Strength Of Materials Miscellaneous
- A rod of length L and diameter D is subjected to a tensile load P. Which of the following is sufficient to calculate the resulting change in diameter?
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Both Young's modulus and shear modulus
Correct Option: D
Both Young's modulus and shear modulus
- A 200 × 100 × 50 mm steel block is subjected to a hydro static pressure of 15 MPa. The Young's modulus and Poisson's ratio of the material are 200 GPa and 0.3 respectively. The change in the volume of the block in mm3 is
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NA
Correct Option: B
NA
- In terms of Poisson's ratio (μ) the ratio of Young's modulus (E) to Shear modulus (G) of elastic materials is
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We know E = 2G[1 + v]
E/G = 2[1 + v].
18. Here, Px = Px = Px = – 15 × 106 Pa
(compression)Now, ex Px = Py = Pz = 3 × 10-5 T2 2π π
where, m = Poisson’s ratio
ex = ey = ez = 3 × 10-5
ex + ey + ez = 9 × 10-5
∴ Change in volume = Total strain × Original = 90 m3Correct Option: A
We know E = 2G[1 + v]
E/G = 2[1 + v].
18. Here, Px = Px = Px = – 15 × 106 Pa
(compression)Now, ex Px = Py = Pz = 3 × 10-5 T2 2π π
where, m = Poisson’s ratio
ex = ey = ez = 3 × 10-5
ex + ey + ez = 9 × 10-5
∴ Change in volume = Total strain × Original = 90 m3
- If the wire diameter of a compressive helical spring is increased by 2%, the change in spring stiffness (in%) is _______ (correct to two decimal places.)
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Stiffness of helical spring
k = Gd4 64R3 n
∴ k ∝ d4k' = d' 4 k d k' = 1.02 d 4 k d
k' = 1.08243 k% increase in k = k' - k × 100% k = 1.08243 k - k × 100% = 8.243% k Correct Option: A
Stiffness of helical spring
k = Gd4 64R3 n
∴ k ∝ d4k' = d' 4 k d k' = 1.02 d 4 k d
k' = 1.08243 k% increase in k = k' - k × 100% k = 1.08243 k - k × 100% = 8.243% k
- A helical compression spring made of a wire of circular cross-section is subjected to a compressive load. The maximum shear stress induced in the cross-section of the wire is 24 MPa. For the same compressive load, if both the wire diameter and the mean coil diameter are doubled, the maximum shear stress (in MPa) induced in the cross-section of the wire is ___.
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τmax = 8WD kw πd2 τmax ∝ 1 d2
[ ∴ W1 = W2 = W ; C1 = C2 ; ( Kw)1 = ( Kw)2 = ( Kw) ]τ2 = d1 2 τ1 d2 τ2 = d 2 24 2d
⇒ τ2 = 6 MPaCorrect Option: A
τmax = 8WD kw πd2 τmax ∝ 1 d2
[ ∴ W1 = W2 = W ; C1 = C2 ; ( Kw)1 = ( Kw)2 = ( Kw) ]τ2 = d1 2 τ1 d2 τ2 = d 2 24 2d
⇒ τ2 = 6 MPa