# Why in a rectangular cross section beam the shear stresses are parallel to the height throughout the surface?

Consider a beam with a rectangular c/s subjected to some arbitrary loading. At any cross section the shear force is V. This shear force is the resultant of all the internal resistive forces which act parallel to the section.

These individual internal resistive forces (which make V) might not necessarily be directed parallel to the y axis all over the surface. For instance, it may be as shown in (a) below, where all the individual internal forces are not parallel to y axis.

The textbook that I follow, states the following -

The textbook uses the word reasonable, to explain why the shear stresses are all parallel to the y axis. However, I'm not able to figure out what is the reason. How did we conclude that shear stresses are all parallel to the y axis in a rectangular section?

When I started studying about shear stresses in beams, I went in with a preconception that no matter what is the c/s shape all the shear stresses (or the individual internal resistive forces) will be parallel to the y axis, throughout the area. Then I got to know that in circular sections near the periphery the stresses are not along y, but tangent to the boundary. That made realize, it's not necessary that shear stresses in all c/s shapes will be parallel to y axis all over the c/s. Then it struck me, that the book never proved the shear stresses in a rectangular beam are all parallel to the y axis, and just covered it in a word reasonable.

Can anyone elaborate on the reason why shear stresses in a rectangular beam direction along y?

• Textbooks usually load in the y axis. If the beam is loaded in the x axis the shear would be parallel to x. Torsion problems usually with round sections cause torsional stresses, max at the surface, a function of r. Feb 23, 2022 at 14:44
• @DKNguyen I'm not sure if I fully understand you, but the forces on the section are parallel to it, so the horizontal component is also parallel and hence when divided by the area will give a shear stress. I don't see it being a normal stress. Feb 23, 2022 at 20:02
• "*but the forces on the section are parallel to it, so the horizontal component is also parallel" I don't know what you mean when you say this because I don't know what direction you mean when you say "parallel to the section". Feb 23, 2022 at 20:17
• @DKNguyen I believe I should've specified the loading type, i.e. the loading is only done in the x-y plane. No loads are applied perpendicular to x-y plane. So if the loading is such, the internal forces will be parallel to the area and both the horizontal and vertical components will be shear stresses. Feb 23, 2022 at 20:21
• @HarshitRajput The component of the force perpendicular to the edge of the cross section. Like how when you push down on a ramp, it produces a sideways component of force. Whereas when you push down on a flat surface it only produces a downward component. Feb 23, 2022 at 21:23

You could have a shear stress distribution like (a) if you had a concentrated load on a deformable material cross-section.

However, because in most textbooks longitudinal members are considered (i.e. long beams where the length is much greater compared to either width or breadth), the load is assumed to be distrubuted uniformly on the top side or along the end face. This is what makes the shear stress distribution parallel in (b).

• So if my beam becomes shorter in length, it might be the case where I get a distribution like the one shown in a? and hence just because a beam is made of rectangular c/s doesnt mean that stresses have to be parallel to y axis, it may be as is shown in a. It is just that for long beams it is reasonable to assume the stresses parallel to y axis? Feb 23, 2022 at 10:11
• Yes, "just because a beam is made of rectangular c/s doesnt mean that stresses have to be parallel to y axis" . However, the length of the beam is not important. It is the distribution of the load whether its uniform or not.
– NMech
Feb 23, 2022 at 11:13

Many things can cause the shear stress not to run parallel to the shear force or parallel to the vertical side of the beam.

Only on beams of isotropic material and if the vertical load is distributed uniformly the shear stress runs evenly parallel to the face.

• In lumber beams, the shear stress varies depending on the grain of the lumber, and it can be locally, depending on the grain, much more than the average calculated shear and cause cracking.
• In concrete beams when there is a large load, say a perpendicular precast beam resting on half of the width of the beam, The shear stress is not parallel to the face of the beam and has to be calculated and dealt with by transverse reinforcement in the beam.
• even in isotropic beams if there is a concentrated load, say a post sitting eccentrically on a beam the shear is not uniform on the vertical surface of the beam.
• In the seismic design of beams and joints question of eccentric loading is critical and has to be carefully analyzed. It also sometimes is taken advantage of to design joints with high energy absorbing capacity during the earthquake.

From the shear equations, $$v = \frac{P}{A}$$ and $$v = \dfrac{VQ}{Ib}$$, we know that the unit shear stress has the same intensity within any horizontal strip of a beam across its width, so apparently, there is only one possible orientation of the shear stress that can claim the expression $$P = \sum v \Delta A$$ is "true" - $$v$$ is parallel to the $$y - axis$$, otherwise, we will have to deal with the terms $$P = v \cos \theta \Delta A + v \Delta A + v....$$ (note, $$v$$ is constant).