Page 141 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
16:18
Brown.cls
Brown˙C02
BEAMS
Therefore, the maximum shear force (V max ) is given by Eq. (2.80).
wL 123
V max = (2.80)
2
The bending moment distribution is given by Eq. (2.81) for values of the distance (x)
equal to zero at the left end of the beam to a value (L) at the right end of the beam.
wx 3
M =− 0 ≤ x ≤ L (2.81)
6L
The bending moment (M) distribution is shown in Fig. 2.116.
M
0 x
L
–
2
–wL /6
FIGURE 2.116 Bending moment diagram.

The bending moment (M) is zero at the left end of the beam, then decreases cubically
2
to a maximum negative value (−wL /6) at the right end. The maximum bending moment
(M max ) occurs at the right end of the beam, given by Eq. (2.82).
wL 2
M max = (2.82)
6

U.S. Customary SI/Metric
Example 2. Calculate the shear force (V ) and Example 2. Calculate the shear force (V ) and
bending moment (M) for a cantilevered beam bending moment (M) for a cantilevered beam
of length (L) with a triangular distributed load of length (L) with a triangular distributed load
(w) acting across its entire length, at a distance (w) acting across its entire length, at a distance
(x) from the left end of the beam, where (x) from the left end of the beam, where
w = 300 lb/ft w = 4,500 N/m
L = 6ft L = 1.8 m
x = 2ft x = 0.6 m
solution solution
Step 1. Determine the shear force (V ) from Step 1. Determine the shear force (V ) from
Eq. (2.79) as Eq. (2.79) as
wx 2 (300 lb/ft)(2ft) 2 wx 2 (4,500 N/m)(0.6m) 2
V =− =− V =− =−
2L 2 (6ft) 2L 2 (1.8m)
1,200 ft · lb 1,620 N · m
=− =−100 lb =− =−450 N
12 ft 3.6m