Page 103 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
16:18
Brown.cls
Brown˙C02
x = distance from left end of beam
L = length between supports BEAMS 85
x 1 = (x – L) = distance past roller support on overhang
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 distance (x) in Eq. (2.47a) must between 0 and (L), and the distance (x) in
Eq. (2.47b) must be between the distance (L) and the total length of the beam (L + a).
U.S. Customary SI/Metric
Example 4. Calculate the deflection ( ) of Example 4. Calculate the deflection ( ) of a
a single overhanging beam of length (L) and singleoverhangingbeamoflength(L)andover-
overhang (a) with a uniformly distributed load hang (a) with a concentrated force (F) acting
(w), at a distance (x), where at the free end, at a distance (x), where
w = 113 lb/ft w = 1,540 N/m
L = 3 ft, a = 1ft L = 1m, a = 0.3 m
x = 2ft x = 0.6 m
6
2
9
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).
9
4
2
4
6
2
EI = (30 × 10 lb/in )(12 in ) EI = (207 × 10 N/m )(491 cm )
1ft 2 1m 4
8 2
= 3.6 × 10 lb · in × ×
144 in 2 (100 cm) 4
6
6
= 2.5 × 10 lb · ft 2 = 1.02 × 10 N · m 2
Step 2. As the distance (x) is less than the Step 2. As the distance (x) is less than the
length (L), the deflection ( ) is determined length (L), the deflection ( ) is determined
from Eq. (2.47a). from Eq. (2.47a).
wx 4 2 2 3 wx 4 2 2 3
= (L − 2L x + Lx = (L − 2L x + Lx
24 EIL 24 EIL
2 2 2 2 2 2 2 2
−2 a L + 2 a x ) ↑ −2 a L + 2 a x ) ↑
(113 lb/ft)(2ft) (1,540 N/m)(0.6m)
= =
2
2
6
6
24 (2.5 × 10 lb · ft )(3ft) 24 (1.02 × 10 N · m )(1m)
4
2
2
4
×[(3ft) − 2 (3ft) (2ft) 2 ×[(1m) − 2 (1m) (0.6m) 2
3
2
2
3
+(3ft)(2ft) − 2 (1ft) (3ft) 2 +(1m)(0.6m) − 2 (0.3m) (1m) 2
2
2
2
2
+2 (1ft) (2ft) ] +2 (0.3m) (0.6m) ]
(226 lb) (924 N)
= = 7 3
3
8
(1.8 × 10 lb · ft ) (2.45 × 10 N · m )
4
4
4
4
×[81 ft − 72 ft + 24 ft 4 ×[1 m − 0.72 m + 0.216 m 4
4
4
4
4
−18 ft + 8ft ] −0.18 m + 0.065 m ]

1 4 −5 1 4
−6
= 1.26 × 10 × [23 ft ] = 3.77 × 10 3 × [0.381 m ]
ft 3 m
12 in 100 cm
= 0.000029 ft × = 0.000014 m ×
ft m
= 0.00035 in ↑ = 0.0014 cm ↑