Hibbeler Structural Analysis 8th Txtbk 130

3. 5

109

THE  M ETHOD OF  S ECTIONS

3 .7

EXAMPLE

Determine the force in members BC and MC of the K-truss shown in Fig. Fi g. 3–28a. State whether the members are in tension or compression. The reactions at the supports have been calculated. La

K

J

I

H

 

10 ft

M

O

P

  10 ft



 A

C 

B a

15 ft  A y

N

 0

 A x

 2900 lb



15 ft

D

15 ft

15 ft

E 15 ft

120 200 0 lb 150 500 0 lb 180 800 0 lb

G

F  15 ft G y

3

 1600 lb



(a) SOLUTION

in Fig. Fig. 3–28a cuts Free-Body Diagram. Although section aa shown in through four members, members, it is possible to solve for the force in member BC using this section.The free-body diagram of the left portion of the truss is shown in Fig. Fig. 3–28b.

Equations of Equilibrium. Summing moments about point L eliminates three of the unknowns unknowns,, so that - 29001152 + FBC1202 = 0 d + © ML = 0; FBC = 2175 lb 1T2  Ans. Free-Body Diagrams. The force in MC  can be obtained indirectly by first obtaining the force in MB from vertical force equilibrium of   joint B, Fig ig.. 3– 3–28 28c, i.e., FMB = 1200 lb (T). Then from the free-body diagram in Fig. Fig. 3–28b. + c © Fy = 0; 2900 - 1200 + 1200 - FML = 0 FML = 2900 lb 1T2 Using these these results, results, the free-body free-body diagram diagram of joint M  is shown in Fig. Fi g. 3–28d. Equations of Equilibrium. 3 + © F = 0; FMC x 13 2 13 :

+ c © Fy = 0;

a

b

FLK  FML

 20 ft

FMB

 A

FBC  B

15 ft  2900 lb

1200 lb (b)

 y

FMB FBA

 x

B

FBC 

1200 lb (c)

 a 2 31313 b 2 2900 - 1200 - a b 13 2 13

FMK = 0

FMC -

FMK = 1532 lb 1C2

L

 a 2 21313 b

 y

FMK = 0

FMC = 1532 lb 1T2

 Ans.

Sometimes, as in this example,application Sometimes, example, application of both both the method of sections and the method of joints leads to the most direct solution to the problem. It is also possible to solve for the force in MC by using the result for FBC. In this case, case, pass a vertical section through LK , MK , MC ,and , and BC , Fig. Fi g. 3–28a. Isolate the left section section and apply © MK = 0.

FMK 

2900 lb 2 3

M 

3

 x 2

FMC 

1200 lb (d)

Fig. 3–28