Page 110 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C02
Brown.cls
92
U.S. Customary 16:18 STRENGTH OF MACHINES SI/Metric
solution solution
Calculate the maximum upward deflection Calculate the maximum upward deflection
( Mid ) from Eq. (2.52b). ( Mid ) from Eq. (2.52b).
2
2
FL a FL a
Mid = ↑ Mid = ↑
8 EI 8 EI
2
2
(1,800 lb)(4ft) (1.5ft) (8,000 N)(1.2m) (0.5m)
= 2 =
5
2
6
8 (1.34 × 10 lb · ft ) 8 (6.48 × 10 N · m )
43,200 lb · ft 3 5,760 N · m 3
= = 6 2
2
7
(1.07 × 10 lb · ft ) (5.18 × 10 N · m )
12 in 100 cm
= 0.004 ft × = 0.0011m ×
ft m
= 0.05 in ↑ = 0.11 cm ↑
2.2.10 Double Overhang: Uniform Load
The simply-supported beam in Fig. 2.72 has double overhangs, each of length (a). The beam
has a uniform distributed load (w) acting vertically downward across the entire length of
the beam (L). The units on this distributed load (w) are force per length. Therefore, the total
force acting on the beam is the uniform load (w) times the length of the beam (L),or(wL).
w
A D
B C
a a
L
FIGURE 2.72 Double overhang: uniform load.
Reactions. The reactions at the supports are shown in Fig. 2.73—the balanced free-body-
diagram. Notice that the total downward force (wL) is split evenly between the vertical
reactions (B y and C y ), and as the uniform load (w) is acting straight down, the horizontal
reaction (B x ) is zero.
w (force/length)
B x = 0
B = wL/2 C = wL/2
y
y
FIGURE 2.73 Free-body-diagram.
U.S. Customary SI/Metric
Example 1. Determine the reactions for a Example 1. Determine the reactions for a
double overhanging beam of length (L) and double overhanging beam of length (L) and
overhangs (a) with a uniform distributed load overhangs (a) with a uniform distributed load
(w), where (w), where
w = 15 lb/ft w = 225 N/m
L = 12 ft L = 4m
a = 2ft a = 0.6 m