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Lecture 27

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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Lecture 27

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  1. ENGR-1100 Introduction to Engineering Analysis Lecture 27

  2. Today Lecture outline • Distributed Loads

  3. Distributed forces Concentrated forces

  4. dR=wdx W=f(x) R d Distributed load

  5. 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

  6. 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

  7. 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

  8. + 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

  9. 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

  10. 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

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