Chapter 13 : Deflection of Beams¶

Example 13.1 Page No : 313¶

In [1]:

# Variables
E = 1.5*10**6
F1 = -100. 			#lb
F2 = -100. 			#lb
x1 = 6 			#in
x2 = 6 			#in
Ina = 64 			#in**4
h1 = -600 			#lb ft
h2 = -1200 			#lb ft
xa1 = 10 			# in
xa2 = 8 			#in

# Calculations
deltamax = ((1./2)*x1*xa1*h1+(1./2)*(x1+x2)*h2*xa2)*(1728)/(E*Ina)

# Results
print  'maximum deflection = %.2f in'%(deltamax)
maximum deflection = -1.36 in

Example 13.2 Page No : 314¶

In [2]:

# Variables
E = 1.5*10**6
I = 50. 	    		#in**4
delta = -1. 			#in
l = 8.       			#ft

# Calculations
w = -delta*8*E*I/(l**4*1728)

# Results
print  'distributed weight = %.1f lb per ft'%(w)
distributed weight = 84.8 lb per ft

Example 13.3 Page No : 315¶

In [2]:

# Variables
W = 50. 			#lb/ft
x = 5.  			#ft
x1 = 2. 			#ft

# Calculations
V = W*x
M = W*((x/2)+x1)*x
M1 = W*x*(x+x1)
M2 = -M
M3 = -W*x*x/2
EIdeltamax = ((1./2)*(x+x1)*M1*((x+x1)/3))+(x+x1)*M2*((x+x1)/2)+(1./3)*x*M3*(x/4)

# Results
print  'maximum value of EIdeltax = %.1f lb ft**3'%(round(EIdeltamax,-2))
maximum value of EIdeltax = -14600.0 lb ft**3

Example 13.5 Page No : 329¶

In [3]:

# Variables
w = 180. 			#lb/ft
l = 8.   			#ft
P = 1200. 			#lb
b = 6.   			#ft
E = 3*10.**6
I = 64. 			#in**4

# Calculations
delta = ((w*l**4)/(8))+((P*b**2)*(3*l-b)/(6))

# Results
print  'deflection of the free end = %.1fbyEI ft'%(round(delta,-3))
deflection of the free end = 222000.0byEI ft

Example 13.6 Page No : 329¶

In [5]:

# Variables
P = 6.  			#kips
w = 3. 	    		#kips/ft
L1 = 8. 			#ft
L2 = 8. 			#ft

# Calculations
delta = (P*(L1+L2)**3/192)+(w*(L1+L2)**4/768)

# Results
print  'midspan value of deflection = %.1f kip ft**3'%(delta)
midspan value of deflection = 384.0 kip ft**3

Example 13.7 Page No : 331¶

In [6]:

# Variables
x1 = 3. 			#ft
x2 = 3. 			#ft
x3 = 3. 			#ft
x4 = 3. 			#ft
W1 = 4. 			#kips
W2 = 8. 			#kips
l = x1+x2+x3+x4

# Calculations
b = x2+x3+x4
b1 = x4
a = x1
x = l/2
P = (((W1*b*(l/b*(x-a)**3+(l**2-b**2)*x-x**3))/(6*l))+((W2*b1*x*(l**2-x**2-b1**2))/(6*l)))*(48/l**3)
R1 = 3+2-(P/2)
R2 = P
R3 = 1+6-(P/2)

# Results
print  'R1 = %.3f kips'%(R1)
print  'R2 =%.2f kips'%(R2)
print  'R3 =%.3f kips'%(R3)
R1 = 0.875 kips
R2 =8.25 kips
R3 =2.875 kips

Example 13.8 Page No : 333¶

In [7]:
from numpy import linalg

# Variables
P = 680. 			#lb
K = 1000. 			#lb/in
L = 6.  			#ft
E = 30.*10**6
Ina = 1.728 			#in**4

# Calculations
A = [[((L*12)**3/(3*E*Ina)),-(1/K)],[1,1]]
b = [0,P]
c = linalg.solve(A,b)
Pb = c[0]
Ps = c[1]

# Results
print  'Force in the spring = %.2f psi'%(Ps)
Force in the spring = 480.00 psi

Example 13.9 Page No : 334¶

In [8]:

# Variables
I = 1.5 			#in**4
Da = 0.5 			#in
E = 30.*10**6
l = 60. 			#in

# Calculations
F = 6*Da*E*I/(l**3)

# Results
print  'F = %.2f lb'%(F)
F = 625.00 lb