# Chapter 11:Design of Beams and Shafts¶

### Example 11.1 Page No 544¶

In [5]:
#Given
sigma_allow = 24    #ksi, stress
tou_allow = 14.5      #ksi, allowable stress

#Shear and Moment Diagrams
V_max = 30         #kip,
M_max = 120         #kip, bending moment

#Bending Stress
S_reqd = (M_max*(10**3))/sigma_allow
#Shear Stress
d = 17.9   #in
tw = 0.315  #inch
tou_avg = (V_max)/(d*tw)

if tou_avg<14.5:
print"Use a W18*14"

Use a W18*14


### Example 11.2 Page No 545¶

In [5]:
#Given
udl = 0.5         #kN/m
h_by_a = 1.5
sigma_allow = 9 #MPa, allowable bending stress
tou_allow = 0.6 #MPa, allowable shear stress

#calculation
#Shear and Moment Diagrams
import math
V_max =20         #kNm
M_max =10.67       #kNm
#Bending Stress
S_reqd = (M_max)/(sigma_allow*1000)
c = h_by_a/2.0
a_cube = (S_reqd*c*12)/(1.5**3) #S_reqd = I/c
a = a_cube**(1/3.0)
A = a*h_by_a*a
tou_max = (1.5*V_max)/(A*1000)
if tou_max>tou_allow:
a_sqr = (3/2.0)*(V_max)/(h_by_a*tou_allow*1000)
a =math.sqrt(a_sqr)
else:
print"not"

#Display
print"The smallest width for the laminated wooden beam = ", round(a,2),"m"

The smallest width for the laminated wooden beam =  0.18 m


### Example 11.3 Page No 546¶

In [1]:
#Given
l = 200/1000.0       #m, length
t = 30/1000.0        #m
sigma_allow = 9     #MPa, shear stress
tou_allow = 0.6      #MPa, bending stress
V_nail = 1.50        #kN
l_bc = 2             #m
l_cd = 2            #m

#Shear and Moment Diagrams
V_max = 20        #kN
M_max =10.67         #kNm
#Bending Stress
y1 = l/2.0
A1 = l*t
y2 = l+(t/2.0)
A2 = t*l
y_dash = (y1*A1 + y2*A2)/(A1 + A2)

I1 = (t*l**3)/12.0 + (t*l*(y_dash - y1)**2)
I2 = (l*t**3)/12.0 + (t*l*(y2 - y_dash)**2)
I =I1 + I2

c = y_dash
sigma = (M_max*c)/(I)
flag1 = 0
sigma_allow = sigma_allow*1000 #kPa

if(sigma<sigma_allow):
flag1 = 1
else:
print"otherwisw not"

#Shear Stress
y3 = y_dash/2.0
A3 = y_dash*t
Q = y3*A3

tou = (V_max*Q)/(I*t)
tou_allow = tou_allow*1000 #kPa
flag2 =0

if(tou<tou_allow):
flag2 = 1
else:
print"flag2 is not equal to one"

#Nail Spacing
y4a = (l+t-y_dash)
y4 = y4a - (t/2.0)
A4 = l*t
Q4 = y4*A4
V_bc = 1.5 #kN
V_cd = 1 #kN

q_bc = (V_bc*Q4)/I
q_cd = (V_cd*Q4)/I

s_bc = (V_nail)/(q_bc)
s_cd = (V_nail)/(q_cd)

chosen_bc = 150 #mm
chosen_cd = 250 #mm

#Result
print'The design is safe in bending and shear.'
print'The calculated nail spacing BC ',round(s_bc,2),"m"
print'The calculated nail spacing CD ',round(s_cd,2),"m"
print'The chosen nail spacing BC     ',chosen_bc,"mm"
print'The chosen nail spacing CD     ',chosen_cd,"mm"

otherwisw not
flag2 is not equal to one
The design is safe in bending and shear.
The calculated nail spacing BC  0.17 m
The calculated nail spacing CD  0.26 m
The chosen nail spacing BC      150 mm
The chosen nail spacing CD      250 mm


### Example 11.6 Page No 560¶

In [3]:
#Find smallest allowable diameter

#Given
tou_allow = 50*10**6 #MPa, shear stress
T = 7.5              #Nm, torque
R_ah = 150           #N, horizontal force
R_av = 475           #N
l_ac = 0.25          #m

#Calculation
import math
mc = R_ah*l_ac
m = R_av*l_ac
M_c = sqrt(m**2 + mc**2)
k = sqrt(M_c**2 + T**2)
c1 = (2*k)/(math.pi*tou_allow)
c = c1**(1/3.0)
d = 2*c*1000

#Display
print"The smallest allowable diameter of the shaft =",round(d,1),"mm"

The smallest allowable diameter of the shaft = 23.3 mm