Page 43 - Marks Calculation for Machine Design

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12:26
January 4, 2005
Brown.cls
Brown˙C01
FUNDAMENTAL LOADINGS
y 25
h Neutral axis
b
FIGURE 1.25 Rectangular beam.

automatically results for positive values of the distance (y). This can be confusing, so this
minus sign is not used in this book. Besides, it is usually obvious in most problems, where
the bending stress is tensile and where it is compressive.
The most common beam cross section is rectangular, as shown in Fig. 1.25.
For the rectangular beam of Fig. 1.25, the maximum value of the distance (y) is half of
the height (h). The moment of inertia (I) for this rectangular cross section about the neutral
axis that passes through the centroid of the area, is given by Eq. (1.36) as
1 3
I = bh (1.36)
12

U.S. Customary SI/Metric
Example1. Determinethemaximumbending Example1. Determinethemaximumbending
stress (σ max ) in a beam with a rectangular cross stress (σ max ) in a beam with a rectangular cross
section, where section, where
M = 20,000 ft · lb = 240,000 in · lb M = 30,000 N · m
b = 2in b = 5cm = 0.05 m
h = 6in = 2y max h = 15 cm = 0.15 m = 2y max
solution solution
Step 1. Calculate the moment of inertia (I) of Step 1. Calculate the moment of inertia (I) of
the rectangular cross section using Eq. (1.36). the rectangular cross section using Eq. (1.36).
1 3 1 3 1 3 1 3
I = bh = (2in)(6in) I = bh = (0.05 m)(0.15 m)
12 12 12 12
= 36 in 4 = 0.000014 m 4
Step 2. Substitute the bending moment (M), Step 2. Substitute the bending moment (M),
the maximum distance (y max ), and the moment the maximum distance (y max ), and the moment
of inertia (I) just found, in Eq. (1.35) to deter- of inertia (I) just found, in Eq. (1.35) to deter-
mine the maximum bending stress (σ max ). mine the maximum bending stress (σ max ).
My max M (h/2) My max M (h/2)
σ max = = σ max = =
I I I I
(240,000 in · lb)(3in) (30,000 N · m)(0.075 m)
= 4 = 4
36 in 0.000014 m
2
2
= 20,000 lb/in = 20 kpsi = 160,000,000 N/m = 160 MPa