(a) Explanation Given information: The beam is subjected to a vertical shear of V . Calculation: Calculate the area ( A ) of the cross section as shown below. A = 2 ( 1 2 b h ) = b h . Here, b is the width of the cross section and h half height of the cross section. Calculate the moment of inertia ( I ) of the thin-walled cross section as shown below. I = 2 × b h 3 12 = b h 3 6 Sketch the cross section at a cut section through y ≤ h as shown in Figure 1. Refer to Figure 1. Area of the cut section A ( y ) = 1 2 × b y h × y = b y 2 2 h Location of the centroid of the cut section y ¯ ( y ) = h − 2 3 y . Calculate the first moment of area of the section as shown below. Q = A y ¯ Here, A is the area of the cross section and y ¯ is the location of centroid. Substitute b y 2 2 h for A and h − 2 3 y for y ¯ . Q = b y 2 2 h × ( h − 2 3 y ) = b y 2 2 − b y 3 3 h Thickness of the cross section is t ( y ) = b y h . Calculate the maximum shear stress as shown below. τ max = V Q I t Here, V is the vertical shear, Q is the first moment of area, I is the moment of inertia, and t is the thickness of the section. Substitute b y 2 2 − b y 3 3 h for Q , b h 3 6 for I and b y h for t (b) To determine The constant k in the expression τ max = k V A .

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Chapter 6.2, Problem 27P

(a)

To determine

The horizontal line along which the shear stress is maximum.

(b)

To determine

The constant k in the expression τmax=kVA.

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Chapter 6.2, Problem 26P

Chapter 6.2, Problem 28P

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The horizontal line along which the shear stress is maximum.

Solution Summary: The author calculates the area of the cross section at a cut section through yle h as shown in Figure 1.

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The horizontal line along which the shear stress is maximum.

The constant k in the expression τmax=kVA.

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Chapter 6 Solutions