Torsion members - The Behaviour and Design of Steel Structures to EC3 - page 479

209.3

15.6

533 × 210 UB 92

12.7
radius

1.60 dm 6

I =

=
=
=

w
I t

y

75.7 cm 4

517.5

533.1

210 000 N/mm 2
81 000 N/mm 2

E
G

10.1

Dimensions in mm

z

Figure 10.37 Examples 1-7.

Solution.

(a) If EI w = 0, then the end twist rotation (see Section 10.2.2.1) is

3 × 10 6 × 5000
81000 × 75.7 × 10 4 = 0.245 radians = 14.0 ◦ .

φ um = TL

GI t =

(b) If GI t = 0, then the end twist rotation is (see Section 10.3.2.2)

3 × 10 6 × 5000 3
3 × 210000 × 1.6 × 10 12 = 0.372 radians = 21.3 ◦ .

φ wm = TL 3 / 3 EI w =

(c) Usingtheresultsof(a)and(b)aboveintheapproximationofequation10.56

φ m = 0.245 × 0.372

0.245 + 0.372 = 0.148 radians = 8.5 ◦

10.9.2 Example 2 - serviceability twist rotation

Problem. Use the solution obtained in Section 10.8.5 to determine an accurate
value of the serviceability end twist rotation of the cantilever of example 1.

Solution. Using equation 10.51,

210000 × 1.6 × 10 12
81000 × 75.7 × 10 4

EI w
GI t =

a = = 2341mm.

Therefore e − 2 L / a = e − 2 × 5000 / 2341 = 0.01396