Page 115 - Shigley's Mechanical Engineering Design

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90 Mechanical Engineering Design
Figure 3–14 y Compression

Bending stresses according to
Eq. (3–24). c Neutral axis, Centroidal axis
y
x

Tension

neutral axes of all cross sections, is called the neutral plane. Elements of the beam
coincident with this plane have zero stress. The location of the neutral axis with
respect to the cross section is coincident with the centroidal axis of the cross section.
The bending stress varies linearly with the distance from the neutral axis, y, and is
given by
My
σ x =− (3–24)
I
where I is the second-area moment about the z axis. That is,

2
I = y dA (3–25)
The stress distribution given by Eq. (3–24) is shown in Fig. 3–14. The maximum magni-
tude of the bending stress will occur where y has the greatest magnitude. Designating σ max
as the maximum magnitude of the bending stress, and c as the maximum magnitude of y
Mc
σ max = (3–26a)
I
Equation (3–24) can still be used to ascertain whether σ max is tensile or compressive.
Equation (3–26a) is often written as
M
σ max = (3–26b)
Z
where Z = I/c is called the section modulus.

EXAMPLE 3–5 A beam having a T section with the dimensions shown in Fig. 3–15 is subjected to a bend-
ing moment of 1600 N · m, about the negative z axis, that causes tension at the top surface.
Locate the neutral axis and find the maximum tensile and compressive bending stresses.
Solution Dividing the T section into two rectangles, numbered 1 and 2, the total area is
2
A 12(75) 12(88) 1956 mm . Summing the area moments of these rectangles
about the top edge, where the moment arms of areas 1 and 2 are 6 mm and (12
88/2) 56 mm respectively, we have
1956c 1 = 12(75)(6) + 12(88)(56)

and hence c 1 = 32.99 mm. Therefore c 2 = 100 − 32.99 = 67.01 mm.
Next we calculate the second moment of area of each rectangle about its own cen-
troidal axis. Using Table A–18, we find for the top rectangle
1 3 1 3 4 4
I 1 = bh = (75)12 = 1.080 × 10 mm
12 12