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b-Ta+6 kN/mA2 kN/m- 4m-

Question

Replace the loading by a single resultant force, specify the location of force measured from point A.  a = 0.75 m and b = 2.25 m

b-
Ta+
6 kN/m
A
2 kN/m
- 4m
-
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b- Ta+ 6 kN/m A 2 kN/m - 4m -

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Step 1

Note that, the center of mass of a uniformly varying loading (UVL) acts at the position of two-third of is spam from its narrow side and one-third of its spam from it’s another side. Also, the center of mass of a uniformly distributed load (UDL) acts its mid position.

 

Next, draw the schematic diagram of the provided structure by considering F1 is the force that is acting on the beam due to the UVL about its centroid (CG1) and F2 is the force that is acting on the beam due to the UDL about its centroid (CG1).

4-0.7Sym =3.25m F
b
a-o7sm
16 Ktm
b22.25
2.25
3
F0.75m
CG2
2.Somm
F2
4 m
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4-0.7Sym =3.25m F b a-o7sm 16 Ktm b22.25 2.25 3 F0.75m CG2 2.Somm F2 4 m

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Step 2

Determine the value of F1 and F2 that will be the corresponding area under the structure of UVL and UDL respectively.

w
2
)(6 kN/m)
1
(2.25 m)(6 kN/m)
2
- 6.75 kN
F wl
(2 kN/m)ab)
=(2 kN/m)(0.75 m+ 2.25 m)
=6 kN
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w 2 )(6 kN/m) 1 (2.25 m)(6 kN/m) 2 - 6.75 kN F wl (2 kN/m)ab) =(2 kN/m)(0.75 m+ 2.25 m) =6 kN

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Step 3

Determine the value of net resultant force FR, Consider, the downwar...

F F-F
= 6.75 kN 6.00 kN
=0.75 kN
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F F-F = 6.75 kN 6.00 kN =0.75 kN

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