6d-ex. Shear Stress Examples • Ex. 6d.1 • Ex. 6d.2 Back | Index | Next

 Example 6d.1 Given: A cantilever beam, is built in at the left end, has length L, and rectangular cross-section of width b, and height h. The beam is loaded by point load P at its free end. Req'd: Determine the shear stress at the top, bottom and neutral axis at a cross-section in the beam. Sol'n: The shear force, V(x) = P, is constant across the entire beam. At the top (or bottom) surface (y = ±h/2), y* = h/2 and A* = 0, therefore t = 0. At the neutral axis (y1 = 0), y* = h/4 and A* = bh/2. Thus: The shear distribution, calculated as a function of y (= y1), is a parabola, given by the function: Note that these equations for t(y) are only valid for beams of rectangular cross-section. Note: shear-stress acts parallel to the beam-face. The parabola is a plot of the magnitude of the shear stress.

 Example 6d.2 Given: The I-beam at right is subjected to shear force, V = 5 kN. The flange and the web both have a thickness of 20 mm, the height of the beam is 150 mm and the width is 100 mm. The beam has a moment of inertia of I = 19x106 mm4. Req'd: Determine the shear stress at a-a' and b-b'. Sol'n: The first moment of area (about the neutral axis of the entire cross-section) of the area to the left of a-a' is: Q = A*y* = [(40 mm)(20 mm)](65 mm) = 5.2x104 mm3 y* =[ (150/2 mm) - (20/2 mm)] = (half the height of the cross section) minus (half the thickness of the flange).
 Then the shear stress is in the flange is: txy = 670 kPa The first moment of area of the area to below b-b' is: Q = A*y* = [(100 mm)(20 mm)](65 mm) = 1.3x105mm3 The shear stress in the web just above (below) the lower (upper) flange, is: txy = 1.68 MPa

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Updated: 05/24/09 DJD