Smart example on shear stress.

 7laha is a smart example to learn more.

 Two plates(5*20) are welded in continuous way with an extruded beam has the shown cross section and a uniform thickness of 3mm and they carry the loads as shown.

Determine the average shear stress at point B.

 

Step 1

Ra+Rc=(6.2+3*0.36)=7.28 KN

 

MOMENT AROUND C =0

SO 3*0.18*RA=7.28*0.18   RA=2.4 KN

 A cording to rod fg and the similar rod the length of the rod from the triangle

 

We divide the trapezoid into 4 rods

Where i=second moment of area.

Q=first moment of area

b=thickness of part is cut.

 

Step 2:

Y`=SUM (A*Y)/SUM A

Look to datum position in section figure.

Y`=[(20*5)*38+2*(34*3)*(30/2  +3/2 + 5/2) +(60*3)*(30+1.5+2.5)+(28*3)*(1.5+2.5)]

÷[(20*5)+(2*34*3)+(60*3)+(28*3)+(20*5)]

=24.88 mm =25mm

Step 3:

Calculation of i=(b(h^3)/12 +A(d^2) )

I for

Upper plate=((20*(5^3)/12  +(20*5)*((38-25)^2))

Upper rod =(60*(3^3)/12  +(60*3)*((34-25)^2))

How we can get I for fg rod? 

b`=b/cosɵ

b`=3/(30/34)=3.4mm and h=30 mm not 34mm(deal with it now as a vertical rectangular)

For fg and its isotope i=2*((3.4*(30^3)/12     +(3.4*30*(25-19)^2).

For lower rod of trapezoid  i=((28*(3^3)/12 +(28*3*(25-4)^2)

For lower plate i=((20*(5^3)/12  +(20*5*(25-2.5)^2)

I total =14.2 *(10^-8) m^4

Q=A*Y`

Q AT b will include the cut rectangular and the upper plate .

For symmetry we will cut at two edges at distance 25 mm from each edge=10mm width for the cut part.

The thickness of cut part=3mm(trapezoid thickness)
Q=(10*3*(34-25))+(20*5*13)=1570*(10^-9) m^3

b=2*3=6mm

sh.stat b=RA*Q/I*b*2

=(2.4*1000)*(1570*10^-9) /(14.2*6*10^-11)*2 

2.21MPA