# Chapter 12:Deflection of Beams and Shaft¶

## Example 12.10 Page No 610¶

In :
#Given
E = 200*10**6   #kN/m**2, stress
I = 17*10**-6   #mm**4, moment of inertia

#The given dimension are
l_ac = 2 #m
l_cF = 4 #m
l_Fb = 2 #m
l_cb = 6 #m
l_aF = 6 #m
l_ab = 8 #m
F = 16 #kN
R_b = (F*l_cb)/l_ab
R_a = F - R_b

#Calculation
mc = R_a*l_ac
mf = R_b*l_Fb
theta_ca = (0.5*l_ac*mc)/(E*I)
A1 = 0.5*l_aF*mf
t1_ba = (l_Fb + l_aF/3)*(A1)
A2 = 0.5*l_Fb*mf
t2_ba = (l_Fb*2*A2)/3
t_ba = (t1_ba+t2_ba)/(E*I)
theta_c = (t_ba/l_ab)-(theta_ca)

#Display
print"The slope at point C of the steel beam    = ",round(theta_c,4),"rad"

The slope at point C of the steel beam    =  0.0094 rad


## Example 12.12 Page No 612¶

In :
#Given
E = 29*10**3    #ksi, stress
I = 125        #inch**4, moment of inertia
l_ab = 12        #ft, dimension
l_bc =12        #ft
l_ac = l_ab+l_bc
R_a = -5     #kip, normal force
R_b = 10      #kip
R_c = 5      #kip

#calculation
mb = R_a*l_ab
#Moment-Area Theorem
t_ca = (l_ab*0.5*l_ac*mb)/(E*I)
t_ba = ((1/3.0)*l_ab*0.5*l_ab*mb)/(E*I)
del_c = (t_ca - 2*t_ba)*1728

#Display
print"The displacement at point C for the steel overhanging beam    =",round(del_c,2),"inch"

The displacement at point C for the steel overhanging beam    = -2.75 inch


## Example 12.13 Page No 620¶

In :
#Given
L = 8 #m, length
P = 8 #kN

#Calculations
EI_theta_A1 = (3*w*L**3)/(128) #ThetaA1 = (3wL**3)/(128EI)
EI_nu_C1 = (5*w*L**4)/(768) #NuC1 = (5wL**4)/(768EI)
EI_theta_A2 = (P*L**2)/(16) #theta_A2 = (PL**2)/(16EI)
EI_nu_C2 = (P*L**3)/(48) #nu_C2 = (PL**3)/(48EI)
theta_A = EI_theta_A1 + EI_theta_A2
nu_C = EI_nu_C1 + EI_nu_C2

#Display
print'The slope at A in terms of EI                =',theta_A,"kNm**2"
print'The displacement at point C in terms of EI   =',nu_C,"kNm**2"

The slope at A in terms of EI                = 56 kNm**2
The displacement at point C in terms of EI   = 138 kNm**2


## Example 12.14 Page No 621¶

In :
#Given
w = 5    #kN/m, force per unit length
l_ab = 4 #m, length
l_bc = 2 #m
M = w*l_ab #kNm

#Calculations
EI_theta_B1 = (w*l_ab**3)/(24) #ThetaB1 = (wL**3)/(24EI)
EI_nu_C1 = l_bc*EI_theta_B1

EI_theta_B2 = (M*l_ab)/(3) #
EI_nu_C2 = l_bc*EI_theta_B2
EI_nu_C3 = (P*l_bc**3)/(3) #nuC3 = (PL**3)/(24EI)
nu_C = -EI_nu_C1 + EI_nu_C2 + EI_nu_C3

#Display
print'The displacement at end C of the overhanging beam, in terms of EI   = ',nu_C,"kNm**3"

The displacement at end C of the overhanging beam, in terms of EI   =  52 kNm**3


## Example 12.15 Page No 622¶

In :
#Given
w = 4    #kN/m, force per unit length
l = 6   #m, length
l_bc =2  #m

#Calculations
EI_theta_B = (w*l**3)/(24.0)  #ThetaB1 = (wL**3)/(24EI)
EI_nu_B = (w*l**4)/(30.0)     #nuB = (wL**4)/(30EI)
nu_C = EI_nu_B + (EI_theta_B*l_bc)

#Display
print'The displacement at end C of the cantilever beam, in terms of EI   = ',nu_C,"kNm**3"

The displacement at end C of the cantilever beam, in terms of EI   =  244.8 kNm**3


## Example 12.16 Page No 623¶

In :
#Given
k = 15.0        #kip/ft, force per unit length
F = 3           #kip,  force
E = 29*10**3  #ksi, stress
l_ab = 3.0       #ft, length
l_ac = 1       #ft
l_cb = 2       #ft
I = 12         #in**4, moment of inertia
R_a = (F*l_cb)/(l_ab)
R_b = F-R_a

#Calculations
mu_a = (R_a)/k
mu_b = (R_b)/k
mu_c1 = mu_b + (l_cb/l_ab)*(mu_a - mu_b)
#From fig b
a=3            #ft
b=6            #ft
L=9            #ft
mu_c2 = ((F*a*b)*(L**2 - a**2 - b**2))/(6*E*144*I*(1/20736.0)*L)
mu_c = mu_c1 + mu_c2

#Display
print'The vertical displacement of the force at C   = ',round(mu_c,3),"ft"

The vertical displacement of the force at C   =  0.126 ft


## Example 12.21 Page No 643¶

In :
#Given
l = 10        #ft, length
w = 2       #kip/ft, force per unit length

#Calculation
#Compatibility Equation
EI_nu_b1 = (w*l**4)/8.0 + (5*P*l**3)/48.0 #nu_b = (wl**4)/8EI + (5Pl**3)/48EI
EI_nu_b2 = (l**3)/3.0
B_y = EI_nu_b1 / EI_nu_b2

#Display
print"The reactions at roller support B    = ",B_y,"kip"

The reactions at roller support B    =  10.0 kip


## Example 12.22 Page No 644¶

In :
#Given:
l = 8         #ft, length
l_ab = 5      #ft
l_bc = 5       #ft
l_af = 1/2.0  #inch
b = 12/1000.0 #m
w = 8        #kip, force per unit length
E = 29*10**3  #Ksi
I = 475.0     # inch**4, moment of inertia

#Compatibility Equation:
import math
A=math.pi/4.0*(l_af**2)
muB__byFbc=l*12/(A*E)
L=l_ab+l_bc
muB=5*w*L**3*12/(48*E*I)
muB_byFbc=L**3*12/(3*E*I)
#From equation muB__=muB-muB_
Fbc=muB/(muB_byFbc+muB__byFbc)

#Display:
print "Force developed in the rod is",round(Fbc,3),"kip"
print"In the book: Calculation mistake"

Force developed in the rod is 0.042 kip
In the book: Calculation mistake


## Example 12.23 Page No: 646¶

In :
#Given:
L = 12       #ft
#E and I are constant  say
E = 29*10**3 #Ksi
I = 475        # inch**4
w = 3         #kip/ft

#calculation
thetab=w*L**3/(48*E*I)
mub=7*w*L**4/(384*E*I)
theta_bbyBy=L**2/(2*E*I)
mu_bbyBy=L**3/(3*E*I)
theta__byMb=L/(E*I)
muB__byMb=L**2/(2*E*I)

#From eq 1 and 2 (solving  by matrix)
#72By+12Mb=-108
#576By+72Mb=-1134
M = array([[576, 72], [72, 12]])
N=([-108,-1134])
X=inv(M)*N #Inverse matrix
a=X[0,0]
b=X[1,0]*2.5

#Display:
print"Moment at B is",b,"kip-ft"

Moment at B is 11.25 kip-ft