JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year STRENGTH OF MATERIALS - I Previous Question Paper (March - 2021 - R18)
Answer any five questions. All questions carry equal marks.
Main
Draw the stress-strain diagram for mild steel and explain salient points.
Answer
The stress-strain diagram for mild steel under tension consists of the following key points:
- Limit of Proportionality: The point up to which stress is directly proportional to strain, obeying Hooke's Law.
- Elastic Limit: The maximum stress up to which the material regains its original shape upon unloading.
- Yield Point: The stress level where the material undergoes significant deformation without an increase in load. It includes Upper and Lower yield points.
- Ultimate Stress Point: The maximum stress the material can withstand before necking begins.
- Breaking Point: The point at which the material finally fractures.
A solid steel bar 1000 mm long and 50 mm diameter, is placed inside an Aluminium tube of 60 mm inside diameter and 75 mm outside diameter. The aluminum tube is 1.5 mm longer than steel bar. An axial load of 500 kN is applied to the bar and tube through rigid cover plates. Assess the stresses developed in the steel bar and Aluminium tube. Adopt E for steel = 2 × 105 N/mm2 and E for Aluminium = 0.7 × 105 N/mm2.
Answer
Given: Steel bar length Ls = 1000 mm, diameter ds = 50 mm, As = π/4 × 502 = 1963.5 mm2, Es = 2 × 105 N/mm2. Aluminium tube inner diameter = 60 mm, outer diameter = 75 mm, Aa = π/4 × (752 - 602) = 1590.4 mm2, Ea = 0.7 × 105 N/mm2. The tube is 1.5 mm longer, so the load P = 500,000 N acts on the steel first until the gap is closed.
1. Load required to compress the tube by 1.5 mm: δ = (Pa × La) / (Aa × Ea) => 1.5 = (Pa × 1001.5) / (1590.4 × 0.7 × 105). Solving gives Pa = 166,740 N.
2. Remaining load P' = 500,000 - 166,740 = 333,260 N. This load is shared by both members: P' = Ps + Pa'. Since both deform equally after contact, (Ps × Ls) / (As × Es) = (Pa' × La) / (Aa × Ea).
3. Solving the compatibility equation: Ps = Pa' × (As × Es × La) / (Aa × Ea × Ls) = Pa' × (1963.5 × 2 × 105 × 1001.5) / (1590.4 × 0.7 × 105 × 1000) = 3.53 Pa'.
4. Substituting into P': 3.53 Pa' + Pa' = 333,260 => Pa' = 73,567 N. Then Ps = 259,693 N.
5. Total load on steel = 259,693 N; Total load on Aluminium = 166,740 + 73,567 = 240,307 N.
6. Stresses: σs = 259,693 / 1963.5 = 132.26 N/mm2; σa = 240,307 / 1590.4 = 151.10 N/mm2.
Build the relation between the modulus of elasticity and modulus of rigidity.
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A vertical steel rod 1200 mm long is fixed at its top and a weight of 500 N is dropped from a height of 600 mm on to a collar at the lower end. The upper 800 mm length of the rod is has a diameter of 25 mm while the reaming portion has 20 mm diameter. Determine the maximum instantaneous stress induced in the rod.
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A simply supported beam subjected to the loading as shown in Figure 1. Draw the shear force and bending moment diagrams.
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Draw the shear force and bending moment diagrams for a beam supported and loaded as shown in Figure 2.
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A simply supported steel beam of span 6 m has an unsymmetrical I-section, top flange 150 mm × 10 mm, bottom flange 100 mm × 12 mm, web 8 mm thick and the overall depth of the beam is 300 mm. Determine the maximum uniformly distributed load the beam can support, if the permissible stresses are 125 N/mm2 in compression and 175 N/mm2 in tension.
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A simply supported beam of span 5 m carries a uniformly distributed load of 60 kN/m over its entire span. The cross-section of the beam has a T-section with flange width 150 mm, flange thickness 12 mm, web thickness 10 mm and overall depth 250 mm. Draw the distribution of shear stress across the depth of the section, subjected to maximum shear force. Also find the ratio of maximum shear stress to the average shear stress.
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Determine the location and magnitude of maximum deflection and the slopes at the supports of a beam supported and loaded as shown in Figure 3.
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Explain the following theories of failure:
Maximum principal strain theory
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Shear strain energy theory.
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