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.
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?
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.
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.
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.
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.
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.
I have an exam tomorrow morning. I wish i had found you before.
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.
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?
Same question here. Why is not negative?
You saved my life!
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?
you're right
6:40
I got 52.8 MPa.
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.
9:22 how did you know that stress in z direction, pls i got 5 hours till my final exam
Why is the stress from the Mx acting in the z-direction?
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.
quick questions, your torque Tz working on -z , does that mean your 22.64mpa should be - 22.64 mpa?
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.
you could've also used the formula Tmax=4V/3A for max shear stress in deflection
You switched the values of Mx and My in your calculations
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.
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.
thaks alot
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