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Solutions to Chapter 10 Problems
SOLUTIONS TO CHAPTER 10 PROBLEMS S.10.1
The area of cross section is given by A ¼ 10 60 þ 40 10 ¼ 1000 mm2 Referring to Fig. S.10.1(a) 1000 y ¼ 10 60 40 þ 40 10 5 10 mm A y y z
G
7.74 N/mm2
y 1.68 N/mm2
10 mm
A 40 mm
(a)
60 mm
6.72 N/mm2
(b)
FIGURE S.10.1
from which y ¼ 26 mm Then Iz ¼
10 603 40 103 þ 10 60 142 þ þ 40 10 212 ¼ 4:77 105 mm4 12 12
First consider the web. The area A0 ¼ 10(44y), b0 ¼ 10 mm and y ¼ ð44 þ yÞ=2 (i.e. the distance to the centroid of A0 ) so that, from Eq. (10.4), the shear stress distribution is given by 4 103 10ð44 yÞ 44þy 2 ¼ 0:004ð442 y 2 Þ τ¼ 10 4:77 105
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Solutions to Chapter 10 Problems
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When y ¼ 44 mm, τ ¼ 0 and when y ¼ 16 mm, τ ¼ 6.72 N/mm2. The maximum value of shear stress will occur when y ¼ 0, i.e. τmax ¼ 7.74 N/mm2. Now consider the flange. The area A0 ¼ 40[26(y)], b0 ¼ 40 mm and y ¼ ½26 þ ðyÞ=2. Then, from Eq. (10.4) 4 103 40ð26 yÞ 26þy 2 ¼ 0:004ð262 y 2 Þ τ¼ 40 4:77 105 When y ¼ 26 mm, τ ¼ 0 and when y ¼ 16 mm, τ ¼ 1.68 N/mm2 The complete distribution is shown in Fig. S.10.1(b). S.10.2
The problem is identical to Ex. 10.2 except that numerical dimensions are given. Therefore, from the dimensions given in Fig. P.10.2 Iz ¼
150 4003 135 3603 ¼ 2:75 108 mm4 12 12
Then, from Eq. (10.8) τðflangeÞ ¼
80 103
4002 2 4 y 8
2 2:75 10
¼ 1:45 104 ð40000 y 2 Þ
When y ¼ 200 mm, τ (flange) ¼ 0 and when y ¼ 180 mm, τ (flange) ¼ 1.1 N/mm2 The shear stress distribution in the web is given by Eq. (10.11) and is 2 360 2 2 2 4 y 3 150ð400 360 Þ 80 10 þ 2 815 τðwebÞ ¼ 8 2:75 10 i.e. τðwebÞ ¼ 15:77 1:45 104 y 2 When y ¼ 180 mm, τ (web) ¼ 11.1 N/mm2. The maximum value of shear stress occurs at y ¼ 0 and is 15.77 N/mm2. The form of the distribution is identical to that shown in Fig. 10.6(b). The shear load carried by the web is given by SðwebÞ ¼ 2
ð 180
15ð15:77 1:45 104 y2 Þdy ¼ 76701:6 N ¼ 76:7 kN
0
Therefore, the percentage of the total load carried by the web is given by 76:7 100 ¼ 95:9%: 80
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Solutions Manual
S.10.3
The area of cross section of the reinforced beam is given by A ¼ 400 40 þ 2 200 30 þ 25 540 ¼ 41 500 mm2 Referring to Fig. S.10.3(a) and taking moments of areas about the top edge of the plate 41 500y ¼ 400 40 20 þ 200 30 55 þ 25 540 340 þ 200 30 625
400 mm y
y z
G
40 mm 30 mm
0.68
1.36
14.22
1.78 N/mm2
25 mm
15.15
600 mm
30 mm (a)
(b)
200 mm
FIGURE S.10.3
i.e. y ¼ 216:6 mm Then 400 403 þ 400 40 196:62 þ Iz ¼ 12
200 6003 175 5403 12 12
!
þ ð200 600 175 540Þ 123:42 i.e. Iz ¼ 2:31 109 mm4 In the plate, A0 ¼ 400(216.6y), y ¼ ð216:6 þ yÞ=2, b0 ¼ 400 mm. Then from Eq. (10.4) τ ¼ 4:33 105 ð216:62 y2 Þ
Solutions to Chapter 10 Problems
e145
When y ¼ 216.6 mm, τ ¼ 0 and when y ¼ 176.6 mm, τ ¼ 0.68 N/mm2. In the flange of the I-beam, A0 y ¼ ½400 40 196:6 þ 200ð176:6 yÞð176:6 þ yÞ=2 and b0 ¼ 200 mm. Substituting in Eq. (10.4) τ ¼ 4:33 107 ð6:26 106 100y2 Þ When y ¼ 176.6 mm, τ ¼ 1.36 N/mm2 and when y ¼ 146.6 mm, τ ¼ 1.78 N/mm2. In the web of the I-beam, A0 y ¼ ½400 40 196:6 þ 200 30 161:6 þ 25ð146:6 yÞ ð146:6 þ yÞ=2 and b0 ¼ 25 mm. Substituting in Eq. (10.4) τ ¼ 3:46 106 ð4:38 106 12:5y 2 Þ When y ¼ 146.6 mm, τ ¼ 14.22 N/mm2 and when y ¼ 0, τ ¼ 15.15 N/mm2. The required distribution is shown in Fig. S.10.3(b). The shear stress at the top of the flange is 1.36 N/mm2 so that the shear force/unit length of beam is 1.36 200 ¼ 272 N/mm or 272 kN/m. S.10.4
From Ex. 10.1 the maximum shear stress in a rectangular section beam is given by τðmax Þ ¼
3Sy 2bd
The maximum shear force on a section of this beam is 90 kN, therefore τðmax Þ ¼ S.10.5
3 90 103 ¼ 3 N=mm2 : 2 150 300
From Fig. P.10.5 the area of cross section of the girder is A ¼ 2ð300 40 þ 2 100 30 þ 390 30Þ ¼ 59400 mm2 Therefore the weight per metre length of the girder is Wt=m ¼ 77 1 59400 106 ¼ 4:6 kN=m From Exs 3.8 and 3.9 the maximum shear force occurs at a support and is S:Fðmax Þ ¼
500 4:6 5 þ ¼ 261:5 kN 2 2
The second moment of area of the girder cross section about a horizontal axis through its centroid is Iz ¼ 2
300 403 200 303 30 3903 þ 300 40 2452 þ þ 200 30 2102 þ 12 12 12
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Solutions Manual which gives Iz ¼ 2270:5 106 mm4 The bolts resist the horizontal shear force between the outer surfaces of the flanges of the channels and the inner surfaces of the plates. We therefore require the distribution of shear stress in the flanges of the channels. Then, in Eq. (10.4) A0 y ¼ 300 40 245 þ 200ð225 yÞð225 þ yÞ=2 b0 ¼ 200 mm Substituting in Eq. (10.4) and simplifying τ¼
261:5 103 ð8:003 106 100y2 Þ 200 2270:5 106
which gives, when y ¼ 225 mm τ ¼ 1:69 N=mm2 The shear force per metre length of the girder is ¼ 1.69 300 1000 103 ¼ 507 kN. This shear force is resisted by bolts in each flange of the channels and since the shear resistance of a single bolt is 150 kN then two bolts in each flange of both channel sections will give a shear resistance/ metre of 4 150 ¼ 600 kN. Therefore two bolts per flange of each channel section per metre length of the girder will be sufficient. S.10.6
The maximum shear force in the beam is given by
W þ 11:5 kN S:Fðmax Þ ¼ 2 The maximum shear stress occurs in the channel section webs at mid-height. The shear stress distribution in the webs is, from Eq. (10.4) τ¼
½ðW =2Þ þ 11:5 103 ½300 40 245 þ 2 100 30 210 þ 60ð195 yÞð195 þ yÞ=2 60 2270:5 106
The maximum value of τ occurs at y ¼ 0 and is then
W τmax ¼ þ 11:5 0:039 ¼ 100 N=mm2 2 which gives W ¼ 5105 kN
Solutions to Chapter 10 Problems
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The maximum bending moment occurs at mid-span and is B:MðmaxÞ ¼
W 5 4:6 52 kNm þ 8 4
Then, from Eq. (9.9), the maximum direct stress due to bending is σ max ¼
ð1:25W þ 14:4Þ 106 265 ¼ 200 N=mm2 2270:5 106
which gives W ¼ 1359 kN Therefore the maximum allowable value of W is 1359 kN and bending is the limiting case. S.10.7
The beam section is singly symmetrical so that the shear centre lies on the axis of symmetry, the z axis. The shear flow distribution is given by Eq. (10.22) which, since Izy ¼ 0 and only Sy is applied, reduces to ð Sy s ty ds (i) qs ¼ Iz 0 where Iz ¼ 2
ða
ts 2 sin 2 α ds þ 2
0
ða
tða sin α þ s sin αÞ2 ds
0
In the first term on the right hand side of the expression for Iz, s is measured in the direction CB from C and in the second term s is measured from B in the direction BA. Then Iz ¼
16a3 t sin 2 α 3
In AB, y ¼ a sin α þ ða sA Þ sin α ¼ ð2a sA Þ sin α Substituting in Eq. (i) qAB ¼
Sy Iz
ðs
tð2a sA Þ sin α ds
0
which gives qAB ¼
s2 3Sy 2as 2A 16a3 sin α
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Solutions Manual When sA ¼ 0, qAB ¼ 0 and when sA ¼ a, qAB ¼ 29Sy/32a sin α. In BC, y ¼ (asB) sin α. Then, from Eq. (i) Sy qBC ¼ Iz
ðs
tða sB Þ sin α ds
0
9Sy 32a sin α
i.e. qBC ¼
3Sy
3 sB sB2 þ 2 a 2a2
16a sin α
When sB ¼ 0, qBC ¼ 29Sy/32a sin α and when sB ¼ a, qBC ¼ 23Sy/8a sin α. Taking moments about C Sy zS ¼ 2
ða qAB a sin 2α ds 0
Substituting for qAB from the above and integrating S y zS ¼
3Sy cos α 2 sA3 a as 6 4a2 0
which gives zS ¼ S.10.8
5a cos α : 8
The centroid of the section coincides with the centre of the circle and since Gz is an axis of symmetry Izy ¼ 0. Also, Sz ¼ 0 so that Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds
(i)
0
The second moment of area, Iz, of the section about Gz may be deduced from the second moment of area (πr3t/2) of the semi-circular section shown in Fig. 9.30 and is πr3t. Then, at any point a distance s from one edge of the narrow slit Eq. (i) becomes qs ¼
Sy πr 3 t
ðs ty ds 0
Referring now to Fig. S.10.8(a) and working with angular coordinates, Eq. (ii) becomes, since t ¼ constant qθ ¼
Sy πr 3
ðθ 0
r sin θ rd θ
(ii)
Solutions to Chapter 10 Problems (a)
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y t
Sy r
z
G
S
s
θ
zs (b)
y
2Sy πr G
z
FIGURE S.10.8
so that
qθ ¼
Sy πr
ðθ
sin θ dθ
(iii)
0
Integrating and substituting the limits in Eq. (iii) gives qθ ¼
Sy ðcos θ 1Þ πr
(iv)
From Eq. (iv) qθ ¼ 0 when θ ¼ 0 and 2π as would be expected at the open edges of the narrow slit. Also qθ is a maximum when θ ¼ π and is equal to 2Sy/πr. Note that qθ is negative and therefore in the opposite sense to increasing values of θ for all values of θ. The complete distribution is shown in Fig. S.10.8(b).
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Solutions Manual
S.10.9
The shear centre S will lie on the horizontal axis of symmetry, Gz, a distance zS, say, from the vertical web 24 (see Fig. S.10.9). Therefore we apply an arbitrary shear force, Sy, through S. Since Gz is an axis of symmetry Izy ¼ 0 and Eq. (9.22) reduces to Sy qs ¼ Iz
ðs ty ds
(i)
0
Referring to Fig. S.10.9 " 2 # th3 td 3 sin 2 α h þ td Iz ¼ þ 2 12 12 2
y α
5
4 6 zs G
z
Sy S
h
t
α
3 2
1
s2 s1 d/2 d/2
FIGURE S.10.9
that is Iz ¼
th3 ð1 þ 6ρ þ 2ρ3 sin 2 αÞ 12
From Eq. (i) q12 ¼
Sy Iz
ð s1 h d t þ s1 sin α ds1 2 2 0
(ii)
Solutions to Chapter 10 Problems
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which gives q12 ¼
Sy t ðhs1 þ ds1 sin α s12 sin αÞ 2Iz
(iii)
q32 ¼
Sy t ðhs2 ds2 sin α þ s22 sin αÞ 2Iz
(iv)
Similarly
Taking moments about G in Fig. S.10.9 Sy zS ¼ 2
ð d =2 0
h q12 cos α ds1 þ 2 2
ð d =2 0
h q32 cos α ds2 2
(v)
Substituting in Eq. (v) for q12 and q32 from Eqs (iii) and (iv) and evaluating gives zS ¼
thd 3 sin α cos α 12Iz
Now substituting for Iz from Eq. (ii) zS ¼ d
ρ2 sin α cos α 1 þ 6ρ þ 2ρ3 sin 2 α
S.10.10 The shear centre, S, lies on the horizontal axis of symmetry a distance zS, say, from the point 2 (see Fig. S.10.10). We therefore apply an arbitrary shear load, Sy, through S and since Izy ¼ 0 and Sz ¼ 0 Eq. (10.22) reduces to Sy qs ¼ Iz
ðs ty ds
(i)
0
The second moment of area, Iz, of the section will have the same value as the section in P.9.24, that is 3πr3t. Then, working with angular coordinates Eq. (i) becomes Sy q12 ¼ Iz
ðθ
tðr þ rcos θÞrdθ
0
which gives, after integration and substitution of the limits and Iz q12 ¼
Sy ðθ þ sin θÞ 3πr
(ii)
Now taking moments about the point 2 Sy zS ¼ 2
ðπ 0
q12 ðr þ rcos θÞrdθ
(iii)
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Solutions Manual 3 t r
Sy
zs S
z
2
q12
θ
1
s
FIGURE S.10.10
Substituting in Eq. (iii) for q12 from Eq. (ii) and expanding Sy zS ¼
2Sy r 3π
ðπ
ðθ þ θ cos θ þ sin θ þ sin θ cos θÞdθ
0
which gives zS ¼
π 2r θ2 cos 2θ þ θ sin θ 3π 2 4 0
from which zS ¼
πr 3
S.10.11 The shear centre S lies on the Gz axis which is an axis of symmetry so that Izy ¼ 0. Therefore we apply an arbitrary shear force, Sy, through S and since Sz ¼ 0 Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds 0
(i)
Solutions to Chapter 10 Problems
t
Sy S
z
h
e153
zs t/β
t 1
2
3
s1
s2 βd
d
FIGURE S.10.11
Referring to Fig. S.10.11 " 2 # th3 h 2 t h 2 h þd þ βd Iz ¼ þ 2 td ¼ th 12 2 β 2 12 From Eq. (i) q12 ¼
Sy Iz
ð s1 h t ds1 2 0
from which q12 ¼
Sy th s1 2Iz
(ii)
q32 ¼
Sy th s2 2βIz
(iii)
Similarly
Taking moments about the mid-point of the web Sy zS ¼ 2
ðd
h q12 ds1 2 2 0
ð βd 0
h q32 ds2 2
Substituting for q12 and q32 from Eqs (ii) and (iii) in Eq. (iv) Sy zS ¼
Sy th2 2Iz
ðd 0
s1 ds1
Sy th2 2βIz
ð βd s2 ds2 0
(iv)
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Solutions Manual which gives zS ¼
th2 d 2 βd 2 2Iz 2 2
Substituting for Iz and for (d/h) ¼ ρ zS 3ρð1 βÞ ¼ d ð1 þ 12ρÞ S.10.12 The shear centre S lies on the horizontal axis of symmetry so that we apply an arbitrary shear load Sy through S to determine its horizontal position. Further since Gz is an axis of symmetry Izy ¼ 0 and Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds
(i)
0
in which, referring to Fig. S.10.12 "ð Iz ¼ 2
r
ty ds1 þ 2
0
ð π=4
# 2
tðr sin θÞ r dθ
0
1 s1 2
Sy
S
q
3
O
r
z
f
4 zS 5
FIGURE S.10.12
so that "ð Iz ¼ 2
r 0
o 2
tð2r sin 45o s1 sin 45 Þ ds1 þ
ð π=4 0
# tr 3 sin 2 θ dθ
Solutions to Chapter 10 Problems
e155
Integrating gives Iz ¼ t
4r 2 s1 2rs12 þ
s1 3 r 3 3 0 þr
h iπ=4 θ sin22θ 0
from which Iz ¼ 2:62 tr 3 From Eq. (i) ð s1
Sy q12 ¼ Iz
tð2r sin 45o s1 sin 45o Þ ds1
0
which gives q12 ¼
0:27Sy s12 2rs 1 r3 2
(ii)
From Eq. (ii) at the point 2 when s1 ¼ r q2 ¼
0:4Sy r
Further Sy q23 ¼ Iz
ðφ
t r sin ð45o φÞ r dφ
0
0:4Sy r
which simplifies to q23 ¼
Sy 0:13Sy cos ð45o φÞ 2:62r r
(iii)
Now taking moments about O "ð Sy zS ¼ 2
r
q12 r ds1 þ
0
ð π=4
# 2
q23 r dφ 0
Substituting for q12 and q23 from Eqs (ii) and (iii) in Eq. (iv) and integrating (
0:27 zS ¼ 2 r2
h
3
S rs12 61
π=4 ) sin ð45o φÞ r þ 0:13φ 2:62 0 0
ir
(iv)
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Solutions Manual which gives zS ¼ 1:2 r
S.10.13 Since we are only required to find the horizontal position of the shear centre we apply an arbitrary shear load Sy through the shear centre S as shown in Fig. S.10.13. Then, since Sz ¼ 0 Eq. (10.22) becomes Sy Izy qs ¼ Iz Iy Izy2
ðs
S y Iy tz ds 2 I I z y Izy 0
ðs ty ds
(i)
0
2r
zs Sy
0.14r t
s2
3
S
4
y G
O z
r
θ
t
2t 1
2 s1 r
FIGURE S.10.13
From Fig. S.10.13 Iz ¼ 2rtr þ 2rtr þ 2
2
ðπ
tr 2 cos 2 θ rdθ ¼ 5:6tr 3
0
tð2rÞ3 2tr 3 Iy ¼ þ 2rtð0:86rÞ2 þ þ 2rtð0:36rÞ2 þ 12 12
ðπ
tðr sin θ þ 0:14 rÞ2 rdθ ¼ 4:8tr 3
0
Izy ¼ 2rtð0:86rÞðþrÞ þ r2tð0:36rÞðrÞ þ 0 ¼ tr 3 Substituting for Iz etc in Eq. (i) 0:04Sy qs ¼ tr 3
ðs
0:19Sy tz ds tr 3 0
ðs ty ds 0
(ii)
Solutions to Chapter 10 Problems
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On the flange 12, z ¼ 0.86 þ s1, y ¼ r and the thickness is 2t. Substituting in Eq. (ii) q12 ¼
Sy ð0:45rs1 0:04s12 Þ r3
(iii)
so that at 2 where s1 ¼ r, q2 ¼
0:41Sy r
In the semicircular portion 23, z ¼ 0.14r þ r sin θ, y ¼ r cos θ and the thickness is t. Then from Eq. (ii) 0:04Sy q23 ¼ tr 3
ðθ
0:19Sy tðþ0:14r þ r sin θÞ rdθ tr 3 0
ðθ
tðr cos θÞ rdθ þ
0
0:41Sy r
which gives q23 ¼
Sy ð0:19 sin θ þ 0:04 cos θ 0:006θ þ 0:37Þ r
(iv)
and q3 ¼
0:31Sy r
In the flange 34, z ¼ s2 þ 0.14r, y ¼ r and the thickness is t. Then 0:04Sy q34 ¼ tr 3
ð s2 0
0:19Sy tðs2 þ 0:14rÞ ds2 tr 3
ð s2 tr ds2 0
which gives q34 ¼
Sy ð0:31 r 2 0:196s2 r þ 0:02s22 Þ r3
(v)
Now taking moments about O Sy zS ¼
ðr 0
q12 rds1 þ
ðπ
q23 r rdθ þ
0
ð 2r q34 r ds2 0
Substituting in Eq. (vi) for q12 etc from Eqs (iii), (iv) and (v) and carrying out the integration gives zS ¼ 2:0r
(vi)
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Solutions Manual
S.10.14 The shear centre is the point in a beam cross section through which shear loads must be applied for there to be no twisting of the section. The z axis is an axis of symmetry so that the shear centre lies on this axis. Its position is found by applying a shear load Sy, through the shear centre, determining the shear flow distribution and then taking moments about some convenient point. Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds
(i)
0
in which, referring to Fig. S.10.14 tr 3 Iz ¼ 2 þ 2rtr 2 þ 3
!
ðπ
2
tr cos θr dθ 2
0
s2
2
2
3 Sy
s1 2r
q
r
1
4 O
z
S zs
t
2r
FIGURE S.10.14
i.e. Iz ¼ 6:22tr 3 In the wall 12, y ¼ s1. Therefore substituting in Eq. (i) q12 ¼
Sy Iz
ð s1
ts1 ds ¼
0
Then q2 ¼
Sy tr 2 Iz 2
Sy ts12 Iz 2
Solutions to Chapter 10 Problems
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In the wall 23, y ¼ r, then q23 ¼
ð s2 Sy tr 2 tr ds þ Iz 2 0
i.e. Sy tr 2 q23 ¼ trs2 þ Iz 2 and q3 ¼ 5
Sy tr 2 Iz 2
In the wall 34, y ¼ r cos θ, then q34 ¼
Sy Iz
ð θ
tr 2 cos θ dθ þ
0
5tr 2 2
i.e. Sy 5 q34 ¼ tr 2 sin θ þ Iz 2 Taking moments about O "ð Sy zS ¼ 2
r
0
q12 2r ds þ
ð 2r
#
ðπ q23 r ds þ
0
2
q34 r 2 dθ
0
The negative sign arises from the fact that the moment of the applied shear load is in the opposite sense to the moments produced by the internal shear flows. Substituting for q12, q23 and q34 from the above Sy S y zS ¼ t Iz
"ð # ð 2r ðπ r 2 4 s12 r2 5 2r ds þ r ds þ r sin θ þ dθ rs2 þ 2 2 0 0 2 0
which gives zS ¼ 2:66r: S.10.15 In this problem the axis of symmetry is the vertical y axis and the shear centre will lie on this axis so that only its vertical position is required. Therefore we apply a horizontal shear load Sz through the shear centre, S, as shown in Fig. S.10.15.
Solutions Manual y S
Sz
mm
4 ys
50
e160
q
3
O 50 mm
s2
s1 1
2 25 mm
100 mm
25 mm
FIGURE S.10.15
The thickness of the section is constant and will not appear in the answer for the shear centre position (see S.10.14), therefore assume the section has unit thickness. Eq. (10.22), since Izy ¼ 0, t ¼ 1 and only Sz is applied, reduces to qs ¼
Sz Iy
ðs z ds
(i)
0
where "
253 þ 25 62:52 þ 50 502 þ Iy ¼ 2 12
#
ðπ
2
ð50 cos θÞ2 50 dθ
0
i.e. Iy ¼ 6:44 105 mm4 In the flange 12, z ¼ 75s1 and Sz q12 ¼ Iy
Sz s12 ð75 s1 Þ ds ¼ 75s1 Iy 2 0
ð s1
and when s1 ¼ 25 mm, q2 ¼ 1562.5Sz/Iy In the wall 23, z ¼ 50 mm, then q23 ¼
Sz Iy
ð s2 0
Sz 50 ds þ 1562:5 ¼ ð50s2 þ 1562:5Þ Iy
Solutions to Chapter 10 Problems
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When s2 ¼ 50 mm, q3 ¼ 4062.5Sz/Iy In the wall 34, z ¼ 50 cos θ, therefore Sz q34 ¼ Iy
ð θ
50 cos θ 50 dθ þ 4062:5 ¼
0
Sz ð2500 sin θ þ 4062:5Þ Iy
Now taking moments about O Sz yS ¼ 2
ð 25
q12 50 ds1
0
ð 50
!
ðπ q23 50 ds2
2
2
q34 50 dθ
0
0
Note that the moments due to the shear flows in the walls 23 and 34 are opposite in sign to the moment produced by the shear flow in the wall 12. Substituting for q12, etc. gives yS ¼ 87:5 mm: S.10.16 The section is singly symmetrical so that the shear centre S lies on the axis of symmetry, the z axis. Then, since only Sy, is applied and Izy ¼ 0, Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds 0
where 2 t0 h3 3t0 h 2 h þ 9d þ 2d ¼ t0 h Iz ¼ 12 2 4 12 In the wall AB, y ¼ h/2, t ¼ 2t0(lsA/2d). Substituting in Eq. (i) and integrating gives Sy s2 qAB ¼ t 0 h sA A Iz 4d When sA ¼ d, qB ¼ 3Syt0hd/4Iz. In the wall BC, y ¼ h/2sB. Then Sy qBC ¼ Iz
ð sB h 3t0 hd t0 sB ds þ 2 4 0
i.e. Sy 3hd 2 qBC ¼ t0 hsB sB þ 2Iz 2
(i)
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Solutions Manual If moments are taken about the mid-point of the web qBC will not contribute to the moment equation, i.e. h ds Sy zS ¼ 2 qAB 2 0 ðd
Substituting for qAB and Iz from the above gives zS ¼
5d 2 : h þ 9d
S.10.17 The section is unsymmetrical so that both horizontal and vertical positions of the shear centre are unknown. To find the horizontal position apply a vertical shear load through the shear centre and then to find the vertical position apply a horizontal shear load through the shear centre. Initially the position of the centroid of area must be found so that the second moments of area can be determined. The cross-sectional area of the flange DC is equal to the cross-sectional area of the flange AB so that the centroid of area is at the mid-depth of the section. The area of cross section is given by A ¼ 10 50 þ 10 100 þ 5 100 ¼ 2000 mm2 Referring to Fig. S.10.17 and taking moments of area about the web BC 2000 z ¼ 10 50 25 þ 5 100 50 from which z ¼ 18:75 mm 50 mm D
C
10 mm y
s A
Sy zs
Sz 5 mm 100 mm
FIGURE S.10.17
z
G
z
10 mm
S B
ys
100 mm
Solutions to Chapter 10 Problems
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Then Iz ¼ 10 5 502 þ 5 100 502 þ Iy ¼
10 1003 ¼ 3:33 106 mm4 12
10 503 5 1003 þ 10 50 6:252 þ þ 5 100 31:252 12 12 þ 10 100 18:752
¼ 1:38 106 mm4 Izy ¼ 10 50ð6:25Þð50Þ þ 5 100ð31:25Þð50Þ ¼ 0:625 106 mm4 The numerator in both terms on the right hand side of Eq. (10.22) is then Iz Iy Izy2 ¼ ð3:33 1:38 0:6252 Þ 1012 ¼ 4:205 1012 Since only the position of the shear centre is required it is not necessary to calculate the shear flow distribution in the complete cross section; if moments are taken about, say, the corner C the shear flows in the walls BC and CD will not contribute to the moment equation so that only the shear flow distribution in the flange AB is needed. First consider the application of Sy. Equation (10.22) becomes
Ðs Ðs Sy Izy 0 tz ds Iy 0 ty ds qs ¼ 4:205 1012
(i)
In the flange AB, z ¼ 81.25s and y ¼ 250 mm. Therefore ðs ðs qAB ¼ Sy 0:149 106 5ð81:25 sÞ ds 0:328 106 5ð50Þ ds 0
0
i.e. qAB ¼ Sy ½21:47 106 s þ 0:37 106 s 2 Taking moments about C Sy zS ¼
ð 100
qAB 100 ds
0
Substituting for qAB from Eq. (ii) gives zS ¼ 23:1 mm
(ii)
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Solutions Manual Now apply Sz through the shear centre. Eq. (10.22) becomes
Ðs Ðs Sz Iz 0 tz ds þ Izy 0 ty ds qAB ¼ 4:205 1012
(iii)
The expressions for z and y in terms of s are identical to those above so that Eq. (iii) simplifies to ðs ðs 6 6 5ð81:25 sÞ ds 0:149 10 5ð50Þ ds qAB ¼ Sz 0:792 10 0
i.e.
0
qAB ¼ Sz 284:5 106 s þ 1:98 106 s 2
Taking moments about C S z yS ¼
ð 100
(iv)
qAB 100 ds
0
Again substituting for qAB from Eq. (iv) and carrying out the integration yS ¼ 76:3 mm: S.10.18 The weight/unit length of girder is given by, referring to Fig. P.10.18, Weight=unit length ¼ ð2 30 400 þ 1400 10Þ 7750 1 9:81 103 ¼ 2:9 kN=m: Therefore, one girder is subjected to the loading system shown in Fig. S.10.18(a). 750 kN
750 kN 2.9 kN/m
B
A 4m
C 4m
D 4m
(a)
FIGURE S.10.18(a)
The total load on a girder ¼ 2 750 þ 2:9 12 ¼ 1534:8 kN: The support reactions are then both ¼ 767.4 kN, which will be the maximum shear force on a girder. Since the girder section is doubly symmetrical and only a single vertical shear force is applied, Eq. (10.22) reduces to ð Sy s ty ds (i) qs ¼ Iz 0
Solutions to Chapter 10 Problems
e165
Referring to Fig. P.10.18 Iz ¼ 2 400 30 7002 þ 10 14003 =12 ¼ 1:4 1010 mm4 Substituting for Sy (¼ 767.4 kN) and Iz in Eq. (i) gives ðs 5 ty ds qs ¼ 5:5 10 0
Now referring to Fig. S.10.18(b) y 30 mm 3
s1
2
1
10 mm 1400 mm
z
5
6
4 400 mm
(b)
FIGURE S.10.18(b)
in the top flange q12 ¼ 5:5 10
5
ð s1
30 700 d s1 ,
0
which gives q12 ¼ 1:16s1 Then, when s1 ¼ 200 mm q2 ¼ 232 N/mm. There will be an identical contribution from the shear flow in the flange 23 so that the total shear flow at the top of the web will be 464 N/mm. This is resisted by the two fillet welds so that the shear force/mm of weld is 232 N. The required weld throat thickness, tW, is given by Eq. (22.10) i.e. tW ¼
232 ¼ 2:32 mm 100
The leg length of the weld (see Fig. 22.12(c)) is then pffiffiffi Leg length ¼ 2 2:32 ¼ 3:3 mm Say, a leg length of 4 mm.
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Solutions Manual
S.10.19 The most straightforward approach for the solution of this problem is to replace the applied shear load by a shear load through the shear centre, which coincides with the centre of symmetry, together with a torque as shown in Fig. S.10.19. The shear flows produced by the two separate loading systems are calculated separately and then superimposed. The torsion of thin-walled closed section beams is covered in Section 11.4. The torque is given by T ¼ 30 103 100 cos 30° ¼ 2:6 106 Nmm sB
y
C
B
sC
sA
30° 30 kN
z
D
G S
A
2.6
106 Nmm E
F
FIGURE S.10.19
The area enclosed by the mid-line of the section wall is given by A ¼ 100 2 100 cos 30° þ 4 0:5 100 cos 30° 100 sin 30° ¼ 2:6 104 mm2 Then, from Eq. (11.20), the shear flow due to the torque is qT ¼
2:6 106 ¼ 50 N=mm 2 2:6 104
The shear flow distribution due to the shear load is given by Eq. (10.24) in which Sz ¼ 30 103 N and Sy ¼ 0. Further, the section is doubly symmetrical so that Izy ¼ 0. If an origin for s is taken at A on the horizontal axis of symmetry, i.e. the section is “cut” at A, then qs,0 ¼ 0 and Eq. (10.24) reduces to ð Sz s tz ds (i) qs ¼ Iy 0 The thickness of the section walls is constant and, as we have seen previously, will disappear from the expressions for shear flow distribution, therefore we assume that the walls are of unit thickness. Then ð 100 2 1003 þ4 ð100 sA sin 30°Þ2 ds ¼ 2:5 106 mm4 Iy ¼ 12 0 Substituting in Eq. (i) qs ¼ 0:012
ðs z ds 0
Solutions to Chapter 10 Problems
e167
In the wall AB, z ¼ 100sA sin 30° ¼ 1000.5sA. Then qAB ðshearÞ ¼ 0:012ð100sA 0:25sA2 Þ ¼ 1:2sA 0:003sA2 When sA ¼ 100 mm, qAB (shear) ¼ 90 N/mm In the wall BC, z ¼ 50sB. Then qBC ðshearÞ ¼ 0:012ð50sB 0:5sB2 Þ þ 90 ¼ 0:6sB 0:006sB2 þ 90 When sB ¼ 100 mm, qBC ¼ 90 N/mm In the wall CD, z ¼ 50sC sin 30° ¼ 500.5sC. Then qCD ðshearÞ ¼ 0:012ð50sC 0:25sC2 Þ þ 90 ¼ 0:6sC 0:003sC2 þ 90 The remaining shear flow distribution due to shear follows from symmetry. Now superimposing the shear flows due to torsion on those due to shear, qAB ¼ 1:2sA 0:003sA2 þ 50 qBC ¼ 0:6sB 0:006sB2 þ 140 qCD ¼ 0:6sC 0:003sC2 þ 140 In the remaining half of the section the shear flow distribution due to shear is symmetrical with that in ABCD but is then reduced by the shear flow due to torsion. S.10.20 In Fig. P.10.20 BO is an axis of symmetry so take this as the z axis. Then Sy ¼ S and Izy ¼ 0. Eq. (10.24) therefore reduces to ð S s ty ds þ qs,0 qs ¼ Iz 0 where 3 sin
Iz ¼ ð2rÞ
2
45° þ2 12
ðπ
4
ðr sin θÞ2 r dθ ¼ 0:62r 3
0
in which the section is assumed to have unit wall thickness. “Cutting” the section at O ðθ S 1:61Sðcos θ 1Þ r sin θr dθ ¼ qb,OA ¼ 3 0:62r 0 r and when θ ¼ π/4, qb,OA ¼ 0.47S/r. Also S qb,AB ¼ 0:62r 3
ðs 0
ðr sÞ sin 45°ds
0:47S r
i.e. qb,AB ¼
Sð1:14rs 0:57s2 þ 0:47Þ r
e168
Solutions Manual Now taking moments about B ðπ
4
Sr ¼ 2
qb, OA rrdθ þ 2
0
2 πr qs, 0 4
Substituting for qb,OA from the above ðπ Sr ¼ 2
4
1:61Sðcos θ 1Þ r dθ þ
0
πr 2 qs,0 2
from which qs,0 ¼ 0:8S=r The complete shear flow distribution is then Sð1:61cos θ 0:81Þ r Sð0:57s2 1:14rs þ 0:33Þ qAB ¼ : r
qOA ¼
S.10.21 The radius BO is an axis of symmetry so that the shear centre will lie on this axis. Therefore we take BO as the z axis and apply an arbitrary shear load, Sy, through the shear centre S as shown in Fig. S.10.21. Further Izy ¼ 0 and Eq. (10.24) reduces to ð Sy s qs ¼ ty ds þ qs, 0 (i) Iz 0
z
A O
Sy
zs s
S θ C
B
FIGURE S.10.21
Iz has been previously calculated in P.10.20 and is 0.62tr3. Referring to Fig. S.10.21 and “cutting” the section at O we note that the qb distribution will be identical to that in P.10.20 and is 1:61Sy ðcos θ 1Þ r
(ii)
Sy ð0:57s2 1:14rs 0:47r 2 Þ r3
(iii)
qb,OA ¼ and qb,AB ¼
Solutions to Chapter 10 Problems
e169
Since the rate of twist of the beam section is zero (shear load is applied through the shear centre) qs,0 is given by Eq. (10.30) in which þ ds ¼ 2r þ ðπr=2Þ ¼ 3:57r and "ð
þ qb ds ¼ 2
π=4
0
1:61Sy ðcos θ 1Þ r dθ þ r
ðr
Sy ð0:57s 2 1:14rs 0:47r 2 Þ ds 3 0r
#
which gives þ qb ds ¼ 1:95Sy Then, from Eq. (10.30) qs,0 ¼
1:95Sy 0:55Sy ¼ 3:57r r
so that qOA ¼
Sy ð1:61cos θ 1:06Þ r
Taking moments about the point B Sy zS ¼ 2
ð π=4
qOA r dθ ¼ 2 2
ð π=4
0
0
Sy ð1:61cos θ 1:06Þ r 2 dθ r
from which zS ¼ 0:61r S.10.22 The shear centre lies on the vertical axis of symmetry therefore apply an horizontal shear load Sz through the shear centre. Since Sy ¼ 0 and Izy ¼ 0 (the y axis is an axis of symmetry) Eq. (10.24) reduces to ð Sz s qs ¼ tz ds þ qs,0 Iy 0 where, referring to Fig. S.10.22 Iy ¼
23 þ2 12
ð3
ðs sin θÞ2 ds
0
assuming that the thickness is unity and s is measured from the point 3. Then Iy ¼ 2:67 m4
e170
Solutions Manual y O
1 s1 s2
2 ys
S
Sz G
z
q s 3
FIGURE S.10.22
Now taking an origin for s at O Sz qb,O1 ¼ 2:67
ð s1 0
s1 ds ¼ 0:19Sz s12
and when s1 ¼ 1 m, qb,O1 ¼ 0.19Sz ð s2 qb,13 ¼ 0:37Sz ð1 s2 sin θÞ ds 0:19Sz ¼ 0:37Sz ð0:5 þ s2 0:17s22 Þ Þ
0
In Eq. (10.30), ds ¼ 8 m. Then h Ð i Ð3 1 2 0:37Sz 0:5 0 s12 ds þ 0 ð0:5 þ s2 0:17s22 Þ ds ¼ 0:44Sz qs,0 ¼ 8 so that qO1 ¼ Sz ð0:44 0:19s12 Þ Take moments about 3 Sz ð3 cos θ yS Þ ¼ 2
ð1
qO1 3 cos θ ds
0
from which yS ¼ 0:7 m: S.10.23 Referring to Fig. P.10.23, the wall DB is 3 m long so that its cross-sectional area, 3 103 8 ¼ 24 103 mm2, is equal to that of the wall EA, 2 103 12 ¼ 24 103 mm2. It follows that the centroid of area of the section lies mid-way between DB and EA on the vertical axis of symmetry. Also since Sy ¼ 500 kN, Sz ¼ 0 and Izy ¼ 0, Eq. (10.24) reduces to ð 500 103 s qs ¼ ty ds þ qs,0 (i) Iz 0
Solutions to Chapter 10 Problems
e171
If the origin for s is taken on the axis of symmetry, say at O, then qs,0 is zero. Also Iz ¼3 103 8 ð0:43 103 Þ2 þ 2 103 12 ð0:43 103 Þ2 þ 2 ð1 103 Þ3 10 sin 2 60°=12 i.e. Iz ¼ 101:25 108 mm4 Eq. (i) then becomes qs ¼ 4:94 105
ðs ty ds 0
In the wall OA, y ¼ 0.43 103 mm. Then ð sA 5 12 0:43 103 ds ¼ 0:25sA qOA ¼ 4:94 10 0
and when sA ¼ 1 10 mm, qOA ¼ 250 N/mm In the wall AB, y ¼ 0.43 103 þ sB cos 30°. Then ð sB qAB ¼ 4:94 105 10ð0:43 103 þ 0:866sB Þ ds þ 250 3
0
i.e. qAB ¼ 0:21sB 2:14 104 sB2 þ 250 When sB ¼ 1 103 mm, qAB ¼ 246 N/mm In the wall BC, y ¼ 0.43 103 mm. Then ð sC qBC ¼ 4:94 105 8 0:43 103 ds þ 246 0
i.e. qBC ¼ 0:17sC þ 246 Note that at C where sC ¼ 1.5 103 mm, qBC should equal zero; the discrepancy, 9 N/mm, is due to rounding off errors. The maximum shear stress will occur in the wall AB (and ED) mid-way along its length (this coincides with the neutral axis of the section) where sB ¼ 500 mm. This gives, from Eq. (ii), qAB(max) ¼ 301.5 N/mm so that the maximum shear stress is equal to 301.5/10 ¼ 30.2 N/mm2.
The area of cross section is given by A ¼ 10 60 þ 40 10 ¼ 1000 mm2 Referring to Fig. S.10.1(a) 1000 y ¼ 10 60 40 þ 40 10 5 10 mm A y y z
G
7.74 N/mm2
y 1.68 N/mm2
10 mm
A 40 mm
(a)
60 mm
6.72 N/mm2
(b)
FIGURE S.10.1
from which y ¼ 26 mm Then Iz ¼
10 603 40 103 þ 10 60 142 þ þ 40 10 212 ¼ 4:77 105 mm4 12 12
First consider the web. The area A0 ¼ 10(44y), b0 ¼ 10 mm and y ¼ ð44 þ yÞ=2 (i.e. the distance to the centroid of A0 ) so that, from Eq. (10.4), the shear stress distribution is given by 4 103 10ð44 yÞ 44þy 2 ¼ 0:004ð442 y 2 Þ τ¼ 10 4:77 105
e142
Solutions to Chapter 10 Problems
e143
When y ¼ 44 mm, τ ¼ 0 and when y ¼ 16 mm, τ ¼ 6.72 N/mm2. The maximum value of shear stress will occur when y ¼ 0, i.e. τmax ¼ 7.74 N/mm2. Now consider the flange. The area A0 ¼ 40[26(y)], b0 ¼ 40 mm and y ¼ ½26 þ ðyÞ=2. Then, from Eq. (10.4) 4 103 40ð26 yÞ 26þy 2 ¼ 0:004ð262 y 2 Þ τ¼ 40 4:77 105 When y ¼ 26 mm, τ ¼ 0 and when y ¼ 16 mm, τ ¼ 1.68 N/mm2 The complete distribution is shown in Fig. S.10.1(b). S.10.2
The problem is identical to Ex. 10.2 except that numerical dimensions are given. Therefore, from the dimensions given in Fig. P.10.2 Iz ¼
150 4003 135 3603 ¼ 2:75 108 mm4 12 12
Then, from Eq. (10.8) τðflangeÞ ¼
80 103
4002 2 4 y 8
2 2:75 10
¼ 1:45 104 ð40000 y 2 Þ
When y ¼ 200 mm, τ (flange) ¼ 0 and when y ¼ 180 mm, τ (flange) ¼ 1.1 N/mm2 The shear stress distribution in the web is given by Eq. (10.11) and is 2 360 2 2 2 4 y 3 150ð400 360 Þ 80 10 þ 2 815 τðwebÞ ¼ 8 2:75 10 i.e. τðwebÞ ¼ 15:77 1:45 104 y 2 When y ¼ 180 mm, τ (web) ¼ 11.1 N/mm2. The maximum value of shear stress occurs at y ¼ 0 and is 15.77 N/mm2. The form of the distribution is identical to that shown in Fig. 10.6(b). The shear load carried by the web is given by SðwebÞ ¼ 2
ð 180
15ð15:77 1:45 104 y2 Þdy ¼ 76701:6 N ¼ 76:7 kN
0
Therefore, the percentage of the total load carried by the web is given by 76:7 100 ¼ 95:9%: 80
e144
Solutions Manual
S.10.3
The area of cross section of the reinforced beam is given by A ¼ 400 40 þ 2 200 30 þ 25 540 ¼ 41 500 mm2 Referring to Fig. S.10.3(a) and taking moments of areas about the top edge of the plate 41 500y ¼ 400 40 20 þ 200 30 55 þ 25 540 340 þ 200 30 625
400 mm y
y z
G
40 mm 30 mm
0.68
1.36
14.22
1.78 N/mm2
25 mm
15.15
600 mm
30 mm (a)
(b)
200 mm
FIGURE S.10.3
i.e. y ¼ 216:6 mm Then 400 403 þ 400 40 196:62 þ Iz ¼ 12
200 6003 175 5403 12 12
!
þ ð200 600 175 540Þ 123:42 i.e. Iz ¼ 2:31 109 mm4 In the plate, A0 ¼ 400(216.6y), y ¼ ð216:6 þ yÞ=2, b0 ¼ 400 mm. Then from Eq. (10.4) τ ¼ 4:33 105 ð216:62 y2 Þ
Solutions to Chapter 10 Problems
e145
When y ¼ 216.6 mm, τ ¼ 0 and when y ¼ 176.6 mm, τ ¼ 0.68 N/mm2. In the flange of the I-beam, A0 y ¼ ½400 40 196:6 þ 200ð176:6 yÞð176:6 þ yÞ=2 and b0 ¼ 200 mm. Substituting in Eq. (10.4) τ ¼ 4:33 107 ð6:26 106 100y2 Þ When y ¼ 176.6 mm, τ ¼ 1.36 N/mm2 and when y ¼ 146.6 mm, τ ¼ 1.78 N/mm2. In the web of the I-beam, A0 y ¼ ½400 40 196:6 þ 200 30 161:6 þ 25ð146:6 yÞ ð146:6 þ yÞ=2 and b0 ¼ 25 mm. Substituting in Eq. (10.4) τ ¼ 3:46 106 ð4:38 106 12:5y 2 Þ When y ¼ 146.6 mm, τ ¼ 14.22 N/mm2 and when y ¼ 0, τ ¼ 15.15 N/mm2. The required distribution is shown in Fig. S.10.3(b). The shear stress at the top of the flange is 1.36 N/mm2 so that the shear force/unit length of beam is 1.36 200 ¼ 272 N/mm or 272 kN/m. S.10.4
From Ex. 10.1 the maximum shear stress in a rectangular section beam is given by τðmax Þ ¼
3Sy 2bd
The maximum shear force on a section of this beam is 90 kN, therefore τðmax Þ ¼ S.10.5
3 90 103 ¼ 3 N=mm2 : 2 150 300
From Fig. P.10.5 the area of cross section of the girder is A ¼ 2ð300 40 þ 2 100 30 þ 390 30Þ ¼ 59400 mm2 Therefore the weight per metre length of the girder is Wt=m ¼ 77 1 59400 106 ¼ 4:6 kN=m From Exs 3.8 and 3.9 the maximum shear force occurs at a support and is S:Fðmax Þ ¼
500 4:6 5 þ ¼ 261:5 kN 2 2
The second moment of area of the girder cross section about a horizontal axis through its centroid is Iz ¼ 2
300 403 200 303 30 3903 þ 300 40 2452 þ þ 200 30 2102 þ 12 12 12
e146
Solutions Manual which gives Iz ¼ 2270:5 106 mm4 The bolts resist the horizontal shear force between the outer surfaces of the flanges of the channels and the inner surfaces of the plates. We therefore require the distribution of shear stress in the flanges of the channels. Then, in Eq. (10.4) A0 y ¼ 300 40 245 þ 200ð225 yÞð225 þ yÞ=2 b0 ¼ 200 mm Substituting in Eq. (10.4) and simplifying τ¼
261:5 103 ð8:003 106 100y2 Þ 200 2270:5 106
which gives, when y ¼ 225 mm τ ¼ 1:69 N=mm2 The shear force per metre length of the girder is ¼ 1.69 300 1000 103 ¼ 507 kN. This shear force is resisted by bolts in each flange of the channels and since the shear resistance of a single bolt is 150 kN then two bolts in each flange of both channel sections will give a shear resistance/ metre of 4 150 ¼ 600 kN. Therefore two bolts per flange of each channel section per metre length of the girder will be sufficient. S.10.6
The maximum shear force in the beam is given by
W þ 11:5 kN S:Fðmax Þ ¼ 2 The maximum shear stress occurs in the channel section webs at mid-height. The shear stress distribution in the webs is, from Eq. (10.4) τ¼
½ðW =2Þ þ 11:5 103 ½300 40 245 þ 2 100 30 210 þ 60ð195 yÞð195 þ yÞ=2 60 2270:5 106
The maximum value of τ occurs at y ¼ 0 and is then
W τmax ¼ þ 11:5 0:039 ¼ 100 N=mm2 2 which gives W ¼ 5105 kN
Solutions to Chapter 10 Problems
e147
The maximum bending moment occurs at mid-span and is B:MðmaxÞ ¼
W 5 4:6 52 kNm þ 8 4
Then, from Eq. (9.9), the maximum direct stress due to bending is σ max ¼
ð1:25W þ 14:4Þ 106 265 ¼ 200 N=mm2 2270:5 106
which gives W ¼ 1359 kN Therefore the maximum allowable value of W is 1359 kN and bending is the limiting case. S.10.7
The beam section is singly symmetrical so that the shear centre lies on the axis of symmetry, the z axis. The shear flow distribution is given by Eq. (10.22) which, since Izy ¼ 0 and only Sy is applied, reduces to ð Sy s ty ds (i) qs ¼ Iz 0 where Iz ¼ 2
ða
ts 2 sin 2 α ds þ 2
0
ða
tða sin α þ s sin αÞ2 ds
0
In the first term on the right hand side of the expression for Iz, s is measured in the direction CB from C and in the second term s is measured from B in the direction BA. Then Iz ¼
16a3 t sin 2 α 3
In AB, y ¼ a sin α þ ða sA Þ sin α ¼ ð2a sA Þ sin α Substituting in Eq. (i) qAB ¼
Sy Iz
ðs
tð2a sA Þ sin α ds
0
which gives qAB ¼
s2 3Sy 2as 2A 16a3 sin α
e148
Solutions Manual When sA ¼ 0, qAB ¼ 0 and when sA ¼ a, qAB ¼ 29Sy/32a sin α. In BC, y ¼ (asB) sin α. Then, from Eq. (i) Sy qBC ¼ Iz
ðs
tða sB Þ sin α ds
0
9Sy 32a sin α
i.e. qBC ¼
3Sy
3 sB sB2 þ 2 a 2a2
16a sin α
When sB ¼ 0, qBC ¼ 29Sy/32a sin α and when sB ¼ a, qBC ¼ 23Sy/8a sin α. Taking moments about C Sy zS ¼ 2
ða qAB a sin 2α ds 0
Substituting for qAB from the above and integrating S y zS ¼
3Sy cos α 2 sA3 a as 6 4a2 0
which gives zS ¼ S.10.8
5a cos α : 8
The centroid of the section coincides with the centre of the circle and since Gz is an axis of symmetry Izy ¼ 0. Also, Sz ¼ 0 so that Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds
(i)
0
The second moment of area, Iz, of the section about Gz may be deduced from the second moment of area (πr3t/2) of the semi-circular section shown in Fig. 9.30 and is πr3t. Then, at any point a distance s from one edge of the narrow slit Eq. (i) becomes qs ¼
Sy πr 3 t
ðs ty ds 0
Referring now to Fig. S.10.8(a) and working with angular coordinates, Eq. (ii) becomes, since t ¼ constant qθ ¼
Sy πr 3
ðθ 0
r sin θ rd θ
(ii)
Solutions to Chapter 10 Problems (a)
e149
y t
Sy r
z
G
S
s
θ
zs (b)
y
2Sy πr G
z
FIGURE S.10.8
so that
qθ ¼
Sy πr
ðθ
sin θ dθ
(iii)
0
Integrating and substituting the limits in Eq. (iii) gives qθ ¼
Sy ðcos θ 1Þ πr
(iv)
From Eq. (iv) qθ ¼ 0 when θ ¼ 0 and 2π as would be expected at the open edges of the narrow slit. Also qθ is a maximum when θ ¼ π and is equal to 2Sy/πr. Note that qθ is negative and therefore in the opposite sense to increasing values of θ for all values of θ. The complete distribution is shown in Fig. S.10.8(b).
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Solutions Manual
S.10.9
The shear centre S will lie on the horizontal axis of symmetry, Gz, a distance zS, say, from the vertical web 24 (see Fig. S.10.9). Therefore we apply an arbitrary shear force, Sy, through S. Since Gz is an axis of symmetry Izy ¼ 0 and Eq. (9.22) reduces to Sy qs ¼ Iz
ðs ty ds
(i)
0
Referring to Fig. S.10.9 " 2 # th3 td 3 sin 2 α h þ td Iz ¼ þ 2 12 12 2
y α
5
4 6 zs G
z
Sy S
h
t
α
3 2
1
s2 s1 d/2 d/2
FIGURE S.10.9
that is Iz ¼
th3 ð1 þ 6ρ þ 2ρ3 sin 2 αÞ 12
From Eq. (i) q12 ¼
Sy Iz
ð s1 h d t þ s1 sin α ds1 2 2 0
(ii)
Solutions to Chapter 10 Problems
e151
which gives q12 ¼
Sy t ðhs1 þ ds1 sin α s12 sin αÞ 2Iz
(iii)
q32 ¼
Sy t ðhs2 ds2 sin α þ s22 sin αÞ 2Iz
(iv)
Similarly
Taking moments about G in Fig. S.10.9 Sy zS ¼ 2
ð d =2 0
h q12 cos α ds1 þ 2 2
ð d =2 0
h q32 cos α ds2 2
(v)
Substituting in Eq. (v) for q12 and q32 from Eqs (iii) and (iv) and evaluating gives zS ¼
thd 3 sin α cos α 12Iz
Now substituting for Iz from Eq. (ii) zS ¼ d
ρ2 sin α cos α 1 þ 6ρ þ 2ρ3 sin 2 α
S.10.10 The shear centre, S, lies on the horizontal axis of symmetry a distance zS, say, from the point 2 (see Fig. S.10.10). We therefore apply an arbitrary shear load, Sy, through S and since Izy ¼ 0 and Sz ¼ 0 Eq. (10.22) reduces to Sy qs ¼ Iz
ðs ty ds
(i)
0
The second moment of area, Iz, of the section will have the same value as the section in P.9.24, that is 3πr3t. Then, working with angular coordinates Eq. (i) becomes Sy q12 ¼ Iz
ðθ
tðr þ rcos θÞrdθ
0
which gives, after integration and substitution of the limits and Iz q12 ¼
Sy ðθ þ sin θÞ 3πr
(ii)
Now taking moments about the point 2 Sy zS ¼ 2
ðπ 0
q12 ðr þ rcos θÞrdθ
(iii)
e152
Solutions Manual 3 t r
Sy
zs S
z
2
q12
θ
1
s
FIGURE S.10.10
Substituting in Eq. (iii) for q12 from Eq. (ii) and expanding Sy zS ¼
2Sy r 3π
ðπ
ðθ þ θ cos θ þ sin θ þ sin θ cos θÞdθ
0
which gives zS ¼
π 2r θ2 cos 2θ þ θ sin θ 3π 2 4 0
from which zS ¼
πr 3
S.10.11 The shear centre S lies on the Gz axis which is an axis of symmetry so that Izy ¼ 0. Therefore we apply an arbitrary shear force, Sy, through S and since Sz ¼ 0 Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds 0
(i)
Solutions to Chapter 10 Problems
t
Sy S
z
h
e153
zs t/β
t 1
2
3
s1
s2 βd
d
FIGURE S.10.11
Referring to Fig. S.10.11 " 2 # th3 h 2 t h 2 h þd þ βd Iz ¼ þ 2 td ¼ th 12 2 β 2 12 From Eq. (i) q12 ¼
Sy Iz
ð s1 h t ds1 2 0
from which q12 ¼
Sy th s1 2Iz
(ii)
q32 ¼
Sy th s2 2βIz
(iii)
Similarly
Taking moments about the mid-point of the web Sy zS ¼ 2
ðd
h q12 ds1 2 2 0
ð βd 0
h q32 ds2 2
Substituting for q12 and q32 from Eqs (ii) and (iii) in Eq. (iv) Sy zS ¼
Sy th2 2Iz
ðd 0
s1 ds1
Sy th2 2βIz
ð βd s2 ds2 0
(iv)
e154
Solutions Manual which gives zS ¼
th2 d 2 βd 2 2Iz 2 2
Substituting for Iz and for (d/h) ¼ ρ zS 3ρð1 βÞ ¼ d ð1 þ 12ρÞ S.10.12 The shear centre S lies on the horizontal axis of symmetry so that we apply an arbitrary shear load Sy through S to determine its horizontal position. Further since Gz is an axis of symmetry Izy ¼ 0 and Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds
(i)
0
in which, referring to Fig. S.10.12 "ð Iz ¼ 2
r
ty ds1 þ 2
0
ð π=4
# 2
tðr sin θÞ r dθ
0
1 s1 2
Sy
S
q
3
O
r
z
f
4 zS 5
FIGURE S.10.12
so that "ð Iz ¼ 2
r 0
o 2
tð2r sin 45o s1 sin 45 Þ ds1 þ
ð π=4 0
# tr 3 sin 2 θ dθ
Solutions to Chapter 10 Problems
e155
Integrating gives Iz ¼ t
4r 2 s1 2rs12 þ
s1 3 r 3 3 0 þr
h iπ=4 θ sin22θ 0
from which Iz ¼ 2:62 tr 3 From Eq. (i) ð s1
Sy q12 ¼ Iz
tð2r sin 45o s1 sin 45o Þ ds1
0
which gives q12 ¼
0:27Sy s12 2rs 1 r3 2
(ii)
From Eq. (ii) at the point 2 when s1 ¼ r q2 ¼
0:4Sy r
Further Sy q23 ¼ Iz
ðφ
t r sin ð45o φÞ r dφ
0
0:4Sy r
which simplifies to q23 ¼
Sy 0:13Sy cos ð45o φÞ 2:62r r
(iii)
Now taking moments about O "ð Sy zS ¼ 2
r
q12 r ds1 þ
0
ð π=4
# 2
q23 r dφ 0
Substituting for q12 and q23 from Eqs (ii) and (iii) in Eq. (iv) and integrating (
0:27 zS ¼ 2 r2
h
3
S rs12 61
π=4 ) sin ð45o φÞ r þ 0:13φ 2:62 0 0
ir
(iv)
e156
Solutions Manual which gives zS ¼ 1:2 r
S.10.13 Since we are only required to find the horizontal position of the shear centre we apply an arbitrary shear load Sy through the shear centre S as shown in Fig. S.10.13. Then, since Sz ¼ 0 Eq. (10.22) becomes Sy Izy qs ¼ Iz Iy Izy2
ðs
S y Iy tz ds 2 I I z y Izy 0
ðs ty ds
(i)
0
2r
zs Sy
0.14r t
s2
3
S
4
y G
O z
r
θ
t
2t 1
2 s1 r
FIGURE S.10.13
From Fig. S.10.13 Iz ¼ 2rtr þ 2rtr þ 2
2
ðπ
tr 2 cos 2 θ rdθ ¼ 5:6tr 3
0
tð2rÞ3 2tr 3 Iy ¼ þ 2rtð0:86rÞ2 þ þ 2rtð0:36rÞ2 þ 12 12
ðπ
tðr sin θ þ 0:14 rÞ2 rdθ ¼ 4:8tr 3
0
Izy ¼ 2rtð0:86rÞðþrÞ þ r2tð0:36rÞðrÞ þ 0 ¼ tr 3 Substituting for Iz etc in Eq. (i) 0:04Sy qs ¼ tr 3
ðs
0:19Sy tz ds tr 3 0
ðs ty ds 0
(ii)
Solutions to Chapter 10 Problems
e157
On the flange 12, z ¼ 0.86 þ s1, y ¼ r and the thickness is 2t. Substituting in Eq. (ii) q12 ¼
Sy ð0:45rs1 0:04s12 Þ r3
(iii)
so that at 2 where s1 ¼ r, q2 ¼
0:41Sy r
In the semicircular portion 23, z ¼ 0.14r þ r sin θ, y ¼ r cos θ and the thickness is t. Then from Eq. (ii) 0:04Sy q23 ¼ tr 3
ðθ
0:19Sy tðþ0:14r þ r sin θÞ rdθ tr 3 0
ðθ
tðr cos θÞ rdθ þ
0
0:41Sy r
which gives q23 ¼
Sy ð0:19 sin θ þ 0:04 cos θ 0:006θ þ 0:37Þ r
(iv)
and q3 ¼
0:31Sy r
In the flange 34, z ¼ s2 þ 0.14r, y ¼ r and the thickness is t. Then 0:04Sy q34 ¼ tr 3
ð s2 0
0:19Sy tðs2 þ 0:14rÞ ds2 tr 3
ð s2 tr ds2 0
which gives q34 ¼
Sy ð0:31 r 2 0:196s2 r þ 0:02s22 Þ r3
(v)
Now taking moments about O Sy zS ¼
ðr 0
q12 rds1 þ
ðπ
q23 r rdθ þ
0
ð 2r q34 r ds2 0
Substituting in Eq. (vi) for q12 etc from Eqs (iii), (iv) and (v) and carrying out the integration gives zS ¼ 2:0r
(vi)
e158
Solutions Manual
S.10.14 The shear centre is the point in a beam cross section through which shear loads must be applied for there to be no twisting of the section. The z axis is an axis of symmetry so that the shear centre lies on this axis. Its position is found by applying a shear load Sy, through the shear centre, determining the shear flow distribution and then taking moments about some convenient point. Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds
(i)
0
in which, referring to Fig. S.10.14 tr 3 Iz ¼ 2 þ 2rtr 2 þ 3
!
ðπ
2
tr cos θr dθ 2
0
s2
2
2
3 Sy
s1 2r
q
r
1
4 O
z
S zs
t
2r
FIGURE S.10.14
i.e. Iz ¼ 6:22tr 3 In the wall 12, y ¼ s1. Therefore substituting in Eq. (i) q12 ¼
Sy Iz
ð s1
ts1 ds ¼
0
Then q2 ¼
Sy tr 2 Iz 2
Sy ts12 Iz 2
Solutions to Chapter 10 Problems
e159
In the wall 23, y ¼ r, then q23 ¼
ð s2 Sy tr 2 tr ds þ Iz 2 0
i.e. Sy tr 2 q23 ¼ trs2 þ Iz 2 and q3 ¼ 5
Sy tr 2 Iz 2
In the wall 34, y ¼ r cos θ, then q34 ¼
Sy Iz
ð θ
tr 2 cos θ dθ þ
0
5tr 2 2
i.e. Sy 5 q34 ¼ tr 2 sin θ þ Iz 2 Taking moments about O "ð Sy zS ¼ 2
r
0
q12 2r ds þ
ð 2r
#
ðπ q23 r ds þ
0
2
q34 r 2 dθ
0
The negative sign arises from the fact that the moment of the applied shear load is in the opposite sense to the moments produced by the internal shear flows. Substituting for q12, q23 and q34 from the above Sy S y zS ¼ t Iz
"ð # ð 2r ðπ r 2 4 s12 r2 5 2r ds þ r ds þ r sin θ þ dθ rs2 þ 2 2 0 0 2 0
which gives zS ¼ 2:66r: S.10.15 In this problem the axis of symmetry is the vertical y axis and the shear centre will lie on this axis so that only its vertical position is required. Therefore we apply a horizontal shear load Sz through the shear centre, S, as shown in Fig. S.10.15.
Solutions Manual y S
Sz
mm
4 ys
50
e160
q
3
O 50 mm
s2
s1 1
2 25 mm
100 mm
25 mm
FIGURE S.10.15
The thickness of the section is constant and will not appear in the answer for the shear centre position (see S.10.14), therefore assume the section has unit thickness. Eq. (10.22), since Izy ¼ 0, t ¼ 1 and only Sz is applied, reduces to qs ¼
Sz Iy
ðs z ds
(i)
0
where "
253 þ 25 62:52 þ 50 502 þ Iy ¼ 2 12
#
ðπ
2
ð50 cos θÞ2 50 dθ
0
i.e. Iy ¼ 6:44 105 mm4 In the flange 12, z ¼ 75s1 and Sz q12 ¼ Iy
Sz s12 ð75 s1 Þ ds ¼ 75s1 Iy 2 0
ð s1
and when s1 ¼ 25 mm, q2 ¼ 1562.5Sz/Iy In the wall 23, z ¼ 50 mm, then q23 ¼
Sz Iy
ð s2 0
Sz 50 ds þ 1562:5 ¼ ð50s2 þ 1562:5Þ Iy
Solutions to Chapter 10 Problems
e161
When s2 ¼ 50 mm, q3 ¼ 4062.5Sz/Iy In the wall 34, z ¼ 50 cos θ, therefore Sz q34 ¼ Iy
ð θ
50 cos θ 50 dθ þ 4062:5 ¼
0
Sz ð2500 sin θ þ 4062:5Þ Iy
Now taking moments about O Sz yS ¼ 2
ð 25
q12 50 ds1
0
ð 50
!
ðπ q23 50 ds2
2
2
q34 50 dθ
0
0
Note that the moments due to the shear flows in the walls 23 and 34 are opposite in sign to the moment produced by the shear flow in the wall 12. Substituting for q12, etc. gives yS ¼ 87:5 mm: S.10.16 The section is singly symmetrical so that the shear centre S lies on the axis of symmetry, the z axis. Then, since only Sy, is applied and Izy ¼ 0, Eq. (10.22) reduces to qs ¼
Sy Iz
ðs ty ds 0
where 2 t0 h3 3t0 h 2 h þ 9d þ 2d ¼ t0 h Iz ¼ 12 2 4 12 In the wall AB, y ¼ h/2, t ¼ 2t0(lsA/2d). Substituting in Eq. (i) and integrating gives Sy s2 qAB ¼ t 0 h sA A Iz 4d When sA ¼ d, qB ¼ 3Syt0hd/4Iz. In the wall BC, y ¼ h/2sB. Then Sy qBC ¼ Iz
ð sB h 3t0 hd t0 sB ds þ 2 4 0
i.e. Sy 3hd 2 qBC ¼ t0 hsB sB þ 2Iz 2
(i)
e162
Solutions Manual If moments are taken about the mid-point of the web qBC will not contribute to the moment equation, i.e. h ds Sy zS ¼ 2 qAB 2 0 ðd
Substituting for qAB and Iz from the above gives zS ¼
5d 2 : h þ 9d
S.10.17 The section is unsymmetrical so that both horizontal and vertical positions of the shear centre are unknown. To find the horizontal position apply a vertical shear load through the shear centre and then to find the vertical position apply a horizontal shear load through the shear centre. Initially the position of the centroid of area must be found so that the second moments of area can be determined. The cross-sectional area of the flange DC is equal to the cross-sectional area of the flange AB so that the centroid of area is at the mid-depth of the section. The area of cross section is given by A ¼ 10 50 þ 10 100 þ 5 100 ¼ 2000 mm2 Referring to Fig. S.10.17 and taking moments of area about the web BC 2000 z ¼ 10 50 25 þ 5 100 50 from which z ¼ 18:75 mm 50 mm D
C
10 mm y
s A
Sy zs
Sz 5 mm 100 mm
FIGURE S.10.17
z
G
z
10 mm
S B
ys
100 mm
Solutions to Chapter 10 Problems
e163
Then Iz ¼ 10 5 502 þ 5 100 502 þ Iy ¼
10 1003 ¼ 3:33 106 mm4 12
10 503 5 1003 þ 10 50 6:252 þ þ 5 100 31:252 12 12 þ 10 100 18:752
¼ 1:38 106 mm4 Izy ¼ 10 50ð6:25Þð50Þ þ 5 100ð31:25Þð50Þ ¼ 0:625 106 mm4 The numerator in both terms on the right hand side of Eq. (10.22) is then Iz Iy Izy2 ¼ ð3:33 1:38 0:6252 Þ 1012 ¼ 4:205 1012 Since only the position of the shear centre is required it is not necessary to calculate the shear flow distribution in the complete cross section; if moments are taken about, say, the corner C the shear flows in the walls BC and CD will not contribute to the moment equation so that only the shear flow distribution in the flange AB is needed. First consider the application of Sy. Equation (10.22) becomes
Ðs Ðs Sy Izy 0 tz ds Iy 0 ty ds qs ¼ 4:205 1012
(i)
In the flange AB, z ¼ 81.25s and y ¼ 250 mm. Therefore ðs ðs qAB ¼ Sy 0:149 106 5ð81:25 sÞ ds 0:328 106 5ð50Þ ds 0
0
i.e. qAB ¼ Sy ½21:47 106 s þ 0:37 106 s 2 Taking moments about C Sy zS ¼
ð 100
qAB 100 ds
0
Substituting for qAB from Eq. (ii) gives zS ¼ 23:1 mm
(ii)
e164
Solutions Manual Now apply Sz through the shear centre. Eq. (10.22) becomes
Ðs Ðs Sz Iz 0 tz ds þ Izy 0 ty ds qAB ¼ 4:205 1012
(iii)
The expressions for z and y in terms of s are identical to those above so that Eq. (iii) simplifies to ðs ðs 6 6 5ð81:25 sÞ ds 0:149 10 5ð50Þ ds qAB ¼ Sz 0:792 10 0
i.e.
0
qAB ¼ Sz 284:5 106 s þ 1:98 106 s 2
Taking moments about C S z yS ¼
ð 100
(iv)
qAB 100 ds
0
Again substituting for qAB from Eq. (iv) and carrying out the integration yS ¼ 76:3 mm: S.10.18 The weight/unit length of girder is given by, referring to Fig. P.10.18, Weight=unit length ¼ ð2 30 400 þ 1400 10Þ 7750 1 9:81 103 ¼ 2:9 kN=m: Therefore, one girder is subjected to the loading system shown in Fig. S.10.18(a). 750 kN
750 kN 2.9 kN/m
B
A 4m
C 4m
D 4m
(a)
FIGURE S.10.18(a)
The total load on a girder ¼ 2 750 þ 2:9 12 ¼ 1534:8 kN: The support reactions are then both ¼ 767.4 kN, which will be the maximum shear force on a girder. Since the girder section is doubly symmetrical and only a single vertical shear force is applied, Eq. (10.22) reduces to ð Sy s ty ds (i) qs ¼ Iz 0
Solutions to Chapter 10 Problems
e165
Referring to Fig. P.10.18 Iz ¼ 2 400 30 7002 þ 10 14003 =12 ¼ 1:4 1010 mm4 Substituting for Sy (¼ 767.4 kN) and Iz in Eq. (i) gives ðs 5 ty ds qs ¼ 5:5 10 0
Now referring to Fig. S.10.18(b) y 30 mm 3
s1
2
1
10 mm 1400 mm
z
5
6
4 400 mm
(b)
FIGURE S.10.18(b)
in the top flange q12 ¼ 5:5 10
5
ð s1
30 700 d s1 ,
0
which gives q12 ¼ 1:16s1 Then, when s1 ¼ 200 mm q2 ¼ 232 N/mm. There will be an identical contribution from the shear flow in the flange 23 so that the total shear flow at the top of the web will be 464 N/mm. This is resisted by the two fillet welds so that the shear force/mm of weld is 232 N. The required weld throat thickness, tW, is given by Eq. (22.10) i.e. tW ¼
232 ¼ 2:32 mm 100
The leg length of the weld (see Fig. 22.12(c)) is then pffiffiffi Leg length ¼ 2 2:32 ¼ 3:3 mm Say, a leg length of 4 mm.
e166
Solutions Manual
S.10.19 The most straightforward approach for the solution of this problem is to replace the applied shear load by a shear load through the shear centre, which coincides with the centre of symmetry, together with a torque as shown in Fig. S.10.19. The shear flows produced by the two separate loading systems are calculated separately and then superimposed. The torsion of thin-walled closed section beams is covered in Section 11.4. The torque is given by T ¼ 30 103 100 cos 30° ¼ 2:6 106 Nmm sB
y
C
B
sC
sA
30° 30 kN
z
D
G S
A
2.6
106 Nmm E
F
FIGURE S.10.19
The area enclosed by the mid-line of the section wall is given by A ¼ 100 2 100 cos 30° þ 4 0:5 100 cos 30° 100 sin 30° ¼ 2:6 104 mm2 Then, from Eq. (11.20), the shear flow due to the torque is qT ¼
2:6 106 ¼ 50 N=mm 2 2:6 104
The shear flow distribution due to the shear load is given by Eq. (10.24) in which Sz ¼ 30 103 N and Sy ¼ 0. Further, the section is doubly symmetrical so that Izy ¼ 0. If an origin for s is taken at A on the horizontal axis of symmetry, i.e. the section is “cut” at A, then qs,0 ¼ 0 and Eq. (10.24) reduces to ð Sz s tz ds (i) qs ¼ Iy 0 The thickness of the section walls is constant and, as we have seen previously, will disappear from the expressions for shear flow distribution, therefore we assume that the walls are of unit thickness. Then ð 100 2 1003 þ4 ð100 sA sin 30°Þ2 ds ¼ 2:5 106 mm4 Iy ¼ 12 0 Substituting in Eq. (i) qs ¼ 0:012
ðs z ds 0
Solutions to Chapter 10 Problems
e167
In the wall AB, z ¼ 100sA sin 30° ¼ 1000.5sA. Then qAB ðshearÞ ¼ 0:012ð100sA 0:25sA2 Þ ¼ 1:2sA 0:003sA2 When sA ¼ 100 mm, qAB (shear) ¼ 90 N/mm In the wall BC, z ¼ 50sB. Then qBC ðshearÞ ¼ 0:012ð50sB 0:5sB2 Þ þ 90 ¼ 0:6sB 0:006sB2 þ 90 When sB ¼ 100 mm, qBC ¼ 90 N/mm In the wall CD, z ¼ 50sC sin 30° ¼ 500.5sC. Then qCD ðshearÞ ¼ 0:012ð50sC 0:25sC2 Þ þ 90 ¼ 0:6sC 0:003sC2 þ 90 The remaining shear flow distribution due to shear follows from symmetry. Now superimposing the shear flows due to torsion on those due to shear, qAB ¼ 1:2sA 0:003sA2 þ 50 qBC ¼ 0:6sB 0:006sB2 þ 140 qCD ¼ 0:6sC 0:003sC2 þ 140 In the remaining half of the section the shear flow distribution due to shear is symmetrical with that in ABCD but is then reduced by the shear flow due to torsion. S.10.20 In Fig. P.10.20 BO is an axis of symmetry so take this as the z axis. Then Sy ¼ S and Izy ¼ 0. Eq. (10.24) therefore reduces to ð S s ty ds þ qs,0 qs ¼ Iz 0 where 3 sin
Iz ¼ ð2rÞ
2
45° þ2 12
ðπ
4
ðr sin θÞ2 r dθ ¼ 0:62r 3
0
in which the section is assumed to have unit wall thickness. “Cutting” the section at O ðθ S 1:61Sðcos θ 1Þ r sin θr dθ ¼ qb,OA ¼ 3 0:62r 0 r and when θ ¼ π/4, qb,OA ¼ 0.47S/r. Also S qb,AB ¼ 0:62r 3
ðs 0
ðr sÞ sin 45°ds
0:47S r
i.e. qb,AB ¼
Sð1:14rs 0:57s2 þ 0:47Þ r
e168
Solutions Manual Now taking moments about B ðπ
4
Sr ¼ 2
qb, OA rrdθ þ 2
0
2 πr qs, 0 4
Substituting for qb,OA from the above ðπ Sr ¼ 2
4
1:61Sðcos θ 1Þ r dθ þ
0
πr 2 qs,0 2
from which qs,0 ¼ 0:8S=r The complete shear flow distribution is then Sð1:61cos θ 0:81Þ r Sð0:57s2 1:14rs þ 0:33Þ qAB ¼ : r
qOA ¼
S.10.21 The radius BO is an axis of symmetry so that the shear centre will lie on this axis. Therefore we take BO as the z axis and apply an arbitrary shear load, Sy, through the shear centre S as shown in Fig. S.10.21. Further Izy ¼ 0 and Eq. (10.24) reduces to ð Sy s qs ¼ ty ds þ qs, 0 (i) Iz 0
z
A O
Sy
zs s
S θ C
B
FIGURE S.10.21
Iz has been previously calculated in P.10.20 and is 0.62tr3. Referring to Fig. S.10.21 and “cutting” the section at O we note that the qb distribution will be identical to that in P.10.20 and is 1:61Sy ðcos θ 1Þ r
(ii)
Sy ð0:57s2 1:14rs 0:47r 2 Þ r3
(iii)
qb,OA ¼ and qb,AB ¼
Solutions to Chapter 10 Problems
e169
Since the rate of twist of the beam section is zero (shear load is applied through the shear centre) qs,0 is given by Eq. (10.30) in which þ ds ¼ 2r þ ðπr=2Þ ¼ 3:57r and "ð
þ qb ds ¼ 2
π=4
0
1:61Sy ðcos θ 1Þ r dθ þ r
ðr
Sy ð0:57s 2 1:14rs 0:47r 2 Þ ds 3 0r
#
which gives þ qb ds ¼ 1:95Sy Then, from Eq. (10.30) qs,0 ¼
1:95Sy 0:55Sy ¼ 3:57r r
so that qOA ¼
Sy ð1:61cos θ 1:06Þ r
Taking moments about the point B Sy zS ¼ 2
ð π=4
qOA r dθ ¼ 2 2
ð π=4
0
0
Sy ð1:61cos θ 1:06Þ r 2 dθ r
from which zS ¼ 0:61r S.10.22 The shear centre lies on the vertical axis of symmetry therefore apply an horizontal shear load Sz through the shear centre. Since Sy ¼ 0 and Izy ¼ 0 (the y axis is an axis of symmetry) Eq. (10.24) reduces to ð Sz s qs ¼ tz ds þ qs,0 Iy 0 where, referring to Fig. S.10.22 Iy ¼
23 þ2 12
ð3
ðs sin θÞ2 ds
0
assuming that the thickness is unity and s is measured from the point 3. Then Iy ¼ 2:67 m4
e170
Solutions Manual y O
1 s1 s2
2 ys
S
Sz G
z
q s 3
FIGURE S.10.22
Now taking an origin for s at O Sz qb,O1 ¼ 2:67
ð s1 0
s1 ds ¼ 0:19Sz s12
and when s1 ¼ 1 m, qb,O1 ¼ 0.19Sz ð s2 qb,13 ¼ 0:37Sz ð1 s2 sin θÞ ds 0:19Sz ¼ 0:37Sz ð0:5 þ s2 0:17s22 Þ Þ
0
In Eq. (10.30), ds ¼ 8 m. Then h Ð i Ð3 1 2 0:37Sz 0:5 0 s12 ds þ 0 ð0:5 þ s2 0:17s22 Þ ds ¼ 0:44Sz qs,0 ¼ 8 so that qO1 ¼ Sz ð0:44 0:19s12 Þ Take moments about 3 Sz ð3 cos θ yS Þ ¼ 2
ð1
qO1 3 cos θ ds
0
from which yS ¼ 0:7 m: S.10.23 Referring to Fig. P.10.23, the wall DB is 3 m long so that its cross-sectional area, 3 103 8 ¼ 24 103 mm2, is equal to that of the wall EA, 2 103 12 ¼ 24 103 mm2. It follows that the centroid of area of the section lies mid-way between DB and EA on the vertical axis of symmetry. Also since Sy ¼ 500 kN, Sz ¼ 0 and Izy ¼ 0, Eq. (10.24) reduces to ð 500 103 s qs ¼ ty ds þ qs,0 (i) Iz 0
Solutions to Chapter 10 Problems
e171
If the origin for s is taken on the axis of symmetry, say at O, then qs,0 is zero. Also Iz ¼3 103 8 ð0:43 103 Þ2 þ 2 103 12 ð0:43 103 Þ2 þ 2 ð1 103 Þ3 10 sin 2 60°=12 i.e. Iz ¼ 101:25 108 mm4 Eq. (i) then becomes qs ¼ 4:94 105
ðs ty ds 0
In the wall OA, y ¼ 0.43 103 mm. Then ð sA 5 12 0:43 103 ds ¼ 0:25sA qOA ¼ 4:94 10 0
and when sA ¼ 1 10 mm, qOA ¼ 250 N/mm In the wall AB, y ¼ 0.43 103 þ sB cos 30°. Then ð sB qAB ¼ 4:94 105 10ð0:43 103 þ 0:866sB Þ ds þ 250 3
0
i.e. qAB ¼ 0:21sB 2:14 104 sB2 þ 250 When sB ¼ 1 103 mm, qAB ¼ 246 N/mm In the wall BC, y ¼ 0.43 103 mm. Then ð sC qBC ¼ 4:94 105 8 0:43 103 ds þ 246 0
i.e. qBC ¼ 0:17sC þ 246 Note that at C where sC ¼ 1.5 103 mm, qBC should equal zero; the discrepancy, 9 N/mm, is due to rounding off errors. The maximum shear stress will occur in the wall AB (and ED) mid-way along its length (this coincides with the neutral axis of the section) where sB ¼ 500 mm. This gives, from Eq. (ii), qAB(max) ¼ 301.5 N/mm so that the maximum shear stress is equal to 301.5/10 ¼ 30.2 N/mm2.