Page 71 - Marks Calculation for Machine Design

P1: Sanjay
16:18
January 4, 2005
Brown˙C02
Brown.cls
BEAMS
where = deflection of beam (positive downward)
C = applied couple at an intermediate point 53
x = distance from left end of beam
L = length of beam
a = location of couple (C) from left end of beam
b = location of couple (C) from right end of beam
E = modulus of elasticity of beam material
I = area moment of inertia of cross-sectional area about axis through centroid
Note that the deflection ( ) is downward for some values of the distance (x) and
upward for other values. Also, the distance (x) in Eq. (2.16a) must be between 0 and
(a), and the distance (x) in Eq. (2.16b) must be between (a) and the length of the beam
(L).
The deflection ( a ) at the point where the couple (C) acts, which is the distance (a),as
given by Eq. (2.17),
Cab
a = (2 a − L) at x = a (2.17)
3 EIL

If the distance (a) is less than half the length of the beam (L), then the deflection
( a ) will be above the axis of the beam. If the distance (a) is greater than half the
length of the beam, then the deflection ( a ) will be below the axis of the beam. If the
distance (a) is equal to half the length of the beam, then the deflection ( a ) will be
zero.

U.S. Customary SI/Metric
Example 4. Calculate the deflection ( ) of Example 4. Calculate the deflection ( ) of
a simply-supported beam with a concentrated a simply-supported beam with a concentrated
couple (C) a distance (x) of (L/6), where couple (C) a distance (x) of (L/4), where
C = 15 ft · kip = 15,000 ft · lb C = 20 kN · m = 20,000 N · m
L = 12 ft, a = 4 ft, b = 8ft L = 4m, a = 1 1 / 2 m, b = 2 1 / 2 m
9
6
2
2
E = 10.3 × 10 lb/in (aluminum) E = 71 × 10 N/m (aluminum)
I = 20 in 4 I = 781 cm 4
solution solution
Step 1. Determine the distance (x). Step 1. Determine the distance (x).
L 12 ft L 4m
x = = = 2ft x = = = 1m
6 6 4 4
Step 2. Calculate the stiffness (EI). Step 2. Calculate the stiffness (EI).
2
4
9
2
6
4
EI = (10.3 × 10 lb/in )(20 in ) EI = (71 × 10 N/m )(781 cm )
2
8
2.06 × 10 lb · in × 1ft 2 1m 4
= × 4
144 in 2 (100 cm)
6
5
= 1.43 × 10 lb · ft 2 = 5.54 × 10 N · m 2