Q21GATE 2014NAT

For the state of stresses (in MPa) shown in the figure below, the maximum shear stress (in MPa) is _____

📖 Explanation

\begin{aligned} \tau_{\max } &=\frac{\sigma_{1}-\sigma_{2}}{2}=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}^{2}} \\ &=5.0 \mathrm{MPa} \end{aligned}

Q22GATE 2013MCQ

The creep strains are

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Q23GATE 2012MCQ

The Poisson's ratio is defined as

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📖 Explanation

$\mu =\left | \frac{\text{Lateral strain}}{\text{Axial strain}} \right |$

Q24GATE 2010MCQ

The number of independent elastic constants for a linear elastic isotropic and homogeneous material is

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Q25GATE 2008MCQ

A mild steel specimen is under uniaxial tensile stress. Young's modulus and yield stress for mild steel are $2 \times 10^{5}$ MPa and 250 MPa respectively. The maximum amount of strain energy per unit volume that can be stored in this specimen without permanent set is

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📖 Explanation

The strain energy per unit volume may be given as

\begin{aligned} u &=\frac{1}{2} \times \frac{\sigma_{y}^{2}}{E}=\frac{1}{2} \times \frac{(250)^{2}}{2 \times 10^{5}} \\ &=0.156 \mathrm{N}\; \mathrm{mm} / \mathrm{mm}^{3} \end{aligned}

Q26GATE 2008MCQ

A vertical rod PQ of length L is fixed at its top end P and has a flange fixed to the bottom end Q. A weight W is dropped vertically from a height h ( $\lt$ L) on to the flange. The axial stress in the rod can be reduced by

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📖 Explanation

The kinetic energy of the weight (W) at flange level will be transformed in the form of strain energy in the rod. We know that $\mathrm{U}=\sigma^{2} / 2 \mathrm{E} \times$ Area of cross-section of rod $\times$ length of rod Since, energy remains constant, hence to reduce the axial stress in the rod, the length of the rod should be increased or area of cross-section of the rod should be increased or modulus of elasticity should be decreased.

Q27GATE 2007MCQ

A metal bar of length 100 mm is inserted between two rigid supports and its temperature is increased by $10^{\circ}C$ . If the coefficient of thermal expansion is $12\times 10^{-6}$ per $^{\circ}C$ and the Young's modulus is $2\times 10^{5}$ MPa, the stress in the bar is

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📖 Explanation

\begin{aligned} \text { Temperature stress } &=\alpha \text { TE } \\ &=12 \times 10^{-6} \times 10 \times 2 \times 10^{5} \\ &=24 \mathrm{MPa} \end{aligned}

Q28GATE 2007MCQ

For an isotropic material, the relationship between the Young's modulus (E), shear modulus (G) and Poisson's ratio ( $\mu$ ) is given by

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Q29GATE 2005MCQ

The components of strain tensor at a point in the plane strain case can be obtained by measuring longitudinal strain in following directions

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📖 Explanation

In case of plane strain condition, the components of strain tensor at a point are $\epsilon_{x}, \in_{y} \text{ and }\phi_{x y} .$ Here we have three unknowns so we require 3 equations to find them and these unknowns can be find by the equations of longitudinal strain, $\epsilon_{1}=\epsilon_{x} \cos ^{2} \theta_{1}+\epsilon_{y} \sin ^{2} \theta_{1}+\frac{\phi_{x y}}{2} \sin 2 \theta_{1}$ $\epsilon_{2}=\epsilon_{x} \cos ^{2} \theta_{2}+\epsilon_{y} \sin ^{2} \theta_{2}+\frac{\phi_{x y}}{2} \sin 2 \theta_{2}$ $\epsilon_{3}=\epsilon_{x} \cos ^{2} \theta_{3}+\epsilon_{y} \sin ^{2} \theta_{3}+\frac{\phi_{x y}}{2} \sin 2 \theta_{3}$ Therefore to find out component of strain tensor in plain strain condition, we measure longitudinal strains $\left(\epsilon_{1}, \epsilon_{2}\text{ and }\epsilon_{3}\right)$ along any three arbitrary direction.

Q30GATE 2005MCQ

The symmetry of stress tensor at a point in the body under equilibrium is obtained from

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📖 Explanation

Taking moment equilibrium about the centre, we get

\begin{aligned} \tau_{y x} \times \frac{d}{2}+\tau_{y x} \times \frac{d}{2}&=\tau_{x y} \times \frac{d}{2}+\tau_{x y} \times \frac{d}{2} \\ \therefore \quad \tau_{x y}&=\tau_{y x} \end{aligned}

Q31GATE 2004MCQ

For linear elastic systems, the type of displacement function for the strain energy is

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📖 Explanation

$\text { Strain Energy }=\frac{1}{2} \times \sigma \times \varepsilon=\frac{1}{2} \mathrm{E} \varepsilon^{2}$ Since strain is directly proportional to displacement so strain energy is directly proportional to quadratic equation of displacement.

Q32GATE 2003MCQ

A bar of varying square cross-section is loaded symmetrically as shown in the figure. Loads shown are placed on one of the axes of symmetry of cross-section. Ignoring self weight, the maximum tensile stress in $N/mm^{2}$ anywhere is

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📖 Explanation

The stress in lower bar $=\frac{50 \times 1000}{50 \times 50}=20 \mathrm{N} / \mathrm{mm}^{2}$ The stress in upper bar $=\frac{250 \times 1000}{100 \times 100}=25 \mathrm{N} / \mathrm{mm}^{2}$ Thus the maximum tensile stress any where in the bar is $25 \mathrm{N} / \mathrm{mm}^{2}.$

Q33GATE 2002MCQ

The shear modulus (G), modulus of elasticity (E) and the Poisson's ratio (v) of a material are related as,

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