# Why the vertical shear stress in the flange neglected?

Consider a thin walled I section beam. At some intermediate c/s the shear force is V. The shear stress direction in flanges and web will be as shown.

The textbook says that, the vertical shear stresses in the flange (as shown in green) can be neglected and

"This is because the flange is thin, and the top and bottom surfaces of the flange are free of stress"

How having zero stresses at the top and bottom surfaces of flange result in negligible vertical shear in flange?

• Because they are thin compared to the distance between them. Mar 1 at 16:40
• The answer should be a no-brainer. You should stare at the shear flow diagram hard and harder (hint, pay attention to the direction of the shear flow in each component of the cross-section.)
– r13
Mar 1 at 17:22
• What text book are you using? Perhaps you should check out another one as you might find the explanations clearer in a different book. Mar 2 at 7:32
• @SolarMike Yes, I did refer to other books too, but wasn't able to find an answer. One of the explanation given in the book, was exactly what kamran told in his answer, but I wasnt able to make sense of it. Mar 2 at 12:02
• "'....cut will never be much more', the last line of the text in the image" based on your own comment... Mar 2 at 12:15

It can be easily demonstrated that the share of vertical shear force in the flange is smaller than that carried by the web, and very importantly - the intensity of shear stress is much lower in the flange sections. The shear diagram below is the quantitative result derived from the shear equation - $$\tau = \dfrac{VQ}{Ib}$$.

Note: - As depicted below, "shear flow" can be understood as the shear friction (similar to glue) in between two cross-sectional segments that prevent the segments from undergoing relative motions/displacements. The shear equation, same as the flexural formulas, also is applicable to all shapes of beams that follow the "Hookie's Law" and satisfy the compatibility requirements, "with the rectangular shape as a special case".

• Um, I'm having trouble with the distribution of shear stress you show. Isnt it that shear formula couldn't be applied for flanges of an I section since, the shear stresses are not all parallel to the shear force and along the width the shear stress varies too? Mar 2 at 11:58
• I don't know where you get the impression that the shear equation does not apply to the flange, but I think it is a misunderstanding resulting from the fact that the vertical shear stress in the flange is ignored for it lacks practical significance in the design of I-beam but complicates the matter. However, it is a significant matter when making a built-up shape to act compositely, such as adding a cover plate to strengthen the I-shape, and/or designing the shear connectors for a composite slab.
– r13
Mar 2 at 14:50
• The shear formula is derived for a rectangular beam originally, with the assumptions 1) All the shear stresses in the c/s must be directed along the direction of total shear force V 2) The shear stress remains constant along the width. Both these assumptions are not met for flange. May be the book I follow, was pointing out to net shear stresses in the flange can't be determined by shear formula. I guess, the distribution you show, is only for vertical component in the flange and that can be calculated using shear formula Mar 2 at 16:18

Shear flow in the flanges of an I beam is as your sketch shows predominantly transversal along the flange, due to the tensile or compressive stresses caused by the moment, dm.

If they were not part of a beam and not integrated by the web then they would carry vertical shear.

'

• "Since the distance between the top and bottom of this element is very small the vertical shear stress on this cut will never be much more" - the text in the image. How did we conclude, it could never be much more because the flange is thin and shear stresses on top and bottom are zero? For instance, consider two points on the x axis, x=0 and x= L, such that L is close to x=0. At x= 0 and L, y= 0. Just because y is zero at x= 0 and L doesn't mean for every x between 0 and L, the value of y will be close to zero. It could happen that there is a sudden peak at say L/2. Mar 2 at 12:19