Page 126 - Marks Calculation for Machine Design

P1: Sanjay
16:18
January 4, 2005
Brown.cls
Brown˙C02
108
STRENGTH OF MACHINES
L = length of beam
a = distance to force (F) from left end of beam
b = distance from force (F) to right end of beam
E = modulus of elasticity of beam material
I = area moment of inertia of cross-sectional area about axis through centroid
The maximum deflection ( max ) occurs at the free end, and is given by Eq. (2.67),
Fb 2
max = (3L − b) at x = 0 (2.67)
6 EI
and deflection ( a ) at the location of the force (F) is given by Eq. (2.68),
Fb 3
a = at x = a (2.68)
3 EI
U.S. Customary SI/Metric
Example 4. Calculate the deflection ( ) for a Example 4. Calculate the deflection ( ) for a
cantilevered beam of length (L) with a concen- cantilevered beam of length (L) with a concen-
trated force (F) acting at an intermediate point, trated force (F) acting at an intermediate point,
at a distance (x) from the left end of the beam, at a distance (x) from the left end of the beam,
where where
F = 150 lb F = 700 N
L = 8ft L = 2.5 m
a = 3 ft, b = 5ft a = 1m, b = 1.5 m
x = 6ft x = 2m
9
6
2
2
E = 1.6 × 10 lb/in (Douglas fir) E = 11 × 10 N/m (Douglas fir)
I = 145 in 4 I = 6,035 cm 4
solution solution
Step 1. Calculate the stiffness (EI). Step 1. Calculate the stiffness (EI).
4
6
2
9
2
4
EI = (1.6 × 10 lb/in )(145 in ) EI = (11 × 10 N/m )(6,035 cm )
1ft 2 1m 4
8 2
= 2.32 × 10 lb · in × ×
144 in 2 (100 cm) 4
6
5
= 1.61 × 10 lb · ft 2 = 6.64 × 10 N · m 2
Step 2. As the distance (x) is greater than the Step 2. As the distance (x) is greater than the
distance (a), determine the deflection ( ) from distance (a), determine the deflection ( ) from
Eq. (2.66b). Eq. (2.66b).
F(L − x) 2 F(L − x) 2
= (3 b − L + x) = (3 b − L + x)
6 (EI) 6 (EI)
(150 lb)(8ft − 6ft) 2 (700 N)(2.5m − 2m) 2
= =
2
2
5
6
6 (1.61 × 10 lb · ft ) 6 (6.64 × 10 N · m )
×[3 (5ft) − (8ft) + (6ft)] ×[3 (1.5m) − (2.5m) + (2m)]
2
2
(600 lb · ft ) (175 N · m )
=
= 2 6 2
6
9.66 × 10 lb · ft (3.98 × 10 N · m )
×[(15 − 8 + 6) ft] ×[(4.5 − 2.5 + 2) m]
= (6.21 × 10 −5 ) × (13 ft) = (4.39 × 10 −5 ) × (4m)
12 in 100 cm
= 0.00081 ft × = 0.00018 m ×
ft m
= 0.010 in ↓ = 0.018 cm ↓