Shear center in beams (Strength of Materials - II, Midterm Exam-42-7 ...

Problem :
Shear center in beams
(Strength of Materials - II, Midterm Exam-42-7)
(a) Locate the shear center S of the hat section by determining the eccentricity, e. (b) If a vertical
shear V =10kips acts through the shear center of this hat what are the values of the shear stresses τ A
and τ B at the locations and directions indicated in Figure.
Solution :
The centroidal principal moment of inertia of the beam section is
µ Ã
3 0.5 × 3 3 × 0.25
Izz = 2× +(0.5 × 3) × 4.52 +2×
12
3
+(3× 0.25) ×
12
Izz = 86. 656 in 4
The first moment areas of the locations shown in figure are
³
(Qz) 1 =(0.5 × s1) × 6 − s1
´
µ
2
(Qz) 2 =(0.25 × s2) × 3+ 0.25
+(3× 0.5) × 4.5
2
µ
3+ 0.25
!
2
+
2
0.5 × 63
12
Dr. M. Kemal Apalak 1

(Qz) 2 =0.781 25s2 +6. 75
µ
(Qz) 3 = (0.5 × 3) × (1.5) + (0.25 × 3) × 3+ 0.25
+(3× 0.5) × 4.5
2
(Qz) 3 = 11. 344 in 3
The shear stresses at these locations are
µ
Vy ×
³
³
Qz Vy
τ 1 =
=
(0.5 × s1) × 6 −
t × Izz 1 0.5 × Izz
s1
´´
2
τ 1 = Vy
³
6 −
Izz
s1
´
s1
2
µ
Vy × Qz Vy
τ 2 =
= (0.781 25s2 +6. 75)
t × Izz 0.25 × Izz
The moment about point A is
the resultant forces are
τ 2 = Vy
Izz
2
(3.125s2 + 27)
τ B =
10 × 103
(3.125 × 3 + 27)
86.66
τ B = 4197. 4 psi
τ 3 =
µ
Vy × Qz
t × Izz
Vy
= (11. 344)
0.5 × Izz
τ A = 22. 688 Vy
τ A = 2618 psi
F1 =
Izz
3
=22. 688 ×
10 × 103
86.66
Vy × e =6× F2 − 3 × (2 × F1)
Z Z 3
τ 1dA1 =
A1
s1=0
Vy
Izz
³
6 − s1
2
´
s1 (0.5 × ds1)
Dr. M. Kemal Apalak 2

F2 =
F1 = 0.5 Vy
Z 3 µ
F1 = 0.5 Vy
Izz
s1=0
Izz
F1 = 11. 25 Vy
6s1 − s2 1
2
µ
3 × 9 − 27
6
Izz
Z Z 3
τ 2dA2 =
A2
s2=0
Vy
Izz
ds1 =0.5 Vy
Izz
(3.125s2 +27)(0.25 × ds2)
F2 = 0.25 Vy
Z 3
(3.125s2 + 27) ds2 =0.25
Izz
s2=0
Vy
µ
3.125
Izz 2
F2 = 0.25 Vy
µ
3.125
× 9+27× 3
2
Izz
F2 = 23. 766 Vy
Izz
Vy × e = 6× 23. 766 Vy
e =
75. 096
86.66
Izz
=0.866 56 in
µ
3s 2 1 − s3 ¯3 ¯¯¯
1
6 0
µ
− 6 × 11. 25 Vy
Izz
s 2 2 +27s2
Dr. M. Kemal Apalak 3
¯3 ¯¯¯
0