Torsion members - The Behaviour and Design of Steel Structures to EC3 - page 482

The corresponding maximum warping normal stress (see Section 10.8.2) is

d 2 φ
d x 2

σ w , max = Ed f b f
2 × 2

max
= 210000 × 517.5 × 209.3 × 5 × 10 6

2 × 2 × 2341 × 81000 × 75.7 × 10 4 × ( 1 − 0.01396 ) ( 1 + 0.01396 ) N / mm 2

= 192.6 N / mm 2 .

10.9.5Example - first yield design torque capacity

Problem. Ifthecantileverofexample2hasayieldstressof275N/mm 2 ,thenuse
Section10.4.7todeterminethetorqueresistancewhichisconsistentwithEC3and
a first yield strength criterion.
Solution. In this case, the maximum uniform torsion shear and warping torsion
shear and normal stresses occur either at different points along the cantilever, or
else at different points around the cross-section. Because of this, each stress may
be considered separately.
For first yield in uniform torsion, equation 10.79 is equivalent to

τ t /τ y ≤ 1
in which τ y ≈ 275 / √ 3 = 159 N/mm 2 .Thus

τ t , max ≤ 159 N/mm 2 .

UsingSection10.9.3,adesigntorqueof T = 5kNmcausesamaximumshearstress
of τ t , max = 79.0 N/mm 2 , if the stress concentration at the flange-web-junction is
ignored.Thus the design uniform torsion torque resistance is

T t = 5 × 159 / 79.0 = 10.05 kNm.

The maximum warping shear stress τ w , max caused by a design torque of
T = 5kNm(seeSection10.9.4)is4.4N/mm 2 , whichismuchlessthanthecorre-
sponding maximum uniform torsion shear stress τ t , max = 79.0 N/mm 2 . Because
of this, the design torque resistance is not limited by the warping shear stress.
For first yield in warping torsion, equation 10.79 reduces to

σ w / f y ≤ 1

Thus

σ w , max ≤ 275 N/mm 2 .

Using Section 10.9.4, a design torque of T = 5 kNm causes a design warping
normal stress of σ w , max = 192.6 N/mm 2 . Thus the design warping torsion torque