07.2-2 Combined loading - EXAMPLE

[@tarik_arikan]

I have an exam tomorrow morning. I wish i had found you before.

[@molopes7386]

You saved my life!

[@_basu_6320]

Sir can you kindly tell me why is the direction of Transverse Shear Stress due to Fx at point A(=3.77MPa) is in X Direction? Shouldn't it be in the in the Positive Z Direction? Please let me know Sir.

[@salifyeo7116]

Great video. Thanks for uploading Sir. Can you please explains how in general do we know that they will be a Torque or a Moment on a PARTICULAR AXIS given some forces? I really get confused to differentiate them when given forces.

[@Alhussainba]

6:40 I got 52.8 MPa.

[@jiungshin2083]

I don't quite get why V_x is 500N. Shouldn't it be in the opposite direction of the load to maintain equilibrium of the force in x direction?

[@sam_the.patriot]

Same question here. Why is not negative?

[@thejoaoprx]

why is Vx and Vy oriented towards positive direction? Shouldn't Vx counter act the 500 N force? So if the 500N force is headed to the positive direction, then Vx should be oriented to the negative direction, right?

[@mohameddrissi1075]

you're right

[@MegaMauricio120]

you could've also used the formula Tmax=4V/3A for max shear stress in deflection

[@memocool4279]

9:22 how did you know that stress in z direction, pls i got 5 hours till my final exam

[@Garebare1]

You switched the values of Mx and My in your calculations

[@rhythmm860818]

quick questions, your torque Tz working on -z , does that mean your 22.64mpa should be - 22.64 mpa?

[@introductoryengineeringmec9114]

The torque Tz is in the negative direction by the right-hand rule, however, the sign on the stress depends on what point you are evaluating on the cross-section. Think of the stress acting clock-wise all around the cross-section (from a top view), like a whirlpool. At point A, the stress is acting in the positive x-direction. At a point opposite A on the cross-section, the stress is acting in the negative x-direction. At 90 degrees from A, the stress is going in the positive or negative y-direction, depending on the point.

[@sinothandomoerane4315]

Why is the stress from the Mx acting in the z-direction?

[@introductoryengineeringmec9114]

Mx is a bending moment on the cross-section. Bending moments cause normal stress defined by the equation sigma = Mc/I. The normal stresses push or pull on the cross-section surface. The direction of the push or pull is perpendicular to the cross-section which here is the z-direction.

[@beleteaynalem4430]

thaks alot

[@abbas2294]

The torsion part is not correct. The sign of torsion is same through the section depending upon its direction whether towards the positive z axis or negative. Once it indicates during the equilibrium that its sign is positive then it has to be taken positive.

[@introductoryengineeringmec9114]

I disagree. The sign on the internal torque at the cross-section is negative by the right-hand rule. The shear stress caused by the internal torque is acting over the cross-section (x-y plane). However, the sign on the shear stress caused by the torque changes over the cross-section, depending on the point where stress is being evaluated. The shear stress acts in the direction of the torque. At point A it is in the positive x-direction. On the opposite side of the cross-section from point A, the direction of the stress is the negative x-direction. Halfway between opposite sides, the stress is in the positive or negative y-directions.

[@wasidsayyadwzr1355]

sir comment me your email