TENSIONS IN THE PLAN - 4 Exercises Solved | Mechanics of Materials

In this video I solve 4 problems from appendix D of the book about review of principal stresses and maximum in-plane shear. Let's learn how to determine the maximum principal stress and the maximum shear stress acting on the element.
00:00 TRUE OR FALSE TEST ON MAIN TENSIONS
01:10 IDENTIFYING PLAN TENSIONS AND ITS SIGNS
03:40 APPLYING THE VOLTAGE TRANSFORMATION EQUATION
04:54 DETERMINING THE MAXIMUM MAIN VOLTAGE
6:10 AM IDENTIFYING PLAN TENSIONS AND ITS SIGNS
06:55 AM DETERMINING THE MAXIMUM SHEAR STRESS IN THE PLANE
07:50 AM IDENTIFYING PLAN TENSIONS AND ITS SIGNS
8:30 AM APPLYING THE VOLTAGE TRANSFORMATION EQUATION
To determine the maximum main voltage we need to follow the following step-by-step:
Identify the tensions acting on the element and its senses, that is, if the normal tension is compressing the element (-) or if it is pulling the element (+). In the case of shear stress, we have to observe whether the stress acts upwards (+) or downwards (-), if the plane to the right of the element is analyzed.
After identifying all the stresses acting on the element and their respective signs, it is now easy, just replace the values ​​in the voltage transformation equation if you want to find the principal stresses or replace it in the maximum in-plane shear stress equation if you want to find the greater shear stress acting on the element.
Note: The highest main voltage that can act on the element is the one whose value in module is the highest between sigma 1 and sigma 2.
🔴 Complete your studies on PLAN TENSIONS by clicking here 👇:
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So that's it folks, I hope you enjoyed today's review of tensions in the plan.
Thanks for your support and until the next video! 😉
📚 Source: Mechanics of Materials 7th Ed. (R.C. Hibbeler)
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[@henriquesantos7217]

Excelente explicação professor!🥳🥳🥳🥳🤝🤝🤝

[@wolfdiogo]

Explêndida didática!

Muito obrigado, Diogo! tmj

[@diegorafael4964]

Boa explicação, parabéns!

Valeu, Diego!

[@tarcisiofernandes4856]

Na D.78 a resposta não seria 100Mpa? Pois ficou (-10)^2 + 0^2 = 100MPa

Tarcísio, você esqueceu que o 100 está dentro da raiz quadrada da fórmula, aí da 10 MPa mesmo.

[@tarcisiofernandes4856]

@ true