Chapter 5: Stresses in Beams Basic Topics¶

Example 5.1, page no. 307¶

In [1]:
import math
import numpy

#initialisation

L = 8.0                       # length of beam in ft
h = 6.0                       # Height of beam in inch
e = 0.00125                   # elongation on the bottom surface of the beam
y = -3.0                      # Dismath.tance of the bottom surface to the neutral surface of the beam in inch

#Calculations
r = -(y/e)                    # Radius of curvature
print "radius of curvature is", round(r), "inch"
k = 1/r                     # curvature in in-1
print "curvature", round(k,5), "ft-1"
theta = numpy.degrees(numpy.arcsin(((L*12.0)/(2.0*r))))         # angle in degree
print "Angle of twist", round(theta,3), "degree"
my_del = r*(1-math.cos(math.radians(theta)))       #Deflection in inch
print "Deflection in the beam is ", round(my_del,4), "inch"

radius of curvature is 2400.0 inch
curvature 0.00042 ft-1
Angle of twist 1.146 degree
Deflection in the beam is  0.48 inch


Example 5.2, page no. 315¶

In [10]:
import math

#initialisation
d = 0.004                       # thickness of wire in m
R0 = 0.5                        # radius of cylinder in m
E = 200e09                      # Modulus of elasticity of steel
s = 1200e06                     # proportional limit of steel

#calculation

M = (math.pi*E*d**4)/(32*(2*R0+d))      # Bending moment in wire in N-m
print "Bending moment in the wire is ", round(M,2), "N-m"
s_max = (E*d)/(2*R0+d)                  # Maximum bending stress in wire in Pa
print "Maximum bending stress in the wire is %e" %(s_max), "Pa"

Bending moment in the wire is  5.01 N-m
Maximum bending stress in the wire is 7.968127e+08 Pa


Example 5.3, page no. 316¶

In [11]:
import math

#initialisation
L = 22                          # Span of beam in ft
q = 1.5                         # Uniform load intensity in k/ft
P = 12                          # Concentrated in k
b = 8.75                        # width of cross section of beam in inch
h = 27                          # height of cross section of beam in inch
Ra = 23.59                      # Reaction at point A
Rb = 21.41                      # Reacyion at point B
Mmax = 151.6                    # Maximum bending moment

#calculation

S = (b*h**2)/6                  # Section modulus
s = (Mmax*12)/S                 # stress in k
st = s*1000                     # Tensile stress
print "Maximum tensile stress in the beam", round(st), "psi"
sc = -s*1000                    # Compressive stress
print "Maximum compressive stress in the beam", round(sc), "psi"

Maximum tensile stress in the beam 1711.0 psi
Maximum compressive stress in the beam -1711.0 psi


Example 5.4, page no. 318¶

In [5]:
import math

#initialisation
q = 3200.0                        # Uniform load intensity in N/m
b = 0.3                           # width of beam in m
h = 0.08                          # Height of the beam in m
t = 0.012                         # thickness of beam in m
Ra = 3600.0                       # Reaction at A in N
Rb = 10800.0                      # Reaction at B in N
Mpos = 2025.0                     # Moment in Nm
Mneg = -3600.0                    # Moment in Nm

#calculation
y1 = t/2.0
A1 = (b-2*t)*t
y2 = h/2
A2 = h*t
A3 = A2
c1 = ((y1*A1)+(2*y2*A2))/((A1)+(2*A2))
c2 = h - c1
Ic1 = (b-2*t)*(t**3)*(1.0/12.0)
d1 = c1-(t/2.0)
Iz1 = (Ic1)+(A1*(d1**2))
Iz2 = 956600e-12
Iz3 = Iz2
Iz = Iz1 + Iz2 + Iz3                # Moment of inertia of the beam cross section

# Section Modulli
S1 = Iz / c1                        # for the top surface
S2 = Iz / c2                        # for the bottom surface

# Maximum stresses for the positive section
st = Mpos / S2
print "Maximum tensile stress in the beam in positive section is", st, "Pa"
sc = -Mpos / S1
print "Maximum compressive stress in the beam in positive section is", sc, "Pa"

# Maximum stresses for the negative section
snt = -Mneg / S1
print "Maximum tensile stress in the beam in negative section is", snt, "Pa"
snc = Mneg / S2
print "Maximum compressive stress in the beam in negative section is", snc, "Pa"

# Conclusion
st_max = st
sc_max = snc

Maximum tensile stress in the beam in positive section is 50468539.6422 Pa
Maximum compressive stress in the beam in positive section is -15157118.8248 Pa
Maximum tensile stress in the beam in negative section is 26945989.0219 Pa
Maximum compressive stress in the beam in negative section is -89721848.2528 Pa


Exampe 5.5, page no. 325¶

In [6]:
import math

#initialisation
L = 12                      # Length of beam in ft
q = 420                     # Uniform load intensity in lb/ft
s = 1800                    # Allowable bending stress in psi
w = 35                      # weight of wood in lb/ft3

#calculation
M = (q*L**2*12)/8           # Bending moment in lb-in
S = M/s                     # Section Modulli in in3

# From Appendix F
q1 = 426.8                  # New uniform load intensity in lb/ft
S1 = S*(q1/q)               # New section modulli in in3

# From reference to appendix F, a beam of cross section 3*12 inch is selected
print ("Beam of crosssection 3*12 is sufficient")

Beam of crosssection 3*12 is sufficient


Example 5.6, page no. 326¶

In [18]:
import math

#initialisation
P = 12000                           # Lataeral load at the upper end in N
h = 2.5                             # Height of post in m
Mmax = P*h                          # Maximum bending moment in Nm

#calculation
# Part (a) : Wood Post
s1 = 15e06                          # Maximum allowable stress in Pa
S1 = Mmax/s1                        # Section Modulli in m3
d1 = ((32.0*S1)/math.pi)**(1.0/3.0)       # diameter in m
print "the minimum required diameter d1 of the wood post is", round(d1,3), "m"

# Part (b) : Alluminium tube
s2 = 50e06                          # Maximum allowable stress in Pa
S2 = Mmax/s2                        # Section Modulli in m3
d2 = (S2/0.06712)**(1.0/3.0)            # diameter in meter.....(1)
print "minimum required outer diameter d2 of the aluminum tube is", round(d2,3),"m"

the minimum required diameter d1 of the wood post is 0.273 m
minimum required outer diameter d2 of the aluminum tube is 0.208 m


Example 5.7, page no. 326¶

In [1]:
import math

#initialisation
q = 2000.0                        # Uniform load intensity in lb/ft
s = 18000.0                       # Maximum allowable load in Psi
Ra = 18860.0                      # Reaction at point  A
Rb = 17140.0                      # Reaction at point  B

#calculation
x1 = Ra/q                         # Distance in ft from left end to the point of zero shear
Mmax = (Ra*x1)-((q*(x1**2))/2.0)  # Maximum bending moment in lb-ft
S = (Mmax*12.0)/s                 # Section Modulli in in3

# Trial Beam
Ra_t = 19380.0                            # Reaction at point  A
Rb_t = 17670.0                            # Reaction at point  B

#in Python the value for x1 differes by some points and hence the subsequent results differ
x1_t = Ra_t/q                             # Distance in ft from left end to the point of zero shear
Mmax_t = (Ra_t*x1_t)-((q*(x1_t**2))/2.0)  # Maximum bending moment in lb-ft
S_t = (Mmax_t*12.0)/s                     # Section Modulli in in3
# From table E beam 12*50 is selected
print "Beam of crosssection 12*50 is selected with section modulli", round(S_t,2), "in^3"

Beam of crosssection 12*50 is selected with section modulli 62.6 in^3


Example 5.8, page 329¶

In [15]:
import math

#initialisation
g = 9810                        # Specific weight of water in N/m3
h = 2                           # Height of dam in m
s = 0.8                         # Dismath.tance between square cross section in m
sa = 8e06                       # Maximum allowable stress in Pa

#Calculations
b = ((g*(h**3)*s)/sa)**(1.0/3.0)    # Dimension of croossection in m

#Result
print "the minimum required dimension b of the posts", round(b,3), "m"

the minimum required dimension b of the posts 0.199 m


Example 5.11, page no. 341¶

In [10]:
import math

#initialisation
L = 3                       # Span of beam in ft
q = 160                     # Uniform load intensity in lb/in
b = 1                       # Width of cross section
h = 4                       # Height of cross section

# Calculations from chapter 4
Mc = 17920                  # Bending moment in ld-in
I = (b*(h**3))/12.0           # Moment of inertia in in4
sc = -(Mc*1)/I              # Compressive stress at point C in psi
Ac = 1*1                    # Area of section C in inch2
yc = 1.5                    # dismath.tance between midlayers od section C and cross section of beam
Qc = Ac*yc                  # First moment of C cross section in inch3
tc = (Vc*Qc)/(I*b)          # Shear stress in Psi
print "Normal stress at C", sc, "psi"
print "Shear stress at C", tc, "psi"

Normal stress at C -3360.0 psi
Shear stress at C -450.0 psi


Example 5.12, page no. 342¶

In [11]:
import math

#initialisation
s = 11e06                       # allowable tensile stress in pa
t = 1.2e06                      # allowable shear stress in pa
b = 0.1                         # Width of cross section in m
h = 0.15                        # Height of cross section in m
a = 0.5                         # in m

#Calculations
P_bending = (s*b*h**2)/(6.0*a)    # Bending stress in N
P_shear = (2*t*b*h)/3.0           # shear stress in N
Pmax = P_bending                # Because bending stress governs the design

#Result
print "the maximum permissible value Pmax of the loads", Pmax, "N"

the maximum permissible value Pmax of the loads 8250.0 N


Example 5.13, page no. 345¶

In [20]:
import math

#initialisation
d2 = 4                      # Outer diameter in inch
d1 = 3.2                    # Inner diameter in inch
r2 = d2/2                   # Outer radius in inch
r1 = d1/2                   # inner radius in inch
P = 1500                    # Horizontal force in lb

#calculation
# Part (a)
t_max = ((r2**2+(r2*r1)+r1**2)*4*P)/(3*math.pi*((r2**4)-(r1**4)))  # Mximum shear stress in Psi
print "Maximum shear stress in the pole is", round(t_max), "psi"

# Part (b)
d0 = math.sqrt((16*P)/(3*math.pi*t_max))  # Diameter of solid circular cross section in meter
print "Diameter of solid circular cross section is ", round(d0,2), "m"

Maximum shear stress in the pole is 658.0 psi
Diameter of solid circular cross section is  1.97 m


Example 5.14, page no. 351¶

In [3]:
import math

#initialisation
b = 0.165                   # in m
h = 0.320                   # in m
h1 = 0.290                  # in m
t = 0.0075                  #  in m
V = 45000.0                   # Vertical force in N

#calculation
I = (1.0/12.0)*((b*(h**3))-(b*(h1**3))+(t*(h1**3))) # Moment of inertia of the cros section
t_max = (V/(8.0*I*t))*((b*(h**2))-(b*(h1**2))+(t*(h1**2))) # Maximum shear stress in Pa
t_min = ((V*b)/(8*I*t))*(h**2-h1**2) # Minimum shear stress in Pa
T = ((t*h1)/3.0)*(2*t_max + t_min) # Total shear force in Pa
t_avg = V/(t*h1)  # Average shear stress in Pa

#Result
print "Maximum shear stress in the web is", round(t_max,2), "Pa"
print "Minimum shear stress in the web is", round(t_min,2), "Pa"
print "Total shear stress in the web is", round(T,2), "N"

Maximum shear stress in the web is 20985785.26 Pa
Minimum shear stress in the web is 17359517.46 Pa
Total shear stress in the web is 43015.04 N


Example 5.15, page no. 352¶

In [24]:
import math

#initialisation
V = 10000                   # Vertical shear force in lb
b = 4                       # in inch
t = 1                       # in inch
h = 8                       # in inch
h1 = 7                      # in inch

#calculation
A = b*(h-h1) + t*h1                             # Area of cross section
Qaa = ((h+h1)/2.0)*b*(h-h1) + (h1/2.0)*(t*h1)       # First moment of cross section
c2 = Qaa/A                                      # Position of neutral axis in inch
c1 = h-c2                                       # Position of neutral axis in inch
Iaa = (b*h**3)/3.0 - ((b-t)*h1**3)/3.0              # Moment of inertia about the line aa
I = Iaa - A*c2**2                               # Moment of inertia of crosssection
Q1 = b*(h-h1)*(c1-((h-h1)/2.0))                   # First moment of area above the line nn
t1 = (V*Q1)/(I*t)                               # Shear stress at the top of web in Psi
Qmax = (t*c2)*(c2/2.0)                            # Maximum first moment of inertia below neutral axis
t_max = (V*Qmax)/(I*t)                          # Maximum Shear stress in Psi

#Result
print "Shear stress at the top of the web is", round(t1), "psi"
print "Maximum Shear stress in the web is", round(t_max), "Psi"

Shear stress at the top of the web is 1462.0 psi
Maximum Shear stress in the web is 1762.0 Psi


Example 5.16, page no. 357¶

In [25]:
import math

#initialisation

Af = 40*180                     # Area of flange in mm2
V = 10500                       # Shear force acting on cross section
F = 800                         # Allowable load in shear
df = 120                        # Dismath.tance between centroid of flange and neutral axis in mm

#calculation
Q = Af*df                       # First moment of cross section of flange
I = (1.0/12.0)*(210*280**3) - (1.0/12.0)*(180*200**3)  # Moment of inertia of entire cross section in mm4
f = (V*Q)/I                     # Shear flow
s = (2*F)/f                     # Spacing between the screw

#Result
print "The maximum permissible longitudinal spacing s of the screws is", round(s,1), "mm"

The maximum permissible longitudinal spacing s of the screws is 46.6 mm


Example 5.17, page no. 362¶

In [26]:
import math

#initialisation
L = 60                           # Length of beam in inch
d = 5.5                          # dismath.tance from the point of application of the load P to the longitudinal axis of the tube in inch
b = 6                            # Outer dimension of tube in inch
A = 20                           # Area of cross section of tube in inch
I = 86.67                        # Moment of inertia in in4
P = 1000                         # in lb
theta = 60                       # in degree
Ph = P*math.sin(math.radians(60)) # Horizontal component
Pv = P*math.cos(math.radians(60)) # Vertical component

#Calculations
M0 = Ph*d                        # Moment in lb-in
y = -3                           # Point at which maximum tensile stress occur in inch
N = Ph                           # Axial force
M = 9870                         # Moment in lb-in
st_max = (N/A)-((M*y)/I)         # Maximum tensile stress in Psi
yc = 3                           # in inch
M1 = 5110                        # moment in lb-in
sc_left = (N/A)-((M*yc)/I)       # Stress at the left of point C in Psi
sc_right = -(M1*yc)/I            # Stress at the right of point C in Psi
sc_max = min(sc_left,sc_right)   # Because both are negative quantities

#Result
print "The maximum compressive stress in the beam is", round(sc_max), "psi"
print "The maximum tensile stress in the beam is", round(st_max), "psi"

The maximum compressive stress in the beam is -298.0 psi
The maximum tensile stress in the beam is 385.0 psi