In-plane bending of beams - The Behaviour and Design of Steel Structures to EC3 - page 215

The torque about the centroid exerted by the shear stresses is equal to the sum
ofthetorquesoftheforceintheflange1234andtheforcesinthewebs62and53.
Thus

τ h t ρ d s = ( 100 × 50 ) + ( 2 × 42.86 × 50 ) = 9286 kNmm.

A

Fortheappliedsheartobestaticallyequivalenttotheshearstresses, itmustexert
an equal torque about the centroid, and so

− 100 × z 0 = 9286

whence z 0 =− 92.9 mm.
Because of symmetry, the shear centre lies on the z axis, and so y 0 = 0.

5.12.11 Example 11 - shear stresses in a box section

Problem. The rectangular box section shown in Figure 5.42c has a shear force of
100 kN acting parallel to the y axis. Determine the maximum shear stress.
Solution. If initially a slit is assumed at point 1 on the axis of symmetry so that
( τ ho t ) 1 = 0, then

s 1

(τ ho t ) 12 = − 100 × 10 3
6 × 10 6

s 1 × 12 × d s 1 =− 0.1 s 1

0

(τ ho t ) 2 =− 0.1 × 50 2 =− 0.1 × 2500 N / mm
(τ ho t ) 23 =− 0.1 × 2500 − 100 × 10 3

s 2

50 × 6 × d s 2

6 × 10 6

0

=− 0.1 ( 2500 + 50 s 2 )
(τ ho t ) 3 =− 0.1 ( 2500 + 50 × 150 ) =− 0.1 × 10000 N / mm
(τ ho t ) 34 =− 0.1 × 10000 − 100 × 10 3

s 3

( 50 − s 3 ) × 6 × d s 3

6 × 10 6

0

=− 0.1 [ 10000 + ( 50 s 3 − s 3 / 2 ) ]
(τ ho t ) 4 =− 0.1 [ 10000 + ( 50 × 100 − 100 2 / 2 ) ]=− 0.1 × 10000N / mm
= (τ ho t ) 3 , which is a symmetry check.

The remainder of the shear stress distribution can therefore be obtained by
symmetry.