Page 65 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
16:18
Brown.cls
Brown˙C02
U.S. Customary BEAMS SI/Metric 47
2
2
(60,000 lb · ft ) (50,625 N · m )
= =
3
3
6
7
(4.00 × 10 lb · ft ) (7.49 × 10 N · m )
2
2
2
2
2
2
×[(64 ft ) − (4ft ) − (9ft )] ×[(9m ) − (1m ) − (1.27 m )]

1 1
2
2
= 1.5 × 10 −3 × [51 ft ] = 6.76 × 10 −3 × [6.73 m ]
ft m
12 in 100 cm
= 0.0765 ft × = 0.92 in ↓ = 0.0455 m × = 4.55 cm ↓
ft m
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
F = 10 kip = 10,000 lb F = 45 kN = 45,000 N
L = 8 ft, a = 6 ft, b = 2ft L = 3m, a = 2m, b = 1m
5
5
EI = 8.33 × 10 lb · ft 2 EI = 4.16 × 10 N · m 2
solution solution
Step 1. Calculate the maximum deflection Step 1. Calculate the maximum deflection
from Eq. (2.10). from Eq. (2.10).
3/2 3/2

Fb a (L + b) Fb a (L + b)
max = max =
3 (EI) L 3 3 EIL 3
(10,000 lb)(2ft) (45,000 N)(1m)
= =
2
5
2
5
3 (8.33 × 10 lb · ft )(8ft) 3 (4.16 × 10 N · m )(3m)
3/2
3/2
(6ft) [(8ft) + (2ft)] (2m) [(3m) + (1m)]
× ×
3 3
20,000 lb · ft 45,000 N · m
max = 3 max =
7
2.00 × 10 lb · ft 3.74 × 10 N · m 3
6
3/2
3/2
2
2
(60 ft ) (8m )
× ×
3 3

1 1
3
3
= 1.00 × 10 −3 2 (89.4ft ) = 1.20 × 10 −2 2 (4.35 m )
ft m
12 in 100 cm
= 0.089 ft × = 1.07 in ↓ = 0.0523 m × = 5.23 cm ↓
ft m
Step 2. Calculate the location of the maximum Step 2. Calculate the location of the maximum
deflection from Eq. (2.10). deflection from Eq. (2.10)

a (L + b) a (L + b)
x max = x max =
3 3

(6ft) [(8ft) + (2ft)] (2m) [(3m) + (1m)]
= =
3 3

(6ft)(10 ft) (2m)(4m)
= =
3 3

2 2
60 ft 8m
= = 4.47 ft = = 1.63 m
3 3