# How do you select the correct area to consiser for first moment of area calculations?

I have just begun studying mechanics of materials and I am struggling to understand intuitively how to select the area in first moment of area calculations. I was hoping someone has a relatively easy explanation.

The problem arises when calculating the shear stress $$\tau$$ at a specific point in a beam due to a specified shear force. For $$\tau_{xy}$$ the calculations seem to be the same:

$$\tau_{xy}$$ due to $$V(x)$$ at point A requires the calculation of the first moment of area, $$Q$$, shaded here:

However, then the problem requires me to find $$\tau_{xz}$$ due to $$V(x)$$ at point B. The shaded area under is the area we have to use, and my question is why?

I know this is probably a very banal question for you guys, but I just really want to understand this, and browsing the web has led me nowhere.

You're looking for $\tau_{xz}$, which pertains to horizontal shear flow along the top flange. (Rather than the vertical shear flow considered in $\tau_{xy}$.) As Mark noted, the shear flow starts from the center of the flange and flows outward/down.

The formulation for the shear stress calculation remains essentially the same. Only the area changes -- we're simply cutting out a different section, and because of the tube shape we have to make two cuts to remove our section. (Versus and I-shape where only one cut is required.)

This presentation has a pretty good explanation of shear stress.

V causes no shear stress at point B. The 1st moment of area refers to the area above the point you're studying, in the negative direction of the shear stress applied. You can see that at point B, that area is inexistent, and so the shear stress due to V is equal to zero.

• Thank you for your answer. I believe $\tau_{xy,B,V}=0$, as you say. However, $\tau_{xz,B,V}=0.96 \frac{N}{mm^2}$ according to the solution. May 28, 2015 at 14:59
• The shear flow will require some shear stress at B
– Mark
Jun 27, 2015 at 12:36

Due to your shear force, the entire beam will attempt to shear. (i.e. change from a nice long rectangular beam (regardless of cross section):

 --------------------------------- -> Pull on the top flange
|<-----L----------------------->| - Due to shear across the cross section
---------------------------------


to a parallelogram:

 ---------------------------------
/                                /
---------------------------------


(see this diagram on Wikipedia).

This pull on the top flange is what causes a shear stress in the horizontal portion of the beam. Since the beam has a giant hole, the shearing force can't grow straight down. Instead, it has to travel out, through the top part, and down the beam. This path of the shearing force is called shear flow. As you can see in the definition of shear flow, it only happens in sections made of thin plates (like yours!).

I beams will have shear flow that starts on the edges and makes it's way to the middle. This lecture from the University of Maryland shows the shear flow really well, especially on page 10.

• You've misunderstood the type of shear applied. The shear is not applied horizontally to the top flange, hence it does not distort the section into a parallelogram. See V(x) as drawn on the top left diagram. Jun 29, 2015 at 12:40
• I understand the shear is vertical. The shear has to flow from the top beam out to the sides and down. That's shear flow.
– Mark
Jun 29, 2015 at 12:48
• I don't disagree with the shear flowing out to the sides. I disagree about the section distorting into a parallelogram. The shear flow will be symmetrical about a vertical line of symmetry - see the slide you've referenced (slide no 19 / page 10). Jun 29, 2015 at 14:39
• Oh, I was trying to show the beam would go from flat on top with vertical sides to slightly tilted sides. I'll rewrite to make that more clear.
– Mark
Jun 29, 2015 at 14:53