Page 95 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
16:18
Brown.cls
Brown˙C02
77
2
2
0 ≤ x ≤ L
= Fax (L − x )↑ BEAMS (2.38a)
6 EIL
F(x − L) 2
= [a (3 x − L) − (x − L) ] ↓ L ≤ x ≤ L + a (2.38b)
6 EI
where = deflection of the beam
F = applied force at free end
x = distance from left end of beam
L = length between supports
a = length of overhang
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 upward between the supports and downward for the
overhang. The distance (x) in Eq. (2.38a) must be between 0 and (L), and the distance (x)
in Eq. (2.38b) must be between (L) and the total length of the beam (L + a).
There is a maximum upward deflection between the supports, given by Eq. (2.39),
FaL 2 L
max = √ ↑ at x = √ (2.39)
9 3 EI 3
and a maximum downward deflection at the free end, where the force (F) acts, as given by
Eq. (2.40),
Fa 2
max = (L + a) ↓ at x = L + a (2.40)
3 EI
Note that the maximum upward deflection does not occur at the midpoint (L/2) between
the supports, but at a point closer to the roller at point B. The magnitude of the deflection
at the free end is usually greater than the magnitude of the deflection between the supports.

U.S. Customary SI/Metric
Example 4. Calculate the deflection ( ) of a Example 4. Calculate the deflection ( ) of a
singleoverhangingbeamoflength(L)andover- singleoverhangingbeamoflength(L)andover-
hang (a) with a concentrated force (F) acting hang (a) with a concentrated force (F) acting
at the free end, at a distance (x), where at the free end, at a distance (x), where
F = 450 lb F = 2,000 N
L = 3 ft, a = 1ft L = 1m, a = 0.3 m
x = 2ft x = 0.6 m
9
2
6
2
E = 30 × 10 lb/in (steel) E = 207 × 10 N/m (steel)
I = 12 in 4 I = 491 cm 4
solution solution
Step 1. Calculate the stiffness (EI). Step 1. Calculate the stiffness (EI).
4
4
2
2
9
6
EI = (30 × 10 lb/in )(12 in ) EI = (207 × 10 N/m )(491 cm )
8
2
3.6 × 10 lb · in × 1ft 2 1m 4
= ×
144 in 2 (100 cm) 4
6
6
= 2.5 × 10 lb · ft 2 = 1.02 × 10 N · m 2