3 kN 30° 0.1 m 0.1 m -0.2 m -0.3 m- 5 kN 4 kN (a)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Replace the force and couple system shown by an
equivalent resultant force and couple moment acting at point O.
Expert Solution
Step 1 Introduction:-
Free body diagram with all representation of values obtained in answer
Step 2 Solution:-
- Forces Calculations:-
The forces are resolved in x and y components.
∑(FR)x= ∑ Fx
(FR)x = 3cos 30 + (3/5)*5 = 5.598 kN
∑(FR)y= ∑ Fy
(FR)y = 3sin30 – (4.5) *5 – 4 = -6.50 kN
Using Pythagorean theorem:-
FR =√(FR)x2 + (FR)y 2
FR=√(5.598)2+(6.50)2
FR = 8.58kN
ϴ = tan-1 ((FR)x /(FR)y)
ϴ = tan-1 ( 6.50 / 5.598) = 49.3
- Calculation of moments:-
(MR)O = ∑ Mo
(MR)O= {3 * sin30 *0.2} –{ 3* cos 30*0.1} +{( 3/5) *5*0.1}-{(4/5) *5*0.5} – ( 4*0.2)
(MR)O = - 2.46kNm
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