# Lecture 27 - R = ∑ F = A 1 A 2 A 3 = 1000 500 875 = 2375...

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ENGR-1100 Introduction to Engineering Analysis Lecture 27 Notes courtesy of: Prof. Yoav Peles

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Homework Assignment (Last HW due next class) Problem 5-69 Problem 5-73 Problem 5-80

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Concentrated forces Distributed forces
Distributed load dR=wdx W=f(x) R d Rd Mo = = L wdx R

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Finding d xdR dM = 0 The moment produced by the distributed load: = = xdR dM M o o (a) The moment produced by the concentrated load: Rd M o = (b) From (a) and (b): R x xdR Rd M c o = = = R M x d o c = = dR=wdx x
R M x d o c = = R d Therefore, resultant R is located at centroid x c of area of distributed load,

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Example 5-71 Determine the resultant 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 R ur
Solution A 1 = 250(4) = 1000 lb; x c1 = 2 ft A 2 = 0.5 (250) (4) = 500 lb; x c2 = 1/3 (4) = 1.333 ft A 3 = 250 (3.5) = 875 lb; x c3 = 5.75 ft

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Unformatted text preview: R = ∑ F = A 1 + A 2 + A 3 = 1000 + 500 + 875 = 2375 lb = 2375 lb ↓ R ur + M A = A 1 x c1 + A 2 x c2 + A 3 x c3 = 1000(2) + 500(1.333) + 875(5.75) = 7698 ft lb Rd = M A d = = 3.24 ft → A M 7698 R 2375 = Class assignment: Exercise set 5-70 Determine the resultant 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 R ur Example 5-75 Determine the resultant of the system of distributed loads and locate its line of action with respect to the left support for the beam shown R ur Class assignment: Exercise set 5-74 Determine the resultant 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 R ur...
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## This note was uploaded on 04/04/2011 for the course ENGR 1100 taught by Professor Anderson during the Fall '06 term at Rensselaer Polytechnic Institute.

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Lecture 27 - R = ∑ F = A 1 A 2 A 3 = 1000 500 875 = 2375...

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