A steel bar of rectangular cross-section (20 x 50 mm) carries a tensile load P and is attached to a support by means of a round pin of diameter 20 mm. The allowable stresses for the bar in tension and the pin in shear are σallow = 150 MPa and τallow = 75 MPa, respectively. What is the maximum permissible value of the load P?
Net area for tensile load
= 20 x (50 - 20) = 600 mm2
∴ P1 (based on tensile stress)
Net area for shearing load
∴ P2 (based on shearing stress)
= 75 x 2 x 0.785 x 400 = 47.1 kN
So, permissible value of Pis 47 kN.
A temperature stress is a function of
1. Coefficient of linear expansion
2. Temperature rise
3. Modulus of elasticity
4. Poisson’s ratio
The correct answer is
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All of the following statements are correct, except.
Each member of a composite bar is subjected to same strain.
A steel rod passes through a brass tube which is closed by thin rigid washers and fastened by nuts screwed to the rod. Which type of stress will be induced in the tube when the nut is tightened on the bolt?
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Two wires of different materials but of same diameter are connected end to end and a force is applied which stretches them by 1 cm. The two wires will have the
Load and area remains same for both cases, so stresses are same.
Stress = E x strain
E is different for both materials, so strains are different.
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A steel rod of length 2 and diameter d, fixed at both ends is uniformly heated to a temperature rise of ΔT. The Young’s modulus is E and the coefficient of linear expansion is α. The thermal stress in the rod is
Thermal stress = E x Thermal strain