# Properties of Metals, Stress and Strain

 Question 1
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 $8 \mathrm{~mm}$, 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 ?

 A 185.2 B 111.3 C 214.5 D 134.3
GATE CE 2023 SET-1   Solid Mechanics
Question 1 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}
 Question 2
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 2 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 3
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 3 Explanation:
$E=2G(1+\mu )$
$G=$ Shear modulas
$\mu =$Poission's ratio
$E=$ Young's modulus
 Question 4
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 4 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 5
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 5 Explanation:

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

There are 5 questions to complete.