Questions and Numerical problems on Bending Stress

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

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

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: 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.