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QUESTION: 1

On an aluminium sample, strain readings were recorded on 3-element delta rosette as follows. ε_{0°} = −100 μs, ε_{120°} = −500 μs and ε_{60°} = +600 μs Then the maximum tensile strain is __________ μs.

(A) 589

(B) 600

Solution:
ε_{0} = ε_{xx} = −100 μs ⋯ ①

ε_{120°} = −500 μs =

−500 = −25 + 0.75 ε_{yy} − 0.433 γ_{xy}

or − 475 = 0.75 ε_{xy} − 0.433 γ_{xy} ⋯ ②

ε_{60°} = 600 =

= 25 + 0.75 ε_{yy} + 0.433 γ_{xy}

575 = 0.75 ε_{yy} + 0.433 γ_{xy} ⋯③

−475 = 0.75 ε_{yy} − 0.433 γ_{xy} ⋯④

100 = 1.5 ε_{yy}

or ε_{yy} = +66.67 μs

Putting the values in equation ③

575 = 0.75 × 66.67 + 0.433 γ_{xy} = 50 + 0.433 γ_{xy}

γ_{xy} = 1212.5 μs

Now = −16.665 μs

= −83.33 μs

=

= √(−83.33)^{2} + (606.25)^{2}

√6943.9 + 367539 = 611.95 μs

Principal strains ε_{p1} = −16.665 + 611.95 = +595.285 μs

ε_{p2} = −16.665 − 611.95 = −628.615 μs

Question_Type: 5

QUESTION: 2

If a small concrete cube is submerged deep in still water in such a way that the pressure exerted on all faces of the cube is P, then the maximum shear stress developed inside the cube is

Solution:
Maximum shear stress

τ_{max} =

σ_{1}, σ_{2} =

=

σ_{1}, σ_{2} = P

τ_{max} = = 0

QUESTION: 3

Which point on the stress strain curve occurs after yield plateau?

Solution:

After the yield plateau the curve will go up to its maximum limit of stress which is its ultimate point.

QUESTION: 4

Two wooden joists 50mm and 100mm are glued together along the joist AB as shown. Determine the normal and tangential stress in the glue if P = 200 kN

Solution:

σ_{x} = = 40N/mm^{2}

Normal stress, σ_{θ}

σ_{θ} =

=

σ_{θ} = 16.5N/mm^{2}

Tangential stress, τ_{θ}

τ_{θ} =

= 19.7MPa

QUESTION: 5

In a strained material the normal stress on one plane is 10t/cm^{2} tensile and the shear stress is 5t/cm^{2}. On a plane perpendicular to this plane there is only shear stress. The center for Mohr’s circle

Solution:
Centre of Mohr’s circle

σ_{1 }=

=

= 12.07 t/cm^{2}

σ_{2} = 5 − √5^{2} + 5^{2} = −2.07 t/cm^{2}

Centre of Mohr′s circle =

=

= 5 t/cm^{2}

QUESTION: 6

In a material under a state of plane strain, a square centered at a point gets deformed as shown in figure.

Find the shear strain γ_{xy}

Solution:
Shear strain is defined as the change in initial right angle.

⇒ Initial angle = 90°

Final angle = 70°

γ_{xy} = 90 − 70 = 20°

=

= 0.35 radian

QUESTION: 7

Assuming E = 160 GPa and G = 100 GPa for a material, a strain tensor is given as

The shear stress, τ_{xy} is

Solution:

τ_{xy} = G γ_{xy} = 100 × 10^{3 }× (0.004 × 2) MPa

= 800 MPa

QUESTION: 8

n a strained material, normal stresses on two mutually perpendicular planes are σx and σy (both alike) accompanied by a shear stress τ_{xy}, one of the principal stresses will be zero, only if,

Solution:

σ_{1.2} =

let σ_{2} = 0

Square both sides

τ_{xy2} =

τ_{xy2} =

τ_{xy2} =

τ_{xy2} =

τ_{xy} = √σ_{x}σ_{y}

τ_{xy} = √σ_{x} × σ_{y}

QUESTION: 9

For which one of the following stress conditions the diameter of Mohr’s stress circle is zero?

Where σ_{1} =major principal stress,

σ_{2} = Minor principal stress and

q = Shear stress

Solution:
Diameter of Mohr’s circle

2(τ_{max}) =

Option (A):

2(τ_{max}) =

= 800 N/mm^{2}

Option (B):

2(τ_{max}) =

= 1,131.37 N/mm^{2}

Option (C):

2(τ_{max}) =

Option (D):

2(τ_{max}) =

= 800 N/mm^{2}

QUESTION: 10

If Normal stresses of same nature px and py and shear stress q is acting on two perpendicular planes and q = (pxpy)^{1/2}, then the major and minor principal stresses respectively are

Solution:

=

=

=

=

∴ σ_{1} =

σ_{2} =

QUESTION: 11

A tensile specimen with 12 mm initial diameter and 50 mm initial length is subjected to a load of 90 kN. After sometime, the diameter is 10mm. Assuming it as in compressible material, calculate true strain along the length. (Give answer upto 3 decimal places)

Solution:

As, it is incompressible, volume = constant Hence, D_{0}^{2}L_{0} = D_{1}^{2}L_{1}

Now, e =

=

= 2 ln(1.2)

= 0.364

Question_Type: 5

QUESTION: 12

A square plate of side 4 units undergoes deformation. The angle between the diagonals becomes 60°. Calculate τ_{45°} (4 decimal places)

(A) 0.5235

(B) 0.5235

Solution:
Before deformation, the angle between diagonals is 90°

τ_{45°} = 90° − 60° = 30°

=

=0.5235

Question_Type: 5

QUESTION: 13

A square plate of side 4 units experiences the following strain components.

ϵ_{x} = 0.001

ϵ_{y} = 0.002

τ_{xy} = 0.003

Find the elongation of the diagonal (4 decimal places)

(A) 0.0169

(B) 0.0169

Solution:

ϵ_{45°} =?

As, ϵ_{θ} =

ΔL = (0.003) × 4√2

ΔL = 0.0169 units

Question_Type: 5

QUESTION: 14

For which of the following cases distortion energy theory is more liberal when compared to shear stress theory.

I. σ_{1} , σ_{2} are equal and are tensile in nature.

II. σ_{1} , σ_{2} are equal and are compressive in nature.

III. σ_{1} is compressive and σ_{2} is tensile and the magnitudes are the same.

IV. σ_{1 }is tensile and σ_{2} is compressive and the magnitudes are same

Solution:

For I, II both the theories will give the same result. For III, IV distortion theory is more liberal than shear stress theory.

QUESTION: 15

Following stress Tensor represents tri-axial state of stress at a point. Determine strain energy per unit volume in kJ/m^{3} .

Consider, Poisson’s ratio = 0.3 and Young’s modulus of elasticity = 200 GPa

τ =

(A) 1.766

(B) 1.766

Solution:
σ_{x} = 10 MPa, τ_{xy} = 5

σ_{y} = 20 MPa, τ_{xz} = 0

σ_{z} = −10 MPa, τ_{yz} = 0

σ_{1,2} =

=

= 15 ± √52 + 52

= 15 ± 5√2

σ_{1} = 22.07 MPa

σ_{2} = 3.82 MPa

σ_{3} = σz = −10 MPa [∵ τ_{yz} = τ_{xz} = 0]

Strain energy per unit volume

=

= = −2(0.3)[(22.07 × 3.82) + (3.82 × −10) + (−10 × 22.07)]}

= 1.766 × 10^{−3} mJ/m^{3}

= 1.766 kJ/m^{3}

Question_Type: 5

QUESTION: 16

Following stress Tensor represents tri-axial state of stress at a point. Determine the distortion energy per unit volume in terms of kJ/m^{3} . Take μ = 0.3, E = 200 GPa

[σ] =

(A) 1.682

(B) 1.682

Solution:
Given,

σ_{x} = 10 MPa

σ_{y} = 20 MPa

σ_{z} = −10 MPa

τ_{xy} = 5 MPa

τ_{xz} = τ_{yz} = 0

σ_{1,2} =

=

= 15 ± √52 + 52

σ_{1} = 22.07 MPa

σ_{2} = 3.82 MPa

σ_{3} = σ_{2} = −10 MPa [∵ τ_{yz} = τ_{xz} = 0]

Distortion energy per unit volume

=

= + +(3.82 + 10)^{2} + (−10 − 22.07)^{2}]

= 1.682 × 10^{−3} mJ/m^{3}

= 1.682 kJ/m^{3}

Question_Type: 5

QUESTION: 17

Consider the below shown loading conditions, which one will fail first according to principal strain energy theory [μ = 0.3]

I.

II.

Solution:
σ_{x} = σ = σ_{1} similarly, σ_{2} = 2σ

σ_{3} = 3σ

S_{yt }=

S_{yt} )_{I} =

=

=

= √7.4 σ2

∵ [σ_{1} = 2σ, σ_{2} = σ, σ_{3} = 4σ]

S_{yt} )_{II} =

=

=

=√12.6σ^{2}

S_{yt} )_{II} > S_{yt} )_{I}

QUESTION: 18

Consider the below shown loading conditions. Which one will fail first according to distortion energy theory? [take, σ/τ = 2]

I.

II.

Solution:
According to distortion Energy Theory

S_{yt }=

S_{yt})_{I}:

S_{yt} =

=

=

= √3 × [4τ^{2} + 2τ^{2}]

= √18 τ^{2}

S_{yt})_{II}:

=

=

=

= √7 × 4 × τ^{2} + 3τ^{2}

= √31 τ^{2}

As, S_{yt })_{II} > S_{yt})_{I}

II will fail first.

QUESTION: 19

An I-beam with flanges of size 200 mm × 200 mm and a web of 600 mm × 12 mm is subjected to a bending moment of 450 kNm and a shear force of 400 kN at a section. The major principal stress at a point 200 mm above the neutral axis is __________ MPa

(A) 114.0

(B) 116.0

Solution:

Moment of inertia of the section = I

=

= = 985.067 × 106 mm^{4}

At a distance of 200 mm above the neutral axis Bending stress = p =

=

= 91.364 N/mm^{2} (compressive)

Shear stress = q =

ay̅ = (200 × 20 × 310) + (12 × 100 × 250) = 1540000 mm^{3}

q = = 52.112 N/mm^{2}

Principal stresses are given by

=

= 45.682 ± 69.300 N/mm^{2}

∴ p_{1} = 45.682 + 69.300

= 114.982 N/mm^{2} (tensile)

and p_{2} = 45.682 − 69.300

= −23.618 N/mm^{2} (compressive)

Question_Type: 5

QUESTION: 20

A circle of diameter 500 mm is inscribed on a square plate of copper, before the application of stresses as shown, σ_{1} = 160 MPa, σ_{2} = 80 MPa, τ = 30 MPa. After the application of stresses, the circle is deformed into an ellipse. Then the major and minor axes of the ellipse will be Given E = 100 GPa, ν = 0.35.

Solution:
= 120 MPa

= 40 MPa

where τ = 30 MPa

= √402 + 302 = 50 MPa

Principal stresses, p_{1} = 120 + 50 = 170 MPa p_{2} = 120 − 50 = 70 MPa

Principal strains,

e_{1} =

=

= 1455 × 10^{−6} = 1455 μ strain

e_{2} =

= 105 × 10^{−6 }= 105 μ strain

Major axis = 500 + 1455 × 10^{−6} × 500

= 500 + 0.7275 = 500.7275 mm

Minor axis = 500 + 105 × 10^{−6} × 500

= 500 + 0.0525 = 500.0525 mm

Principal angles, θ_{1} =

=

θ_{1} = −18.44 ° as shear stress on reference plane is positive

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