Page 85 - Marks Calculation for Machine Design

P1: Sanjay
16:18
January 4, 2005
Brown˙C02
Brown.cls
U.S. Customary BEAMS SI/Metric 67
solution solution
Step 1. Calculate the maximum deflection Step 1. Calculate the maximum deflection
from Eq. (2.29). from Eq. (2.29).
wL 4 wL 4
max = 0.00652 max = 0.00652
(EI) (EI)
(750 lb/ft)(6ft) 4 (10,000 N/m)(1.8m) 4
= (0.00652) = (0.00652)
6
2
2
6
(5.83 × 10 lb · ft ) (2.27 × 10 N · m )
5
5
9.72 × 10 lb · ft 3 1.05 × 10 N · m 3
= (0.00652) = (0.00652) 6 2
6
5.83 × 10 lb · ft 2 2.27 × 10 N · m
= (0.00652)(0.1667 ft) = (0.00652)(0.0462 m)
12 in 100 cm
= 0.00109 ft × = 0.013 in ↓ = 0.0003 m × = 0.030 cm ↓
ft m
The location of this maximum deflection The location of this maximum deflection
occurs just to the right of the midpoint of the occurs just to the right of the midpoint of the
beam, approximately at (0.52 L). beam, approximately at (0.52 L).
= 0.52 L = (0.52)(6ft) = 3.12 ft = 0.52 L = (0.52)(1.8m) = 0.94 m
x max x max
2.2.6 Twin Concentrated Forces
The simply-supported beam in Fig. 2.44 has twin concentrated forces, each of magnitude
(F), acting directly downward and located equidistant from each end of the beam. The
distance these two forces are from each support is labeled (a), and the distance between
the supports is labeled (L). Therefore, the distance between the two concentrated forces is
a distance (L – 2a).
F F
a a
A B
L
FIGURE 2.44 Twin concentrated forces.

Reactions. The reactions at the end supports are shown in Fig. 2.45—the balanced free-
body-diagram. The vertical reactions (A y and B y ) are equal, each with magnitude (F).As
both forces are acting directly downward, the horizontal reaction (A x ) is zero.

F F
A = 0
x

A = F B = F
y
y
FIGURE 2.45 Free-body-diagram.