Engineering College

problem 07.071 - determine the maximum shearing stress for the given general state of stress. skip to question consider the given state of stress. take x = 38 mpa and y = 18 mpa.

Answers

Answer 1
:

Given: x = 38 MPa, y = 18 MPa

To find: The maximum shearing stress for the given general state of stress

Solution:

The maximum shearing stress occurs on a plane that is oriented at 45 degrees to the x-axis. The equation to calculate the maximum shearing stress is:

τmax = (σx - σy) / 2 * sin(2θ)

where σx and σy are the normal stresses in the x and y directions, respectively, and θ is the angle between the plane of interest and the x-axis.

Substituting the given values into the equation, we get:

τmax = (38 - 18) / 2 * sin(2 * 45)
τmax = 10 / 2 * sin(90)
τmax = 10 MPa

Therefore, the maximum shearing stress for the given general state of stress is 10 MPa.
Answer 2

The maximum shearing stress for the given general state of stress, considering the given state of stress, taking x = 38 MPa and y = 18 MPa is 20 MPa.

The maximum shearing stress for the given general state of stress, considering the given state of stress, taking x = 38 MPa and y = 18 MPa is 20 MPa.

The general state of stress is given as:σx = 38 MPa, σy = 18 MPa, τxy = -12 MPa

The normal and shear stresses on an inclined plane with respect to x-axis is given by the following equation:

σn = (σx + σy)/2 + [(σx - σy)/2]cos2θ + τxy sin2θσs = [(σx - σy)/2]sin2θ + τxy cos2θ

where, σn = normal stress,σs = shear stress,θ = angle made by the plane with the x-axis

In this case, we need to find the maximum shear stress, which occurs when θ is such that the second term in σs expression is maximum.

To obtain maximum value of σs, we equate the derivative of the second term with respect to θ to zero.

τxy cos2θ - [(σx - σy)/2]sin2θ = 0τxy cos2θ = [(σx - σy)/2]sin2θtan2θ = 2τxy/(σx - σy)

Substituting the given values, we have:tan2θ = 2(-12)/20 = -1.2

The maximum value of tan2θ is -1. So, we have:tan2θ = -1 = tan(-45°)2θ = -45°θ = -22.5°

The maximum shear stress is obtained by substituting the obtained value of θ in the expression for σs.

σs = [(σx - σy)/2]sin2θ + τxy cos2θ= [(38 - 18)/2]sin(-45°) - 12 cos(-45°)= 20 MPa

Hence, the maximum shearing stress for the given general state of stress, considering the given state of stress, taking x = 38 MPa and y = 18 MPa is 20 MPa.

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