# Properties of Metals, Stress and Strain

 Question 1
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)
 A 85 B 23 C 65 D 50
GATE CE 2022 SET-2   Solid Mechanics
Question 1 Explanation:

$R_A+R_B=P$
$\because$ 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}
 Question 2
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
 A $G=\frac{E}{2(1+v)}$ B $G=\frac{E}{(1+2v)}$ C $E=\frac{G}{2(1+v)}$ D $E=\frac{G}{(1+2v)}$
GATE CE 2022 SET-2   Solid Mechanics
Question 2 Explanation:
$E=2G(1+\mu )$
$G=$ Shear modulas
$\mu =$Poission's ratio
$E=$ Young's modulus
 Question 3
Strain hardening of structural steel means
 A experiencing higher stress than yield stress with increased deformation B strengthening steel member externally for reducing strain experienced C strain occurring before plastic flow of steel material D decrease in the stress experienced with increasing strain
GATE CE 2021 SET-2   Solid Mechanics
Question 3 Explanation:
Strain hardening is experiencing higher stress than yield stress with increased deformation
In the figure AB = Strain hardening zone
OA = Linear elastic zone
Stress corresponding to point 'A' is yield stress.

 Question 4
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 ________
 A 1.2 B 6.3 C 5.2 D 2.5
GATE CE 2021 SET-1   Solid Mechanics
Question 4 Explanation:

So horizontal displacement of the point (0.5 m, 0.5 m)
$=-2.5 \mathrm{~mm}+5 \mathrm{~mm}=2.5 \mathrm{~mm}$
 Question 5
The state of stress represented by Mohr's circle shown in the figure is
 A uniaxial tension B biaxial tension of equal magnitude C hydrostatic stress D pure shear
GATE CE 2020 SET-2   Solid Mechanics
Question 5 Explanation:
In pure shear condition, Mohr's circle has its center at origin.
 Question 6
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 ______
 A 110 B 150 C 130 D 160
GATE CE 2020 SET-1   Solid Mechanics
Question 6 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}
 Question 7
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
 A OP-I, OQ-II, OR-IV, OS-III B OP-IV, OQ-III, OR-I, OS-II C OP-III, OQ-II, OR-I, OS-IV D OP-I, OQ-III, OR-II, OS-IV
GATE CE 2020 SET-1   Solid Mechanics
Question 7 Explanation:

 Question 8
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
 A $L \alpha (T_t-T_b)$ B $E \alpha (T_t-T_b)$ C $\frac{E \alpha (T_t-T_b)}{2}$ D Zero
GATE CE 2019 SET-1   Solid Mechanics
Question 8 Explanation:
Due to frictionless, thermal stress developed in concrete pavement slab is zero
$\sigma _{th}=0$
 Question 9
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
 A Zero B 0.001 C 0.0025 D 0.005
GATE CE 2019 SET-1   Solid Mechanics
Question 9 Explanation:
$\varepsilon _x=0.0030$
$\varepsilon _y=0.0020$
Normal strain in the plane of maximum shear strain
$\varepsilon _{avg}=\frac{\varepsilon _x+\varepsilon _y}{2}=\frac{0.0030+0.0020}{2}=0.0025$
 Question 10
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
 A $-120\times 10^{-6}$ B $-60\times 10^{-6}$ C 0 D $120\times 10^{-6}$
GATE CE 2018 SET-1   Solid Mechanics
Question 10 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}
There are 10 questions to complete.