Page 59 - Marks Calculation for Machine Design

P1: Sanjay
16:18
January 4, 2005
Brown.cls
Brown˙C02
U.S. Customary BEAMS SI/Metric 41
solution solution
Step 1. Calculate the maximum deflection Step 1. Calculate the maximum deflection
from Eq. (2.5). from Eq. (2.5).
FL 3 FL 3
max = max =
48 (EI) 48 (EI)
(12,000 lb)(6ft) 3 (55,000 N)(2m) 3
= =
2
5
2
5
48 (4.69 × 10 lb · ft ) 48 (1.82 × 10 N · m )
5
6
2.59 × 10 lb · ft 3 4.40 × 10 N · m 3
= = 6 2
7
2.25 × 10 lb · ft 2 8.74 × 10 N · m
12 in 100 cm
= 0.115 ft × = 1.38 in ↓ = 0.0504 m × = 5.04 cm ↓
ft m
Notice that the maximum deflection ( max ) found at the midpoint of the beam (L/2) in
Example 5 is not twice the deflection ( ) at a distance one quarter the length of the beam
(L/4) found in Example 4. This is because the shape of the deflection curve is parabolic,
not linear.
2.2.2 Concentrated Force at Intermediate
Point
The simply-supported beam in Fig. 2.16 has a concentrated force (F) acting vertically
downward at an intermediate point, meaning not at its midpoint. The distance between the
supports is labeled (L), so the force (F) is located at a distance (a) from the left end of the
beam and a distance (b) from the right end of the beam, where the sum of distances (a) and
(b) equal the length of the beam (L).
F
a b

A B
L
FIGURE 2.16 Concentrated force at intermediate point.

Reactions. The reactions at the end supports are shown in Fig. 2.17, the balanced free-
body-diagram. Notice that the force (F) is unevenly divided between the vertical reactions
(A y and B y ), and because the force (F) is acting straight down, the horizontal reaction

F
a b
A = 0
x

A = Fb/L B = Fa/L
y
y
FIGURE 2.17 Free-body-diagram.