2. Problems on bending stress
1. A cantilever beam of length 2m fails when a load of
2KN is applied at the free end. If the section is
40mmx60mm, find the stress at the failure.
Solution:
Step 1: Given Data:
Length of beam = 2m or 2000mm
load at failure = 2KN
Section dimensions = 40mm X 60mm
Step 2: Calculation of moment of inertia
I = bd3/12
= (40) (603)/12
= 7.2X105 mm4
Step 3: Calculation of bending moment about
fixed end
M = WL
= (2)(2)
= 4KN-m
Step 4:
Calculation of bending stress
M /I= σ / y
Substitute for above (where y = depth /2= 60/2
= 30mm)
There fore
σ = 166.67N/mm2
L=2000mm
B=40mm D=60mm
3. 2.A rectangular beam 200mm deep and 300mm wide is simply supported over the
span of 8m. What uniformly distributed load per metre the beam may carry, if the
bending stress is not exceed 120N/mm2.
Solution:
Step 1: Given Data:
Length of beam = 8m or 8000mm
Section dimensions = 300mm X 200mm
maximum bending stress = σ = 120N/mm2.
condition: uniformly distributed load for
simply supported beam
Step 2: Calculation of bending moment for
the above condition
M = wL2/8
= w (8)2/8
= 8wX106
step 3: Calculation of moment of inertia
I = bd3/12
= (300) (2003) /12
= 2X10 8 mm4
Step 4: Calculation of Udl
M /I= σ / y
Substitute for above (where y = depth /2= 200/2 =100mm)
8wX106 /2X10 8= 120 / 100
w =3X10 4N/m or 30 N/mm
4. 3.A beam is simply supported and carries a uniformly distributed load of 40KN/m
run over the whole span. The section of the beam is rectangular having depth as
500mm.If the maximum stress in the material of the beam is 120N/mm2and
moment of inertia of the section is 7x108mm4, find the span of the beam.
5. 4. Calculate the maximum stress induced in a cast iron pipe of external
diameter 40mm,of internal diameter 20mm and length 4m when the
pipe is supported at its ends and carries a point load of 80N at its centre.
4m
80N
6. 5. A rectangular beam 300mm deep is simply supported over a span of 4m.
Determine the uniformly distributed load per meter which the beam may carry, if
the bending stress should not exceed 120N/mm2.Take I=8x106mm4.
Solution:
Step 1: Given Data:
Length of beam = 4m or 4000mm
Depth of the beam = 300mm
maximum bending stress = σ =120N/mm2
condition: udl for simply supported beam
I=8x106mm4
Step 2: Calculation of maximum bending moment
M= W L2 /8
M= W (4000)2 /8
M= 2 X106 W
Step 3: Calculation of udl
M /I= σ / y
2 X106 W /8x106= 120 / 150
W = 3.2N/mm
7. 6. A square beam 20mmx20mm in section and 2m long is supported at the ends.
The beam fails when a point load of 400N is applied at the centre of the beam.
What uniformly distributed load per meter length will break a cantilever of the same
material 40mm wide,60mm deep and 3m long?
Solution:
Step 1: Given Data:
case 1: point load application at centre of the beam
Length of beam = 2m or 2000mm
Cross section of the beam = 20mmx20mm
condition: simply supported beam
Step 2: Calculation of maximum bending moment
M= W L /4
M= (400) (2000) /4
M= 200x103 N-mm
Step 3: Calculation of moment of inertia
I = bd3/12
= (20) (203)/12
= 13333.33mm4
Step 4: Calculation of bending stress
M /I= σ / y
2 X105 /13333.33= σ / 10
σ = 150N/mm2W=400N
8. 6. A square beam 20mmx20mm in section and 2m long is supported at the ends.
The beam fails when a point load of 400N is applied at the centre of the beam. What
uniformly distributed load per meter length will break a cantilever of the same
material 40mm wide,60mm deep and 3m long?
Step 5: Case 2: calculation of magnitude of
udl when dimensions of the beam is
changed
Length of beam =3m or 3000mm
width of beam = 40mm
depth of beam = 60mm
condition: cantilever beam
Step 6: Calculation of maximum bending
moment
M= W L2 /2
M= W (3000)2 /2
Step 7: Calculation of moment of inertia
I = bd3/12
= (40) (603 ) /12
= 72x104mm4
Step 8: Calculation of load
M /I= σ / y
W (3000)2 /2 /72x104= 150 / 30
W = 800N/m
