Where did-53.13 and -26.6 and 63.4 come from please show the work 4:10ExpianationGiven information:The stress component along x direction as150 MPа.The stress component along y direction asOy = 30 MPaThe shear stress component as 'æy= -80 MPaCalculation:Calculate the principal plane(0,)as shown below.2Tzytan (20,)Substitute 150 MPa for Jx, 30 MPa for Cy, and-80 MPa for Txy.2x(-80)tan (20,)150-30tan (20,) = –1.33320,= -26.6° and 63.4°= -53.13°OpHence, the principal planes of the state of stress is-26.6° and 63.4°To determineThe principal stresses of the state of stress.Answer

Given:- Principle stress question with solution, To find:- To clear the doubt how -53.13, 26.6, and…

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Engineering

Mechanical Engineering

The principal stresses of the state of stress.

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter7: Analysis Of Stress And Strain

Section: Chapter Questions

Problem 7.3.8P: An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of...

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Forming and Shaping

Metal forming is a category of metalworking process through which the material can be given the desired shape and size by subjecting it to plastic deformation, where the component undergoes bulk deformation process upon application of loads. Unlike the shearing and machining process, the material does not undergo up to a failure point as indicated in the stress-strain curve. Hence, the stresses induced during the forming process are greater than yield strength but less than the failure strength of the material. The process does not lead to the loss of material, so it can be said to be an economical process. During forming, the material is subjected to tensile forces, compressive forces, bending, and shearing loading conditions. The material should possess the property of ductility to undergo the forming process.

Question

Where did-53.13 and -26.6 and 63.4 come from please show the work

4:10
Expianation
Given information:
The stress component along x direction as
150 MPа.
The stress component along y direction as
Oy = 30 MPa
The shear stress component as 'æy
= -80 MPa
Calculation:
Calculate the principal plane
(0,)
as shown below.
2Tzy
tan (20,)
Substitute 150 MPa for Jx, 30 MPa for Cy, and
-80 MPa for Txy.
2x(-80)
tan (20,)
150-30
tan (20,) = –1.333
20,
= -26.6° and 63.4°
= -53.13°
Op
Hence, the principal planes of the state of stress is
-26.6° and 63.4°
To determine
The principal stresses of the state of stress.
Answer

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