction at a, Since the system is symmetry
R_b = l*p/2  #KN -The reaction at b
l_s = 10     #mm - The length of the screw
shear_al = 2 #KN - The maximum load the screw can take
I = 2.36*(10**9) #mm2 The moment of inertia of the whole system
#We will divide this into two parts
l_1 = 50.0   #mm
l_2 = 50.0   #mm
b_1 = 100.0  #mm
b_2 = 200.0  #mm
A_1 = l_1* b_1   #in2 - area of part_1
y_1 = 200.0      #mm com distance
A_2 =l_2*b_2     #mm2 - area of part_1
y_2 = 225.0      #in com distance
Q = 2*A_1*y_1 + A_2*y_2 # mm3 For the whole system
q = R_a*Q*(10**3)/I     #N/mm The shear flow
d = shear_al*(10**3)/q  #mm The space between the nails
print "The minimal space between the nails ",round(d,0),"mm"

The minimal space between the nails  123.0 mm


## Example 7.6 page number 376¶

In [3]:
#Given
#we will divide this into two equal parts and other part
l = 10.0     # in - The height
t  = 0.1     # in - The width
b = 5.0      #mm- The width of the above part
A = t* b     #in2 - area of part
y_net = l/2  # The com of the system
y_1 = l      # The position of teh com of part_2
I_1 = t*(l**3)/12          #in4 The moment of inertia of part 1
I_2 = 2*A*((y_1-y_net)**2) #in4 The moment of inertia of part 2
I = I_1 + I_2              #in4 The moment of inertia
e = (b**2)*(l**2)*t/(4*I)  #in the formula of channels
l_sc = e - t/2             #in- The shear centre
print "The shear centre from outside vertical face is ",l_sc ,"in"

The shear centre from outside vertical face is  1.825 in


## Example 7.8 page number 387¶

In [4]:
#Given
dia = 10.0           #mm - The diameter of the cylinder
c = dia/2            #mm - the radius of the cylinder
A = 3.14*(c**2)      #mm2 The area of the crossection
y = 4*c/(3*3.14)     #mm The com of cylinder
I = 3.14*(c**4)/4    #mm4 - The moment of inertia of the cylinder
j = 3.14*(dia**4)/32 #mm4
T = 20.0             #N.m - The torque
V = 250.0            #N - The shear
M = 25.0             #N-m The bending moment
Q = A*y/2  #mm
stress_dmax = 4*V/(3*A)                   #V*Q/(I*d)  #Mpa The direct maximum stress
stress_tmax = T*c*(10**3)/j               #-Mpa The torsion  maximum  stress
stress_total = stress_dmax + stress_tmax  #Mpa The total stress
print "The direct maximum stress",round(stress_dmax,2),"Mpa"
print "The torsion  maximum  stress",round(stress_tmax,2),"Mpa"
print "The total stress",round(stress_total,2),"Mpa"

The direct maximum stress 4.25 Mpa
The torsion  maximum  stress 101.91 Mpa
The total stress 106.16 Mpa


## Example 7.9 page number 393¶

In [1]:
#Given
dia = 15      #mm - The diameter of the rod
h = 0.5       #mt - The freely falling height
A = 3.14*(dia**2)/4 #mm2 The area of the crossection
E = 200             #Gpa -Youngs modulus
L = 750             #mm - The total length of the rod
G = 80              #gpa - Shear modulus
N = 10              #number of live coils
d = 5               #mm the diameter of live coil
m = 3               # the mass of freely falling body
H = 500             #mm -from mass to spring
F= m*9.81           #Kg the force due to that mass
p = 3        #KN-m _ the load applied
#e = e_rod + e_spr
#e_rod
e_rod = p*L*(10**-3)/(A*E)                  #mm The elongation due to freely falling body
#e_spr
e_spr = 64*F*(dia**3)*N*(10**-3)/(G*(d**4)) #mm The elongation due to spring
e = e_rod + e_spr                           #mm The total elongation
p_dyn =F*(1+pow((1+(2*H/e)),0.5))
Stress_max = p_dyn/A                        #MPa - The maximum stress in the system
print "The maximum stress in the system ",round(Stress_max,2),"Mpa"

The maximum stress in the system  4.84 Mpa