JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II STRUCTURAL ANALYSIS Previous Question Paper (July/August - 2021 - R16)
Answer any five questions. All questions carry equal marks.
Main
Analyse a propped cantilever beam shown in figure 1 and draw the shear force and bending moment diagrams.
Answer
A propped cantilever beam of length L, fixed at A and propped at B, subjected to a point load W at the center, is a statically indeterminate structure of the first degree. Let RB be the reaction at the prop B.
Using the principle of superposition, the deflection at B due to the load W is δ1 = (W L3) / (48 EI) downwards, and the deflection due to the prop reaction RB is δ2 = (RB L3) / (3 EI) upwards. Equating these for zero deflection at the prop:
Solving for RB gives RB = W / 16. The vertical reaction at the fixed end A is RA = W - W/16 = 15W / 16. The fixed end moment MA is calculated by taking moments about A: MA = (W * L/2) - (RB * L) = (W L / 2) - (W L / 16) = 7WL / 16.
For the Shear Force Diagram (SFD):
- From x = 0 to L/2, SF = RA = 15W/16.
- From x = L/2 to L, SF = RA - W = -W/16.
For the Bending Moment Diagram (BMD):
- At x = 0 (fixed end), M = -7WL/16.
- At x = L/2 (under load), M = -7WL/16 + (15W/16 * L/2) = 3WL/32.
- At x = L (prop), M = 0.
The SFD is a rectangle of height 15W/16 up to the center, dropping to -W/16 thereafter. The BMD is a trapezoidal shape with a negative peak at the fixed end, crossing zero at x = 14L/15, and reaching a positive peak of 3WL/32 at the center.
Analyse a fixed beam of span 6 m subjected to loads as shown in figure 2 and draw the shear force and bending moment diagrams.
Answer
For a fixed beam of span L = 6 m, the analysis depends on the loading configuration. Assuming a standard point load W at the center (mid-span), the fixed end moments are calculated as:
If the load is a uniformly distributed load (UDL) of intensity w over the entire span, the fixed end moments are:
To determine the shear force and bending moment diagrams, follow these steps:
- Calculate the fixed end moments (FEMs) using the standard formulas based on the specific loading provided in the figure.
- Determine the support reactions (RA and RB) by applying equilibrium equations (ΣV = 0 and ΣM = 0).
- Calculate the bending moment at any section x using the expression: Mx = RAx - MA - (load effects).
- Determine the points of contraflexure where the bending moment changes sign.
- Construct the Shear Force Diagram (SFD) by plotting the vertical forces along the span, noting that the shear force changes abruptly at point loads and linearly under UDLs.
- Construct the Bending Moment Diagram (BMD) by plotting the calculated moments, ensuring the curve is parabolic for UDLs and linear for point loads, with the fixed end moments acting as hogging moments.
The final diagrams will show the beam is in a state of hogging at the supports and sagging in the central region, with the maximum bending moment occurring at the supports for a point load or at the center for a UDL.
A plane truss is made up of equilateral triangles of side equal to 3 m is simply supported and loaded as shown in figure 3. Analyse the truss using the method of joints.
Answer
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Explain about different types of frames with the help of figures.
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Determine the horizontal deflection at the point of application of the inclined load of a steel plane truss supported as shown in figure 4. Assume the axial rigidity is the same for all members of the truss. Use Unit load method.
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A three-hinged parabolic arch of span 30 m and central rise of 6 m is subjected to two concentrated loads 150 kN and 100 kN at a distance of 6 m and 12 m from the right hinge respectively and uniformly distributed load of 25 kN/m over the left half of the span. Find the resultant support reactions and the horizontal thrust, bending moment and the radial shear at a section 9 m from the left support.
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Using slope-deflection method, analyse the continuous beam shown in figure 5, if the support ‘B’ sinks by 20 mm. Assume the flexural rigidity is 2750 kNm2.
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Two point loads 125 kN and 200 kN spaced at 3.6 m apart crosses a simply supported girder of span 20 m from left to right with the 125 kN load leading. Determine the maximum positive and negative shear forces and also maximum bending moment at a section 8 m from the right support.
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