Page 44 - Marks Calculation for Machine Design

P1: Shibu
January 4, 2005
12:26
Brown.cls
Brown˙C01
STRENGTH OF MACHINES
26
y max = c
h Neutral axis
y min = c
b
FIGURE 1.26 Limiting values of the distance (y).
The most difficult fact about calculating the bending stress (σ) using Eq. (1.35) is de-
termining the bending moment (M). This is why Chap. 2 is devoted entirely to finding the
bending moment and shear force distributions for the most common beam configurations
and loadings.
Before moving on to shear stress owing to bending, there is a quantity associated with the
limiting values of the distance (y) in Eq. (1.35). If the maximum value (y max ) is considered
positive upward from the neutral axis, then the minimum value (y min ) is considered negative
downward from the neutral axis. For the rectangular cross section of Fig. 1.25, these two
limiting values are equal in magnitude but opposite in sign. Figure 1.26 shows these limiting
values.
For other cross-sectional areas, these limiting values may be different. In either case, if the
distance (y) in Eq. (1.35) is moved from the numerator to the denominator, then a quantity
called the section modulus (S) is defined. This algebraic process is shown in Eq. (1.37).

My max M M My min M M
σ max = = = or σ min = = = (1.37)
I I/y max S max I I/y min S min
As mentioned earlier, a rectangular cross section has equal maximum and minimum val-
ues of the distance (y), only their signs are opposite, and which are typically labeled (c).
The section modulus for a rectangular cross section becomes that given in Eq. (1.38). The
3
3
units of section modulus are length cubed, that is, in or m .
I I I
S = = = (1.38)
y max y min c

U.S. Customary SI/Metric
Example 2. Calculate the section modulus Example 2. Calculate the section modulus
(S) for the beam with the rectangular cross sec- (S) for the beam with the rectangular cross sec-
tion in Example 1, where tion in Example 1, where
4
4
I = 36 in (from Example 1) I = 0.000014 m (from Example 1)
h = 6in = 2 c h = 15 cm = 0.15 m = 2c
solution solution
Step 1. Substituting the moment of inertia (I) Step 1. Substituting the moment of inertia (I)
and the maximum distance (c) into Eq. (1.38) and the maximum distance (c) into Eq. (1.38)
gives gives
I I I I I I
S = = = S = = =
y max y min c y max y min c
36 in 4 0.000014 m 4 3
= = 12 in 3 = = 0.00019 m
3in 0.075 m