Page 70 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C02
Brown.cls
52
U.S. Customary 16:18 STRENGTH OF MACHINES SI/Metric
Example 3. Calculate and locate the max- Example 3. Calculate and locate the max-
imum shear force (V max ) and the maximum imum shear force (V max ) and the maximum
bending moment (M max ) for the beam of bending moment (M max ) for the beam of
Examples 1 and 2, where Examples 1 and 2, where
C = 15 ft · kip = 15,000 ft · lb C = 20 kN · m = 20,000 N · m
1
1
L = 12 ft, a = 4 ft, b = 8ft L = 4m, a = 1 m, b = 2 m
2 2
solution solution
Step 1. Calculate the maximum shear force Step 1. Calculate the maximum shear force
(V max ) from Eq. (2.12) as (V max ) from Eq. (2.12) as
C 15,000 ft · lb C 20,000 N · m
V max = = V max = =
L 12 ft L 4m
= 1,250 lb = 5,000 N
Step 2. As shown in Fig. 2.27 the maximum Step 2. As shown in Fig. 2.27 the maximum
shear force (V max ) of 1,250 lb does not have a shear force (V max ) of 5,000 N · lb does not have
specific location. a specific location.
Step 3. Calculate the maximum bending Step 3. Calculate the maximum bending
moment (M max ) from Eq. (2.14a), as the dis- moment (M max ) from Eq. (2.14a), as the dis-
tance (a), the location of the couple (C), is less tance (a), the location of the couple (C), is less
than the distance (b). than the distance (b)
Cb (15,000 ft · 1b)(8ft) Cb (20,000 N · m)(2.5m)
M max = = M max = =
L 12 ft L 4m
2
120,000 ft · lb 50,000 N · m 2
= = 10,000 ft · lb = = 12,500 N · m
12 ft 4m
Step 4. Figure 2.28 shows that the maximum Step 4. Figure 2.28 shows that the maximum
bending moment (M max ) of 10,000 ft · lb occurs bending moment (M max ) of 12,500 N · m occurs
where the couple (C) acts. where the couple (C) acts.
a b
C ∆
A B
∆
L
FIGURE 2.29 Beam deflection diagram.

Deflection. For this loading configuration, the deflection ( ) along the beam is shown in
Fig. 2.29, and given by Eq. (2.16a) for the values of the distance (x) from the left end of the
beam to where the couple (C) acts, and given by Eq. (2.16b) for the values of the distance
(x) from where the couple (C) acts to the right end of the beam.
Cx 2 2 2
= [6 aL − x − 3a − 2L ] 0 ≤ x ≤ a (2.16a)
6 EIL
C 2 2 2 2 3
= [3a L + 3Lx − x(2L + 3a ) − x ] a ≤ x ≤ L (2.16b)
6 EIL