Page 132 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C02
Brown.cls
114
U.S. Customary 16:18 STRENGTH OF MACHINES SI/Metric
solution solution
Step 1. Calculate the stiffness (EI). Step 1. Calculate the stiffness (EI).
4
2
6
9
2
4
EI = (29 × 10 lb/in )(13 in ) EI = (207 × 10 N/m )(541 cm )
1ft 2 1m 4
8 2
= 3.77 × 10 lb · in × ×
144 in 2 (100 cm) 4
6
6
= 2.62 × 10 lb · ft 2 = 1.12 × 10 N · m 2
Step 2. As the distance (x) is less than the Step 2. As the distance (x) is less than the
distance (a), determine the deflection ( ) from distance (a), determine the deflection ( ) from
Eq. (2.70a). Eq. (2.70a).
Cx 2 Cx 2
= =
2 (EI) 2 (EI)
(1,500 ft · lb)(2ft) 2 (2,000 N · m)(0.6m) 2
= =
2
2
6
6
2 (2.62 × 10 lb · ft ) 2 (1.12 × 10 N · m )
3
3
(6,000 lb · ft ) (720 N · m )
= = 6 2
2
6
(5.24 × 10 lb · ft ) (2.24 × 10 N · m )
12 in 100 cm
= 0.00115 ft × = 0.00032 m ×
ft m
= 0.014 in ↓ = 0.032 cm ↓
Example 5. Calculate the maximum deflec- Example 5. Calculate the maximum deflec-
tion ( max ) and its location for the beam con- tion ( max ) and its location for the beam con-
figuration in Example 4, where figuration in Example 4, where
C = 1,500 ft · lb C = 2,000 N · m
L = 4ft L = 1.2 m
a = 3 ft, b = 1ft a = 0.9 m, b = 0.3 m
6
6
EI = 2.62 × 10 lb · ft 2 EI = 1.12 × 10 N · m 2
solution solution
Step 1. Calculate the maximum deflection at Step 1. Calculate the maximum deflection at
the free end from Eq. (2.71). the free end from Eq. (2.71).
Ca Ca
max = (2L − a) max = (2L − a)
2 (EI) 2 (EI)
(1,500 ft · lb)(3ft) (2,000 N · m)(0.9m)
= =
2
6
2
6
2 (2.62 × 10 lb · ft ) 2 (1.12 × 10 N · m )
×[2 (4ft) − 3ft)] ×[2 (1.2m) − 0.9m)]
3
3
4.50 × 10 lb · ft 2 1.80 × 10 N · m 2
= = 6 2
6
5.24 × 10 lb · ft 2 2.24 × 10 N · m
×[(8 − 3) ft] ×[(2.4 − 0.9) m]
= (8.59 × 10 −4 ) × (5ft) = (8.04 × 10 −4 ) × (1.5m)
12 in 100 cm
= 0.0043 ft × = 0.0012 m ×
ft m
= 0.052 in ↓ = 0.12 cm ↓