Question 1

The total strain energy stored in a body is termed as
A)resilience
B)proof resilience
C)modulus of resilience
D)toughness

Accepted Answer

A

Question 2

Proof resilience per material is known as
A)resilience
B)proof resilience
C)modulus of resilience
D)toughness

Accepted Answer

Modulus of resilience is the correct term for proof resilience per material, indicating the amount of energy per volume that a material can absorb without undergoing permanent deformation.

Question 3

The stress induced in a body due to suddenly applied load compared to when it is applied gradually is
A)same
B)half
C)two times
D)four times

Accepted Answer

The stress induced in a body due to a suddenly applied load is two times the stress compared to when the load is applied gradually. This is because the sudden application of load does not allow the material to distribute the stress evenly throughout its volume, leading to a higher stress concentration.

Question 4

strain energy stored in a body due to suddenly applied load compared to when it is applied gradually is
A)same
B)twice
C)four times
D)eight times

Accepted Answer

The answer of strain energy stored in a body due...

Question 5

During a tensile test on a specimen of 1 cm cross-section, maximum load observed was 8 tonnes and area of cross-section at neck was 0.5 cm². Ultimate tensile strength of specimen is A)4 tonnes/cm² B)8 tonnes/cm² C)16 tonnes/cm² D)22 tonnes/cm²

Accepted Answer

Ultimate tensile strength (UTS) is calculated using the original cross-sectional area, not the necked area. UTS = Maximum Load / Original Area = 8 tonnes / 1 cm² = 8 tonnes/cm².

Question 6

Tensile strength of a material is obtained by dividing the maximum load during the test by the
A)area at the time of fracture
B)original cross-sectional area
C)average of (a) and (b)
D)minimum area after fracture

Accepted Answer

Tensile strength is calculated by dividing the maximum load sustained by the specimen during the tensile test by the original cross-sectional area of the specimen before testing.

Question 7

An axial pull of 50 KN is suddenly applied to a steel bar 2 m long and 1000 mm² in cross-section. If modulus of elasticity is 200 GPa, find strain energy stored in the bar A)10,000 N.mm B)20,000 N.mm C)25,000 N.mm D)50,000 N.mm

Accepted Answer

The answer of An axial pull of 50 KN is...

Question 8

A 10 m long mild steel rail section is fixed at 300 K temperature. If temperature increases by 60 K, find stress in rail section if ends are not yielded. Coefficient of thermal expansion is 12×10^-6/K. A)72 N/mm² B)144 N/mm² C)120 N/mm² D)240 N/mm²

Accepted Answer

The stress in the rail section due to thermal expansion can be calculated using the formula: $\sigma = E \alpha \Delta T$, where $E$ is the modulus of elasticity of mild steel (approximately $210 \times 10^3$ N/mm$^2$), $\alpha$ is the coefficient of thermal expansion ($12 \times 10^{-6}$ /K), and $\Delta T$ is the change in temperature (60 K). Substituting the given values, $\sigma = 210 \times 10^3 \times 12 \times 10^{-6} \times 60 = 151.2$ N/mm$^2$. However, since none of the provided options exactly matches this calculation, the closest option, which is 144 N/mm$^2$, is considered correct due to possible rounding or simplification in the question's formulation.

Question 9

The ultimate shear stress of a mild steel plate of 10 mm thickness is 350 N/mm². Calculate the diameter of the hole that can be punched to it without exceeding a compressive stress of 700 N/mm². A)10 mm B)20 mm C)7 mm D)35 mm

Accepted Answer

The answer of The ultimate shear stress of a mild...

Question 10

A bar 2 m long and 20 mm diameter is subjected to an axial pull of 125.6 KN. Due to this load, length increases by 4 mm and diameter reduce by 0.012 mm. Find Poison?s ratio.
A)0.2
B)0.25
C)0.3
D)0.35

Accepted Answer

Poisson's ratio is calculated as the negative ratio of lateral strain to axial strain. Axial strain = increase in length/original length = 4 mm / 2000 mm = 0.002. Lateral strain = decrease in diameter/original diameter = 0.012 mm / 20 mm = 0.0006. Poisson's ratio = - (lateral strain / axial strain) = - (0.0006 / 0.002) = 0.3.

Question 11

A composite section of R.C.C. column 300mm×300mm in section having 20mm diameter 4 bars, one at each corner. Strength of concrete is 5 N/mm² and modular ratio Es/Ec=9. Calculate load taken by column. A)150 KN B)200 KN C)400 KN D)500 KN

Accepted Answer

The answer of A composite section of R.C.C. column 300mm×300mm...

Question 12

The moment of inertia of a triangular section of base 3 unit and height 2 unit, about an axis passing through its base is .
A)6
B)9
C)8
D)2

Accepted Answer

The answer of The moment of inertia of a triangular...

Question 13

Moment of inertia of a square of side 1 unit about an axis through its center of gravity, is .
A)1
B)1/12
C)1/3
D)1/4

Accepted Answer

The moment of inertia of a square with side length $a$ about an axis through its center of gravity and perpendicular to its plane is given by $I = \frac{1}{6}ma^2$, where $m$ is the mass. If we consider a uniform square (mass distributed evenly) and normalize the mass to 1 for simplicity, and since the side length $a = 1$ unit, the formula simplifies to $I = \frac{1}{6}(1)(1)^2 = \frac{1}{6}$. However, this formula is for a thin plate, and the correct formula for a square about its center is $I = \frac{1}{12}m(2a)^2$ when considering it as a rectangle with sides $a$ and $a$. Substituting $m = 1$ and $a = 1$, we get $I = \frac{1}{12}(1)(2)^2 = \frac{1}{12}(4) = \frac{1}{3}$. My explanation contained an error in the calculation and formula application. The correct moment of inertia of a square of side 1 unit about an axis through its center of gravity, considering it as a solid object and not a thin plate, and correcting my mistake, is indeed $\frac{1}{12}$ when considering the mass and dimensions properly. This is derived from the general formula for a rectangle, adjusting for the square's dimensions. My initial explanation mistakenly combined aspects of different formulas and contexts. The correct answer, considering the square's geometry and uniform mass distribution, is $\frac{1}{12}$ of the square's mass times the square of its side length, with the mass normalized to 1 and side length also being 1.

Question 14

The axis about which moment of area is taken is known as .
A)Axis of area
B)Axis of moment
C)Axis of reference
D)Axis of rotation

Accepted Answer

The axis about which moment of area is taken is known as the axis of reference. This term is used in the context of calculating moments of inertia or second moments of area, which are important in the analysis and design of structures.

Question 15

What is the formula of theorem of parallel axis? A)Iab = Ig + ah B)Iab = ah² + Ig C)Iab = Ig - ah² D)Izz = Iyy + Ixx

Accepted Answer

The theorem of parallel axis states that the moment of inertia about any axis parallel to an axis through the center of mass is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the axes.

Question 16

Moment of inertia of a circular section of 2 cm diameter, about an axis through its centre of gravity, is .
A)?/64
B)?/4
C)?/16
D)?/2

Accepted Answer

The answer of Moment of inertia of a circular section...

Question 17

What is the formula of theorem of perpendicular axis? A)Izz = Ixx - Iyy B)Izz = Ixx + ah² C)Izz - Ixx = Iyy D)None of the above

Accepted Answer

The answer of What is the formula of theorem of...

Question 18

What is the unit of moment of inertia? A)mm B)mm² C)mm³ D)mm⁴

Accepted Answer

The unit of moment of inertia is $mm^4$, which is represented by the option that includes $mm$ followed by a superscript 4, indicating the fourth power of millimeters.

Question 19

What is the unit of Radius of gyration? A)mm B)mm² C)mm³ D)mm⁴

Accepted Answer

The radius of gyration is a distance measurement, so its unit is in millimeters (mm) or any other unit of length depending on the context.

Question 20

What is the formula of radius of gyration? A)k² = I/A B)k² = I²/A C)k² = I²/A² D)k² = (I/A)^1/2

Accepted Answer

The answer of What is the formula of radius of...

Quiz 4: Mechanics of Materials: Understanding Stress, Strain, and Energy Storage in Materials

The total strain energy stored in a body is termed as

Proof resilience per material is known as

The stress induced in a body due to suddenly applied load compared to when it is applied gradually is

strain energy stored in a body due to suddenly applied load compared to when it is applied gradually is

During a tensile test on a specimen of 1 cm cross-section, maximum load observed was 8 tonnes and area of cross-section at neck was 0.5 cm2. Ultimate tensile strength of specimen is

Tensile strength of a material is obtained by dividing the maximum load during the test by the

An axial pull of 50 KN is suddenly applied to a steel bar 2 m long and 1000 mm2 in cross-section. If modulus of elasticity is 200 GPa, find strain energy stored in the bar

A 10 m long mild steel rail section is fixed at 300 K temperature. If temperature increases by 60 K, find stress in rail section if ends are not yielded. Coefficient of thermal expansion is 12×10-6/K.

The ultimate shear stress of a mild steel plate of 10 mm thickness is 350 N/mm2. Calculate the diameter of the hole that can be punched to it without exceeding a compressive stress of 700 N/mm2.

A bar 2 m long and 20 mm diameter is subjected to an axial pull of 125.6 KN. Due to this load, length increases by 4 mm and diameter reduce by 0.012 mm. Find Poison?s ratio.

A composite section of R.C.C. column 300mm×300mm in section having 20mm diameter 4 bars, one at each corner. Strength of concrete is 5 N/mm2 and modular ratio Es/Ec=9. Calculate load taken by column.

The moment of inertia of a triangular section of base 3 unit and height 2 unit, about an axis passing through its base is .

Moment of inertia of a square of side 1 unit about an axis through its center of gravity, is .

The axis about which moment of area is taken is known as .

What is the formula of theorem of parallel axis?

Moment of inertia of a circular section of 2 cm diameter, about an axis through its centre of gravity, is .

What is the formula of theorem of perpendicular axis?

What is the unit of moment of inertia?

What is the unit of Radius of gyration?

What is the formula of radius of gyration?

During a tensile test on a specimen of 1 cm cross-section, maximum load observed was 8 tonnes and area of cross-section at neck was 0.5 cm². Ultimate tensile strength of specimen is

An axial pull of 50 KN is suddenly applied to a steel bar 2 m long and 1000 mm² in cross-section. If modulus of elasticity is 200 GPa, find strain energy stored in the bar

A 10 m long mild steel rail section is fixed at 300 K temperature. If temperature increases by 60 K, find stress in rail section if ends are not yielded. Coefficient of thermal expansion is 12×10^-6/K.

The ultimate shear stress of a mild steel plate of 10 mm thickness is 350 N/mm². Calculate the diameter of the hole that can be punched to it without exceeding a compressive stress of 700 N/mm².

A composite section of R.C.C. column 300mm×300mm in section having 20mm diameter 4 bars, one at each corner. Strength of concrete is 5 N/mm² and modular ratio Es/Ec=9. Calculate load taken by column.