Test: Analysis of Stress & Strain Level - 3


20 Questions MCQ Test Engineering Mechanics | Test: Analysis of Stress & Strain Level - 3


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QUESTION: 1

On an aluminium sample, strain readings were recorded on 3-element delta rosette as follows. ε = −100 μs, ε120° = −500 μs and ε60° = +600 μs Then the maximum tensile strain is __________ μs.

(A) 589

(B) 600

Solution: ε0 = εxx = −100 μs ⋯ ①

ε120° = −500 μs =

−500 = −25 + 0.75 εyy − 0.433 γxy

or − 475 = 0.75 εxy − 0.433 γxy ⋯ ②

ε60° = 600 =

= 25 + 0.75 εyy + 0.433 γxy

575 = 0.75 εyy + 0.433 γxy ⋯③

−475 = 0.75 εyy − 0.433 γxy ⋯④

100 = 1.5 εyy

or εyy = +66.67 μs

Putting the values in equation ③

575 = 0.75 × 66.67 + 0.433 γxy = 50 + 0.433 γxy

γxy = 1212.5 μs

Now = −16.665 μs

= −83.33 μs

=

= √(−83.33)2 + (606.25)2

√6943.9 + 367539 = 611.95 μs

Principal strains εp1 = −16.665 + 611.95 = +595.285 μs

εp2 = −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

τmax =

σ1, σ2 =

=

σ1, σ2 = P

τmax = = 0

QUESTION: 3

The data obtained from a rectangular strain gage rosette attached to a stressed steel member are ε0 = −220 × 10−6, ε45° = 120 × 10−6, ε90° = +220 × 10−6.Given E = 2 × 105 N⁄mm2 and Poisson’s ratio = 0.3. The principal stress acting at the point is ___________ MPa

(A) 37.5

(B) 39.5

Solution: εxx = ε0 = −220 μs

εyy = ε90o = +220 μs

Principal strain,

ε1 , ε2 =

ε1, ε2 = 0 ± √(−220)2 + (120)2

= ±250.6 μs

p1 =

=

= 38.55 N/mm2

p2 = −38.55 N/mm2

Question_Type: 5

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:

σx = = 40N/mm2

Normal stress, σθ

σθ =

=

σθ = 16.5N/mm2

Tangential stress, τθ

τθ =

= 19.7MPa

QUESTION: 5

In a strained material the normal stress on one plane is 10t/cm2 tensile and the shear stress is 5t/cm2. On a plane perpendicular to this plane there is only shear stress. The center for Mohr’s circle

Solution: Centre of Mohr’s circle

σ1 =

=

= 12.07 t/cm2

σ2 = 5 − √52 + 52 = −2.07 t/cm2

Centre of Mohr′s circle =

=

= 5 t/cm2

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 γxy

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

⇒ Initial angle = 90°

Final angle = 70°

γxy = 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, τxy is

Solution:

τxy = G γxy = 100 × 103 × (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 τxy, one of the principal stresses will be zero, only if,

Solution:

σ1.2 =

let σ2 = 0

Square both sides

τxy2 =

τxy2 =

τxy2 =

τxy2 =

τxy = √σxσy

τxy = √σx × σy

QUESTION: 9

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

Where σ1 =major principal stress,

σ2 = Minor principal stress and

q = Shear stress

Solution: Diameter of Mohr’s circle

2(τmax) =

Option (A):

2(τmax) =

= 800 N/mm2

Option (B):

2(τmax) =

= 1,131.37 N/mm2

Option (C):

2(τmax) =

Option (D):

2(τmax) =

= 800 N/mm2

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:

=

=

=

=

∴ σ1 =

σ2 =

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)

(A) 0.364

(B) 0.364

Solution: As, it is incompressible, volume = constant Hence, D02L0 = D12L1

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 τ45° (4 decimal places)

(A) 0.5235

(B) 0.5235

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

τ45° = 90° − 60° = 30°

=

=0.5235

Question_Type: 5

QUESTION: 13

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

ϵx = 0.001

ϵy = 0.002

τxy = 0.003

Find the elongation of the diagonal (4 decimal places)

(A) 0.0169

(B) 0.0169

Solution:

ϵ45° =?

As, ϵθ =

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. σ1 , σ2 are equal and are tensile in nature.

II. σ1 , σ2 are equal and are compressive in nature.

III. σ1 is compressive and σ2 is tensile and the magnitudes are the same.

IV. σ1 is tensile and σ2 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/m3 .

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

τ =

(A) 1.766

(B) 1.766

Solution: σx = 10 MPa, τxy = 5

σy = 20 MPa, τxz = 0

σz = −10 MPa, τyz = 0

σ1,2 =

=

= 15 ± √52 + 52

= 15 ± 5√2

σ1 = 22.07 MPa

σ2 = 3.82 MPa

σ3 = σz = −10 MPa [∵ τyz = τxz = 0]

Strain energy per unit volume

=

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

= 1.766 × 10−3 mJ/m3

= 1.766 kJ/m3

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/m3 . Take μ = 0.3, E = 200 GPa

[σ] =

(A) 1.682

(B) 1.682

Solution: Given,

σx = 10 MPa

σy = 20 MPa

σz = −10 MPa

τxy = 5 MPa

τxz = τyz = 0

σ1,2 =

=

= 15 ± √52 + 52

σ1 = 22.07 MPa

σ2 = 3.82 MPa

σ3 = σ2 = −10 MPa [∵ τyz = τxz = 0]

Distortion energy per unit volume

=

= + +(3.82 + 10)2 + (−10 − 22.07)2]

= 1.682 × 10−3 mJ/m3

= 1.682 kJ/m3

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: σx = σ = σ1 similarly, σ2 = 2σ

σ3 = 3σ

Syt =

Syt )I =

=

=

= √7.4 σ2

∵ [σ1 = 2σ, σ2 = σ, σ3 = 4σ]

Syt )II =

=

=

=√12.6σ2

Syt )II > Syt )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

Syt =

Syt)I:

Syt =

=

=

= √3 × [4τ2 + 2τ2]

= √18 τ2

Syt)II:

=

=

=

= √7 × 4 × τ2 + 3τ2

= √31 τ2

As, Syt )II > Syt)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 mm4

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

=

= 91.364 N/mm2 (compressive)

Shear stress = q =

ay̅ = (200 × 20 × 310) + (12 × 100 × 250) = 1540000 mm3

q = = 52.112 N/mm2

Principal stresses are given by

=

= 45.682 ± 69.300 N/mm2

∴ p1 = 45.682 + 69.300

= 114.982 N/mm2 (tensile)

and p2 = 45.682 − 69.300

= −23.618 N/mm2 (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, σ1 = 160 MPa, σ2 = 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, p1 = 120 + 50 = 170 MPa p2 = 120 − 50 = 70 MPa

Principal strains,

e1 =

=

= 1455 × 10−6 = 1455 μ strain

e2 =

= 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, θ1 =

=

θ1 = −18.44 ° as shear stress on reference plane is positive

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