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ENGR-1100 Introduction to Engineering Analysis. Lecture 27. Today Lecture outline. Distributed Loads. Distributed forces. Concentrated forces. dR=wdx. W=f(x). R. d. Distributed load. x. The moment produced by the distributed load:. (a). The moment produced by the concentrated load:.
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Today Lecture outline • Distributed Loads
Distributed forces Concentrated forces
dR=wdx W=f(x) R d Distributed load
x The moment produced by the distributed load: (a) The moment produced by the concentrated load: dR=wdx (b) From (a) and (b): Finding d
Determine the resultant force of the system of distributed loads and locate its line of action with respect to the left support for the beam shown in Fig. P5-71 Example 5-71
Solution A1 = 250(4) = 1000 lb; xc1 = 2 ft A2 = 0.5 (250) (4) = 500 lb; xc2 = 1/3 (4) = 1.333 ft A3 = 250 (3.5) = 875 lb; xc3 = 5.75 ft Resultant Force R =F = A1 + A2 + A3 = 1000 + 500 + 875 = 2375 lb
+ MA = A1xc1 + A2xc2 + A3xc3 = 1000(2) + 500(1.333) + 875(5.75) = 7698 ft lb Resultant Moment of the Force Rd = MA d = MA/R d = 7698 ft lb / 2375 lb = 3.24 ft
Determine the resultant force of the system of distributed loads and locate its line of action with respect to the left support for the beam shown in Fig. P5-70 Class Assignment: Exercise set 5-70 Answer: d = 2.67m R = 2.25KN
Determine the resultant force of the system of distributed loads and locate its line of action with respect to the left support for the beam shown in Fig. P5-74 Class Assignment: Exercise set 5-74 Answer: d = 4.35m R = 4.5KN