Math Problem Statement

whats the answer for 1 and 2

For the state of stress shown, determine the maximum principal stress and the shear stress component after rotating the element counterclockwise through 30°.

Solution

Let's start by addressing each question based on the provided image.

Question 1: Maximum Principal Stress

The given stress state involves normal stresses σx=100\sigma_x = 100 MPa, σy=60\sigma_y = 60 MPa, and a shear stress τxy=48\tau_{xy} = 48 MPa. To determine the maximum principal stress, we use the principal stress formula:

σ1,2=σx+σy2±(σxσy2)2+τxy2\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2}

Step-by-Step Calculation:

  1. Calculate the average normal stress: σx+σy2=100+602=80MPa\frac{\sigma_x + \sigma_y}{2} = \frac{100 + 60}{2} = 80 \, \text{MPa}

  2. Calculate the difference in normal stresses: σxσy2=100602=20MPa\frac{\sigma_x - \sigma_y}{2} = \frac{100 - 60}{2} = 20 \, \text{MPa}

  3. Calculate the principal stress difference: σ1,2=80±202+482=80±400+2304=80±2704=80±52MPa\sigma_{1,2} = 80 \pm \sqrt{20^2 + 48^2} = 80 \pm \sqrt{400 + 2304} = 80 \pm \sqrt{2704} = 80 \pm 52 \, \text{MPa}

Thus, the two principal stresses are:

σ1=80+52=132MPaandσ2=8052=28MPa\sigma_1 = 80 + 52 = 132 \, \text{MPa} \quad \text{and} \quad \sigma_2 = 80 - 52 = 28 \, \text{MPa}

The maximum principal stress is 132 MPa.

Answer for Question 1: 132 MPa


Question 2: Shear Stress After Rotation

To determine the shear stress after a counterclockwise rotation of 3030^\circ, we use the transformation equations for stresses, particularly the shear stress transformation equation:

τ=σxσy2sin(2θ)+τxycos(2θ)\tau' = \frac{\sigma_x - \sigma_y}{2} \sin(2\theta) + \tau_{xy} \cos(2\theta)

where:

  • θ=30\theta = 30^\circ
  • σx=100MPa\sigma_x = 100 \, \text{MPa}, σy=60MPa\sigma_y = 60 \, \text{MPa}, τxy=48MPa\tau_{xy} = 48 \, \text{MPa}

Step-by-Step Calculation:

  1. Calculate sin(2θ)\sin(2\theta) and cos(2θ)\cos(2\theta) for θ=30\theta = 30^\circ: sin(60)=320.866andcos(60)=0.5\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \quad \text{and} \quad \cos(60^\circ) = 0.5

  2. Apply the values to the shear stress equation: τ=100602×0.866+48×0.5\tau' = \frac{100 - 60}{2} \times 0.866 + 48 \times 0.5 τ=20×0.866+24=17.32+24=41.32MPa\tau' = 20 \times 0.866 + 24 = 17.32 + 24 = 41.32 \, \text{MPa}

So, the shear stress after the rotation is approximately 41.3 MPa.

Answer for Question 2: 41.3 MPa


Do you want more details on any part of the calculations?

Here are 5 related questions that might interest you:

  1. How do you calculate the minimum principal stress in a stress state?
  2. What is the physical meaning of principal stresses?
  3. How does the direction of rotation affect the stress transformation?
  4. Can we always assume the same transformation formulas for any rotation angle?
  5. How does shear stress vary with different angles of rotation?

Tip: Always double-check angle conventions (clockwise vs. counterclockwise) when applying stress transformation equations to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Stress Transformation
Principal Stress
Shear Stress
Trigonometry

Formulas

Principal Stress Formula: σ1,2 = (σx + σy) / 2 ± √[(σx - σy) / 2]^2 + τxy^2
Shear Stress Transformation Formula: τ' = (σx - σy) / 2 sin(2θ) + τxy cos(2θ)

Theorems

Stress Transformation Equations
Principal Stress Theory

Suitable Grade Level

Undergraduate Engineering