Question
Asked Oct 5, 2019

Replace the loading by a single resultant force, specify the location of force measured from point O.

8kN/m
5 kN/m
-0.75 m---- 0.75 m
-1.5 m-
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8kN/m 5 kN/m -0.75 m---- 0.75 m -1.5 m-

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Step 1
HAingr
8KN
0-75M
1.5M
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HAingr 8KN 0-75M 1.5M

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Step 2

Divide the whole loading into three sections of loading OA, AB, and BC.

Consider the loading section BC.

Convert this uniform variable loading into a single point load say FBC.

The area of the triangle NBC is equal to the magnitude of the single point load. Thus,

The position of the single point load of section BC is equal to the 1/3rd from the base. Thus,

From the point O, the position of the point load is

Fc =(5
kN/m)(0.75 m)
1.875 kN
(0.75 m)
3
0.25 m
2.5 m
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Fc =(5 kN/m)(0.75 m) 1.875 kN (0.75 m) 3 0.25 m 2.5 m

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Step 3

Consider the loading section AB.

Convert this uniform variable loading into a single point load say FAB.

The area of the triangle MNBC is equal to the magnitude of the single point load. Thus,

The position of the single point load for section AB is equal to the 1/2rd from the base. Thus,

From the point O, the position of the point load is 1.875 m.

...
FAB (5 kN/m(0.75 m)
=3.75 kN
0.75 m
2
0.375 m
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FAB (5 kN/m(0.75 m) =3.75 kN 0.75 m 2 0.375 m

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