Strength Of Materials Miscellaneous
- A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and cross-sectional area of 100 mm2. Its elastic modulus varies along its length as given by E(x) = 100e–x GPa, where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x = 1). The axial displacement of the free end is _____ mm.
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Change in length of small stripδ = Px - x(dx) Ax - xEx - x
Total change in length of bar
δ = 1.718 mmCorrect Option: C
Change in length of small stripδ = Px - x(dx) Ax - xEx - x
Total change in length of bar
δ = 1.718 mm
- A rod is subjected to a uni-axial load within linear elastic limit. When the change in the stress is 200 MPa, the change in the strain is 0.001. If the Poisson's ratio of the rod is 0.3, the modulus of rigidity (in GPa) is
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E σ + 200 × 106 = 200GPa t 0.001 G = 200 = 76.92GPa 2(1 + 0.3)
Correct Option: A
E σ + 200 × 106 = 200GPa t 0.001 G = 200 = 76.92GPa 2(1 + 0.3)
- The number of independent elastic constants required to define the stress-strain relationship for an isotropic elastic solid is _____.
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Elastic constants are required to obtain relationship between stress & strain, There are 4 elastic constants.
E, μ, G, & E
But E & μ are independent.Correct Option: B
Elastic constants are required to obtain relationship between stress & strain, There are 4 elastic constants.
E, μ, G, & E
But E & μ are independent.
Direction: A cylindrical container of radius R = 1 m, wall thickness 1 mm is filled with water up to a depth of 2 m and suspended along its upper rim. The density of water is 1000 kg/m3 and acceleration due to gravity is 10 m/s2. The self-weight of the cylinder is negligible. The formula for hoop stress in a thin-walled cylinder can be used at all points along the height of the cylindrical container.
- If the Young's modulus and Poisson's ratio of the container material' are 100 GPa and 0.3, respectively, the axial strain in the cylinder wall at mid-depth is
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Axial strain in the cylinder would be produced due to the axial (longitudinal) stress and due to the lateral stress
i.e.hoop stress, e = 1 σu - σc y m = 1 (5 - 10 × 0.3) 100 × 103 Correct Option: A
Axial strain in the cylinder would be produced due to the axial (longitudinal) stress and due to the lateral stress
i.e.hoop stress, e = 1 σu - σc y m = 1 (5 - 10 × 0.3) 100 × 103
- The axial and circumferential stress (σa, σc) experienced by the cylinder wall at mid-depth (m as shown) are
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Pressure at mid – depth = ρgh = 103 × 10 × 1 = 104 N/m2
∴ σa = pd + 104 × 2 4t 4 × 10-3
= 5 × 106 N/mm2 = 5 MPaσa = pd + 104 × 2 = 10MPa 2t 2 × 10-3 Correct Option: B
Pressure at mid – depth = ρgh = 103 × 10 × 1 = 104 N/m2
∴ σa = pd + 104 × 2 4t 4 × 10-3
= 5 × 106 N/mm2 = 5 MPaσa = pd + 104 × 2 = 10MPa 2t 2 × 10-3