JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II STRUCTURAL ANALYSIS Previous Question Paper (August/September - 2022 - R16)
Answer any five questions
All questions carry equal marks
Main
Draw SFD and BMD and locate point of contra flexure for a rigidly fixed beam AB of span 5 meters carrying a UDL of 14 kN/m over the entire span.
Answer
For a fixed beam of span L = 5 m carrying a uniformly distributed load (UDL) of w = 14 kN/m over the entire span, the fixed end moments (FEM) at both supports A and B are calculated using the formula:
The vertical reactions at the supports are equal due to symmetry:
The Shear Force (SF) at any distance x from support A is given by:
At x = 0, V = 35 kN; at x = 2.5 m, V = 0; at x = 5 m, V = -35 kN. The SFD is a straight line sloping downwards from 35 kN to -35 kN.
The Bending Moment (BM) at any distance x is given by:
At the supports (x=0, x=5), M = -29.17 kNm. At the center (x=2.5), M = 35(2.5) - 29.17 - 7(2.52) = 14.58 kNm. The BMD is a parabola with peak at the center.
The point of contraflexure occurs where the bending moment is zero:
Solving this quadratic equation using the formula x = [-b ± sqrt(b2 - 4ac)] / 2a:
The points of contraflexure are located at 1.057 m from support A and 1.057 m from support B.
Draw SFD and BMD for the propped cantilever beam of length 10m. It carries a point load of 100 kN at 4m from the fixed end.
Answer
For a propped cantilever beam of length L = 10 m with a point load W = 100 kN at a distance a = 4 m from the fixed end (A), the reaction at the prop (B) is calculated using the compatibility condition that the deflection at B is zero.
Using the standard formula for a propped cantilever with a point load at distance a from the fixed end:
Substituting the values: W = 100 kN, a = 4 m, L = 10 m:
The reaction at the fixed end (RA) is:
The fixed end moment (MA) is:
Shear Force Diagram (SFD):
- From x = 0 to 4 m: SF = RA = +79.2 kN.
- From x = 4 to 10 m: SF = RA - 100 = -20.8 kN.
Bending Moment Diagram (BMD):
- At fixed end (x = 0): M = -144 kNm.
- At load point (x = 4 m): M = -144 + (79.2 * 4) = -144 + 316.8 = +172.8 kNm.
- At prop (x = 10 m): M = 0.
- The BMD is linear between these points, changing sign at the point of contraflexure.
A truss of 10m span is loaded as shown in Figure 1. Find the forces in the members of the truss by using the method of sections.
Answer
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The Figure 2 shows a warren type cantilever truss along with the imposed loads. Determine the forces in the members using the method of tension coefficients.
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Derive the strain energy expressions for body subjected to normal stresses, bending stresses and shear stresses.
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The equation of a three-hinged arch, with origin at its left support is
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Analyze the continuous beam by the slope deflection method. Draw the bending moment diagram. As shown in the figure 4.
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Two wheel loads of 160 kN and 400 kN spaced 2 m apart move on a simply supported beam girder of span 16 m. Find the maximum positive and negative shear force at a section (a) 4 m from the left end and (b) 6 m from the left end.
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