# In static equilibrium body, how can an applied moment about an axis contribute to reaction moment to another point away from the axis?

Determine the resultant internal loadings acting on the cross section at B of the pipe shown in Fig. 1–7a. End A is subjected to a vertical force of 50 N, a horizontal force of 30 N, and a couple moment of 70 N m. Neglect the pipe’s mass.

Well, I can't feel the solution. I would add upward reaction force (70/0.5=140 N) at B due to moment. But I wouldn't add 70 N m to moment equation at point B. can you explain the reason?

You cannot replace a couple moment with a single force. That would introduce an erroneous term to the vertical force balance.

There is a reason why it is called a couple moment. If you want to replace it with forces you can do the following. Add 2 forces that are equal in magnitude, opposite in direction and do not share the same line of action such that their net moment along the axis of interest gives you the same value.

Additionally, as long as you do not change the direction of axis, you can move the point of application of moments and even combine them all with one along each orthogonal direction. Therefore, you can imagine moving the moment at point A to point B and performing the moment balance to find your reaction forces.

Be careful, however, when you try to do the same with forces. In that case, you might have to add additional moments to your equations in order to include the effect of forces on moment balance.

• if I apply CW moment (about axis perpendicular to the face) on a point in one face of a cube shaped body, then the body can be balanced by another CCW same magnitude moment on any other point on that face/opposite face? (no gravity) – Shihabus Sakib Rad Jul 10 '19 at 17:28
• @ShihabusSakibRad That is correct – Elruz Rahimli Jul 11 '19 at 8:04

The counterclockwise moment applied at A will be transferred as torque to the 90 elbow and from there it will impart a moment of 70N.m all the way along the length of that section of the pipe, in similar way as when you apply this moment to the free end of a cantilever beam.

Therefore the moment of 70N.m is the constant along the length of that section of the pipe starting from the elbow to and including the point B.

The solution is correct and you must include 70N.m in the sum of the moments.

• if there was no 50 N and 30 N force. and point B was supported by roller support, then it would have only reaction moment? – Shihabus Sakib Rad Jul 10 '19 at 17:18
• Yes. It would have only moment reaction. No shear. However the reaction would be the negative sign to oppose the applied moment foe equilibrium. What you have is a free body diagram. – kamran Jul 10 '19 at 18:25