520 CHAPTER 17 Torsion of Beams
Fig.17.14
Determination of points of zero warping.
and
A 3 =312.5−
1
2
×8.04× 50 =111.5mm^2
Finally,
A 34 =111.5+
1
2
× 25 s 3 (iii)
SubstitutingforA 12 ,A 23 ,andA 34 fromEqs.(i)to(iii)inEq.(17.21),wehave
2 A′R=
1
200
⎡
⎣
∫^25
0
25 ×1.15s 1 ds 1 +
∫^50
0
2 (312.5−4.02s 2 )2.5ds 2 +
∫^25
0
2 (111.5+12.5s 3 )1.5ds 3
⎤
⎦ (iv)
EvaluationofEq.(iv)gives
2 A′R=424mm^2
Wenowexamineeachwallofthesectioninturntodeterminepointsofzerowarping.Supposethatin
thewall12apointofzerowarpingoccursatavalueofs 1 equaltos1,0.Then
2 ×
1
2
× 25 s1,0= 424