Page 78 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C02
Brown.cls
60
U.S. Customary 16:18 STRENGTH OF MACHINES SI/Metric
Step 3. Determine the deflection ( ) from Step 3. Determine the deflection ( ) from
Eq. (2.22). Eq. (2.22).
wx 3 2 3 wx 3 2 3
= (L − 2Lx + x ) = (L − 2Lx + x )
24 (EI) 24 (EI)
(400 lb/ft)(10 ft) (6,000 N/m)(3m)
= =
2
5
5
24 (8.58 × 10 lb · ft ) 24 3.33 × 10 N · m 2
3
3
2
2
3
3
×[(15 ft) − 2 (15 ft)(10 ft) + (10 ft) ] ×[(5m) − 2 (5m)(3m) + (3m) ]
(4,000 lb) (18,000 N)
= 2 =
7
2
6
(2.06 × 10 lb · ft ) (8.00 × 10 N · m )
3
3
×[(3,375) − (3,000) + (1,000) ft ] ×[(125) − (90) + (27) ft ]
1 1
3
3
= 1.94 × 10 −4 2 × [1,375 ft ] = 2.25 × 10 −3 × [62 m ]
ft m 2
12 in 100 cm
= 0.27 ft × = 3.2in ↓ = 0.14 m × = 14.0cm ↓
ft m
Example 5. Calculate the maximum deflec- Example 5. Calculate the maximum deflec-
tion ( max ) and its location for the beam tion ( max ) and its location for the beam
configuration in Example 4, where configuration in Example 4, where
w = 400 lb/ft w = 6,000 N/m
L = 15 ft L = 5m
5
5
EI = 8.58 × 10 lb · ft 2 EI = 3.33 × 10 N · m 2
solution solution
Step 1. Calculate the maximum deflection Step 1. Calculate the maximum deflection
from Eq. (2.23). from Eq. (2.23).
5 wL 4 5 wL 4
max = max =
384 (EI) 384 (EI)
5 (400 lb/ft)(15 ft) 4 5 (6,000 N/m)(5m) 4
= =
2
2
5
5
384 (8.58 × 10 lb · ft ) 384 (3.33 × 10 N · m )
7
8
1.0125 × 10 lb · ft 3 1.875 × 10 N · m 3
= 2 = 8 2
8
3.2947 × 10 lb · ft 1.279 × 10 N · m
12 in 100 cm
= 0.31 ft × = 3.7in ↓ = 0.147 m × m = 14.7cm ↓
ft
The location of this maximum deflection is The location of this maximum deflection is
at the midpoint of the beam, (L/2). at the midpoint of the beam, (L/2).
2.2.5 Triangular Load
The simply-supported beam shown in Fig. 2.37 has a triangular distributed load acting ver-
tically downward across the length of the beam (L), where the magnitude of the distributed
load is zero at the left pin support and increases linearly to a magnitude (w) at the right
roller support. The units on the distributed load (w) are force per length. Therefore, the total
force acting on the beam is the area under this triangle, which is the distributed load (w)
times the length of the beam (L) divided by two, that is (wL/2).