Torsion members - The Behaviour and Design of Steel Structures to EC3 - page 480

Using Section 10.8.5,

( 1 − e − 2 L / a )
( 1 + e − 2 L / a )

φ L = TL
GI t

1 − a
L

3 × 10 6 × 5000
81000 × 75.7 × 10 4

1 − 2341
5000

( 1 − 0.01396 )
( 1 + 0.01396 )

= 0.133 radians = 7.6 ◦ .

=

which is about 10% less than the approximate value of 8.5 ◦ obtained in
Section 10.9.1c.

10.9.3 Example 3 - elastic uniform torsion shear stress

Problem. Forthecantileverofexample2,determinethemaximumelasticuniform
torsion shear stress caused by a strength design end torque of 5 kNm.

Solution. An expression for the nominal maximum elastic uniform torsion shear
stress can be obtained by combining equations 10.8 and 10.18, whence

d φ
d x

τ t , max ≈ G t max .

max

An expression for d φ /d x can be obtained from the solution for φ in
Section 10.8.5, whence

1 − [ e ( x − 2 L )/ a + e − x / a ]
( 1 + e − 2 L / a )

d φ

d x = T

.

GI t

The maximum value of this occurs at the free end x = L and is given by

d φ
d x

2e − L / a
( 1 + e − 2 L / a )

= T
GI t

1 − .

max

Adapting the solution of example 2,

d φ
d x

1 − 2 × √ ( 0.01396 )
1.01396

= T
GI t

= 0.767 T

GI t .

max

Thus

≈ 0.767 × 5 × 10 6 × 15.6
75.7 × 10 4

τ t , max ≈ 0.767 Tt max
I t

N / mm 2 ≈ 79.0 N / mm 2 .