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Consider the thin-walled section shown below. The section is fixed on one end, and an axial moment (T=12 kNm) is applied to the free end. The section is 6.9 m long. Do the following: Determine the shear flow in each section. What is the maximum shear stress in the section. Determine the angle of rotation of the free end (G = 26 GPa).
More likely the figure on top with shear flow universally going from left to right is due to direct horizontal shear, like the shear under a centered force or due to a support reaction. The one with the shear flow on the flanges going to the right and left and on the web going down is the shear flow due to the moment stresses in the beam. Edit
Therefore the shear flow diagram which is drawn cannot be right. The key thing is that shear flow (like bending stress) must be calculated with respect to the principal axis directions of the cross section. Unlike I beams, the principal axes of an angle section do not line up wit the horizontal and vertical directions.
As preamble, I note that the equation for this shear stress / force / flow is something that seems to suffer from almost everyone adopting a different nomenclature. I learnt it as longitudinal shear = Q A y[bar] / I. In that form, the longitudinal shear is a force per unit length along the section, and Q is the shear force in the section.
217 2 14. Add a comment. Yes, shear flow in a cross sectional element will always be in the same direction as the shear force in that same cross sectional element because the shear flow is essentially the distribution of the shear force. Confusion may arise due to the distinction between the "shear force" in an element and the externally ...
The shear flow in Fig. 7.23-a is the equilibrium shear flow distribution provided one where to cut where the distribution is drawn. Also, if you actually cut there, that is the shear flow distribution on the piece that you remove from the wall.
In an I-beam, the shear flow is usually shown drawn as on the left, not as drawn on the right. I do not understand why. I do understand: The total horizontal shear force must be zero, consistent with statics.This is satisfied in both images. The vertical direction matches the direction of the external load.
Modified 7 years, 11 months ago. Viewed 892 times. 1. for shear flow at D , the Q = Ay , i tried another method , i gt 30000 , instead of 29750, why the ans is different from 29750? why cant i use this method ( i break the shaded area into 3 parts ( as shown in figure) my working is Q = Ay = (60x10x35) + (2x30x10x150) = 30000.
A more quantifiable procedure is to visualize the shear flows and estimate the point about which the net moment produced by the shear flow will be zero. For the Mathematically Inclined. Well, I could go bonanzas with the LaTex (and I may do so a bit later), but since my lunch break is coming to an end, I'll refer you to this explanation.
Taking B for example; the nail is in one side; but nail C isn't doing the same thing as nail B. Instead, the shear is being resisted by the wood at B', so they take the area of 7.5"x1.5" knowing that the wood should be able to resist the shear as well. The same idea is used for C and C'. If they used your section, both Qb and Qc would be the same.