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