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QUESTION: 1

The maximum shear due to an axial compression of 100 MPa is

Solution:
Maximum shear stress,

τ =

=

τ = 50 MPa

QUESTION: 2

When a rod of cross-sectional area 100 mm2is subjected to an tensile force 6kN, the maximum shear stress in the rod will be

Solution:

σ = = 60 MPa

τ_{max} = σ/2 = 30 MPa

QUESTION: 3

A body is subjected to two normal stresses 20 kN/m^{2} (tensile) and 10 kN/m^{2} (compressive) acting mutually perpendicular to each other. The maximum shear stress is

Solution:
τ_{max} = = 15 kN/m^{2}

QUESTION: 4

A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. If is also subjected to shear stress of 400 MPa on the same planes. The maximum shear stress will be

Solution:

maximum Shear Stress , τ_{max}

τ_{max} =

=

= 500 MPa

QUESTION: 5

A Plane stressed element is subjected to the state of stress given by σ_{x} = τ_{xy} = 100 kgf/cm^{2} and σ_{y} = 0. Maximum shear stress in the element is equal to

Solution:
Maximum shear stress,

τ_{max} =

=

= √12, 500

= 50√5 kgf/cm2

QUESTION: 6

The radius of Mohr’s circle for two unlike principal stresses of magnitude σ is

Solution:
Radius of Mohr’s circle, τ_{max} =

τ_{max} = σ

QUESTION: 7

At a point in a stressed material, the stress system is σ_{x} = + 800N/mm^{2} , σ_{y} = 0 and q = 0.

The radius of the Mohr’s stress circle for this case is__________ units.

Solution:
σ_{x} = 800 N/mm^{2}, σ_{y} = 0, q = τ_{xy} = 0

Radius of Mohr’s circle = τ_{max}

τ_{max} = =400 N/mm^{2}

QUESTION: 8

The principal stresses at a point are 150.0 MPa and − 50.0 MPa; the radius of the corresponding Mohr’s circle will be

Solution:
Radius of Mohr’s circle

τ_{max} =

=

= 100 MPa

QUESTION: 9

A spherical ball of volume 10^{6} mm^{3} is subjected to a hydrostatic pressure of 90 MPa. If the bulk modulus for the material is 180 GPa, the change in the volume of the ball is

Solution:
Volume = 10^{6} mm^{3} Hydrostatic pressure, P = 90 MPa

Bulk modulus, k = 180 kN/m^{3} + 2

k =

180 × 103 =

δ_{v} = 500 mm^{3}

QUESTION: 10

An element is subjected to p_{x }= 35 MPa, p_{y} = 20 MPa and shear stress q = 7.5 MPa. Then the direction of principal stresses is

Solution:

tan 2α =

tan 2α = 1 2α = 45°

α = 45°/2

= 22.5°

QUESTION: 11

For an inclined plane in a rectangular block subjected to two mutually perpendicular normal stresses 1000 MPa and 400 MPa and shear stress 400 MPa, the maximum normal stress will be

Solution:

σ1 =

= 700 + √9 × 10^{4} + 16 × 10^{4}

= 700 + 500 = 1200MPa

QUESTION: 12

The normal stresses of 55 N/mm^{2} tensile and 45 N/mm^{2} compressive on two mutually perpendicular planes are acting at a point in a piece of elastic material. These planes also carry shear stress of 50 N/mm^{2}. The minor principal stress will be

Solution:
Minor principal stress, σ_{2}

σ_{2} =

=

σ_{2} = 65.71 N/mm^{2} compressive

σy = 45 N/mm^{2}

QUESTION: 13

A rectangular strain rosette, shown in figure give following reading in a strain measurement task ϵ_{P} = 100 × 10^{−6}, ϵ_{Q} = 150 × 10^{−6}, ϵR = 200 × 10^{−6}. The major principal strain is

Solution:
ϵ_{1,2} =

ϵ_{x} = ϵ_{P} = 100 × 10^{−6}

ϵ_{y} = ϵ_{R} = 200 × 10^{−6}

= 150 × 10^{−6} − 150 × 10^{−6}

= 0

ϵ_{1,2} =

ϵ_{1} =

ϵ_{2} =

ϵ_{1} = ϵ_{y} = 200 × 10^{−6}

QUESTION: 14

If major and minor principal strains are given as 500 × 10^{−6} and − 200 × 10^{−6}. Then calculate the maximum shear strain.

Solution:

⇒ γ_{max} = (ϵ_{1} − ϵ_{2}) = [500 − (−200)]10^{−6}

= 700 × 10^{−6}

QUESTION: 15

A rectangular strain rosette, shown in figure gives following reading in a strain measurement task ϵ_{1} = 1000 × 10^{−6}, ϵ_{2} = 800 × 10^{−6} and ϵ_{3} = 600 × 10^{−6}. The direction of the major principal strain with respect to gauge 1 is

Solution:

tan 2θ_{p} =

=

= 0

QUESTION: 16

Consider the following failure criteria and match them with their corresponding graphical representations.

P - Maximum principal stress Theory

Q - Maximum distortion energy Theory

R - Maximum principal strain Theory

I.

II.

III.

Common Data for Question No: 17 and 18 If the principal stress at a point in an elastic material are 2σ, σ and − σ/2 the properties are S_{yt} = 200 MPa and μ = 0.3

Solution:

QUESTION: 17

The value of σ at failure according to maximum principal stress theory

Solution:
According to MPST

σ_{1} ≤

[N = 1 at failure]

σ = 100 MPa

QUESTION: 18

The value of σ at failure according to strain energy theory

Solution:
According to strain energy theory

σ_{1}^{2} + σ_{2}^{2} + σ_{3}^{2} − 2μ(σ_{1}σ_{2} + σ_{2}σ_{3} + σ_{3}σ_{1}) ≤

4.95 σ^{2} = (200)^{2}

σ = 89.89 MPa ≈ 90 MPa

QUESTION: 19

A spherical pressure shell 3.8 meters diameter and 3 mm thick is subjected to an internal pressure ‘P’. The elastic limit of the shell material in simple tension is 260 MPa. The factor of safety is to 2.8. The internal pressure, if the failure of shell is to be prevented, according to maximum shear stress theory is

Solution:

σ_{noop} = σ_{1} = σ_{2} =

S_{yt }= 260 MPa

N = 2.8

According to MSST

Abs τ_{max} ≤

Abs τ_{max} =

P = 2.932 bar

QUESTION: 20

The force acting on a bolt consists of two components, an axial pull of 14 kN and a transverse shear load of 7 kN. The bolt material yield strength is 340 MPa and the factor of safety is 3 then, the cross sectional area of bolt required according to maximum shear stress theory is

Solution:

σ_{x} =

τ_{xy} =

A= 174.69 mm^{2}

A ≈ 175 mm^{2}

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