The principal stresses of the state of stress.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Where did-53.13 and -26.6 and 63.4 come from please show the work
# Explanation

## Given Information:
- The stress component along the x direction, σₓ = 150 MPa.
- The stress component along the y direction, σᵧ = 30 MPa.
- The shear stress component, τₓᵧ = -80 MPa.

## Calculation:
Calculate the principal plane (θₚ) as shown below.

\[
\tan(2θₚ) = \frac{2τₓᵧ}{σₓ - σᵧ}
\]

Substitute 150 MPa for σₓ, 30 MPa for σᵧ, and -80 MPa for τₓᵧ.

\[
\tan(2θₚ) = \frac{2 \times (-80)}{150 - 30}
\]

\[
\tan(2θₚ) = -1.333
\]

\[
2θₚ = -53.13°
\]

\[
θₚ = -26.6° \quad \text{and} \quad 63.4°
\]

Hence, the principal planes of the state of stress are -26.6° and 63.4°.

## To Determine:
The principal stresses of the state of stress.

## Answer:
Transcribed Image Text:# Explanation ## Given Information: - The stress component along the x direction, σₓ = 150 MPa. - The stress component along the y direction, σᵧ = 30 MPa. - The shear stress component, τₓᵧ = -80 MPa. ## Calculation: Calculate the principal plane (θₚ) as shown below. \[ \tan(2θₚ) = \frac{2τₓᵧ}{σₓ - σᵧ} \] Substitute 150 MPa for σₓ, 30 MPa for σᵧ, and -80 MPa for τₓᵧ. \[ \tan(2θₚ) = \frac{2 \times (-80)}{150 - 30} \] \[ \tan(2θₚ) = -1.333 \] \[ 2θₚ = -53.13° \] \[ θₚ = -26.6° \quad \text{and} \quad 63.4° \] Hence, the principal planes of the state of stress are -26.6° and 63.4°. ## To Determine: The principal stresses of the state of stress. ## Answer:
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