GATE CE · Solid Mechanics

Properties Of Metals Stress And Strain – GATE CE Solid MechanicsPractice Questions & PYQs

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33 questions · 20 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2024MCQ

A homogeneous, prismatic, linearly elastic steel bar fixed at both the ends has a slenderness ratio $(\mathrm{I} / \mathrm{r})$ of 105 , where $I$ is the bar length and $r$ is the radius of gyration. The coefficient of thermal expansion of steel is $12 \times 10^{-6} /{ }^{\circ} \mathrm{C}$ . Consider the effective length of the steel bar as 0.51 and neglect the self-weight of the bar. The differential increase in temperature (rounded off to the nearest integer) at which the bar buckles is

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📖 Explanation

$\sigma_{\text {temp }}=E \alpha \Delta T$ $\sigma_{\mathrm{cr}}=\frac{\pi^{2} \mathrm{E}}{\lambda_{\text {eff }}^{2}}=\frac{\pi^{2} E}{\left(\ell_{\text {eff }} / r\right)^{2}}$ $\Rightarrow$ For given support $\rightarrow \ell_{\text {eff }}=\frac{1}{2}$ Hence,

\begin{aligned} \mathrm{E} \alpha \Delta \mathrm{T} & =\frac{\pi^{2} \mathrm{E}}{\left(\frac{\mathrm{I}}{2 \mathrm{r}}\right)^{2}} \\ \Rightarrow \quad \Delta \mathrm{T} & =\frac{\pi^{2} \times 4}{\alpha 5\left(\frac{1}{2}\right)^{2}}=\frac{\pi^{2} \times 4}{12 \times 10^{-6} \times 105^{2}} \\ \Rightarrow \quad \Delta \mathrm{T} & =298^{\circ} \mathrm{C} \end{aligned}

Q2GATE 2023NAT

Consider the fillet-welded lap joint shown in the figure (Not to scale). The length of the weld shown in the effective length. The welded surface meet at right angle. The weld size is $\mathrm{kN}$ , and the permissible stress in the weld is $120 \mathrm{MPa}$ . What is the safe load $P$ (in $\mathrm{kN}$ , rounded off to one decimal place) that can be transmitted by this welded joint ?

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📖 Explanation

Given data, Weld size $=8 \mathrm{~mm}$ Permissible stress $=120 \mathrm{MPa}$ Effective throat thickness, $\mathrm{t}_{\mathrm{e}}=0.7 \times Weld \; size =0.7 \times 8=5.6 \mathrm{~mm}$ Total effective length of weld, $\ell_{\mathrm{e}}=2 \times 75+50 =200 mm$ $\Rightarrow$ Safe load ' $P$ ' in kN rounded upto one decimal palce

\begin{aligned} & =5.6 \times 200 \times 120 \times 10^{-3} \\ & =134.4 \mathrm{kN} \end{aligned}

Q3GATE 2022NAT

Consider two linearly elastic rods HI and IJ, each of length b, as shown in the figure. The rods are co-linear, and confined between two fixed supports at H and J. Both the rods are initially stress free. The coefficient of linear thermal expansion is $\alpha$ for both the rods. The temperature of the rod IJ is raised by $\Delta T$ , whereas the temperature of rod HI remains unchanged. An external horizontal force P is now applied at node I. It is given that $\alpha ={10^{-6} } {\circ} C^{-1},\Delta T=50^{\circ}C,b = 2 m, AE = 10^6N.$ . The axial rigidities of the rods HI and IJ are 2AE and AE, respectively. To make the axial force in rod HI equal to zero, the value of the external force P (in N) is _________. (round off to the nearest integer)

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📖 Explanation

$R_A+R_B=P$ $\therefore R_B=P$ There is no axial force in rod HI $\therefore R_A=0$ Now check for rod IJ $\therefore R_B=P$ Now as rod IJ is fixed from both end, so net deflection due to increase in temperature will be zero.

\begin{aligned} b\alpha T +\left[ -\frac{Pb}{AE} \right]&= 0\\ b\alpha T &=\frac{Pb}{AE} \\ P &= AE\alpha T\\ P&= 10^6 \times 10^{-6} \times 50\\ P&=50N \end{aligned}

Explanation Figure 2

Explanation Figure 3

Q4GATE 2022MCQ

For a linear elastic and isotropic material, the correct relationship among Young's modulus of elasticity ( $E$ ), Poisson's ratio ( $v$ ), and shear modulus ( $G$ ) is

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📖 Explanation

$E=2G(1+\mu )$ $G=$ Shear modulas $\mu =$ Poission's ratio $E=$ Young's modulus

Q5GATE 2021NAT

A square plate O-P-Q-R of a linear elastic material with sides 1.0 m is loaded in a state of plane stress. Under a given stress condition, the plate deforms to a new configuration O-P'-Q'-R' as shown in the figure (not to scale). Under the given deformation, the edges of the plate remain straight. The horizontal displacement of the point (0.5 m, 0.5 m) in the plate O-P-Q-R (in mm,round off to one decimal place) is ________

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📖 Explanation

So horizontal displacement of the point (0.5 m, 0.5 m) $=-2.5 \mathrm{~mm}+5 \mathrm{~mm}=2.5 \mathrm{~mm}$

Explanation Figure 2

Q6GATE 2021MCQ

Strain hardening of structural steel means

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Q7GATE 2020MCQ

The state of stress represented by Mohr's circle shown in the figure is

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Q8GATE 2020MCQ

The total stress paths corresponding to different loading conditions, for a soil specimen under the isotropically consolidated stress state (O), are shown below: The correct match between the stress paths and the listed loading conditions, is

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Q9GATE 2020NAT

A rigid, uniform, weightless, horizontal bar is connected to three vertical members P, Q and R as shown in the figure. All three members have identical axial stiffness of 10 kN/mm. The lower ends of bars P and R rest on a rigid horizontal surface. When NO laod is applied, a gap of 2 mm exist between the lower end of the bar Q and the rigid horizontal surface. When a vertical load W is placed on the horizontal bar in the downward direction, the bar still remains horizontal and gets displayed by 5 mm in the vertically downward direction. The magnitude of the load W (in kN, round off to the nearst integer), is ______

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📖 Explanation

\begin{aligned} P_1 +P_1+P_2&=W \\ P_1&=P_3 \\ \frac{AE}{L}&=10kN/mm\\ \delta _1&=5mm=\frac{P_1L}{AE} \\ \delta _2&=3 mm = \frac{P_2L}{AE} \\ P_1 &=10 \times 5 =50 kN \\ P_2 &=10 \times 3 =30kN \\ W&=2(20)+30=130 kN \end{aligned}

Explanation Figure 2

Q10GATE 2019MCQ

An isolated concrete pavement slab of length L is resting on a frictionless base. The temperature of the top and bottom fibre of the slab are $T_t \; and \; T_b$ , respectively. Given: the coefficient of thermal expansion $=\alpha$ and the elastic modulus =E. Assuming $T_t \gt T_b$ and the unit weight of concrete as zero, the maximum thermal stress is calculated as

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📖 Explanation

Due to frictionless, thermal stress developed in concrete pavement slab is zero $\sigma _{th}=0$

Q11GATE 2019MCQ

An element is subjected to biaxial normal tensile strains of 0.0030 and 0.0020. The normal strain in the plane of maximum shear strain is

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📖 Explanation

$\varepsilon _x=0.0030$ $\varepsilon _y=0.0020$ Normal strain in the plane of maximum shear strain $\varepsilon _{avg}=\frac{\varepsilon _x+\varepsilon _y}{2}=\frac{0.0030+0.0020}{2}=0.0025$

Q12GATE 2018MCQ

A plate in equilibrium is subjected to uniform stresses along its edges with magnitude $\sigma_{xx}$ = 30 MPa and $\sigma_{yy}$ = 50 MPa as shown in the figure. The Young's modulus of the material is $2 \times 10^{11} N/m^{2}$ and the Poisson's ratio is 0.3. If $\sigma_{zz}$ is negligibly small and assumed to be zero, then the strain $\varepsilon _{zz}$ is

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📖 Explanation

\begin{aligned} \sigma _{xx} &=30MPa \\ \sigma _{yy} &=50MPa \\ \sigma _{zz} &=0 \\ \varepsilon _{zz} &=\frac{\sigma _{zz}}{E}-\mu \frac{\sigma _{xx}}{E}-\mu \frac{\sigma _{yy}}{E} \\ &= -\frac{\mu }{E} (\sigma _{xx}+\sigma _{yy})\\ &= -\frac{0.3}{2 \times 10^5}(30+50)\\ &=-120 \times 10^{-6} \end{aligned}

Q13GATE 2017MCQ

An elastic bar of length L, uniform cross sectional area A, coefficient of thermal expansion $\alpha$ , and Young's modulus E is fixed at the two ends. The temperature of the bar is increased by T, resulting in an axial stress $\sigma$ . Keeping all other parameters unchanged, if the length of the bar is doubled, the axial stress would be

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📖 Explanation

$\sigma=\alpha T E$ $\therefore \;$ Length have no effect on thermal stress. $\therefore \;$ Axial stress is only $\sigma$ .

Explanation Figure 1

Q14GATE 2017NAT

Consider the stepped bar made with a linear elastic material and subjected to an axial load of 1 kN, as shown in the figure. Segments 1 and 2 have cross-sectional are of $100mm^{2}$ and $60mm^{2}$ , Young's modulus of $2 \times 10^{5}$ MPa and $3 \times 10^{5}$ MPa, and length of 400mm and 900mm, respectively. The strain energy (in N-mm, up to one decimal place) in the bar due to the axial load is __________

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📖 Explanation

\begin{aligned} A_{1}=100 \mathrm{mm}^{2}, & E I=2 \times 10^{5} \mathrm{MPa} \\ A_{2}=60 \mathrm{mm}^{2}, &E I=3 \times 10^{5} \mathrm{MPa} \\ \Delta_{A C}=&\Delta_{A B}+\Delta_{B C} \\ =& \frac{1 \times 10^{3} \times 400}{100 \times 2 \times 10^{5}}+\frac{1 \times 10^{3} \times 900}{60 \times 3 \times 10^{5}} \\ =& 0.02+0.05=0.07 \mathrm{mm} \\ U=\frac{1}{2} \times P \times \Delta&=\frac{1}{2} \times 1 \times 1000 \times 0.07=35 \mathrm{N}-\mathrm{mm} \end{aligned}

Explanation Figure 2

Q15GATE 2017NAT

A 2 m long, axially loaded mild steel rod of 8 mm diameter exhibits the load-displacement $(P-\delta )$ behavior as shown in the figure. Assume the yield stress of steel as 250 Mpa. The complementary energy (in N-mm) stored in the bar up to its linear elastic behavior will be____

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📖 Explanation

Elastic strain, $\epsilon_{E}=\frac{\Delta L}{L}=\frac{2.5}{2000}=1.25 \times 10^{-2}$ Elastic strain energy $=\frac{1}{2} \sigma_{y} \in_{E} A L$ $=\frac{1}{2} \times 250 \times 1.25 \times 10^{-3} \times \frac{\pi}{4} \times 8^{2} \times 2000$ $=15707.96 \mathrm{Nmm}$ Note: For linear elastic material both complementary energy and strain energy is same.

Explanation Figure 2

Q16GATE 2017MCQ

In a material under a state of plane strain, a 10x10 mm square centered at a point gets deformed as shown in the figure. If the Shear strain $\gamma _{xy}$ at this point is expressed as 0.001k(in rad),the value of k is

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📖 Explanation

According to the sign convention. In question since angle has been increase Therefore, shear strain should be negative.

\begin{aligned} \therefore \; \gamma_{x y} &=-0.0005 \mathrm{rad} \\ &=0.001 k \\ -0.0005 &=0.001 k \\ \Rightarrow \;\; k &=-0.50 \end{aligned}

Explanation Figure 2

Q17GATE 2016MCQ

An elastic isotropic body is in a hydrostatic state of stress as shown in the figure. For no change in the volume to occur, what should be its Poisson's ratio?

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📖 Explanation

Volumetric strain,

\begin{aligned} \epsilon_{\mathrm{v}}&=\left(\frac{\sigma_{x}+\sigma_{y}+\sigma_{z}}{3}\right)(1-2 \mu)\\ \Rightarrow \quad \frac{\delta V}{V}&=\left(\frac{\sigma_{x}+\sigma_{y}+\sigma_{z}}{3}\right)(1-2 \mu) \\ \text{As }\quad \Delta V&=0 \\ \Rightarrow\text{ Either }\sigma_{x}&+\sigma_{y}+\sigma_{z}=0 or 1-2 \mu=0 \\ \Rightarrow \quad 1-2 \mu&=0 \\ \mu&=0.5 \end{aligned}

Q18GATE 2015NAT

A tapered circular rod of diameter varying from 20 mm to 10 mm is connected to another uniform circular rod of diameter 10 mm as shown in the following figure. Both bars are made of same material with the modulus of elasticity, $E=2\times 10^{5}$ MPa. When subjected to a load $P=30\pi kN$ , the deflection at point A is ______ mm.

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📖 Explanation

Total elongation,

\begin{aligned} \Delta &=\Delta_{A B}+\Delta_{B C} \\ &=\frac{P L}{A E}+\frac{4 P L}{\pi d_{1} d_{2} E} \\ &=\frac{30 \pi \times 1.5}{\frac{\pi \times (0.01)^{2}}{4} \times 2 \times 10^{5}} \\ &+\frac{30 \pi \times 2}{\frac{\pi \times (0.01) \times (0.02)}{4} \times 2 \times 10^{5}} \\ &=9+6=15 \mathrm{mm} \end{aligned}

Explanation Figure 2

Q19GATE 2014MCQ

A box of weight 100 kN shown in the figure is to be lifted without swinging. If all forces are coplanar, the magnitude and direction ( $\theta$ ) of the force (F) with respect to x-axis should be

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📖 Explanation

\begin{aligned} \Sigma F_{\text {horizontal }} &=0 \\ 90 \mathrm{cos} 30^{\circ} &=40 \mathrm{cos} 45^{\circ}+\mathrm{F} \cos \theta \\ 77.94 &=28.28+\mathrm{F} \cos \theta \\ \text{or}\quad \mathrm{F} \cos \theta &=49.66 \mathrm{kN}&\ldots(i)\\ \Sigma F_{\text {vertical }}&=0\\ 90 \sin 30^{\circ}&+40 \sin 45^{\circ}+F \sin \theta=W \\ 45+28.28+F \sin \theta&=100 \\ \text{or}\quad \quad F \sin \theta&=26.72 \mathrm{kN} &\ldots(ii) \end{aligned}

Dividing eq. (ii) by eq. (i), we get

\begin{aligned} \tan \theta &=\frac{26.72}{49.66}=0.5380 \\ \text{or}\quad\theta &=\tan ^{-1}(0.5380)=28.28^{\circ}\\ \text{Substituting }\theta&=28.28 \text{ \in eq. (i), we get}\\ F \cos 28.28^{\circ} &=49.66 \\ \text{or}\quad F &=\frac{49.66}{\cos 28.28^{\circ}}\\ &=\frac{49.66}{0.88}=56.43 \mathrm{kN} \end{aligned}

Explanation Figure 2

Explanation Figure 3

Q20GATE 2014MCQ

The values of axial stress ( $\sigma$ ) in kN/ $m^{2}$ , bending moment (M) in kNm, and shear force (V) in kN acting at point P for the arrangement shown in the figure are respectively

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📖 Explanation

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\begin{array}{l} \text { Axial stress }=\frac{50}{0.2 \times 0.2}=1250 \mathrm{kN} / \mathrm{m}^{2} \\ \text { Bending moment }=50 \times 3=150 \mathrm{kNm} \end{array}

Explanation Figure 2

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