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QUESTION: 1

The state of stress at a point in a loaded member is shown in figure. The magnitude of maximum shear stress is

Solution:

QUESTION: 2

At a point in a strained body carrying two unequal unlike principal stresses p_{1} and p_{2} (p_{1} > p_{2}), the maximum shear stress is given by

Solution:

The maximum shear stress is given by,

Now, σ_{1} = +p_{1}

σ_{2} = -p_{2}

∴

QUESTION: 3

The radius of Mohr’s circle of stress of a strained element is 20 N/mm^{2} and minor principal tensile stress is 10 N/mm^{2}; The major principal stress is

Solution:

Minor principal stress = C - r = 10

Centre of circle (C) = 10 + 20 =30 N/mm^{2}

Major principal stress = C + r = 30 + 20 = 50 N/mm^{2}

QUESTION: 4

The principal stresses σ_{1} , σ_{2} and σ_{3} at a point respectively are 80 MPa, 30 MPa and -40 MPa. The maximum shear stress is

Solution:

∴

QUESTION: 5

Plane stress at a point in a body is defined by principal stress 3σ and σ. The ratio of the normal stress to the maximum shear stress on the plane of maximum shear stress is

Solution:

∴

QUESTION: 6

A shaft subjected to torsion experiences a pure shear stress τ on the surface. The maximum principal stress on the surface which is at 45° to the axis will have a value

Solution:

QUESTION: 7

Principal strains at a point are 100 x 10^{-6} and -200 x 10^{-6}. What is the maximum shear strain at the point?

Solution:

= 150 x 10^{-6 }

∴ φ_{max} = 300 x 10^{-6}

QUESTION: 8

In a plane stress problem there are normal tensile stresses σ_{x} and σ_{y} accompanied by shear stress τ_{xy} at a point along orthogonal Cartesian coordinates x and y respectively. If it is observed that the minimum principal stress on a certain plane is zero the

Solution:

Principal stresses,

Since minimum principal stress is zero, therefore

QUESTION: 9

A point in a strained body is subjected to a tensile stress of 100 MPa on one plane and a tensile stress of 50 MPa on a plane at right angle to it. If these planes are carrying shear stresses of 50 MPa, then the principal stresses are inclined to the larger normal stress at an angle of

Solution:

The angle of inclination of principal stress to the larger normal stress is given by,

⇒ 2θ = tan^{-1}(2)

∴

QUESTION: 10

If the normal cross-section A of a member is subjected to a tensile force P, the resulting normal stress on an oblique plane inclined at angle θ to transverse plane will be

Solution:

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