Question 1 |

Strain hardening of structural steel means

experiencing higher stress than yield stress with increased deformation | |

strengthening steel member externally for reducing strain experienced | |

strain occurring before plastic flow of steel material | |

decrease in the stress experienced with increasing strain |

Question 1 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.

In the figure AB = Strain hardening zone

OA = Linear elastic zone

Stress corresponding to point 'A' is yield stress.

Question 2 |

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 ________

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 ________

1.2 | |

6.3 | |

5.2 | |

2.5 |

Question 2 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 3 |

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

uniaxial tension | |

biaxial tension of equal magnitude | |

hydrostatic stress | |

pure shear |

Question 3 Explanation:

In pure shear condition, Mohr's circle has its center at origin.

Question 4 |

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 ______

The magnitude of the load W (in kN, round off to the nearst integer), is ______

110 | |

150 | |

130 | |

160 |

Question 4 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 5 |

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

The correct match between the stress paths and the listed loading conditions, is

OP-I, OQ-II, OR-IV, OS-III | |

OP-IV, OQ-III, OR-I, OS-II | |

OP-III, OQ-II, OR-I, OS-IV | |

OP-I, OQ-III, OR-II, OS-IV |

Question 5 Explanation:

Question 6 |

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

L \alpha (T_t-T_b) | |

E \alpha (T_t-T_b) | |

\frac{E \alpha (T_t-T_b)}{2} | |

Zero |

Question 6 Explanation:

Due to frictionless, thermal stress developed in concrete pavement slab is zero

\sigma _{th}=0

\sigma _{th}=0

Question 7 |

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

Zero | |

0.001 | |

0.0025 | |

0.005 |

Question 7 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

\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 8 |

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

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

-120\times 10^{-6} | |

-60\times 10^{-6} | |

0 | |

120\times 10^{-6} |

Question 8 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}

Question 9 |

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____

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____

7500 | |

15000 | |

15707.9 | |

15007.9 |

Question 9 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.

Question 10 |

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

If the Shear strain \gamma _{xy} at this point is expressed as 0.001k(in rad),the value of k is

0.5 | |

0.25 | |

-0.25 | |

-0.5 |

Question 10 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}

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}

There are 10 questions to complete.