7laha - Force system Resultants

"7laha - Force system Resultants"

Smart structure problems for never stop thinking

Source:  Mechanics of material by

Ferdinand p. Beer and E .Russell Johnston

We all see a rigid body on surface of something .this body has material with special density so at least we study the weight force of the body on the surface of the beam ;……etc.
We start our study for the force of weight  from :
1#choosing axis for our problem at lowest height
2#determining weight force at the beginning and the end of our body/our axis(at X=0   at X has value)
3#determining the force density for area
4#divid the shape of the body into known shapes
5#determinig centers of mass to all shapes
6#determining the resultant of force according to the direction of forces and taking the moment around  any point .
Remember this rule (moment of the resultant =summation of all forces moments around point
Notice :this steps to conclusive problem .some problems does not need all these steps

Solved example on force system resultant :

The granular material exerts the distributed loading 50(1+x)K pa  on the top surface of the beam as shown in the figure .determine the magnitude and location of the equivalent resultant of this load .

solution :

The loading is distributed on the width of the beam so the force distribution on the beam equal   to 50(x+1)*0.2=10(x+1)KN/m .
The trapezoid can be divided into rectangular and triangle .

The loading is distributed on the
area of the shape
F1=9*10=90KN(length #width)
F2=0.5*9*90=405KN
Parallel forces  so the resultant =495KN=Fr   

Note that : X`2=3m X`1=4.5 m so he last thing to know  X`r
MA=Fr*X`r=405*3 +90*4.5
So  the location of resultant force =3.27m
 

Try this problem :

if the distribution of the ground reaction on the pipe (the diameter=1.6m  )per meter of length can be approximated as (0.5(1+cosα)KN/m  )shown figure .determine the magnitude of the resultant force due to this loading .

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