In-plane bending of beams - The Behaviour and Design of Steel Structures to EC3 - page 214

Forhorizontalequilibriumofthelongitudinalshearforcesatthejunction2,the
shear flows in the segments 12, 62, and 23 must balance (equation 5.20), and so

[ (τ h × 12 ) 23 ] s 3 = 0 =[ (τ h × 12 ) 12 ] s 1 = 50 +[ (τ h × 6 ) 62 ] s 6 = 200

100 × 50 − 50 2
2

= 0.007143 × 12 + 50 × 200 × 6

= 0.007143 × 8750.

The shear flow in the segment 23 is

(τ h × 12 ) 23 = 0.007143 × 12 × 8750 − 100 × 10 3
1400 × 10 4

s 2

× ( − 50 + s 2 ) × 12 × d s 2

0

whence (τ h ) 23 = 0.007143 ( 8750 + 50 s 2 − s 2 / 2 ) N / mm 2 .
As a check, the resultant of the shear stresses in the segments 12, 23, and 34 is

50

( 100 s 1 − s 1 / 2 ) d s 1

τ h t d s = 2 × 0.007143 × 12

0

100

( 8750 + 50 s 2 − s 2 / 2 ) d s 2

+ 0.007143 × 12 ×

0

= 0.007143 × 12 { 2 × ( 100 × 50 2 / 2 − 50 3 / 6 )
+ ( 8750 × 100 + 50 × 100 2 / 2 − 100 3 / 6 ) } N
= 100 kN

which is equal to the shear force.

5.12.10 Example 10 - shear centre of a pi-section

Problem. Determine the position of the shear centre of the pi section shown in
Figure 5.42b.
Solution. The resultant of the shear stresses in the segment 62 is

200

τ h t d s = 0.007143 × 6 ( 50 s 6 ) × d s 6

0

= 0.007143 × 6 × 50 × 200 2 / 2N
= 42.86 kN.

The resultant of the shear stresses in the segment 53 is equal and opposite to this.