Chapter 9: Deflections of beams

Example 9.17, Page number 654

In [8]:
#Variable declaration
P = 5           #load(k)
L = 8*12        #beam length(in)
E = 30*10**6    #modulus of elasticity(psi)
I = 75.0        #moment of inertia(in^4)
q = 1.5*(1./12.)#load intensity(k/in)

#Calculations
Sc = ((P*L**3)/(48*E*I))+((5.*q*L**4)/(384.*E*I))

#Result
print "The download deflection is",Sc,"in(Calculation mistake in textbook)"

The download deflection is 0.0001024 in(Calculation mistake in textbook)

Example 9.23, Page number 682

In [19]:
#Variable declaration
E = 30*10**6*144   #modulus of elasticity(lb/ft^2)
I = 25.92/12       #moment of inertia(ft^4)
x = 12             #ft

#Calculations
'''The equivalent load q(x) for the beam is given by the following equation
q(x) = -700(x)^-1+800(x-6)^0 - 800(x-12)^0 - 5600(x-12)^-1 + 1500(x-16)^-1
Since the equation equals zero at all points except where x=16, we omit the last term
Taking 4th order integration, we obtain the following expression,'''
C1 = (-((350*x**3)/3)+((100*(x-6)**4)/3)-((100*(x-12)**4)/3)+((2800*(x-12)**3)/3))/12
print "C1 =",C1

#For Deflection at point C
x = 6
Elv = ((350*x**3)/3)-(100*((x-6)**4)/3)+(100*((x-12)**4)/3)+(2800*((x-12)**3)/3)+(C1*x)
Sc = Elv/(E*I)    #ft

#For deflection at point D
x = 16
Elv_16 = ((350*x**3)/3)-(100*((x-6)**4)/3)+(100*((x-12)**4)/3)+(2800*((x-12)**3)/3)+(C1*x)
Sd = Elv_16/(E*I)

#Results
print "Deflection at point C is",round((-Sc*12),8),"in"
print "Deflection at point D is",round((Sd*12),8),"in"

print "\n Please note that there is a calculation mistake in textbook while calculating Elv. Hence, the difference in solution"

C1 = -13200
Deflection at point C is 0.00027315 in
Deflection at point D is 2.06e-06 in

 Please note that there is a calculation mistake in textbook while calculating Elv. Hence, the difference in solution