Page 109 - Marks Calculation for Machine Design

P1: Sanjay
16:18
January 4, 2005
Brown.cls
Brown˙C02
BEAMS
91
Unless the length (a) is less than about 22 percent of the length (L), the max-
imum downward deflection ( Tip ) is greater than the maximum upward deflection
( Mid ).
U.S. Customary SI/Metric
Example 4. Calculate the maximum down- Example 4. Calculate the maximum down-
ward deflection ( Tip ) for a double overhanging ward deflection ( Tip ) for a double overhanging
beam, with concentrated forces (F) at the free beam, with concentrated forces (F) at the free
ends, where ends, where
F = 1,800 lb F = 8,000 N
L = 4 ft, a = 1.5 ft L = 1.2 m, a = 0.5 m
2
6
9
2
E = 27.6 × 10 lb/in (stainless steel) E = 190 × 10 N/m (stainless steel)
I = 7in 4 I = 341 cm 4
solution solution
Step 1. Calculate the stiffness (EI). Step 1. Calculate the stiffness (EI).
2
4
4
2
9
6
EI = (27.6 × 10 lb/in )(7in ) EI = (190 × 10 N/m )(341 cm )
1ft 2 1m 4
8 2
= 1.93 × 10 lb · in × ×
144 in 2 (100 cm) 4
6
5
= 1.34 × 10 lb · ft 2 = 6.48 × 10 N · m 2
Step 2. Determine the deflection ( Tip ) from Step 2. Determine the deflection ( Tip ) from
Eq. (2.52a). Eq. (2.52a).
Fa 2 Fa 2
Tip = (3L + 2 a) ↓ Tip = (3L + 2 a) ↓
6 EI 6 EI
(1,800 lb)(1.5ft) 2 (8,000 N)(0.5m) 2
= =
2
2
5
6
6 (1.34 × 10 lb · ft ) 6 (6.48 × 10 lb · ft )
×[3 (4ft) + 2 (1.5ft)] × [3 (1.2m) + 2 (0.5m)]
2
2
(4,050 lb · ft ) (2,000 N · m )
= [12 ft + 3 ft] = [3.6m + 1m]
2
6
2
6
(8.04 × 10 lb · ft ) (3.89 × 10 N · m )
= (5.04 × 10 −4 ) × [15 ft] = (5.14 × 10 −4 ) × [4.6m]
12 in 100 cm
= 0.0076 ft × = 0.0024 ft ×
ft m
= 0.09 in ↓ = 0.24 cm ↓
Example 5. Calculate the maximum upward Example 5. Calculate the maximum upward
deflection ( Mid ) for a double overhanging deflection ( Mid ) for a double overhanging
beam, with concentrated forces (F) at the free beam, with concentrated forces (F) at the free
ends, where ends, where
F = 1,800 lb F = 8,000 N
L = 4 ft, a = 1.5 ft L = 1.2 m, a = 0.5 m
5
6
EI = 1.34 × 10 lb · ft 2 EI = 6.48 ×10 N · m 2