Bending Stress
1. Define pure bending along with neat sketch
2. State the assumptions of simple bending
3. Define section modulus and also derive the equation for the same for rectangular, circular, triangular, hollow rectangular and hallow circular sections.
4. Derive the expression M/I=f/y=E/R
5. Define moment of resistance and neutral axis.
Problems on bending stress
step 3: Calculation of moment of inertia
1. Define pure bending along with neat sketch
2. State the assumptions of simple bending
3. Define section modulus and also derive the equation for the same for rectangular, circular, triangular, hollow rectangular and hallow circular sections.
4. Derive the expression M/I=f/y=E/R
5. Define moment of resistance and neutral axis.
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: 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
Solution:
Step 1:
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
=
2X108 mm4
Step 4:
Calculation
of Udl
M /I= σ /
y
Substitute
for above (where y = depth /2= 200/2 =100mm)
8wX106 /2X108= 120 / 100
w =3X104 N/m or 30 N/mm
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.
Solution:
Step 1: Data:
Depth of beam = 500mm
maximum bending stress = σ = 120N/mm2.
moment of inertia =7x108mm4
Step 2:
Calculation of bending moment for the above condition
M = wL2/8
= 40(L)2/8
= 5L2
Step 3:
Calculation of length of beam
M /I= σ /
y
Substitute
for above (where y = depth /2= 500/2 =250mm)
5L2 /7x108= 120 / 250
L=8197.56
mm
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.
Solution:
Step 1: Data:
Length of beam = 4m or 4000mm
Internal diameter = 20mm
External diameter = 40mm
condition: point load for simply supported
beam
Step 2:
Calculation of maximum bending moment
M= W L /4
M = 80 X
4000 /4
M = 80
KN-m
Step 3:
Calculation of moment of inertia
I = π
(D 4 –d 4)/64
I = π
(40 4 –20 4)/64
I =
117809.7mm4
Step 4:
Calculation of bending stress
M /I= σ /
y
Substitute
for above (where y = depth /2= 40/2 =20mm)
80X1000/117809.7
= σ / 20
σ =
13.58 N/mm2
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: 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/mm2
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: 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
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/mm2
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
7. A timber beam
of rectangular section is to support a load of 20KN uniformly distributed over
a span of 3.6m when beam is simply supported. If the depth is to be twice the
breadth, and the stress in timber is not exceed 7N/mm2, find the
dimensions of the cross section. How could you modify the dimensions with 20KN
of concentrated load is present at centre with same breadth and depth ratio.
Step 1: case 1: when simply supported beam of
length 3.6m carries udl of 20KN and depth is twice the width
We know that W = w L
= 20 X 1000X3.6
= 5.56N
Moment = WL/8
M = 5.56 X 1000X 3.6 /8
M = 2499.75 N-mm
Step 2: Calculation of cross sectional dimensions of the
beam
σ = 7N/mm2
M /I= σ / y
2499.75/(bd3/12) = 7/(d/2)
b = 8.12mm
d = 2b = 16.24mm
Step 3 :Case 2: when simply supported beam of length 3.6m carries
point load of 20KN and depth is twice the width
Moment = WL/4
M = 20 X 106X 3.6 /4
M = 18X 106 N-mm
Step 4: Calculation of cross sectional dimensions of the
beam
σ = 7N/mm2
M /I= σ / y
18X 106 /(bd3/12) = 7/(d/2)
b = 156.82mm
d = 2b = 313.65mm
8. A
steel plate of width 120mm and thickness 20mm is bent into a circular arc of
radius 10m.Determine the maximum stress induced and bending moment which will
produce the maximum stress. Take E=2x105N/mm2.
9. A
timber beam of rectangular section 8m is simply supported. The beam carries a
UDL of 12KN/m run over entire length and point load of 10KN at 3m from the left
support. If the depth is two times the width and the stress in the timber is not
to exceed 8N/mm2.Find the suitable dimensions of the section.
10. A
water main of 500mm internal diameter and 20mm thick is running full. The water
main is of cast iron and is supported at two points 10m apart. Find the maximum
stress in the metal. The cast iron and water weight 72000N/m3 and
10000N/m3 respectively.
11. Prove
that the ratio of depth to width of the strongest beam that can be cut from a
circular log of diameter d is 1. 414.Hence calculate the depth and width of the
strongest beam that can cut of a cylindrical log of wood whose diameter is
300mm.
12. A
cast iron test beam 25mmx25mm cross section and 1m long, supported at its ends
fail when a central load of 800N is applied on it. What UDL will break a
cantilever of same material 50mm wide, 100mm deep and 2m long?
13. A
cast iron pipe has 300mm bore and 10mm metal thickness and is supported at its
ends 10m apart. If the weight of cast iron is 70N/mm3and that of water is 9.81
KN/m3, determine the maximum stress in the metal.
14. Three
beams have same length; same allowable stress and same bending moment. The
cross section of the beams are square, rectangle with depth twice the width and
circle. Find the ratios of the weights of circular and rectangular beams with
respect to the square beams.
15. A
beam of rectangular cross section is to support a load of 20KN over the span of
4m. If the depth of the section is to be twice the width and the stress in the
beam is not to exceed 69.4N/mm2, find the dimensions of the cross
section. What change in the cross section is required, if the given load is a
concentrated load placed at the centre with same width to depth ratio.
16. A
beam of symmetrical section and 200mm deep is simply supported over span of 4m.
Find i) UDL it may carry if the bending stress is not to exceed 100N/mm2.ii)
Maximum bending stress if the beam carries a central load of 40KN.Take I=10x106mm4.
17. A
wooden beam 200mm x200mm is simply supported on a span of 6m. when the beam is
loaded with a 14KNload at each one third span point, it failed. Find the
modulus of rupture.
18. A
simply supported beam of span 10m is 350mm deep. The section of the beam is
symmetrical. The moment of inertia of the section is 9. 5x107mm4.If
the permissible bending stress is 120N/mm2, find
a) The
safe point load that can be applied at the centre of the span.
b) The
safe UDL that can be applied on the span.
19. A 3m
high pole stands as a vertical cantilever fixed at its base. It has to support
a horizontal load of 10KN at its top.
a) Find
the minimum diameter required if the post is of wood, if the permissible
bending stress is 15N/mm2.
b)
Alternatively, if a hallow aluminium tube whose thickness is one eighth of the
external diameter is provided, what should be the external and internal
diameters of the tube? Permissible bending stress for aluminium is 50 N/mm2.
20. The
moment of inertia of the beam section 500mm deep is 69. 49x107mm4.Find
the longest span over which a beam of this section, when simply supported,
could carry a uniformly distributed load of 50KN per meter run. The flange
stress is not to exceed 110 N/mm2.
21. Find
the maximum stress produced in a round steel bar 50mm in diameter and 9m long
due to its own weight when it is simply supported at its ends. Steel weighs at
77000N/cum.
22. A
cast iron of internal diameter 450mm is 15mm thick and is supported on a span
of 8m. Find the maximum stress in the pipe when it is full of water. Take
specific weight of cast iron =71600N/m3.and that of water =9810N/m3.
23. A
timber beam is freely supported on the supports 6m apart. It carries a
uniformly distributed load of 12KN.per meter run and concentrated load of 9KN
at 2.5 m from the left support. If the stress in the timber is not to exceed 8
N/mm2 design the suitable section making the depth twice the width.
24. A
timber beam is 160mm wide and 300mm deep and is simply supported on a span of 5m.
It carries a uniformly distributed load of 3000Nper meter run over the whole
span and three equal concentrated loads of W N each placed at midspan and
quarter span points. If the stress in the timber is not to exceed 8 N/mm2
find the maximum value of W.
25.
Compare the section modulus of two beams of same weight and length if the first
beam is solid circular beam of diameter d and second is the circular tube of
outer diameter D1 and inner diameter D2.
