Page 73 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
16:18
Brown.cls
Brown˙C02
55
BEAMS
2.2.4 Uniform Load
The simply-supported beam shown in Fig. 2.30 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), that is, (wL).
w
A B
L
FIGURE 2.30 Uniform load.

Reactions. The reactions at the end supports are shown in Fig. 2.31, the balanced free-
body-diagram. Notice that the total downward force (wL) is split evenly between the vertical
reactions (A y and B y ), and because the uniform load (w) is acting straight down, the
horizontal reaction (A x ) is zero.
w (force/length)
A = 0
x

A = wL/2 B = wL/2
y
y
FIGURE 2.31 Free-body-diagram.

U.S. Customary SI/Metric
Example 1. Determine the reactions at the Example 1. Determine the reactions at the
ends of a simply-supported beam of length (L) ends of a simply-supported beam of length (L)
with a uniform load (w) acting across the entire with a uniform load (w) acting across the entire
beam, where beam, where
w = 400 lb/ft w = 6,000 N/m
L = 15 ft L = 5m
solution solution
Step 1. From Fig. 2.31 calculate the pin reac- Step 1. From Fig. 2.31 calculate the pin reac-
tions (A x and A y ) at the left end of the beam. tions (A x and A y ) at the left end of the beam.
As the uniform load (w) is vertical, As the uniform load (w) is vertical,
A x = 0 A x = 0
and with the uniform load (w) acting across the and as the uniform load (w) acting across the
entire beam, entire beam,
wL (400 lb/ft)(15 ft) wL (6,000 N/m)(5m)
A y = = A y = =
2 2 2 2
6,000 lb 30,000 N
= = 3,000 lb = = 15,000 N
2 2