chapter 16 shear stresses in bending and shear
CHAPTER 16 SHEAR STRESSES IN BENDING AND SHEAR DEFLECTIONS EXERCISE 60, Page 352
1. A beam of length 3 m is simply supported at its ends and subjected to a uniformly distributed load of 200 kN/m, spread over its entire length. If the beam has a uniform cross-section of depth 0.2 m and width 0.1 m, determine the position and value of the maximum shearing stress due to bending. What will be the value of the maximum shear stress at mid-span?
Maximum shearing force occurs at the ends,
wl 200 3 2 2
where
Fmax =
i.e.
Fmax = 300 kN
Fmax 1.5 300 103 1.5 Maximum shearing stress = 12 bd 0.1 0.2
= 22.5 MPa
i.e. At mid-span τ = F = 0
2. Determine the maximum values of shear stress due to bending in the web and flanges of the sections shown when they are subjected to vertical shearing forces of 100 kN.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
I
(a)
0.1 0.123 0.09 0.13 6.9 106 m 4 12 12
Flange
F
0.045 0.01 0.055 100 103 6 0.01 6.9 10
= 35.87 MN/m 2 = 35.87 MPa Web
W
(0.1 0.01 0.055 0.05 0.01 0.025) 100 103 0.01 6.9 106
= 97.83 MN/m 2 = 97.83 MPa (b) Section
a
y
ay
ay2
i
1 2
1 103 1 103
0.105 0.05
1.05 104 5 105
1.03 105 2.5 106
8.33 109 8.33 107
2 103
–
1.55 104
1.353 105
8.416 107
y = 0.0775 m
IXX = 1.437 105 m4 INA = 2.358 106 m4
F
100 103 0.045 0.01 0.0275 0.01 2.358 106
= 52.48 MPa
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
W
100 103 (0.1 0.01 0.0275 0.02252 / 2 0.01) 100 103 0.01 2.358 106
= 127.36 MPa
3. Determine an expression for the maximum shearing stress due to bending for the section shown, assuming that it is subjected to a shearing force of 0.5 MN acting through its centroid and in a perpendicular direction to NA.
I NA
0.1 0.13 2 = 1.667 10 5 m 4 12
b = 0.1 (1 – 10y)
y dA bdy y
0.1
y
0.11 10 y y dy
0.1
0.1 y 2 10 y 3 0.1 3 y 2 20 y 3 = 0.1 y 3 y 6 2
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
=
0.1 0.01 3 y 2 20 y 3 6
0.01 3 y 2 20 y 3 0.5 0.1 τ 5 6 1.667 10 0.11 10 y
=
For τ,
5000 0.01 3 y 2 20 y 3
1 10 y
d 0 3 dy (1 – 10y)[– (6y – 60y2] – (–10) [0.01 – (3y2 – 20y3)] = 0
i.e.
– 6y + 60y2 + 60y2 – 600y3 + 0.1 – 30y2 + 200y3 = 0
i.e.
– 400y3 + 90y2 – 6y + 0.1 = 0
and
+ 400y3 – 90y2 + 6y – 0.1 = 0 = Ψ dΨ 1200y2 – 180y + 6 = dy = 0
y1 0
0.1 0.017 6
y2 0.017
0.022 = 0.024 3.29
y3 0.024
2.31103 = 0.025 2.37
y4 = 0.025 – 0 i.e.
y = ± 0.025 m perpendicular to NA ( ± because it is a symmetrical beam)
5000 8.4375 103 = ± 56.25 MN/m 2 0.75
4. Determine the value of the maximum shear stress for the cross-section shown, assuming that it is subjected to a shearing force of magnitude 0.5 MN acting through its centroid and in a perpendicular direction to NA.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
b = 2R cos θ y = R sin θ dy = R cos θ dθ
y dA b dy ( y) I NA
0.2 0.43 0.14 1.062 103 m 4 12 64
y dA 0.2 0.2 0.1
/2
0
= 4 103 2 R3
/2
0
2 R cos R sin R cos d
cos 2 d (cos ) /2
cos3 = 4 10 2.5 10 3 0 3
4
1 = 4 103 2.5 104 0 3
y dA 3.917 10
Hence,
Maximum shear stress,
Proof of
3
m
3
0.5 3.917 10 3 = 18.44 MPa 0.11.062 10 3
y dA © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
b = 2R cos θ y = R sin θ dy = R cos θ dθ
y dA
/2
0
2R cos R sin R cos d
2
= 2 R3 cos2 sin d 0
2
= 2R3 cos2 d cos 0
/2
cos3 = 2 R 3 0 3
=
2 R3 3
dA
Therefore,
y
R2 2
2 R3 2 4R 2 3 R 3
5. A simply supported beam, with a cross-section as shown, is subjected to a centrally placed concentrated load of 100 MN, acting through its centroid and perpendicular to NA. Determine the values of the vertical shearing stress at intervals of 0.1 m from NA.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
I
113 0.64 0.08333 6.362 103 12 64
I = 0.077 m4
i.e. At y = 0.5, τ = 0 At y = 0.4 m,
0.4
50 1 0.1 0.45 = 29.22 MPa 1 0.077
At y = 0.3 m,
0.3
50 1 0.2 0.4 = 51.95 MPa 1 0.077
b = 2R cos ϕ
y = R sin ϕ
y dA y b dy 2R = At y = 0.2 m,
dy = R cos ϕ d ϕ 3
cos2 d cos
2 R3 cos 3 0 3
ϕ = sin–1 (0.2/0.3) = 41.81º
2 y dA 1 0.3 0.35 3 0.3 cos 41.81 3
3
= 0.105 – 7.454 103 = 0.0975 m 3 b = 0.553 m,
τ = 114.49 MPa
At y = 0.1 m,
ϕ = sin–1 (0.1/0.3) = 19.47º
2 y dA 1 0.4 0.3 3 0.3 cos 19.47 3
3
= 0.12 – 0.0151 = 0.1049 m 3 b = 0.434 m,
τ = 156.95 MPa
At y = 0, ϕ = 0,
2 y dA 1 0.5 0.25 3 0.3 cos 0 3
3
= 0.107 m 3 b = 1 – 2R = 0.4 m,
τ = 173.70 MPa
Summarising, at y = 0, τo = 173.7 MPa, τ0.1 = 156.95 MPa, τ0.2 = 114.49 MPa, τ0.3 = 51.95 MPa, © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
τ0.4 = 29.22 MPa; at y = 0.5, τ0.5 = 0
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
EXERCISE 61, Page 356
1. Determine the position of the shear centre for the thin-walled section shown.
2
I
i.e.
i.e.
t B3 B Bt 2 12 2
I = 0.5833 t B3
F B 0.5 t B 2 F 0.5B 2 F tI 2 tI 0.5833 t B3
0.8572F Bt
0.429 F FF Bt 0.429F Bt
Taking moments about the centre of the web gives: FΔ = 0.429 F
B 2 2
from which, shear centre, Δ = 0.429B
2. Determine the position of the shear centre for the thin-walled section shown. © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
I NA
= t R3
=
from which,
2 0
t R3 2
2 0
t R d R sin
2
1 cos 2 d 2 2
sin 2 2 0
I NA t R3
0
y dA t R d R sin t R 2 cos 0
0
= t R2 cos 1 t R2 1 cos
F 1 cos F t R 2 1 cos 2 3 t R tR 2
F t R 2 d 0
=
F t R2 2 sin 0 tR
F t R2 2 2FR tR
Hence, the shear centre, Δ =
2FR = 2R F
3. Determine the shear centre position for the thin-walled section shown.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
I NA 0.2t 0.22 2
=
t 0.13 2 0.1 t 0.152 2 12
0
2
0.2 cos t 0.2 d
1 cos 2 = 0.016t 1.667 10 4 t 4.5 10 3 t 8 10 3 t d 0 2
= 0.0207 t + 0.01257t i.e.
I NA 0.0333 t 0.2
AB
F t y3 FAB t dy 5 103 y 0.1 I 6 0.1 0.2
=
Ft 3.333 104 3.333 104 I 6.6667 10 4 F t = 0.02 F 0.0333 t
=
F
y
0.1
dy / t y tI
F y2 3 5 10 I 2
BC
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
At B,
τB = 0.015F/I
At C, τc = 0.015F/I + 0.2 i.e.
t 0.2 F tI
4
τc = 0.055F/I F F 1 FBC 0.015 0.055 0.2t I I 2
7 103 F t 7 103 F t = = 0.21 F I 0.0333t CD
=
0.055F F I tI
0.2d t 0.2cos
0.055 F F 0.04 sin 0 I I
FCD
F I
0.055 0.04sin 0.2 d t 0
=
0.2 Ft 0.055 0.04cos 0 I
=
0.2 Ft 0.055 0.04 0 0.04 0.0333t FCD 1.518F
i.e. Taking moments about O gives:
F 0.02F 0.2 0.21F 0.2 2 1.518F 0.2 from which, shear centre position, Δ = 0.396 m
4. Determine the shear centre position for the thin-walled section of shown. © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
INA = 0.016t + 1.667 104 t + 0.1t × 0.252 × 2 + 0.01257t i.e.
INA = 0.0412 t
AB
y dA
y
0.2
t dy y y
y2 t = t y 2 0.09 2 0.3 2
F y dA tI
0.5 F t 2 y 0.09 tI 0.2
FAB
= i.e. At B,
FAB
0.5 Ft y 3 t dy 0.09 y I 3 0.3
0.5Ft 1.53 102 1.8 102 I 3 1.35 10 Ft I
y dA = 0.1t × 0.25= 0.025t © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
At C,
y dA = 0.025t + 0.04t = 0.065t τB = 0.025F/I τC = 0.065F/I FBC = 9 103 Ft/I
CB
y dA 0.065t 0.04t sin
F 0.065 0.04sin I
FCD 0.2 d t 0
i.e.
=
0.2 Ft 0.065 0.04cos 0 I
=
0.2 Ft 0.065 0.04 0 0.04 I
FCD
0.0568Ft I
Taking moments about 0 gives:
F
=
Ft 1.35 103 0.2 2 9 103 0.2 2 0.0568 2 I
Ft 5.4 104 3.6 103 0.1136 0.0412t
from which, shear centre position, Δ = 2.83 m
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
EXERCISE 62, Page 359
1. Determine the shear centre position for the thin-walled closed tube shown which is of uniform thickness.
tan 1 (0.05 / 0.2) 14.04 I 2
0.206
0
t ds1 s1 sin 0.1 t 0.052 2 t R d R cos 2
2
0
0.206
s13 = 0.1176t 3 0
5 10 4 t t R3 cos 2 d 0
1 cos 2 d t = 3.427 10 4 5 10 4 R3 0 2
0.053 sin 2 4 = 8.427 10 t 2 2 i.e.
I 1.039 103 t
AB
s 1
F tI
t ds s sin 1
1
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
s12 F 0.2426 1.039 10 3 t 2
=
= 116.75 F s12 t
B
4.95F t
BC As ‘t’ is uniform,
s 2
=
c
4.95F F t tI
t ds 0.05 s2
2
0
4.95 F 48.12 F s2 t t 9.76F t
CD As ‘t’ is uniform,
9.76 F F t tI
t R d R cos 0
9.76 F FR 2 sin 0 3 t 1.04 10 t i.e.
9.76F F 2.404sin t t
0
ds ds
However,
ds = 0.206 × 2 + 0.1 × 0.2 + π × 0.05 = 0.769 m and
ds 2
0.206
0
0.1 4.9 F 116.8F s12 48.12F s2 ds1 2 ds2 0 t t t
+
0
F 9.76 F 2.4sin 0.05 d t t
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
0.206
233.6 F s13 = 3 t 0 =
=
i.e.
0.1
s2 48.12Fs2 2 0.488F 0.12F cos 9.9 F t t t t 0 0
F 0.681 0.09 0.481 1.533 0.24 t
3.925F t
0
3.925F 1 t 0.769
0
5.104F t
Taking moments about ‘0’ gives:
F R 2
0.206
0
116.8Fs12 0 t ds1 0.3cos tan t +
0.1 4.95F 48.12 Fs2 2 0 t ds2 0.05 0 t t
=
0
F 9.76 F 2 0 t 2.4 t sin 0.05 d t 0.206
or
116.8s13 F 0.6 F 5.104s1 3 0
0.243 0.1
s 2 0.1F 5.104s2 4.95s2 48.12 2 2 0 Hence,
0.052 F 5.1041 9.76 2.4cos 0
= 0.146(– 1.051 + 0.34) + 0.1(– 0.51 + 0.495 + 0.241)
+0.05 2 (– 16.03 + 30.66 + 4.8) = – 0.104 + 0.023 + 0.049 = – 0.032 m i.e. the shear centre position, Δ = – 0.032 m
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
2. Determine the shear centre position for the thin-walled closed tube shown, which is of uniform thickness.
OB
tan 1 (0.05 / 0.2) 14.04 I 2
0.206
0
t ds1 s1 sin
2
ds1 2 0.2 t 0.052 t R d R cos
2
0
0.206
s 3 2 = 0.05886 1 3 3 0
t 1.963 10
4
t
= 3.43 10 4 t 1 10 3 t 1.963 10 4 t i.e.
I 1.539 103 t
F s1 tI
s1
0
78.82Fs12 s1 t
i.e.
and
s1
F s12 s sin t ds sin 1 1 I 2 0
B
3.345F t
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
BC As ‘t’ is constant,
s 2
3.345F F t tI
s2
0
t ds2 0.05
3.345 F 32.49 F s2 t t
=
c
9.84F t
9.84 F F t tI
CD
=
R d t R cos 0
9.84 F 1.62 F sin t t ds
ds
ds 0.206 2 0.2 2 0.05 = 0.969 m
s ds 2
0.206
0
78.82 Fs12 2F ds1 t t +
3.345 32.49s ds 0.2
2
0
F t
2
9.84 1.62sin 0.05 d 0
0.2
32.49s2 2 157.64 F 3 0.206 2 F s1 3.345s2 = 0 t t 2 0 +
=
0
0.05 F 9.84 1.62cos 0 t
F 4.8 F 0.46 2.636 1.71 t t
4.8F 1 t 0.969
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
0
i.e.
4.96F t
Taking moments about 0 gives:
F 0.05 2F
0.2
0
4.96 3.35 32.49s2 ds2 0.05 +
F 4.96 9.84 1.62sin 0.05 2 d 0
or
0.05 0.1 1.61s2 16.2s22 0
0.2
0.0122 4.05 103 cos
= 0.1 0.322 0.65 0.0383 8.1 10
0
3
= 0.0328 + 0.0464 from which, the shear centre position, Δ = 0.0792 m
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
EXERCISE 63, Page 361
1. Determine the maximum deflection due to shear for the simply supported beam shown. It may be assumed that the beam cross-section is rectangular, of constant width b and of constant depth d.
and
F
W 2
3W d 2 y2 3 bd 4
SSE =
2
2G d (vol ) 2
2 9W 2 d 2 = 2 6 y 2 l b dy 2 Gb d 4
i.e.
SSE =
3W 2l 20Gbd
WD =
1 Ws 2
from which, maximum deflection, s
3Wl 10Gbd
2. Determine the maximum deflection due to shear for the simply supported beam shown. It may be assumed that the beam cross-section is rectangular, of constant width b and of constant depth d.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
F1
W l2 l
and F2
W l1 l 2
2 d /2 36W 2l12 d 2 SSE = y 2 bl1 dy 2 6 2 2 0 Gb d l 4 d /2
W2 d 4 y d 2 y3 y5 = 36 2 6 2 b l2 2l1 l12l2 6 5 0 Gb d l 16
W 2 l2 2l1 l12l2 W 2 l1 l2 l1 l2 = 36 = 36 60 b d l 2 60 b d l 2 i.e.
SSE =
3W 2l1 l2 5G b d l
WD =
1 Ws 2
from which, maximum deflection, s
since l1 l2 l
6Wl1l2 5Gbd
3. Determine the maximum deflection due to shear for the simply supported beam shown. It may be assumed that the beam cross-section is rectangular, of constant width b and of constant depth d.
At x,
F
wl l wx w x 2 2
6F d 2 3 y2 bd 4 2 2 l /2 y 36 F 2 d 2 y 2 b dx dy SSE = 2 6 0 0 2 Gb d 4
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
2
=
l /2 l 72bd 5 w2 x dx 2 6 0 Gb d 60 2
6 w2 = 5G b d
l /2
0
l2 2 lx x dx 4 l /2
6w2 l 2 x l x 2 x3 = 5G b d 4 2 3 0 i.e.
and
SSE =
6 w2 l 3 120G b d
WD =
1 2 wl s 2 3
s
3 WD wl
i.e. the maximum deflection, s
3Wl 2 20Gbd
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
1. A beam of length 3 m is simply supported at its ends and subjected to a uniformly distributed load of 200 kN/m, spread over its entire length. If the beam has a uniform cross-section of depth 0.2 m and width 0.1 m, determine the position and value of the maximum shearing stress due to bending. What will be the value of the maximum shear stress at mid-span?
Maximum shearing force occurs at the ends,
wl 200 3 2 2
where
Fmax =
i.e.
Fmax = 300 kN
Fmax 1.5 300 103 1.5 Maximum shearing stress = 12 bd 0.1 0.2
= 22.5 MPa
i.e. At mid-span τ = F = 0
2. Determine the maximum values of shear stress due to bending in the web and flanges of the sections shown when they are subjected to vertical shearing forces of 100 kN.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
I
(a)
0.1 0.123 0.09 0.13 6.9 106 m 4 12 12
Flange
F
0.045 0.01 0.055 100 103 6 0.01 6.9 10
= 35.87 MN/m 2 = 35.87 MPa Web
W
(0.1 0.01 0.055 0.05 0.01 0.025) 100 103 0.01 6.9 106
= 97.83 MN/m 2 = 97.83 MPa (b) Section
a
y
ay
ay2
i
1 2
1 103 1 103
0.105 0.05
1.05 104 5 105
1.03 105 2.5 106
8.33 109 8.33 107
2 103
–
1.55 104
1.353 105
8.416 107
y = 0.0775 m
IXX = 1.437 105 m4 INA = 2.358 106 m4
F
100 103 0.045 0.01 0.0275 0.01 2.358 106
= 52.48 MPa
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
W
100 103 (0.1 0.01 0.0275 0.02252 / 2 0.01) 100 103 0.01 2.358 106
= 127.36 MPa
3. Determine an expression for the maximum shearing stress due to bending for the section shown, assuming that it is subjected to a shearing force of 0.5 MN acting through its centroid and in a perpendicular direction to NA.
I NA
0.1 0.13 2 = 1.667 10 5 m 4 12
b = 0.1 (1 – 10y)
y dA bdy y
0.1
y
0.11 10 y y dy
0.1
0.1 y 2 10 y 3 0.1 3 y 2 20 y 3 = 0.1 y 3 y 6 2
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
=
0.1 0.01 3 y 2 20 y 3 6
0.01 3 y 2 20 y 3 0.5 0.1 τ 5 6 1.667 10 0.11 10 y
=
For τ,
5000 0.01 3 y 2 20 y 3
1 10 y
d 0 3 dy (1 – 10y)[– (6y – 60y2] – (–10) [0.01 – (3y2 – 20y3)] = 0
i.e.
– 6y + 60y2 + 60y2 – 600y3 + 0.1 – 30y2 + 200y3 = 0
i.e.
– 400y3 + 90y2 – 6y + 0.1 = 0
and
+ 400y3 – 90y2 + 6y – 0.1 = 0 = Ψ dΨ 1200y2 – 180y + 6 = dy = 0
y1 0
0.1 0.017 6
y2 0.017
0.022 = 0.024 3.29
y3 0.024
2.31103 = 0.025 2.37
y4 = 0.025 – 0 i.e.
y = ± 0.025 m perpendicular to NA ( ± because it is a symmetrical beam)
5000 8.4375 103 = ± 56.25 MN/m 2 0.75
4. Determine the value of the maximum shear stress for the cross-section shown, assuming that it is subjected to a shearing force of magnitude 0.5 MN acting through its centroid and in a perpendicular direction to NA.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
b = 2R cos θ y = R sin θ dy = R cos θ dθ
y dA b dy ( y) I NA
0.2 0.43 0.14 1.062 103 m 4 12 64
y dA 0.2 0.2 0.1
/2
0
= 4 103 2 R3
/2
0
2 R cos R sin R cos d
cos 2 d (cos ) /2
cos3 = 4 10 2.5 10 3 0 3
4
1 = 4 103 2.5 104 0 3
y dA 3.917 10
Hence,
Maximum shear stress,
Proof of
3
m
3
0.5 3.917 10 3 = 18.44 MPa 0.11.062 10 3
y dA © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
b = 2R cos θ y = R sin θ dy = R cos θ dθ
y dA
/2
0
2R cos R sin R cos d
2
= 2 R3 cos2 sin d 0
2
= 2R3 cos2 d cos 0
/2
cos3 = 2 R 3 0 3
=
2 R3 3
dA
Therefore,
y
R2 2
2 R3 2 4R 2 3 R 3
5. A simply supported beam, with a cross-section as shown, is subjected to a centrally placed concentrated load of 100 MN, acting through its centroid and perpendicular to NA. Determine the values of the vertical shearing stress at intervals of 0.1 m from NA.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
I
113 0.64 0.08333 6.362 103 12 64
I = 0.077 m4
i.e. At y = 0.5, τ = 0 At y = 0.4 m,
0.4
50 1 0.1 0.45 = 29.22 MPa 1 0.077
At y = 0.3 m,
0.3
50 1 0.2 0.4 = 51.95 MPa 1 0.077
b = 2R cos ϕ
y = R sin ϕ
y dA y b dy 2R = At y = 0.2 m,
dy = R cos ϕ d ϕ 3
cos2 d cos
2 R3 cos 3 0 3
ϕ = sin–1 (0.2/0.3) = 41.81º
2 y dA 1 0.3 0.35 3 0.3 cos 41.81 3
3
= 0.105 – 7.454 103 = 0.0975 m 3 b = 0.553 m,
τ = 114.49 MPa
At y = 0.1 m,
ϕ = sin–1 (0.1/0.3) = 19.47º
2 y dA 1 0.4 0.3 3 0.3 cos 19.47 3
3
= 0.12 – 0.0151 = 0.1049 m 3 b = 0.434 m,
τ = 156.95 MPa
At y = 0, ϕ = 0,
2 y dA 1 0.5 0.25 3 0.3 cos 0 3
3
= 0.107 m 3 b = 1 – 2R = 0.4 m,
τ = 173.70 MPa
Summarising, at y = 0, τo = 173.7 MPa, τ0.1 = 156.95 MPa, τ0.2 = 114.49 MPa, τ0.3 = 51.95 MPa, © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
τ0.4 = 29.22 MPa; at y = 0.5, τ0.5 = 0
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
EXERCISE 61, Page 356
1. Determine the position of the shear centre for the thin-walled section shown.
2
I
i.e.
i.e.
t B3 B Bt 2 12 2
I = 0.5833 t B3
F B 0.5 t B 2 F 0.5B 2 F tI 2 tI 0.5833 t B3
0.8572F Bt
0.429 F FF Bt 0.429F Bt
Taking moments about the centre of the web gives: FΔ = 0.429 F
B 2 2
from which, shear centre, Δ = 0.429B
2. Determine the position of the shear centre for the thin-walled section shown. © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
I NA
= t R3
=
from which,
2 0
t R3 2
2 0
t R d R sin
2
1 cos 2 d 2 2
sin 2 2 0
I NA t R3
0
y dA t R d R sin t R 2 cos 0
0
= t R2 cos 1 t R2 1 cos
F 1 cos F t R 2 1 cos 2 3 t R tR 2
F t R 2 d 0
=
F t R2 2 sin 0 tR
F t R2 2 2FR tR
Hence, the shear centre, Δ =
2FR = 2R F
3. Determine the shear centre position for the thin-walled section shown.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
I NA 0.2t 0.22 2
=
t 0.13 2 0.1 t 0.152 2 12
0
2
0.2 cos t 0.2 d
1 cos 2 = 0.016t 1.667 10 4 t 4.5 10 3 t 8 10 3 t d 0 2
= 0.0207 t + 0.01257t i.e.
I NA 0.0333 t 0.2
AB
F t y3 FAB t dy 5 103 y 0.1 I 6 0.1 0.2
=
Ft 3.333 104 3.333 104 I 6.6667 10 4 F t = 0.02 F 0.0333 t
=
F
y
0.1
dy / t y tI
F y2 3 5 10 I 2
BC
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
At B,
τB = 0.015F/I
At C, τc = 0.015F/I + 0.2 i.e.
t 0.2 F tI
4
τc = 0.055F/I F F 1 FBC 0.015 0.055 0.2t I I 2
7 103 F t 7 103 F t = = 0.21 F I 0.0333t CD
=
0.055F F I tI
0.2d t 0.2cos
0.055 F F 0.04 sin 0 I I
FCD
F I
0.055 0.04sin 0.2 d t 0
=
0.2 Ft 0.055 0.04cos 0 I
=
0.2 Ft 0.055 0.04 0 0.04 0.0333t FCD 1.518F
i.e. Taking moments about O gives:
F 0.02F 0.2 0.21F 0.2 2 1.518F 0.2 from which, shear centre position, Δ = 0.396 m
4. Determine the shear centre position for the thin-walled section of shown. © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
INA = 0.016t + 1.667 104 t + 0.1t × 0.252 × 2 + 0.01257t i.e.
INA = 0.0412 t
AB
y dA
y
0.2
t dy y y
y2 t = t y 2 0.09 2 0.3 2
F y dA tI
0.5 F t 2 y 0.09 tI 0.2
FAB
= i.e. At B,
FAB
0.5 Ft y 3 t dy 0.09 y I 3 0.3
0.5Ft 1.53 102 1.8 102 I 3 1.35 10 Ft I
y dA = 0.1t × 0.25= 0.025t © Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
At C,
y dA = 0.025t + 0.04t = 0.065t τB = 0.025F/I τC = 0.065F/I FBC = 9 103 Ft/I
CB
y dA 0.065t 0.04t sin
F 0.065 0.04sin I
FCD 0.2 d t 0
i.e.
=
0.2 Ft 0.065 0.04cos 0 I
=
0.2 Ft 0.065 0.04 0 0.04 I
FCD
0.0568Ft I
Taking moments about 0 gives:
F
=
Ft 1.35 103 0.2 2 9 103 0.2 2 0.0568 2 I
Ft 5.4 104 3.6 103 0.1136 0.0412t
from which, shear centre position, Δ = 2.83 m
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
EXERCISE 62, Page 359
1. Determine the shear centre position for the thin-walled closed tube shown which is of uniform thickness.
tan 1 (0.05 / 0.2) 14.04 I 2
0.206
0
t ds1 s1 sin 0.1 t 0.052 2 t R d R cos 2
2
0
0.206
s13 = 0.1176t 3 0
5 10 4 t t R3 cos 2 d 0
1 cos 2 d t = 3.427 10 4 5 10 4 R3 0 2
0.053 sin 2 4 = 8.427 10 t 2 2 i.e.
I 1.039 103 t
AB
s 1
F tI
t ds s sin 1
1
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
s12 F 0.2426 1.039 10 3 t 2
=
= 116.75 F s12 t
B
4.95F t
BC As ‘t’ is uniform,
s 2
=
c
4.95F F t tI
t ds 0.05 s2
2
0
4.95 F 48.12 F s2 t t 9.76F t
CD As ‘t’ is uniform,
9.76 F F t tI
t R d R cos 0
9.76 F FR 2 sin 0 3 t 1.04 10 t i.e.
9.76F F 2.404sin t t
0
ds ds
However,
ds = 0.206 × 2 + 0.1 × 0.2 + π × 0.05 = 0.769 m and
ds 2
0.206
0
0.1 4.9 F 116.8F s12 48.12F s2 ds1 2 ds2 0 t t t
+
0
F 9.76 F 2.4sin 0.05 d t t
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
0.206
233.6 F s13 = 3 t 0 =
=
i.e.
0.1
s2 48.12Fs2 2 0.488F 0.12F cos 9.9 F t t t t 0 0
F 0.681 0.09 0.481 1.533 0.24 t
3.925F t
0
3.925F 1 t 0.769
0
5.104F t
Taking moments about ‘0’ gives:
F R 2
0.206
0
116.8Fs12 0 t ds1 0.3cos tan t +
0.1 4.95F 48.12 Fs2 2 0 t ds2 0.05 0 t t
=
0
F 9.76 F 2 0 t 2.4 t sin 0.05 d t 0.206
or
116.8s13 F 0.6 F 5.104s1 3 0
0.243 0.1
s 2 0.1F 5.104s2 4.95s2 48.12 2 2 0 Hence,
0.052 F 5.1041 9.76 2.4cos 0
= 0.146(– 1.051 + 0.34) + 0.1(– 0.51 + 0.495 + 0.241)
+0.05 2 (– 16.03 + 30.66 + 4.8) = – 0.104 + 0.023 + 0.049 = – 0.032 m i.e. the shear centre position, Δ = – 0.032 m
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
2. Determine the shear centre position for the thin-walled closed tube shown, which is of uniform thickness.
OB
tan 1 (0.05 / 0.2) 14.04 I 2
0.206
0
t ds1 s1 sin
2
ds1 2 0.2 t 0.052 t R d R cos
2
0
0.206
s 3 2 = 0.05886 1 3 3 0
t 1.963 10
4
t
= 3.43 10 4 t 1 10 3 t 1.963 10 4 t i.e.
I 1.539 103 t
F s1 tI
s1
0
78.82Fs12 s1 t
i.e.
and
s1
F s12 s sin t ds sin 1 1 I 2 0
B
3.345F t
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
BC As ‘t’ is constant,
s 2
3.345F F t tI
s2
0
t ds2 0.05
3.345 F 32.49 F s2 t t
=
c
9.84F t
9.84 F F t tI
CD
=
R d t R cos 0
9.84 F 1.62 F sin t t ds
ds
ds 0.206 2 0.2 2 0.05 = 0.969 m
s ds 2
0.206
0
78.82 Fs12 2F ds1 t t +
3.345 32.49s ds 0.2
2
0
F t
2
9.84 1.62sin 0.05 d 0
0.2
32.49s2 2 157.64 F 3 0.206 2 F s1 3.345s2 = 0 t t 2 0 +
=
0
0.05 F 9.84 1.62cos 0 t
F 4.8 F 0.46 2.636 1.71 t t
4.8F 1 t 0.969
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
0
i.e.
4.96F t
Taking moments about 0 gives:
F 0.05 2F
0.2
0
4.96 3.35 32.49s2 ds2 0.05 +
F 4.96 9.84 1.62sin 0.05 2 d 0
or
0.05 0.1 1.61s2 16.2s22 0
0.2
0.0122 4.05 103 cos
= 0.1 0.322 0.65 0.0383 8.1 10
0
3
= 0.0328 + 0.0464 from which, the shear centre position, Δ = 0.0792 m
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
EXERCISE 63, Page 361
1. Determine the maximum deflection due to shear for the simply supported beam shown. It may be assumed that the beam cross-section is rectangular, of constant width b and of constant depth d.
and
F
W 2
3W d 2 y2 3 bd 4
SSE =
2
2G d (vol ) 2
2 9W 2 d 2 = 2 6 y 2 l b dy 2 Gb d 4
i.e.
SSE =
3W 2l 20Gbd
WD =
1 Ws 2
from which, maximum deflection, s
3Wl 10Gbd
2. Determine the maximum deflection due to shear for the simply supported beam shown. It may be assumed that the beam cross-section is rectangular, of constant width b and of constant depth d.
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
F1
W l2 l
and F2
W l1 l 2
2 d /2 36W 2l12 d 2 SSE = y 2 bl1 dy 2 6 2 2 0 Gb d l 4 d /2
W2 d 4 y d 2 y3 y5 = 36 2 6 2 b l2 2l1 l12l2 6 5 0 Gb d l 16
W 2 l2 2l1 l12l2 W 2 l1 l2 l1 l2 = 36 = 36 60 b d l 2 60 b d l 2 i.e.
SSE =
3W 2l1 l2 5G b d l
WD =
1 Ws 2
from which, maximum deflection, s
since l1 l2 l
6Wl1l2 5Gbd
3. Determine the maximum deflection due to shear for the simply supported beam shown. It may be assumed that the beam cross-section is rectangular, of constant width b and of constant depth d.
At x,
F
wl l wx w x 2 2
6F d 2 3 y2 bd 4 2 2 l /2 y 36 F 2 d 2 y 2 b dx dy SSE = 2 6 0 0 2 Gb d 4
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
2
=
l /2 l 72bd 5 w2 x dx 2 6 0 Gb d 60 2
6 w2 = 5G b d
l /2
0
l2 2 lx x dx 4 l /2
6w2 l 2 x l x 2 x3 = 5G b d 4 2 3 0 i.e.
and
SSE =
6 w2 l 3 120G b d
WD =
1 2 wl s 2 3
s
3 WD wl
i.e. the maximum deflection, s
3Wl 2 20Gbd
© Carl Ross, John Bird & Andrew Little Published by Taylor and Francis
