Problems on Combined stress

Short Answer Questions:

1.What do you mean by direct stress and bending stress?

2 Draw stress distribution across the section due to Bi axial bending stress and direct stress.

3 Find core diameter of a solid circular section, if diameter is ‘d’

4 Explain about the term kernel and determine the size of kernel for a rectangular section 200 mm x 300mm

5 Explain the conditions for stability of dam.

6 Find core diameter of a hollow section, if external and internal diameter are ‘D’ and ‘d’.

Assignment :

1 A hollow rectangular column of external depth of 1 m and external width 1 m is 10 cm thick.. Calculate the maximum and minimum stress in the section of the column, if vertical load of 200 kN is acting with an eccentricity of 20 cm.

2 A short column of external diameter 40 cm and internal diameter 20 cm carries an eccentric load of 80 kN. Find the greatest eccentricity which the load can have without producing tension on the cross –section.

3 Masonry dam 9m high, 1.5m wide at top and 6m wide at the base retains water to a depth of 7.5m, the water face of the dam being vertical. Find maximum and minimum stress intensities at the base. The weight of water is 9.81kN/m3 and weight of masonry is 24kN/m3

4 A column is rectangular in cross section 300 x 400 mm .The column carries an eccentric loading of 360kN on one diagonal at a distance of quarter diagonal length from a corner. Calculate the stresses at all four corners. Also draw stress distribution diagram for any side

5 A square chimney, 30 m high, has a flue opening of size 1.5 m x 1.5 m. Find the minimum width required at the base for no tension if the masonry weights 20 kN/m2 and the wind pressure is 1.5 kN/m2. The permissible stress in the masonry is 1kN/m2.

6 A solid round bar 3 m long and 5 cm in diameter is used as a strut with one end is fixed and other is hinged. Determine the crippling load.

Take E = 2x105 N/mm2.

Comments

Relation between Modulus of Elasticity and Modulus of Rigidity

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Shearing Stresses Distribution in Circular Section

Show that the shearing stress developed at the neutral axis of a beam with circular cross section is τ max = (4/3)(F/πr2). Assume that the shearing stress is uniformly distributed across the neutral axis. Solution : Let us consider the circular section of a beam as displayed in following figure. We have assumed one layer EF at a distance y1 from the neutral axis of the circular section of the beam Shear stress at a section will be given by following formula as mentioned here Where, F = Shear force (N) τ = Shear stress (N/mm2) A = Area of section, where shear stress is to be determined (mm2) ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section (m) Q = A. ȳ = Moment of the whole shaded area about the neutral axis I = Moment of inertia of the given section about the neutral axis (mm4) For circular cross-section, Moment of inertia, I = ПR4/4 b = Width of the given section where shear stress is to be determined. Let us consider on

Prof.Kodali Srinivas

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