i
i
i
i
Chapter 6
6.1
EIδ max = −EIv|x=L/2
" #
4
3
w0
L
L
L
3
= −
2L
−
−L
24
2
2
2
(b)
(a) Utilize symmetry of displacements about midspan.
Note that due to symmetry v ′ = 0 at x = L/2.
δ max =
Left half (0 ≤ x ≤ L/2)
M=
Px
2
EIv ′ =
P x2
+ C1
4
EIv =
5w0 L4
↓ ◭
384EI
6.3
P x3
+ C1 x + C2
12
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2
P (L/2)
+ C1 = 0
4
2. v ′ |x=L/2 = 0
C1 = −
PL
16
w = w0
2
P 4(L − x)3 − 3L2 (L − x) ◭
48
" #
3
P L3
L
P
2L
4
=
− 3L
↓◭
δ max = −v|x=L/2 = −
48EI
2
2
48EI
Boundary conditions:
1. v|x=0 = 0
′
2. v |x=0 = 0
6.2
EIv =
(a)
M = M0
C1 = 0
w0 x2
−20L3 + 10L2 x − x3 ◭
120L
EIv ′ = M0 x + C1
EIv =
1
M0 x2 + C1 x + C2
2
Boundary conditions:
1. v|x=0 = 0
Boundary conditions:
2. v|x=L = 0
C2 = 0
6.4
x w
w0 L
0
=
Lx − x2
x − w0 x
M =
2
2
2
2
3
w
Lx
x
0
EIv ′ =
+ C1
−
2
2
3
w0 Lx3
x4
EIv =
+ C1 x + C2
−
2
6
12
1. v|x=0 = 0
1
w0 x2
wx =
2
2L
M = −
Right half (L/2 ≤ x ≤ L)
Replace x by L − x in the last expression:
(b)
R=
w0 L2
w0 L
w0 x2 x +
x−
3
2
2L
3
w0
−2L3 + 3L2 x − x3
=
6L
w0
3L2 x2
x4
′
3
EIv =
−2L x +
+ C1
−
6L
2
4
L2 x3
x5
w0
−L3 x2 +
+ C1 x + C2
−
EIv =
6L
2
20
P
4x3 − 3L2 x ◭
EIv =
48
EIv =
x
L
C2 = 0
1
M0 L
M0 L2 + C1 L
C1 = −
2
2
By symmetry, δ max occurs at x = L/2 ◭
"
2
#
1 1
L
M0 L L
δ max = −v|x=L/2 = −
−
M0
EI 2
2
2
2
2. v|x=L = 0
C2 = 0
L4
w0 L4
+ C1 L = 0
−
2
6
12
C1 = −
w0 L3
24
=
w0
2Lx3 − x4 − L3 x ◭
EIv =
24
M0 L2
↓ ◭
8EI
129
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
6.5
6.7
M0
x
L
M =
EIv ′ =
M0 2
x + C1
2L
M = 180x − 30x2 N · m
EIv ′ = 90x2 − 10x3 + C1 N · m2
M0 3
x + C1 x + C2
6L
Boundary conditions:
EIv =
1. v|x=0 = 0
EIv = 30x3 − 2.5x4 + C1 x + C2 N · m3
Boundary conditions:
C2 = 0
M0 3
L + C1 L = 0
6L
δ max occurs where v ′ = 0:
2. v|x=L = 0
M0 2 M0 L
x −
=0
2L
6
δ max = −v|x=L/√3 = −
1
EI
"
C1 = −
1. v|x=0 = 0
M0 L
6
2. v|x=10 m = 0
30(10)3 − 2.5(10)4 + C1 (10) = 0
L
x= √ ◭
3
M0
6L
L
√
3
3
−
M0 L L
√
6
3
C1 = −500 N · m2
EIv|x=5 m = 30(5)3 − 2.5(5)4 − 500(5)
#
= −312.5 N · m3
EIδ mid = −EIv|x=5 m = 313 N · m3 ↓ ◭
2
=
C2 = 0
M0 L
√
↓ ◭
9 3EI
6.8
6.6
EIv ′′ = M = −9.6 + 4.5x kN · m
For segment AB:
M = −
M0
x
L
EIv ′ = −
EIv ′ = −9.6x + 2.25x2 + C1 kN · m2
M0 2
x + C1
2L
EIv = −4.8x2 + 0.75x3 + C1 x + C2 kN · m3
Boundary condiions:
1. v|x=0 = 0
C2 = 0
2. v|x=3.6 m = 0
M0 3
x + C1 x + C2
6L
Boundary conditions:
EIv = −
1. v|x=0 = 0
C2 = 0
2. v|x=L/2 = 0
M0
−
6L
′
−4.8(3.6)2 + 0.75(3.6)3 + C1 (3.6) = 0
3
L
L
+ C1 = 0
2
2
δ max occurs where v = 0:
−
M0 2 M0 L
x +
=0
2L
24
x=
C1 = 7.56 kN · m2
M0 L
C1 =
24
∴ EIv ′ = −9.6x + 2.25x2 + 7.56 kN · m2
EIv = −4.8x + 0.75x2 + 7.56 x kN · m3
√
3L
◭
6
δ max occurs where v ′ = 0:
−9.6x + 2.25x2 + 7.56 = 0
EIδ max = EIv|x=√3L/6
M0
= −
6L
√ !3
3L
M0 L
+
6
24
x = 1.0420 m or x = 3.225 m
√ !
3L
6
At x = 1.0420 m:
√
3M0 L2
( ↑ for segment AB) ◭
=
216
EIv =
−4.8(1.0420) + 0.75(1.0420)2 + 7.56 (1.0420)
= 3.514 kN · m3
130
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
At x = 3.225 m:
6.10
−4.8(3.225) + 0.75(3.225)2 + 7.56 (3.225)
EIv =
= −0.385 kN · m3
I=
∴ δ max =
Displacements are symmetric about the midspan.
Therefore, v ′ = 0 at x = 0.
For right half of beam:
w0 L L
w0 L 2 L
w0 L2
M0 =
−
=
4
2
4
32
24
bh3
160(120)3
=
= 23.04 × 106 mm4
12
12
3.514 × 103
3.514 × 103
=
EI
(12 × 109 )(23.04 × 10−6 )
= 0.01271 m = 12.71 mm (at x = 1.042 m) ◭
w = w0
2x
L
R=
1
w0 x2
wx =
2
L
w0
w0 x2 x w0 L2
=
L3 − 8x3
−
24
L
3
24L
w0
′
3
4
EIv =
L x − 2x + C1
24L
2x5
w0 L3 x2
+ C1 x + C2
−
EIv =
24L
2
5
6.9
M =
(a)
w = w0
x
L
R=
1
w0 x2
w0 x =
2
2L
Boundary conditions:
w0 L
w0 x2 x w0 2
M =
=
L x − x3
x−
6
2L
3
6L
w0 L2 x2
x4
EIv ′ =
+ C1
−
6L
2
4
x5
w0 L2 x3
+ C1 x + C2
−
EIv =
6L
6
20
1. v ′ |x=0 = 0
C1 = 0
w0 L3 (L/2)2
2(L/2)5
2. v|x=L/2 = 0
+ C2 = 0
−
24L
2
5
w0 9L5
3w0 L4
+ C2 = 0
C2 = −
24L 80
640
C2
3w0 L4
δ max = −v|x=0 = −
=
↓ ◭
EI
640EI
Boundary conditions:
1. v|x=0 = 0
C2 = 0
w0 L5
L5
′
+ C1 L = 0
2. v |x=L = 0
−
6L 6
20
7w0 L3
C1 = −
360
2 3
1 w0 L x
7w0 L3
x5
v =
−
−
x
EI 6L
6
20
360
=
w0 x
−3x4 + 10L2 x2 − 7L
360EIL
4
6.11
(a) The displacements are symmetric about midspan, so
that we have to consider the only left half of the beam.
Note that v ′ = 0 at x = L/2.
0≤x≤a
Px
P x2
EIv ′
+ C1
23
Px
EIv
+ C1 x + C2
6
Boundary conditions:
M
◭
(b) δ max occurs where v ′ = 0:
w0 L2 x2
7w0 L3
x4
−
−
= 0
6L
2
4
360
15x4 − 30L2 x2 + 7L4 = 0
x = 0.519L ◭
1. v|x=0 = 0
C2 = 0
2. v ′ |x=L/2 = 0
Pa
a ≤ x ≤ a + L/2
Pa
P ax + C3
P ax2
+ C3 x + C4
2
L
+ C3 = 0
2
C3 = −
P aL
2
131
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
4. v|x=a− = v|x=a+
3
a
P a2
a3
P a3
P
=−
−
a + C4
C4 =
6
2
2
6
1
(C3 L + C4 )
δ max = −v|x=L = −
EI
P a2
P a2
P a3
1
−
=
L+
(3L − a) ↓ ◭
=−
EI
2
6
6EI
P a2
P aL
+ C1 = P a2 −
2
2
3. v ′ |x=a− = v ′ |x=a+
C1 =
Pa
(a − L)
2
4. v|x=a− = v|x=a+
P a2
P a3
P a2 L
P a3
+
(a − L) =
−
+ C4
6
2
2
2
P a3
C4 =
6
6.13
δ max = −v|x=L/2
P aL L P a3
1 P a(L/2)2
+ −
+
= −
EI
2
2
2
6
Pa
=
3L2 − 4a2 ↓ ◭
24EI
Because the displacement is symmetric about the midspan,
we consider only the left half of the beam. Due to
symmetry we have v ′ = 0 at x = L/2.
0≤x≤a
(b) For a W200×100 section (see Appendix B-7):
I = 113 × 106 mm4
M
S = 990 × 103
EIv ′
6000(6000)[3(20000)2 − 4(6000)2 ]
24(200 × 103 )(113 × 106 )
= 70 mm ◭
δ max =
EIv
1. v|x=0 = 0
Mmax = P a = 6000(6000) = 36 × 106 N · mm
36 × 10
Mmax
= 36.4 MPa ◭
=
S
990 × 103
6.12
w0 b3
w0 b(L/2)2
−
+ C3 = 0
2
6
w0 b
C3 = −
3L2 − 4b2
24
w0 ba2
w0 ba2
+ C1 =
+ C3
2
2
w0 b
C1 = C3 = −
3L2 − 4b2
24
3. v ′ |x=a− = v ′ |x=a+
M
EIv
C2 = 0
2. v ′ |x=L/2 = 0
6
EIv
w0 bx2
+ C1
2
w0 bx3
+ C1 x + C2
6
Boundary conditions:
Mmax occurs in a ≤ x ≤ L − a:
σ max =
w0 bx
a ≤ x ≤ L/2
w0 (x − a)2
w0 bx −
2
3
w0 bx2
w0 (x − a)
−
+ C3
2 3
6
4
w0 (x − a)
w0 bx
−
6
24
+ C3 x + C4
′
0≤x≤a
P x− P a = P
(x − a)
x2
P
− ax + C1
23
ax2
x
+ C1 x + C2
−
P
6
2
a≤x≤L
0
4. v|x=a− = v|x=a+
C3
w0 ba3
w0 ba3
+ C1 a =
+ C3 a + C4
6
6
C3 x + C4
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2. v ′ |x=0 = 0
C1 = 0
3. v ′ |x=a− = v ′ |x=a+
2
a
− a2 = C3
P
2
C4 = 0
EIδ max = −EIv|x=L/2
w0 b(L/2)3
w0 b4
w0 b
2
2 L
= −
3L − 4b
−
+−
6
24
24
2
C3 = −
= −
P a2
2
=
w0 b 3
L − 2b3 − 3L3 + 4b2 L
48
w0 b 3
(L − 2b2 L + b3 ) ↓ ◭
24
132
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
(b) For a W200×22.5 section (see Appendix B-2):
I = 20 × 10
−6
4
m
S = 193 × 10
−6
(b) For δ = L/360 = 8/360 = 22.22 × 10−3 m we must
have
3
m
I=
3.6 × 103 (2)
(83 − 2(2)2 (8) + 23 )
δ max =
24(200 × 109 )(20 × 10−6 )
= 0.0342 m = 34.2 mm ◭
6.15
Mmax occurs at midspan (x = 4 m):
22
Mmax = (3.6 × 103 ) (2)(4) −
= 21.6 × 103 N · m
2
M
Mmax
21.6 × 103
= 111.9 × 106 Pa ◭
=
S
193 × 10−6
σ max =
16 000
1600
=
= 7.20×10−6 m4 ◭
Eδ
(10 × 109 )(22.22 × 10−3 )
EIv ′
EIv
6.14
0 ≤ x ≤ 2b
x
−P
2
x2
−P
+ C1
4
−P
2b ≤ x ≤ 3b
x3
+ C1 x + C2
12
−P (3b − x)
x2
−P 3bx −
+ C3
2 3
2
x
3bx
−
−P
2
6
+ C3 x + C4
Boundary conditions:
1. v|x=0 = 0
(a)
M
(N · m)
EIv ′
(N · m2 )
EIv
(N · m3 )
0≤x≤4m
2
1800x − 300x
900x2 − 100x3
+ C1
300x3 − 25x4
+ C1 x + C2
P b2
(2b)3
+ C1 (2b) = 0
C1 =
2. v|x=2b− = 0
−P
12
3
3. v ′ |x=2b− = v ′ |x=2b+
P b2
(2b)2
(2b)2
+ C3
+
= −P 3b(2b) −
−P
4
3
2
4m≤x≤8m
−600(x − 8)
−300(x − 8)2
+ C3
−100(x − 8)3
+ C3 x + C4
C3 =
10P b2
3
4. v|x=2b+ = 0
10P b2
3b(2b)2 (2b)3
+
−
(2b) + C4 = 0
−P
2
6
3
Boundary conditions:
1. v|x=0 = 0
C2 = 0
C2 = 0
C4 = −2P b3
2. v ′ |x=4 m− = v ′ |x=4 m+
900(4)2 − 100(4)3 + C1 = −300(−4)2 + C3
(a) Under the load δ = −v|x=3b
P 3b(3b)2
(3b)3
∴δ = −
−
EI
2
6
2
C1 − C3 = −12 800 N · m
3. v|x=4 m− = v|4 m+
3
4
+
3
300(4) − 25(4) + C1 (4) = −100(−4) + C3 (4) + C4
4 (C1 − C3 ) − C4 = −6400
(b)
π(22.5/2)4
πR4
=
= 112580.6 mm4
4
4
Under the load:
I=
4(−12 800) − C4 = −6400
C4 = −44 800 N · m3
δ=
4. v|x=8 m = 0
C3 (8) + (−44 800) = 0
10P b2
2P b3
P b3
(3b) −
=−
↓ ◭
3EI
EI
EI
C3 = 5600 N · m2
150(720)3
= 22.3 mm ◭
(200 × 103 ) (12580.6)
Mmax occurs at the right support (x = 2b):
EIδ = −EIv|x=4m
= − −100(−4)3 + 5600(4) + (−44 800)
Mmax = P b = 150(720) = 108000 N · mm
σ max =
= 16 000 N · m3 ↓ ◭
(108000)( 22.5
Mmax R
2 )
=
= 96.6 MPa ◭
I
12580.6
133
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
6.16
1. v ′ |x=0 = 0
C1 = 0
2. v|x=0 = 0
= 1.6 × 103 kN · m2
0≤x≤6m
6 m ≤ x ≤ 10 m
M (kN·m)
−2x
−2x + 20
EIv ′ (kN·m2 ) −x2 + C1
−x2 + 20x + C3
3
x
x3
EIv (kN·m3 ) − + C1 x + C2 − + 10x2 + C3 x + C4
3
3
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2. v ′ |x=6 m− = v ′ |x=6 m+
−(6)2 + C1 = −(6)2 + 20(6) + C3
C3 = −
5w0 a4
C4 =
8
1
δ max = −v|x=2a = −
[C3 (2a) + C4 ]
EI
7w0 a3
41w0 a4
5w0 a4
1
−
=
(2a) +
↓ ◭
= −
EI
6
8
24EI
(a)
6.18
Because the displacements are symmetric about the
midspan, we consider only the right half of the beam. Due
to symmetry we have v ′ = 0 at x = 0.
(c)
Solution of Eqs. (a)-(c) is
2
2
C1 = 17.33 kN · m C3 = −102.67 kN · m C4 = 360 kN · m
M
(kN · m)
EIv ′
(kN · m2 )
EIv
(kN · m3 )
3
At the point where the couple acts:
−(6)2 + (17.33)
= −1.167 × 10−2 ◭
v ′ |x=6 m =
1.6 × 103
v|x=6 m =
−
7w0 a3
6
4. v|x=a− = v|x=a+
3a3 a3
w0
7w0 a3
w0 a −
= − (−a)4 −
+
a + C4
4
6
24
6
3. v|x=6 m− = v|x=6 m+
(6)3
(6)3
−
+ C1 (6) = −
+ 10(6)2 + C3 (6) + C4
3
3
6 (C1 − C3 ) − C4 = 360 kN · m3
(b)
4. v|x=10 m = 0
3
(10)
−
+ 10(10)2 + C3 (10) + C4 = 0
3
10C3 + C4 = −666.7
′
3. v |x=a− = v |x=a+
3a2 a2
w0
w0 a −
= − (−a)3 + C3
+
2
2
6
EI = (200 × 109 )(8 × 10−6 ) = 1.6 × 106 N · m2
C1 − C3 = 120 kN · m2
C2 = 0
′
(6)3
+ (17.33) (6)
3
= 0.0200 m = 20.0 mm ◭
1.6 × 103
0≤x≤1m
1m≤x≤5m
2.4
2.4 − 0.15(x − 1)2
2.4x + C1
2.4x − 0.05(x − 1)3 + C3
1.2x2 + C1 x
+C2
1.2x2 − 0.0125(x − 1)4
+ C3 x + C4
Boundary conditions:
1. v ′ |x=0 = 0
′
C1 = 0
′
2. v |x=1 m− = v |x=1 m+
6.17
2.4(1) = 2.4(1.0) − 0.05(1 − 1)3 + C3
C3 = 0
M
EIv ′
EIv
0≤x
≤ a
3a
w0 a − + x
2
3ax x2
+
w0 a −
2
2
+ C1
3ax2
x3
w0 a −
+
4
6
+ C1 x + C2
3. v|x=5 m = 0
a ≤ x ≤ 2a
w0
− (x − 2a)2
2
w0
− (x − 2a)3
6
+ C3
w0
− (x − 2a)4
24
+ C3 x + C4
1.2(5)2 − 0.0125(5 − 1)4 + C4 = 0
C4 = −26.8 kN · m3
4. v|x=1 m− = v|x=1 m+
1.2(1)2 + C2 = 1.2(1)2 − 0.0125(1 − 1)4 − 26.8
C2 = −26.8 kN · m3
EIδ max = −EIv|x=0 = −C2 = 26.8 kN · m3 ↓ ◭
134
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
6.19
1. v ′ |x=0 = 0
0 ≤ x ≤ 900 mm
π
EI
200 × 103 64
(50)4
10
2
(N · mm ) = 6.14 × 10
M
−2100x
(N · mm)
EIv ′
−1050x2 + C1
(N · mm2 )
EIv
−350x3 + C1 x + C2
(N · mm3 )
Boundary conditions:
2. v ′ |x=a− = v ′ |x=a+
900 mm ≤ x ≤ 2400 mm
4
200 × 103 π(75)
64
= 31 × 1010
a3
1
w0
3
3a −
E(3I/2) 2
3
1
a2
=
w0 a 2a2 −
+ C3
EI
2
−2100x
−1050x2 + C3
C3 = −
−350x3 + C3 x + C4
C3 = 6.048 × 109 N · mm2
2. v|x=2400 mm = 0 −350(2400)3 +(6048×109)(2400)+C4
C4 = −9.68 × 1012 N · mm3
′
′
4. v|x=a− = v|x=a+
3. v |
=v|
−1050(900)2 + 6.048 × 109
−1050(900)2 + C1
=
6.14 × 1010
31 × 1010
x=900 mm−
x=900 mm+
w0 3 4 a4
1
+ C2
a −
E(3I/2) 2 2
12
3
1
a
11w0 a4
13w0 a4
3
=
w0 a a −
−
−
EI
6
18
9
C1 = 1.029 × 109 N · mm2
, Solution is :
C1 = 1.029 × 109 N · mm2 = 1.029 × 103 N · m2
4. v|x=900 mm− = v|x=900 mm+
−350(900)3 + 1.88 × 109 (900) + C2
6.14 × 1010
−350(900)3 + 6.048 × 109 (900) − 9.68 × 1012
=
31 × 1010
C2 = −
Moment of inertia varies as I = I0
0≤x≤a
3
1
2
2
2 w0 a − 2 w0 x w0
2
2
=
3a − x
2
x3
w0
2
3a x −
2
3
+ C
1
w0 3 2 2 x4
a x −
2 2
12
+ C1 x + C2
x
L
M = −P x
Px
PL
M
=−
=
v ′′ = −
EI
EI0 (x/L)
EI0
For the right half of the beam:
EIv
C2
61w0 a4
61w0 a4
=
=
↓ ◭
E(3I/2)
24E(3I/2)
36EI
6.21
6.20
EIv ′
61w0 a4
24
δ mid = −v|x=0 =
C2 = −2.33 × 1012 N · mm3
2.33 × 1012
C2
= 37.9 mm ↓ ◭
=
δ max = −v|x=0 = −
EI
6.14 × 1010
M
11w0 a3
18
3. v|x=2a = 0
(2a)3
11w0 a3
w0 a a(2a)2 −
−
(2a) + C4 = 0
6
18
13w0 a4
C4 = −
9
− 1050(2400)2 + C3 = 0
1. v ′ |x=2400 mm = 0
C1 = 0
a ≤ x ≤ 2a
w0 a (2a − x)
x2
w0 a 2ax −
2
+ C3
x3
w0 a ax2 −
6
+ C3 x + C4
v′ = −
PL
x + C1
EI0
v = −
P L x2
+ C1 x + C2
EI0 2
Boundary conditions:
1. v ′ |x=L = 0
−
PL
L + C1 = 0
EI0
C1 =
P L2
EI0
135
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary condition: v ′ |x=0 = 0
P L L2
P L2
−
L + C2 = 0
+
EI0 2
EI0
2. v|x=L = 0
C2 = −
−
P L3
2EI0
P L3
↓ ◭
δ = −v|x=0 = −C2 =
2EI0
w0
6
3
L
−
+ C1 = 0
2
EIv ′ |x=L = C1 = −
C1 = −
w0 L3
48
w0 L3
48
EIθ|x=L = −EIv ′ |x=L =
w0 L3
◭
48
6.22
6.24
M = −4x − 2 hx − 2i kN · m
2
EIv ′ = −2x2 − hx − 2i + C1 kN · m2
2
1
3
EIv = − x3 − hx − 2i + C1 x + C2 kN · m3
3
3
M =
Boundary conditions:
1. v ′ |x=4 m = 0
− 2(4)2 − (4 − 2)2 + C1 = 0
EIv ′ =
470 2
2
2
x − 50 hx − 2i − 40 hx − 6i + C1 kN · m2
9
EIv =
40
470 3 50
3
3
x −
hx − 2i −
hx − 6i
27
3
3
C1 = 36.0 kN · m2
2. v|x=4 m = 0
940
x − 100 hx − 2i − 80 hx − 6i kN · m
9
+ C1 x + C2 kN · m3
1
2
− (4)3 − (4 − 2)3 + 36.0(4) + C2 = 0
3
3
Boundary conditions:
1. v|x=0 = 0
C2 = −98.67 kN · m3
C2 = 0
470
50
40
(9)3 − (7)3 − (3)3 + C1 (9) = 0
27
3
3
2. v|x=9 m = 0
3
EIδ max = −EIv|x=0 = −C2 = 98.67 kN · m
C1 = −734.8 kN · m2
3
δ max =
98.67 × 10
= 0.01 m (given)
(10 × 109 )(0.05h3 /12)
EIv|x=4.5 m =
h = 0.619 m = 619 mm ◭
50
470
(4.5)3 − (2.5)3 − 734.8(4.5)
27
3
= −1981 kN · m3
6.23
EIδ mid = −EIv|x=4.5 m = 1981 kN · m3 ↓ ◭
6.25
2
L
w0 L
w0 x2
w0
w0 L2
x−
+
x−
+
8
2
2
2
2
2
2
w0
L
L
w0
= −
x−
x−
+
2
2
2
2
3
3
w0
L
L
w0
EIv ′ = −
x−
x−
+
+ C1
6
2
6
2
M = −
2
2
M = 600x − 120 hx − 4i + 120 hx − 10i kN · m
EIv ′ = 300x2 − 40 hx − 4i3 + 40 hx − 10i3 + C1 kN · m2
4
4
EIv = 100x3 − 10 hx − 4i + 10 hx − 10i
+ C1 x + C2 kN · m3
136
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
1. v|x=0 = 0
6.27
C2 = 0
2. v|x=12 m = 0
100(12)3−10(8)4 +10(2)4 +C1 (12) = 0
C1 = −11 000 kN · m
(a)
2
M = 220x − 1000 hx − 6i + 1980 hx − 10i
2
−200 hx − 10i N · m
EIv|x=6 m = 100(6)4 − 10(2)3 − 11 000(6)
2
= −44 600 kN · m3
EIδ mid = −EIv|x=6 m = 44 600 kN · m3 ↓ ◭
6.26
1. v|x=0 = 0
C2 = 0
(a)
2. v|x=10 m = 0
0
M0
L
x + M0 x −
L
3
M0 2
L
′
EIv = −
+ C1
x + M0 x −
2L
3
M = −
M0 3 M0
x +
EIv = −
6L
2
500 3
110 3
(10 ) −
(4) + C1 (10) = 0
3
3
C1 = −2600 N · m2
EIv =
2
L
+ C1 x + C2
x−
3
110 3 500
3
3
x −
hx − 6i + 330 hx − 10i
3
3
−
50
4
hx − 10i − 2600x N · m3 ◭
3
(b) At right end
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2
L
M0 3 M0
L−
+ C1 L = 0
L +
2. v|x=L = 0
−
6L
2
3
1
C1 = − M0 L
18
#
"
2
L
Lx
1
x3
x−
◭
−
+
EIv = M0 −
6L 2
3
18
EIδ = −EIv|x=13 m = −
500 3
110
(13)3 +
(7) − 330(3)3
3
3
50 4
(3) + 2600(13)
3
= 2850 N · m3 ↓ ◭
+
6.28
(b) At point of application of M0
(L/3)3
L(L/3)
EIδ = −EIv|x=L/3 = M0
+
6L
18
=
2
EIv ′ = 110x2 − 500 hx − 6i + 990 hx − 10i
200
3
−
hx − 10i + C1 N · m2
3
110 3 500
3
3
EIv =
x −
hx − 6i + 330 hx − 10i
3
3
50
4
− hx − 10i + C1 x + C2 N · m3
3
Boundary conditions:
(a)
2
M0 L2 ↓ ◭
81
M = −300x + 2700 hx − 3i0 − 150 hx − 6i2
+ 2100 hx − 9i N · m
EIv ′ = −150x2 + 2700 hx − 3i − 50 hx − 6i3
2
+ 1050 hx − 9i + C1 N · m2
25
EIv = −50x3 + 1350 hx − 3i2 −
hx − 6i4
2
+ 350 hx − 9i3 + C1 x + C2 N · m3
137
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
1. v|x=0 = 0
6.30
C2 = 0
2. v|x=9 m = 0 − 50(9)3 + 1350(6)2 −
25 4
(3) + C1 (9) = 0
2
C1 = −1238 N · m2
2
EIv = −50x3 + 1350 hx − 3i −
(a)
M = 8x − 4 hx − 3i − 6 hx − 6i − 8 hx − 9i kN · m
25
4
hx − 6i
2
EIv ′ = 4x2 − 2 hx − 3i2 − 3 hx − 6i2 − 4 hx − 9i2
+ C1 kN · m2
4
2
4
EIv = x3 − hx − 3i3 − hx − 6i3 − hx − 9i3
3
3
3
3
+ C1 x + C2 kN · m
3
+ 350 hx − 9i − 1238x N · m3 ◭
(b) Midway between supports: δ = −v|x=4.5 m
∴ EIδ = 50(4.5)3 − 1350(4.5 − 3)2 + 1238(4.5)
= 7090 N · m3 ↓ ◭
Boundary conditions:
1. v|x=0 = 0
6.29
2. v|x=12 m = 0
M =
=
2w0 x
L
w0 L
x−
4
w2 =
4w0 (x − L/2)
L
1
x
1
hx − L/2i
w1 x
+ w2 hx − L/2i
2
3
2
3
EIv =
w0 3 2w0
w0 L
3
x−
x +
hx − L/2i
4
3L
3L
w0 L 2
w0 4 w0
x −
x +
hx − L/2i4 + C1
8
12L
6L
EIv =
w0 5
w0
w0 L 3
x −
x +
hx − L/2i5 + C1 x + C2
24
60L
30L
2. v|x=L = 0
1
1
(1/2)5
−
+
24 60
30
5
3
w0 L
C1 = −
192
w0 L4
4 3 2
4
3
3
3
x − hx − 3i − hx − 6i − hx − 9i
3
3
3
−130.5x kN · m3 ◭
(b) At midspan (x = 6 m):
EIv ′ =
Boundary conditions:
1. v|x=0 = 0
C2 = 0
4
2
4
(12)3 − (9)3 − (6)3 − (3)3
3
3
3
+ C1 (12) = 0
C1 = −130.5 kN · m3
(a)
w1 =
C2 = 0
4
2
3
3
EIδ = −EIv|x=6 m = − (6) + (3) + 130.5 (6)
3
3
= 513 kN · m3 ↓ ◭
6.31
+ C1 L = 0
2
M = −15x + 39 hx − 3i − 3 hx − 3i kN · m
w0 L 3
w0 5
w0
5w0 L3
5
EIv =
x −
x +
hx − L/2i −
x ◭
24
60L
30L
192
EIv ′ = −7.5x2 + 19.5 hx − 3i2 − hx − 3i3 + C1 kN · m2
EIv = −2.5x3 + 6.5 hx − 3i3 − 0.25 hx − 3i4
(b) Maximum displacement occurs at midspan
+ C1 x + C2 kN · m3
EIδ max = −EIv|x=L/2
(1/2)3 (1/2)5
w0 L4
5 (1/2)
4
= w0 L −
=
+
+
◭
24
60
192
120
138
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
6.33
1. v|x=3 m = 0
−2.5(3)3 + C1 (3) + C2 = 0
(a)
M = −W x + W hx − ai + W hx − 3ai −
2. v|x=8 m = 0
W 2 W
W
x +
hx − ai2 +
hx − 3ai2
2
2
2
W
hx − 3ai3 + C1
−
12a
W
W
W
EIv = − x3 +
hx − ai3 +
hx − 3ai3
6
6
6
W
hx − 3ai4 + C1 x + C2
−
48a
Boundary conditions:
EIv ′ = −
−2.5(8)3 + 6.5(5)3 − 0.25(5)4 + C1 (8) + C2 = 0 (b)
Solving Eqs. (a) and (b) yields
C1 = 111.25 kN · m2
C2 = −266.25 kN · m3
Under 15-kN load (at x = 0):
EIδ = −EIv|x=0 = −C2 = 266 kN · m3 ◭
1. v|x=a = 0
6.32
W 3
a + C1 a + C2 = 0
6
W
W
3
− (3a)3 +
(2a) + C1 (3a)+ C2 = 0
6
6
−
2. v|x=3a = 0
b
M = P 1−
x − 2P hx − ai
2a
b
hx − 2ai
+P 2+
2a
P
b
2
′
EIv =
1−
x2 − P hx − ai
2
2a
b
P
2
2+
hx − 2ai + C1
+
2
2a
P
b
P
EIv =
1−
x3 − hx − ai3
6
2a
3
b
P
3
2+
hx − 2ai + C1 x + C2
+
6
2a
C1 =
2. v|x=2a = 0
Eδ = −Ev|x=0 = −
C2
4W a3
=
↓ ◭
EI
3EI
M = 8x − 24 hx − 3i + 16 hx − 6i kN · m
2
2
EIv ′ = 4x2 − 12 hx − 3i + 8 hx − 6i + C1 kN · m2
8
4 3
x − 4 hx − 3i3 + hx − 6i3 + C1 x + C2 kN · m3
3
3
Boundary conditions:
EIv =
1. v|x=0 = 0
2. v|x=6 m = 0
Pa
(3a − 2b)
6
C2 = 0
4 3
(6) − 4(3)3 + C1 (6) = 0
3
C1 = −30.0 kN · m2
(a) Under 24-kN load:
4
EIδ = −EIv|x=3 m = − (3)3 + 30(3) = 54.0 kN · m3 ↓ ◭
3
(b) δ max occurs in 0 < x < 3 m where v ′ = 0:
√
4x2 − 30 = 0
x = 7.5 = 2.739 m
Midway between supports (x = a):
b
Pa
P
1−
a3 +
EIδ = −EIv|x=a = −
(3a − 2b)a
6
2a
6
δ =
4
C2 = − W a3
3
6.34
C2 = 0
b
P
P
1−
(2a)3 − a3 + C1 (2a) = 0
6
2a
3
C1 = −
3
W a2
2
At left end (x = 0):
Boundary conditions:
1. v|x=0 = 0
W
hx − 3ai2
4a
4
EIδ max = −EIv|x=2.739 m = − (2.739)3 + 30(2.739)
3
P a2
(4a − 3b) ↓ ◭
12EI
= 54.8 kN · m3 ↓ ◭
139
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
6.35
6.37
2
L
x−
2
3
L
w0
w0
x−
+ C1
EIv ′ = − x3 +
6
6
2
4
L
w0 4 w0
x−
+ C1 x + C2
EIv = − x +
24
24
2
0
M = −250x + 2000 hx − 2i N · m
M = −
EIv ′ = −125x2 + 2000 hx − 2i + C1 N · m2
EIv = −
125 3
2
x + 1000 hx − 2i + C1 x + C2 N · m3
3
Boundary conditions:
− 125(4)2 + 2000(2) + C1 = 0
1. v ′ |x=4 m = 0
Boundary conditions:
C1 = −2000 N · m2
2. v|x=4 m = 0
−
At left end (x = 0):
w0 2 w0
x +
2
2
′
1. v |x=L = 0
125 3
(4) +1000(2)2 −2000(4)+C2 = 0
3
w0 3 w0
−
L +
6
6
C1 =
C2 = 6667 N · m3
3
EIδ = EIv|x=0 = C2 = 6670 N · m ↑ ◭
2. v|x=L = 0
w0
w0
− L4 +
24
24
7
w0 L3
48
4
L
7
+ w0 L3 (L)+C2 = 0
2
48
C2 = −
6.36
δ max = −v|x=0 = −
M = Rx − P hx − 6i + P hx − 7i
C2
41w0 L4
=
↓ ◭
EI
384EI
M0 = (2 × 3) (10.5) + (2 × 12) (6) − (2 × 6) (9)
= 99 kN · m
Boundary conditions:
2. v ′ |x=12 m = 0
41
w0 L4
384
6.38
R 2 P
P
2
2
EIv ′ =
x − hx − 6i + hx − 7i + C1
2
2
2
P
R 3 P
3
3
x − hx − 6i + hx − 7i + C1 x − C2
EIv =
6
6
6
1. v|x=0 = 0
3
L
+ C1 = 0
2
C2 = 0
2
R
P
P
(12)2 − (6)2 + (5)2 + C1 = 0
2
2
2
C1 =
1
1
3
EIv ′ = −99x + 9x2 − x3 + hx − 6i
3
3
1
− hx − 9i3 + C1 kN · m2
3
1
1
4
hx − 6i
EIv = −49.5x2 + 3x3 − x4 +
12
12
1
4
− hx − 9i + C1 x + C2 kN · m3
12
11
P − 72R
2
3. v|x=12 m = 0
R
P
P
(12)3 − (6)3 + (5)3 +
6
6
6
2
M = −99 + 18x − x2 + hx − 6i − hx − 9i kN · m
11
P − 72R (12) = 0
2
R = 0.08825P = 0.08825(2000) = 176.5 N ◭
140
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
1. v ′ |x=0 = 0
C1 = 0
2. v|x=0 = 0
C2 = 0
6.40
At right end (x = 12 m):
2
2
M = 960x − 200 hx − 2i + 200 hx − 6i N · m
EIδ = −EIv|x=12 m
200
200
3
3
hx − 2i +
hx − 6i + C1 N · m2
3
3
50
50
4
4
hx − 2i +
hx − 6i + C1 x
EIv = 160x3 −
3
3
EIv ′ = 480x2 −
1
1
1
= 49.5(12)2 − 3(12)3 + (12)4 − (6)4 + (3)4
12
12
12
= 3571 kN · m3 ↓ ◭
+ C2 N · m3
6.39
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2. v|x=10 m = 0
50
50
160(10)3 − (8)4 + (4)4 + C1 (10) = 0
3
3
M2 − M1
M2 − M1
hx − ai −
hx − a − Li
L
L
M2 − M1
2
EIv ′ = M1 x +
hx − ai
2L
M2 − M1
2
−
hx − a − Li + C1
2L
M1 2 M2 − M1
3
x +
hx − ai
EIv =
2
6L
M2 − M1
3
−
hx − a − Li + C1 x + C2
6L
M = M1 +
C1 = −9600 N · m2
At midspan:
EIδ = −EIv|x=5 m = −160(5)3 +
50 4
(3) + 9600 (5)
3
= 29.4 × 103 N · m3 = 29.4 kN · m3 ↓ ◭
6.41
Boundary conditions:
1. v|x=a = 0
M1 2
a + C1 a + C2 = 0
2
2
M = 240x − 50x2 + 50 hx − 4i + 400 hx − 8i N · m
2. v|x=a+L = 0
EIv ′ = 120x2 −
M2 − M1
M1
(a + L)2 +
(L)3 + C1 (a + L) + C2 = 0
2
6L
+ C1 N · m2
50
50
200
4
3
EIv = 40x3 − x4 +
hx − 4i −
hx − 8i
12
12
3
Subtracting first equation from second:
M2 − M1 2
M1
(2aL + L2 ) +
L + C1 L = 0
2
6
1
1
C1 = − M1 a + M1 L + M2 L
3
6
+ C1 x + C2 N · m3
Boundary conditions:
1
1
EIθ1 = −EIv|x=0 = −EIC1 = M1 a + M1 L + M2 L ◭
3
6
M2 − M1
(a + L)2
2L
1
1
M2 − M1 2
a − M1 a + M1 L + M2 L
−
2L
3
6
EIθ2 = EIv|x=2a+L = M1 (2a + L) +
=
1
1
M1 L + M2 a + M2 L
6
3
50 3 50
3
2
x +
hx − 4i − 200 hx − 8i
3
3
1. v|x=0 = 0
C2 = 0
2. vx=8 m = 0
40(8)3 −
At right support:
50 4 50 4
(8) + (4) + C1 (8) = 0
12
12
C1 = −560 N · m2
EIθ = −EIv ′ |x=8 m
= −120(8)2 +
50 3 50 3
(8) − (4) + 560
3
3
= 347 N · m2 ◭
◭
141
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
6.42
1. v|x=0 = 0
C2 = 0
10(12)3 −
2. v|x=12 m = 0
x − L/2
w = w0
L/2
2
1
L
w0
L
R = w x−
=
x−
2
2
L
2
3
2
L
w0 L
w0
5w0 L
x−
+
x−
M = −
24
4
3L
2
4
L
5w0 L2
w0 L 2
w0
′
x−
EIv = −
+ C1
x+
x −
24
8
12L
2
5
L
w0
5w0 L2 2 w0 L 3
x−
x +
x −
EIv = −
48
24
60L
2
C1 = 288 N · m2
At right end:
EIδ = −EIv|x=18 m
= −10(18)3 +
1
1
(18)5 − 170(6)3 − (6)5
12
12
−5(6)4 − 288(18)
= 50.1 × 103 N · m3 ↓ ◭
6.44
14400 N
y
4500 N
1500 N/m
2400 N/m
x
MA
+ C1 x + C2
6m
3m
Boundary conditions:
1. v ′ |x=0 = 0
1
(12)5 + C1 (12) = 0
12
R2
C1 = 0
R1
x− 6
150 9 N/m
240 N/m
30780 N . m
2. v|x=0 = 0
C2 = 0
δ max = −v|x=L
"
5 #
1
5w0 L2 2 w0 L 3
w0 L
= −
−
L +
L −
EI
48
24
60L 2
M
2835 N
6m
x− 6
RA = 14400 + 4500 = 18900 N
MA = 4500(7) + 14400(6) = 117900 N · m
1
x−3
R1 =
1500
(x − 3) = 125(x − 3)2 N
2
6
w0L4
121w0 L4
= 0.0630
↓ ◭
=
1920EI
EI
R2 = 2400(x − 3)
6.43
hx − 3i
hx − 3i
− R2
3
2
2400
125
(x − 3)3 −
(x − 3)2
= −117900 + 18900x −
3
3
M = −MA − RA x − R1
EIv ′ = −117900x + 9450x2 − 10.42(x − 3)4 − 400(x − 3)3
+C1 N · m2
5
5
3
M = 60x − x3 + 1020 hx − 12i + hx − 12i
3
3
EIv = −58950x2 + 3150x3 − 2.084(x − 3)5
−100(x − 3)4 N · m3
+ 60 hx − 12i2 N · m
EIv ′ = 30x2 −
Bounday conditions v ′ |x=0 = v|x=0 = 0 yield C1 = C2 = 0
5 4
5
2
4
x + 510 hx − 12i +
hx − 12i
12
12
δ max = −v|x=9 m
+ 20 hx − 12i3 + C1 N · m2
EIv = 10x3 −
=
1
1 5
3
5
x + 170 hx − 12i +
hx − 12i
12
12
58950(9)2 − 3150(9)3 + 2.084(9 − 3)5 + 100(9 − 3)4
(200 × 109 )(126 × 10−6 )
= 0.1041 m
= 104.1 mm ◭
4
+ 5 hx − 12i + C1 x + C2 N · m3
142
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
6.47
6.45
EItC/A = area of M diag.|C
A · x̄/C
1
4
3
1
− (2 × 2)(2) − (2 × 1)
= − (8 × 2) 1 +
2
3
3
4
= −27.17 kN · m3 = 27.17 kN · m3 ↓
I=
EItD/A = area of M diag.|D
A · x̄/D
1
20
28
1
− (420 × 14)
= (1020 × 10) 4 +
2
3
2
3
1
8
− (600 × 4) 10 +
2
3
0.05(0.150)3
bh3
=
= 14.063 × 10−6 m4
12
12
δ C = tC/A =
27.17 × 103
= 0.0280 m
(69 × 109 ) (14.063 × 10−6 )
= 28.0 mm ↓ ◭
6.48
= 11 760 N · m3
EItC/A = area of M diag.|C
A · x̄/C
=
1
20
10
(1020 × 10) − (120 × 10)
2
3
2
1
20 1
8
− (300 × 10) − (600 × 4) 6 +
2
3
2
3
EItA/C = area of M diag.|C
A · x̄/A
1
3
1
× 8 + (5400 × 6)(6)
= − (6400 × 8)
3
4
2
= 7600 N · m3
tC/A
7600
= 760 N · m2
EIθA = EI
=
10
AC
EIδ D = EI tD/A − θA AD = 11 760 − 760(14)
= −5200 N · m3 = 5200 N · m3 ↓
δ A = tA/C =
5200 × 109
= 65 mm ↓ ◭
(10 × 103 )(8 × 106 )
6.49
= 1120 N · m3 ↑ ◭
6.46
EItB/C =
=
EIδ C = EItC/A = area of M diag.|C
A · x̄/C
1
1
2
× 8 − (6 × 24)(2 + 4)
= (8 × 32)
2
3
2
1
(3000 × 6)(4) − (400 × 6)(3)
2
1
− (3600 × 6)(4.5)
3
= −3600 N · m3 = 3600 N · m3 ↓
= 251 kN · m3 ↑ ◭
δ B = tB/C =
3600 × 109
= 45 mm ↓ ◭
(10 × 103 )(8 × 106 )
143
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
6.52
6.50
(a)EItA/B = area of M diag.|B
A · x̄/A = 0:
2L 1
1
(P L × L)
−
2
3
4
P =
w0 L2
×L
6
4L
=0
5
δ B = θC BC − tB/C =
1
w0 L ◭
10
AC
BC − tB/C
EItA/C = area of M diag.|C
A · x̄/A
=
(b)
1
1
(640 × 6)(2) − (640 × 4)(1.0)
2
3
= 2987 N · m3
EIθA = area of M diag.|B
A
1
1 w0 L2
= (P L × L) −
×L
2
4
6
1 1
1 w0 L2
=
w0 L2 × L −
×L
2 10
4
6
=
tA/C
EItB/C = area of M diag.|C
B · x̄/B
2
1
(213.3 × 2) = 142.20 N · m3
2
3
2987
EIδ B =
(2) − 142.20 = 853 N · m3 ↓ ◭
6
=
1
w0 L3 ◭
120
6.53
6.51
A
C
12 kN
M0
B
A
3m
6 +M0 /6 kN
M (kN . m)
C
3m
36 +M0
0
(a)
−M0
δ B = θA AB − tB/A =
−36 − M0
tC/A = 0
tD/A
AD
AB − tB/A
EItD/A = area of M diag.|D
A · x̄/D
area of M -diagram|C
A · x̄/C = 0
=
1
1
(36 + M0 )(6)(2) − M0 (6)(3) − (36)(3)(1.0)
2
2
M0 = 13.50 kN · m ◭
0 =
1
8 1
5 1
1
(1400 × 8) − (1000 × 5) − (400 × 1.0)
2
3 2
3 2
3
= 10 700 N · m3
EItB/A = area of M diag.|B
A · x̄/B
1
(525 × 3)(1.0) = 787.5 N · m3
2
10 700
(3) − 787.5 = 3225 N · m3 ↓ ◭
EIδ B =
8
=
144
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
(b)
δ C = θA AC − tC/A =
tD/A
AD
6.55
AC − tC/A
EItC/A = area of M diag.|C
A · x̄/C
=
1
7 1
4
(1225 × 7) − (800 × 4) = 7871 N · m3
2
3 2
3
EIδ C =
10 700
(7) − 7871 = 1492 N · m3 ↓ ◭
8
6.54
δ C = θA AC − tC/A =
tD/A
AD
AC − tC/A
EItD/A = area of M diag.|D
A · x̄/D
=
Due to skew-symmetry, δ B = 0. Consider segment AB
only.
EItB/A
M0 L
=
L/2
24
x M x3
x
1
0
M0 × x
=
=
2
L
3
6L
=
6.56
A
0.6L
θΑ
δΒ
B
tB/A
√
3L
6
0.6L
area of M diag.|D
A · x̄/D
1
2
= 1530 N · m3
2655
(4.5) − 1530 = 461 N · m3 ↓ ◭
EIδ C =
6
Let δ max occur at D. Thus θD = 0, or
EItD/A =
4.5
3
EItC/A = area of M diag.|C
A · x̄/C
1
1
4.5
=
− (600 × 3) (1.0)
(4.5 × 720)
2
3
2
EIθA =
b=
= 2655 N · m3
EItB/A = area of M diag.|B
A · x̄/B
M0 L2
L
L
1 M0
=
×
=
2
2
2
6
48
EIθ A + area of M diag.|D
A · x̄/D = 0
b
M0 L 1
M0 × b = 0
−
24
2
L
1
1
(960 × 6) (2) − (900 × 4.5)
2
2
1.5
−(60 × 1.5)
2
M0
0.4L
C
tC/A
0.4L
M0/L
b
M0 b3
b
M0 × b
=
L
3
6L
M
M0 L
M0 b3
b−
EIδ D = EI θA x − tD/A =
24
6L
√ !
√ !3
√
M0 L
3L
3L
3M0 L2
M0
=
−
=
24
6
6L
6
216
√
3M0 L2
↑ ◭
δ max = δ D =
216EI
0.6M0
M0
M0
EItC/A = area of M -diagram|C
A · x̄/C
=
1
L
(M0 L) − M0 (0.4L)(0.2L) = 0.08667M0L2
2
3
EItB/A = area of M -diagram|B
A · x̄/B
=
1
(0.6M0 ) (0.6L)(0.2L) = 0.036M0L2
2
145
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
δ B = θA (0.6L) − tB/A =
tC/A
(0.6L) − tB/A
L
EIδ B = 0.08667M0L(0.6L) − 0.036M0L
6.59
2
= 0.01600M0L2 ↓ ◭
6.57
EItA/C = area of M diag.|C
A · x̄/A = 0
0 =
EItA/B = area of M -diagram|B
A · x̄/B
1
1
3
4
× 3 − (1.8 × 3)
×3
= − (3.6 × 3)
3
4
4
5
P = 2700 N ◭
= −11.340 kN · m3
6.60
0.075(0.15)2
bh2
=
= 281.3 × 10−6 m3
S =
6
6
δ A = tA/B =
1
8
1
(4P − 2400) (6) (4) − (4P × 4) 2 +
2
2
3
11.340 × 103
= 4.03 × 10−3 m
(10 × 109 )(281.3 × 10−6 )
= 4.03 mm ◭
6.58
EItB/A = area of M diag.|B
A · x̄/B
=
Consider the left half only, with E being the midpoint of
the beam.
1
(7.2 × 3)(1) − (6.4 × 3)
2
3
1
− (1.35 × 3)
4
5
3
2
= −18.608 kN · m3 = 18.608 kN · m3 ↓
EItA/E = area of M diag.|E
A · x̄/A
1
4
8
1
+ (12 000 × 2) 2 +
= − (24000 × 4)
2
3
2
3
δ B = tB/A =
18.608 × 103
(10 × 109 )(30 × 10−6 )
= 62.0 × 10−3 m = 62.0 mm ◭
= −88 000 N · m3 = 88 000 N · m3 ↓
EItB/E = area of M diag.|E
B · x̄/C = − (12 000 × 2) (1)
= −24 000 N · m3 = 24 000 N · m3 ↓
EIδ A = EI tA/E − tB/E = 88 000 − 24 000
= 64000 N · m3 ↓ ◭
146
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
6.61
6.63
(a) Due to zero slope at C we have
δ B = θA AB − tB/A =
EItA/C = area of M diag.|C
A · x̄/A = 0
1
20 1
[(600 − 3P ) × 10]
− (600 × 6) (8)
2
3
2
P = 56.0 N ◭
0 =
AC
AB − tB/A
EItC/A = area of M diag.|C
A · x̄/C
1
4
1
8
− (1600 × 4) 4 +
= (3200 × 8)
2
3
3
4
(b)
EIδ B = EItB/C = area of M diag.|C
B · x̄/B
=
tC/A
− (1600 × 4)(2)
1
− [(4800 − 1600) × 4]
2
1
[(432 − 172.8) × 6] (4) + (172.8 × 6)(3)
2
1
− (600 × 6)(4)
2
4
3
= 2133 N · m3
= −979 N · m3 = 979 N · m3 ↓ ◭
EItB/A = area of M diag.|B
A · x̄/B
4
1
= (1600 × 4)
2
3
6.62
1
− (1600 × 4) (1)
3
= 2133 N · m3
EIδ B =
2133
(4) − 2133
8
= −1067 N · m3 = 1067 N · m3 ↑ ◭
δ D = θC CD + tD/C =
tA/C
AC
CD + tD/C
6.64
EItA/C = area of M diag.|C
A · x̄/A
=
1
20 1
(360 × 10) − (600 × 6)(8)
2
3
2
= −2400 N · m3 = 2400 N · m3 ↓
EItD/C = area of M diag.|D
C · x̄/D
1
= − (240 × 3)(2) = −720 N · m3 = 720 N · m3 ↓
2
2400
EIδ D =
(3) + 720 = 1440 N · m3 ↓ ◭
10
δ A = θB AB + tA/B =
tC/B
BC
AB + tA/B
147
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
EItC/B =
=
tC/A = area of M/EI diag.|C
A · x̄/C
M0
L 2L
1
×
=
2 4EI0
4 34
1
M0
L 1 3L
3L
−
×
+
2 4EI0
4
4
3 4
area of M diag.|C
B · x̄/C
1
3
(1950 × 3)(1.0) − (600 × 3)
2
2
3
1
− [(1950 − 600) × 3]
4
5
= −382.5 N · m3 = 382.5 N · m3 ↓
= −
EItA/B = area of M diag.|B
A · x̄/A
= − (600 × 1.0)
EIδ A =
M0 L2
M0 L2
=
↓
24EI0
24EI0
tB/A = area of M/EI diag.|B
A · x̄/B
M0
3L 1 3L
1
×
= −
2 4EI0
4 3 4
1
= −300 N · m3 = 300 N · m3 ↓
2
382.5
(1.0) + 300 = 428 N · m3 ↓ ◭
3
= −
6.65
δB =
3 M0 L2
3 M0 L2
=
↓
128 EI0
128 EI0
M0 L2
3 M0 L2
M0 L2 3L/4
=
↑ ◭
−
24EI0 L
128 EI0
128EI0
6.67
δ max = δ A = tA/C = area of M/EI diag.|C
A · x̄/A
3b
3b
w0 b2
1 w0 b2
×b
×b
−
= −
3 2EI0
4
4EI0
2
3w0 b2
w0 b2
5b
1
−
−
×b
2
4EI0
4EI0
3
= −
EItC/A = area of M diag.|C
A · x̄/C
1
1
12
12
− (14 400 × 12)
= (16 800 × 12)
2
3
3
4
6
1
− (2400 × 6)
4
5
11 w0 b4
11 w0 b4
=
↓ ◭
12 EI0
12 EI0
6.66
= 226 100 N · m3
EItC/A
226 100
EIθA =
= 18 842 N · m2
=
12
AC
EItB/A = area of M diag.|B
A · x̄/B
1
6
6
1
− (3600 × 6)
= (8400 × 6)
2
3
3
4
= 39 600 N · m3
δ B = θ A AB − tB/A =
tC/A
AC
EIδ B = EIθA AB − EItB/A = 18 842(6) − 39600
= 73 500 N · m3 ◭
AB − tB/A
148
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
6.68
6.70
Due to the 100 N load alone (a = 2 m, b = 7 m):
Due to dist. load alone:
P bx 2
L − x2 − b2
EIδ B = EIδ|x=2 m =
6L
100(7)(2) 2
9 − 22 − 72 = 725.9 N · m3
=
6(9)
EIδ|x=2.5 m =
Due to conc. load alone (a = 1 m, b = 4 m, L = 5 m):
Pb L
3
3
2
2
EIδ|x=2.5 m =
(x − a) + L − b x − x
6L b
2000(4) 5
3
2
2
3
(2.5 − 1) + (5 − 4 )2.5 − 2.5
=
6(5)
4
Pb L
(x − a)3 + L2 − b2 x − x3
6L b
100(7) 9
(6 − 2)3 + (92 − 72 )6 − 63
=
6(9) 7
EIδ C = EIδ|x=6 m =
= 2958 N · m3
= 755.6 N · m3
Due to both loads acting simultaneously:
Due to the 80 N load alone (a = 6 m, b = 3 m):
δ|x=2.5 m =
P bx 2
EIδ B = EIδ|x=2m
L − x2 − b2
6L
80(3)(2) 2
=
9 − 22 − 32 = 604.4 N · m3
6(9)
EIδ C
5(1000)(5)4
5w0 L4
=
= 8138 N · m3
384
384
8138 + 2958
(10 × 109 )(20 × 10−6 )
= 0.0555 m = 55.5 mm ↓ ◭
6.71
P bx 2
L − x2 − b2
= EIδ|x=6 m =
6L
80(3)(6) 2
9 − 62 − 32 = 960.0 N · m3
=
6(9)
Due to loading on left side:
Due to both loads acting simultaneously:
δ x=L/2 =
EIδ B = 725.9 + 604.4 = 1330 N · m3 ↓ ◭
w0 a2
(3L2 − 2a2 )
96EI
Loading on right side contributes the same amount. Thus
the effect of both loads is
EIδ C = 755.6 + 960.0 = 1716 N · m3 ↓ ◭
δ x=L/2 = 2
6.69
w0 a2
w0 a2
(3L2 − 2a2 ) =
(3L2 − 2a2 ) ↓ ◭
96EI
48EI
6.72
EIδ center =
X Pb
48
where b is the shorter segment.
EIδ center =
(3L2 − 4b2 )
100(4) 3(10)2 − 4(4)2
48
100(2) 3(10)2 − 4(2)2
+
48
= 3150 N · m3 ◭
149
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
From segment BD (a = 3 m, L = 6 m):
6.75
M0 L w0 a2
−
(a − 2L)2
EIθ B =
3
24L
=
900(6) 200(3)2
−
[3 − 2(6)]2 = 787.5 N · m2
3
24(6)
Due to the 500-N load alone (a = 2 m, b = 3 m):
P bx 2
L − x2 − b2
6L
500(3)(2) 2
=
5 − 22 − 32 = 1200 N · m3
6(5)
EIδ B = EIδ|x=2 m =
From segment AB (L = 3 m):
EIδ ′A =
200(3)4
w0 L4
=
= 2025 N · m3
8
8
EIδ C
′
EIδ A = EI θB AB + δ A = 787.5(3) + 2025
= 4390 N · m3 ↓ ◭
Pb L
3
3
2
2
= EIδ|x=4 m =
(x − a) + L − b x − x
6L b
500(3) 5
(4 − 2)3 + (52 − 32 )4 − 43
=
6(5) 3
= 666.7 N · m3
6.73
Due to the 800-N load alone (a = 4 m, b = 1 m):
EIδ center =
X Pb
48
where b is the shorter segment.
EIδ center =
P bx 2
L − x2 − b2
6L
800(1)(2) 2
=
5 − 22 − 12 = 1066.7 N · m3
6(5)
EIδ B = EIδ|x=2 m =
(3L2 − 4b2 )
80(3) 100(2) 3(9)2 − 4(2)2 +
3(9)2 − 4(3)2
48
48
P bx 2
L − x2 − b2
6L
800(1)(4) 2
=
5 − 42 − 12 = 853.3 N · m3
6(5)
EIδ C = EIδ|x=4 m =
3
= 1981 kN · m ◭
6.74
Due to both loads acting simultaneously:
EIδ B = 1200 + 1066.7 = 2270 N · m3 ◭
EIδ C = 666.7 + 853.3 = 1520 N · m3 ◭
I=
bh3
120(240)3
=
= 138.24 × 106 mm4
12
12
EIδ center =
=
5w0 L4
−2
384
M0 L2
16
6.76
E = 10 GPa
EIδ center =
X Pb
48
where b is the shorter segment.
5(2)(10)4
(2P + 4)(10)2
−
384
8
EIδ center =
= 210.4 − 25P kN · m3
= (210.4 − 25P ) × 1012 N · mm3
(3L2 − 4b2 )
600(2) 400(1) 3(5)2 − 4(1)2 +
3(5)2 − 4(2)2
48
48
= 2067 N · m3 ◭
0.05h3
9
10 × 10
(0.02) = 2067
12
(10 × 103 )(138.24 × 106 )(15) = (210.4 − 25P ) × 1012
P = 7.6 MPa ◭
h = 0.1354 m = 135.4 mm ◭
150
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
6.77
6.80
P ab
M0 L
(2L − b) −
6L
3
1000(6)(4)
1800(10)
=
(2(10) − 4) −
6(10)
3
EIθC =
= 400 kN · m2
δ A = δ ′A + θB AB
From segment BC (L = 15 m):
◭
EIθ B =
6.78
M0 L w0 L3
27000(15) 400(153)
−
=
−
3
24
3
24
= 78.75 × 103 N · m2
From segment AB (L = 9 m):
EIδ ′A =
P L3
3000(9)3
=
= 729.0 × 103 N · m3
3
3
EIδ A = 729.0×103 +78.75×103(9) = 1438×103 N·m3 ↓ ◭
From segment AB:
EIθB =
6.81
M0 L
w0 L3
−
24
3
w0
A
140(6)3 2P (6)
−
= 1260 − 4P N · m2
=
24
3
w0
From segment BC:
EIδ ′A =
EIδ A = EI
B
δ C = δ ′ − θB a
P L3
P (2)3
=
= 2.667P N · m3
3
3
θB BC − δ ′A
B
12 m
w0a2
2
θB
θB =
w0 L3
(w0 a2 /2)L
w0 L 2
−
=
(L − 4a2 )
24EI
3EI
24EI
δ′ =
w0 a4
8EI
δC =
w0 L 2
w0 a4
−
(L − 4a2 )a = 0
8EI
24EI
0 = (1260 − 4P )(2) − 2.667P
P = 236 N ◭
a
C
δ'
0 = 3a3 + 4La2 − L3
0 = 3a3 + 4(12)a2 − 123
6.79
0 = 3a3 + 48a2 − 1728
Refer to figure in solution of Prob. 6.78.
EIθ B =
a = 5.21 m ◭
M0 L
w0 L3
−
24
3
140(6)3 2P (6)
−
=0
24
3
P = 315 N ◭
=
151
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
δ = δ1 − δ2 + δ3
6.82
π
π
I = (R4 − r4 ) = (0.084 − 0.074 ) = 13.313 × 10−6 m4
4
4
9
EI = (200 × 10 ) 13.313 × 10−6 = 2.663 × 106 N · m2
6.85
3
w0 a3
60(4)3
(4L − a) =
(4(8) − 4)
24
24
EIδ = (30.72 − 14.04 + 4.48)×103 = 21.2 ×103 kN·m3 ↓ ◭
Due to the triangular loading:
M0
4
(15 × 10 )(3)
w0L
= 0.006510
= 0.006510
EI
2.663 × 106
M0
L
A
B
The maximum dislacement occurs at the center of the
beam. The contribution of each couple to this
displacement is
M0 L2
δ center =
16EI
Hence the maximum displacement is
= 2.970 × 10−3 m
θA =
60(6)3
w0 a3
(4L − a) =
[4(8) − 6]
24
24
= 4.48 × 103 kN · m3
w0 L3
(15 × 103 )(3)3
θA =
=
= 6.337 × 10−3 rad
24EI
24(2.663 × 106 )
δC
EIδ 2 =
EIδ 3 =
5w0 L4
5(15 × 103 )(3)4
=
=
= 5.941 × 10−3 m
384EI
384(2.663 × 106 )
4
w0 L4
60(8)4
=
= 30.72 × 103 kN · m3
8
8
= 14.04 × 103 kN · m3
Due to the uniform loading:
δC
EIδ 1 =
w0 L3
(15 × 103 )(3)3
=
= 3.380 × 10−3 rad
45EI
45(2.663 × 106 )
Due to both loadings acting simulaneously:
δ C = (5.941 + 2.970) × 10−3 = 8.91 × 10−3 m = 8.91 mm ◭
δ max = 2δcenter =
θA = (6.337 + 3.380) × 10−3 = 9.72 × 10−3 rad = 0.557◦ ◭
M0 L2
◭
8EI
6.86
6.83
EIδ =
P L3
M0 L2
6000(12)3 8000(12)2
−
=
−
48
16
48
16
= 144.0 × 103 N · m3 ↓ ◭
′
δ C = δ B + θB (3) + δ C
From segment AB (L = 3 m):
6.84
EIδ B =
M0 L2
P L3
+
3
2
3
=
EIθ B =
2
1800 (3)
900 (3)
+
= 16 200 N · m3
3
2
P L2
+ M0 L
2
2
=
900 (3)
+ 1800 (3) = 9450 N · m2
2
152
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Due to the load at C:
From segment BC (L = 3 m):
′
EIδ C =
w0 L4
w0 L4
−
8
30
4
=
600(3)4
600 (3)
−
= 4455 N · m3
8
30
EIδ C =
P L3
1500(16000)3
=
= 2.048 × 1015 N · mm3
3
3
EIθC =
1500(16000)2
P L2
=
= 1.92 × 1011 N · mm2
2
2
Due to both loads acting simultaneously:
EIδ C = 16 200 + 9450 (3) + 4455 = 49.0 × 103 N · m3 ↓ ◭
EIδ C = (1.024 + 2.048) × 1015 = 3.072 × 1015 N · mm3
EIθC = (7.68 + 19.2) × 1010 = 26.88 × 1010 N · mm2
6.87
δC =
θC =
3.072 × 1015
= 121.9 mm ◭
25.2 × 1012
1.92 × 1011
= 7.619 × 10−3 rad = 0.437◦ ◭
25.2 × 1012
6.89
C
δ 'C
300 N
′
δ C = θ B BC + δ c
720 mm
From segment AB (L = 4 m):
EIθB = −
B
M0 L
w0 L3
+
45
3
A
1800(4)3 3200(4)
= −
+
= 1707 N · m2
45
3
I =
From segment BC (L = 2 m):
θB
1440 mm
216000 N . mm
B
πd4
π(30)4
=
= 3.976 × 104 mm4
64
64
EI = (200 × 103 )(3.976 × 104 ) = 7.952 × 109 N · mm2
1600(2)3
P L3
=
= 4267 N · m3
3
3
EIδ C = 1707(2) + 4267 = 7680 N · m3 ↓ ◭
′
EIδ C =
δ C = δ ′C + LBC θ B
δ ′C =
6.88
θB =
For W10 × 33 section I = 126 × 106 mm4 . Therefore,
3
6
EI = (200 × 10 )(126 × 10 ) = 25.2 × 10
12
P L3BC
300(720)3
=
= 4.69 mm
3EI
3(7.952 × 109 )
21600(1440)
MB LAB
=
= 0.0130 rad
3EI
3(7.952 × 109 )
δ C = 4.69 + 720(0.0130) = 14.05 mm ◭
2
N · mm
6.90
Due to the load at B (a = 8000 mm, L = 16000 mm):
EIδ C =
EIθA =
2400(8000)2
P a2
(3L − a) =
(3 × 16000 − 8000)
6
6
P (L/2)2
− M0 L
2
PL
◭
M0 =
8
0 =
= 1.024 × 1015 N · mm3
EIθ C =
P a2
− M0 L
2
P a2
2400(8000)2
=
= 7.68 × 1010 N · mm2
2
2
153
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
6.94
6.91
25w0 L4
1 5(w0 /2)L4
5w0 L4
=
+
δ = δ1 + δ2 =
384EI
2
384EI
1536EI
δ C = θB a + δ ′C
4
25w0 (4)4
=
360
1536(70 × 109 )(30 × 10−6 )
w0 = 5600 N/m ◭
θB =
M0 L
(w0 a2 /2)(2a)
w0 a3
=
=
EI
E(2I0 )
2EI0
δ ′C =
w0 L4
w0 a4
=
8EI
8EI0
δC =
w0 a4
5 w0 a4
w0 a3
a+
=
↓◭
2EI0
8EI0
8 EI0
6.92
6.95
θB =
δC =
=
M0 b
EI
θB a + δ ′C =
δ ′C =
1
EI
M 0 a2
2EI
M 0 a2
M0 ba +
2
(a)
M0 a
(2b + a) ↓ ◭
2EI
M = − (w0 x)
EIv ′ = −
6.93
w0 x2
x
=−
2
2
w0 x3
+ C1
6
EIv = −
w0 x4
+ C1 x + C2
24
Boundary conditions:
θB =
δC =
=
(P a) b
EI
θB a + δ ′C =
δ ′C =
1
EI
1. v ′ |x=L = 0
−
w0 L3
+ C1 = 0
6
2. v|x=L = 0
−
w0 L3
w0 L4
+
L + C2 = 0
24
6
P a3
3EI
C2 = −
P a3
2
Pa b +
3
C1 =
w0 L3
6
w0 L4
8
w0 L3
w0 L4
w0 x4
+
x−
24
6
8
w0
4
3
4
−x + 4L x − 3L
◭
=
24
EIv = −
P a2
(3b + a) ↓ ◭
3EI
(b)
EIδ max = −EIv|x=0 =
w0 L4
Checks with Table 6.2
8
154
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
6.96
M = w0 Lx − (w0 x)
1. v|x=0 = 0
C2 = 0
2. v|x=L = 0
L6
L6
−
+ C1 L = 0
6
30
w0 Lx2
w0 x3
−
+ C1
2
6
EIv =
w0 x4
w0 Lx3
−
+ C1 x + C2
6
24
2L5
15
3 3
w0
L x
x6
2L5
EIv =
−
−
x
12L2
6
30
15
w0 x
−4L5 + 5L3 x2 − x5 ◭
=
2
360L
x
w0 x2
= w0 Lx −
2
2
EIv ′ =
C1 = −
6.98
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2. v ′ |x=L = 0
w0 L3
w0 L3
−
+ C1 = 0
2
6
C1 = −
w0 L3
3
w0 x4
w0 L3 x
w0 Lx3
−
−
6
24
3
w0 x
3
2
=
−x + 4Lx − 8L3 ◭
24
EIv =
ΣMB = 0
RA =
M =
=
EIv ′ =
EIv =
+
1
Lw0
3
+
1
w0 L
2
L
− RA L = 0
3
2
w0 L
3
x2 x
x
2
w0 L x − (w0 x) − w0
M =
3
2
2L 3
w0
=
−x3 − 3Lx2 + 4L2 x
6L
4
x
w0
3
2 2
′
− − Lx + 2L x + C1
EIv =
6L
4
5
w0
x
Lx4
2L2 x3
EIv =
− −
+
+ C1 x + C2
6L
20
4
3
RA =
6.97
ΣMB = 0
L
(w0 L) +
2
L
− RA L = 0
4
Boundary conditions:
1. v|x=0 = 0
w0 L
12
C2 = 0
2. v|x=L = 0
2L5
11L4
L5 L5
−
+
+ C1 L = 0
C1 = −
20
4
3
30
5
4
2 3
4
w0
x
Lx
2L x
11L x
EIv =
− −
+
−
6L
20
4
3
30
w0 x
−3x4 − 15Lx3 + 40L2x2 − 22L4 ◭
=
360L
We could also solve this problem by superimposing the
equations of the elastic curves for uniform and triangular
loads in Table 6.2.
w0 L
x
w0 L
1
x
1 x2
x−
wx
=
x−
w0 2 x
12
3
4
12
3 L
4
w0
L3 x − x4
12L2
3 2
L x
x5
w0
−
+
C
1
12L2
2
5
3 3
w0
L x
x6
−
+ C1 x + C2
12L2
6
30
−
155
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
3. v ′ |x=a− = v ′ |x=a+
6.99
w0 a3
w0 a3
w0 a3
−
+ C1 =
− w0 a3
2
6
2
M
(kN · m)
EIv ′
(kN · m2 )
EIv
(kN · m3 )
0≤x≤6m
6 m ≤ x ≤ 10 m
2x
12
2
x + C1
12x + C3
x3
+ C1 x + C2
3
6x2 + C3 x + C4
C1 = −
4. v|x=a− = v|x=a+
w0 a4
5w0 a4
w0 a4
w0 a4
−
−
=
− w0 a4 + C4
6
24
6
4
C4 =
Boundary conditions:
1. v|x=0 = 0
w0 a4
24
w0 x4
5w0 a3
w0 ax3
−
−
x
6
24
6
w0 x
=
−x3 + 4ax2 − 20a3 in AB ◭
24
EIv =
C2 = 0
2. v|x=6 m− = 0
(6)3
+ C1 (6) = 0
3
5w0 a3
6
C1 = −12 kN · m3
w0 a4
w0 a2 x2
− w0 a3 x +
4
24
w0 a2
6x2 − 24ax + a2 in BC ◭
=
24
EIv =
3. v ′ |x=6 m− = v ′ |x=6 m+
(6)2 − 12 = 12(6) + C3
C3 = −48 kN · m2
6.101
4. v|x= 6m+ = 0
6(6)2 − 48(6) + C4 = 0
For 0 ≤ x ≤ 6 m:
EIv =
C4 = 72 kN · m3
x
x3
− 12x = (x2 − 36) kN · m3 ◭
3
3
By superposition (L = 5 m, a = 2 m, b = 3 m):
For 6 m ≤ x ≤ 10 m:
EIθ B =
EIv = 6x2 − 48x + 72 = 6 (x − 6) (x − 2) kN · m3 ◭
=
P ab
M0 L
w0 L3
+
(2L − b) −
24
6L
3
300(5)
150(5)3 800(2)(3)
+
(2 × 5 − 3) −
24
6(5)
3
= 1401 N · m2
6.100
◭
6.102
Segment AB
w0 x2
M
w0 ax −
2
w0 x3
w0 ax2
′
−
+ C1
EIv
2 3
64
w0 x
w0 ax
−
+ C1 x + C2
EIv
6
24
Boundary conditions:
1. v|x=0 = 0
2. v ′ |x=2a = 0
Segment BC
w0 a2
2
w0 a2 x
+ C3
22 2
w0 a x
+ C3 x + C4
4
By superposition (L = 12 m, a = 6 m):
EIδ =
=
C2 = 0
M0 L2
w0 a2
(3L2 − 2a2 ) −
96
16
2160(12)2
120(6)2
(3 × 122 − 2 × 62 ) −
96
16
= −3240 N · m3 = 3240 N · m3 ↑◭
C3 = −w0 a3
156
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
6.103
6.106
(a) By superposition (L = 10 m, a = 4 m, b = 6 m):
M0 L
P ab
(2L − b) −
6L
3
100(4)(6)
(3P )(10)
=
(2 × 10 − 6) −
=0
6(10)
3
EIθB =
δ = θB b + δ′
P = 56.0 N ◭
EIθB =
M0 L
(w0 b2 /2)a
=
3
3
EIδ ′ =
w0 b4
w0 L4
=
8
8
(b) By superposition (L = 10 m, b = 6 m, x = 4 m):
M0 x 2
P bx 2
(L − x2 − b2 ) −
(L − x2 )
6L
6L
100(6)(4)
(56.0 × 3)(4)
=
(102 − 42 − 62 ) −
(102 − 42 )
6(10)
6(10)
EIδ =
EIδ =
w0 b4
w0 b3
w0 b2 a
b+
=
(3b + 4a) ↓ ◭
6
8
24
6.107
= 979.2 N · m3 ↓ ◭
Due to 60 N acting alone (L = 18 m, a = 6 m, b = 12 m):
6.104
P bx 2
(L − x2 − b2 )
6L
60(12)(6) 2
(18 − 63 − 122 ) = 5760 N · m3
=
6(18)
Pb L
(x − a)3 + (L2 − b2 )x − x3
EIδ|x=14m =
6L b
60(12) 18
3
2
2
3
=
(14 − 6) + (18 − 12 ) (14) − 14
6(18) 12
EIδ|x=6m =
δ = δ1 − δ2
EIδ 1 =
w0 L4
8
EIδ 2 =
w0 (L/2)3
w0 a3
(4L − a) =
24
24
EIδ =
w0 L4
7w0 L4
41
−
=
w0 L4 ↓ ◭
8
384
384
L
7w0 L4
4L −
=
2
384
= 3627 N · m3
Due to 90 N acting alone (L = 18 m, a = 14 m, b = 4 m):
P bx 2
(L − x2 − b2 )
6L
90(4)(6)
(182 − 62 − 42 ) = 5440 N · m3
=
6(18)
EIδ|x=6m =
6.105
P bx 2
(L − x2 − b2 )
6L
90(4)(14)
=
(182 − 142 − 42 ) = 5227 N · m3
6(18)
EIδ|x=14m =
I=
120(180)3
bh3
=
= 58.32 × 106 mm4
12
12
Due to both loads acting simultaneously:
M0 L2
δ = 2
16EI
EIδ|x=6m = 5760 + 5440 = 11 200 N · m3 ↓ ◭
2
15 =
2(4P )(20000)
16 × (10 × 103 )(58.32 × 106 )
EIδ|x=14m = 3627 + 5227 = 8854 N · m3 ↓ ◭
P = 43740 N ◭
157
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Use superposition.
6.108
Due to 45 N load acting alone (L = 18 m, b = 6 m):
Pb
(3L2 − 4b2 )
48
45(6) =
3(18)2 − 4(6)2 = 4658 N · m3
48
EIδ =
EIAB = (200 × 103 )
π(30)4
= 7.952 × 109 N · mm2
64
EIBC = (200 × 103 )
π(22.5)4
= 2.516 × 109 N · mm2
64
δ C = δ 1 + LθB + δ 2
δ1 =
MB L2
P L3
+
3EIAB
2EIAB
270(1080)3
97200(1080)2
+
= 21.4 mm
9
3(7.952 × 10 ) 2(7.952 × 109 )
MB L
P L2
+
LθB = L
2EIAB
EIAB
Due to 90 N load acting alone (L = 18 m, b = 8 m):
=
Pb
(3L2 − 4b2 )
48
90(8) 3(18)2 − 4(8)2 = 10 740 N · m3
=
48
EIδ =
=
Due to both loads acting simltaneously:
δ2 =
3
EIδ = 4658 + 10 740 = 15 400 N · m ↓ ◭
270(1080)3
97200(1080)2
= 35.6 mm
+
2(7.952 × 109 )
7.952 × 109
0.5(1080)4
w0 L4
=
= 9.01 mm
30EIBC
30(2.516 × 109 )
∴ δ C = 21.4 + 35.6 + 9.01 = 66.01 mm ◭
6.109
C6.1
The deflection formula for a concentrated load P in
Table 6.2 is

2

 P x (3a − x) if 0 ≤ x ≤ a

6EI
δ(x) =
2


 P a (3x − a) if a ≤ x ≤ L
6EI
We can apply this formula for the load element wdξ by
substituting P = wdξ and a = ξ. The result is

(wdξ)x2


(3ξ − x) if 0 ≤ x ≤ ξ

6EI
dδ(x) =
2


 (wdξ)ξ (3x − ξ) if ξ ≤ x ≤ L
6EI
δ C = θ B BC + δ ′C
From member AB (L = 3 m):
EIθB =
120(3)
M0 L
=
= 120 kN · m2
3
3
From member BC (L = 2 m):
EIδ ′C =
P L3
60(2)3
=
= 160 kN · m3
3
3
EIδ C = 120(2) + 160 = 400 kN · m3 → ◭
6.110
The displacement at x is obtained by adding the
contributions of all the load elements:
Z x
1
δ(x) =
w(ξ)ξ 2 (3x − ξ)dξ
6EI 0
Z L
1
w(ξ)x2 (3ξ − x)dξ
+
6EI x
.
1080 mm 97200 N mm
A
B
270 N
δ1
θB
0.5 N/mm
C
B 1080 mm
δ2
158
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
1.4
C6.1 MathCad worksheet for
Part (a)
1.2
Displacement (mm)
1
Given:
6
4
L := 6 m E := 200 GPa I := 58.3 × 10 mm
x
w(x) := (1200 Nm−1 ) ·
L
Computations:


Z x
1
2
w(ξ)
·
ξ
·
(3
·
x
−
ξ)dξ
.
.
.
·
 6·E·I

0Z

δ(x) := − 
L


1
2
·
w(ξ) · x · (3 · ξ − x)dξ
+
6·E ·I x
x := 0, 0.01 · L..L
0.8
0.6
0.4
0.2
0
–0.2
–0.4
0
1
2
3
x (m)
4
5
6
C6.2
Plotting range and interval
0
Displacement (mm)
–2
–4
–6
The deflection formula for a concentrated load P in
Table 6.3 is

P bx
2
2
2


 6LEI (L − x − b ) if 0 ≤ x ≤ a


Pb
L
δ(x) =
3
2
2
3
(x
−
a)
+
(L
−
b
)x
−
x



6LEI b


if a ≤ x ≤ L
–8
–10
–12
–14
0
1
2
3
x (m)
4
5
6
We can apply this formula for the load element wdξ by
substituting P = wdξ, a = ξ and b = L − ξ. The result is

(wdξ)(L − ξ)x 2


[L − x2 − (L − ξ)2 ]


6LEI



if 0 ≤ x ≤ ξ



 (wdξ)(L − ξ)
dδ(x) =
6LEI




L

3
2
2
3


(x
−
ξ)
+
[L
−
(L
−
ξ)
]x
−
x
×


L−ξ


if ξ ≤ x ≤ L
C6.1 MathCad worksheet for
Part (b)
Given:
L := 6 m E := 200 GPa I := 102 × 106 mm4
w(x) :=
(4000 Nm−1 ) if
x ≤ 3.9 m
(−2000 Nm−1 ) otherwise
Computations:


w(L − ξ)x


(2Lξ − ξ 2 )dξ if 0 ≤ x ≤ ξ


6LEI

w(L − ξ)x L(x − ξ)3
dδ(x) =
2
3
+
(2Lξ
−
ξ
)x
−
x
dξ



6LEI
L−ξ


if ξ ≤ x ≤ L
Z x

1
2
·
w(ξ)
·
ξ
·
(3
·
x
−
ξ)dξ
.
.
.
 6·E·I

0Z

δ(x) := − 
L


1
+
·
w(ξ) · x2 · (3 · ξ − x)dξ
6·E ·I x
x := 0, 0.01 · L..L
Plotting range and interval
159
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
The displacement at x as is obtained by adding the
contributions of all the load elements:
C6.2 MathCad worksheet for
Part (b)
δ(x)
1
=
6LEI
Z x
0
1
+
6LEI
L(x − ξ)3
2
3
+ (2Lξ − ξ )x − x dξ
w(L − ξ)
L−ξ
Z L
x
2
Given:
L := 12 · m E := 200 · 109 · Pa I := 101.2 × 106 mm4
2
w(L − ξ)x(2Lξ − ξ − x )dξ
Given:
L := 12 · m E := 200 · 109 · Pa I := 58.3 × 106 mm4
x
w(x) := 1200 · N/m
L
x
I1 (x) + I2 (x)
6·L·E ·I
x := 0, 0.01 · L..L
δ(x) :=
Computations:
Z x
w(ξ) · (L − ξ) ·
I1 (x) :=
0
L · (x − ξ)3
+ (2 · L · ξ − ξ 2 ) · x − x3 dξ
L−ξ
Z L
w(ξ) · (L − ξ) · x · (2 · L · ξ − ξ 2 − x2 )dξ
I2 (x) :=
Plotting range and interval
Displacement (mm)
0
x
Plotting range and interval
0
Displacement (mm)
if x ≤ 7.8 · m
otherwise
Computations:
Z x
L · (x − ξ)3
I1 (x) :=
w(ξ) · (L − ξ) ·
L−ξ
0
2
+ (2 · L · ξ − ξ ) · x − x3 dξ
Z L
I2 (x) :=
w(ξ) · (L − ξ) · x · (2 · L · ξ − ξ 2 − x2 )dξ
C6.2 MathCad worksheet for
Part (a)
I1 (x) + I2 (x)
δ(x) :=
6·L·E·I
x := 0, 0.01 · L..L
(400 · N/m)
(−200 · N/m)
w(x) :=
–0.01
–0.02
–0.03
–0.04
0
2
4
6
x (m)
8
10
12
–0.005
C6.3 MathCad worksheet for
Part (a)
–0.01
Given:
–0.015
0
2
4
6
x (m)
8
10
L := 10 · m
P1 := 12000 · N
E := 70 · 109 · Pa
12
b := 3 · m
P2 := 6000 · N
I := 250 · 106 · mm4
Computations:
P1 · (L − x) · x 2
[L − x2 − (L − x)2 ]
6·L·E·I
P2 · (L − b − x) · x 2
[L − x2 − (L − b − x)2 ]
δ 2 (x) :=
6·L·E ·I
δ 1 (x) :=
160
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
(−δ 1 (x) − δ 2 (x))
(−δ 1 (x))
δ(x) :=
x := 0, 0.01 · L..L
C6.4
if x ≤ L − b
otherwise
Plotting range and interval
Displacement (mm)
0
–5
For segment AB the displacement is (use Table 6.3):
–10
w0 x 3
MB x 2
(b − 2bx2 + x3 ) −
(b − x2 )
24EI
6LEI
MB b
w0 b3
−
θB =
24EI
3EI
δ AB =
–15
–20
0
2
4
6
8
For segment BC the displacement relative to the tangent
at B is (use Table 6.2):
10
x (m)
δ BC =
C6.3 MathCad worksheet for
Part (b)
Thus the displacement of the beam is
(
δ AB
if 0 ≤ x ≤ b
δ=
δ BC − θ B (x − b)
if x > b
Given:
L := 10 · m
P1 := 12000 · N
E := 70 · 109 · Pa
w0 (x − b)2 2
[6a − 4a(x − b) + (x − b)2 ]
24EI
b := 3 · m
P2 := −6000 · N
I := 250 · 106 · mm4
C6.4 MathCad worksheet
Computations:
P1 · (L − x) · x 2
[L − x2 − (L − x)2 ]
6·L·E ·I
P2 · (L − b − x) · x 2
δ 2 (x) :=
[L − x2 − (L − b − x)2 ]
6·L·E·I
Given:
δ(x) :=
Computations:
δ 1 (x) :=
(−δ 1 (x) − δ 2 (x))
(−δ 1 (x))
x := 0, 0.01 · L..L
b := 4.04 · m
L := 6 · m
E := 200 · 109 · Pa
if x ≤ L − b
otherwise
Plotting range and interval
a := L − b MB :=
Displacement (mm)
θB :=
w0 · b3
MB · b
−
24 · E · I
3·E·I
w0 · x
(b3 − 2 · b · x2 + x3 )
24 · E · I
MB · x
(b2 − x2 )
−
6·b·E·I
5
δ AB (x) :=
0
–5
δ BC (x) :=
–10
δ(x) :=
–15
w0 · a2
2
w0 := 12000 · N · m−1
I := 95 · 106 · mm4
0
2
4
6
8
w0 · (x − b)2
[6 · a2 − 4 · a · (x − b) + (x − b)2 ]
24 · E · I
(−δAB (x))
if x ≤ b
[−δBC (x) + θB · (x − b)]
otherwise
10
x (m)
161
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
x := 0, 0.01 · L..L
C6.5 MathCad worksheet
Plotting range and interval
Given:
Displacement (mm)
0.5
0
h1 := 60 mm
h2 := 300 mm
b := 60 mm
L := 3 m
P := 10000 N
E := 200GPa
h(x) := h1 + (h2 − h1 ) ·
–0.5
–1
X
L
Computations:
I(x) :=
0
1
2
3
x (m)
4
5
6
b · h(x)3
12
δ(x) :=
Z L
x
P · (x − ξ) · ξ
dξ
E · I(ξ)
x := 0, 0.01 · L..L Plotting range and interval
Note that δ max is minimized when δ max in AB equals δ C
0
–1
C6.5
Displacement (mm)
–2
–3
–4
–5
–6
–7
–8
By second moment-area theorem,
–9
δ = tC/B
–10
M
diagram between C and B about C
= moment of
EI
Z L
Pξ
(ξ − x)dξ
=
EI
x
0
0.5
1
1.5
x (m)
2
2.5
3
C6.6
By second moment-area theorem,
M
t = moment of
diagram between A and C about C
EI
Z x
M (ξ)
(x − ξ)dξ
=
EI(ξ)
0
The displacement at C is
δ = −(θA x − t) ↑
162
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
C6.6 MathCad Worksheet
Given:
L := 8 m
b := 5 m
h1 := 240 mm
h2 := 600 mm
Af := 7200 mm2
E := 200 GPa
P := 150000 N
X
h1 + 2(h2 − h1 ) ·
L
h(x) := L−x
h1 + 2(h2 − h1 ) ·
L
if x ≤
L
2
otherwise
Given:
M (x) :=
t(x) :=
P · (L − b)
·x
L
P ·b
· (L − x)
L
Z x
0
M (ξ) ·
if x ≤ b
I(x) := 2 · Af ·
otherwise
x−ξ
E · I(ξ)
θA :=
h(x)
2
2
t(L)
L
δ(x) := t(x) − θ A · x
x := 0, 0.05 · L..L
Plotting range and interval
0
–1
Displacement (mm)
–2
–3
–4
–5
–6
–7
–8
–9
–10
0
1
2
3
4
x (m)
5
6
7
8
163
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
Chapter 7
7.1
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2. v|x=L = 0
1
1
RA L3 −
w0 L4 + C1 L
6
120
1
1
C1 = − RA L2 +
w0 L3
6
120
0 =
From FBD of beam:
ΣMB = 0
ΣFy = 0
24 + MB − 6RA = 0
RB − RA = 0
3.v ′ |x=L = 0
(a)
(b)
1
1
RA L2 − w0 L3 + C1
2
24
1
1
1
1
2
3
2
3
w0 L
0 = RA L − w0 L + − RA L +
2
24
6
120
1
RA =
w0 L ◭
10
0 =
From FBD of segment:
M = 24 − RA x kN · m
RA x2
EIv ′ = 24x −
+ C1 kN · m2
2
RA x3
+ C1 x + C2 kN · m3
EIv = 12x2 −
6
7.3
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2
2. v ′ |x=6m = 0
RA (6)
2
= −144 + 18RA kN·m2
C1 = −24 (6) +
Due to skew-symmetry about the midspan RA = RB ,
MA = MB and v|x=4 m = 0.
From FBD of beam:
3. v|x=6m = 0
2
12 (6) −
3
RA (6)
+ (−144 + 18RA ) (6) = 0
6
−432 + 72RA = 0 (c)
ΣMB = 0
(a)
From FBD of segment 0 ≤ x ≤ 4 m:
Equations (a)-(c) yield
RA = RB = 6.00 kN ◭
1200 + 2MA − 8RA = 0
M = RA x − MA
RA x2
− MA x + C1
EIv ′ =
2
RA x3
MA x2
EIv =
−
+ C1 x + C2
6
2
Boundary conditions:
MB = 12 kN · m ◭
7.2
1. v|x=0 = 0
C2 = 0
2. v ′ |x=0 = 0
C1 = 0
3. v|x=4m = 0
From the FBD of the segment:
1 w0 3
1 w0 x2 x M = RA x −
= RA x −
x
2 L
3
6 L
3
2
RA (4)
MA (4)
−
= 0
6
2
10.667RA − 8MA = 0
1
1 w0 4
∴ EIv ′ = RA x2 −
x + C1
2
24 L
1 w0 5
1
x + C1 x + C2
EIv = RA x3 −
6
120 L
(b)
Solving Eqs. (a) and (b):
RA = RB = 225 N ◭
MA = MB = 300 N · m ◭
165
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
7.4
1. v|x=0 = 0
C2 = 0
2
2. v ′ |x=8 m = 0
RA (8)
3
+ 30 (8)
2
= −32RA + 15 360 N · m2
C1 = −
3. v|x=8m = 0
(a) Due to symmetry about midspan
P
RA = RB =
◭
MA = MB
2
From FBD of segment 0 ≤ x ≤ L/2:
RA (8)3
4
− 7.5 (8) + (−32RA + 15 360) (8) = 0
6
v ′ |x=L/2 = 0
−170.67RA + 92 160 = 0
Px
2
P x2
+ C1
EIv ′ = −MA x +
4
MA x2
P x3
EIv = −
+
+ C1 x + C2
2
12
Boundary conditions:
Solution of Eqs. (a) and (b) is
M = −MA +
1. v|x=0 = 0
′
2. v |x=0 = 0
(b)
RA = RC = 540 N ◭
RB = 1800 N ◭
7.6
C2 = 0
C1 = 0
3. v ′ |x=L/2 = 0
−MA
L P (L/2)2
+
=0
2
4
MA = MB =
PL
◭
8
(b)
(a) Due to symmetry
RA = RB =
w0 L
◭
2
MA = MB
v ′ |x=L/2 = 0
From FBD of segment:
w0 Lx w0 x2
−
2
2
w0 x3
w0 Lx2
′
−
+ C1
EIv = −MA x +
4
6
M = −MA +
7.5
Boundary conditions:
1. v ′ |x=0 = 0
2. v ′ |x=L/2 = 0
Due to symmetry RA = RC and v ′ |x=8m = 0.
From FBD of beam:
ΣFy = 0
2RA + RB − 2880 = 0
C1 = 0
−MA
L w0 L(L/2)2 w0 (L/2)3
+
−
=0
2
4
6
MA = MB =
(a)
From FBD of segment 0 ≤ x ≤ 8 m:
w0 L2
◭
12
(b)
M = RA x − 90x2 N · m
RA x2
− 30x3 + C1 N · m2
EIv ′ =
2
RA x3
− 7.5x4 + C1 x + C2 N · m3
EIv =
6
166
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
7.7
1. v|x=0 = 0
C2 = 0
2. v ′ |x=0 = 0
C1 = 0
3. v ′ |x=L = 0
RA L2
w0 L3
−
=0
2
24
(c)
MA L2
RA L3
w0 L4
+
−
=0
2
6
120
(d)
−MA L +
From FBD of beam:
ΣFy = 0
ΣMA = 0
4. v|x=L = 0
RA − 3 = 0
RA = 3 kN ◭
MA + MB − 3 (5) = 0
−
(a)
Solution of Eqs. (a)-(d) is
From FBD of segment:
3w0 L
◭
20
w0 L2
◭
MA =
30
M = −MA + 3x kN · m
7w0 L
◭
20
w0 L2
MB =
◭
20
RA =
EIv ′ = −MA x + 1.5x2 + C1 kN · m2
Boundary conditions:
1. v ′ |x=0 = 0
C1 = 0
2. v ′ |x=5m = 0
− MA (5) + 1.5 (5) = 0
RB =
7.9
2
MA = 7.50 kN · m ◭
From Eq. (a):
M = −RA x + M0 hx − ai0
RA x2
+ M0 hx − ai + C1
EIv ′ = −
2
M0
RA x3
2
+
hx − ai + C1 x + C2
EIv = −
6
2
Boundary conditions:
MB = 15 − MA = 15 − 7.50 = 7.50 kN · m ◭
7.8
1. v|x=0 = 0
C2 = 0
2. v ′ |x=L = 0
−
RA L2
+ M0 b + C1 = 0
2
C1 =
RA L3
M0 2
+
b + C1 L = 0
6
2
M0 2
RA L2
RA L3
+
b +
− M0 b L = 0
−
6
2
2
From FBD of beam:
ΣFy = 0
ΣMB = 0
RA L2
− M0 b
2
3. v|x=L = 0
w0 L
=0
(a)
2
w0 L L
= 0 (b)
MA − MB − RA L +
2
3
RA + RB −
−
RA = RB =
From FBD of segment:
3M0 b(2L − b)
◭
2L3
From FBD of beam:
w0 x2 x w0 x3
M = −MA + RA x −
= −MA + RA x −
2L
3
6L
w0 x4
RA x2
′
−
+ C1
EIv = −MA x +
2
24L
RA x3
w0 x5
MA x2
+
−
+ C1 x + C2
EIv = −
2
6
120L
ΣMB = 0
RA L − M0 − MB = 0
3M0 b(2L − b)
L − M0
MB = RA L − M0 =
2L3
M0 (−3b2 + 6bL − 2L2 )
=
◭
2L2
167
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2. v|x=8 m = 0
7.10
83
− 4(8 − 3)3 − 5(8 − 6)3 + 8C1
6
256
RA + 8C1 − 540
0 =
3
3. v ′ |x=8 m = 0
0 = RA
(a) Due to symmetry MA = MB , v|x=3a/2 = 0 and
RA = RB = P ◭
For the left half of beam:
82
− 12(8 − 3)2 − 15(8 − 6)2 + C1
2
0 = 32RA + C1 − 360
(a)
0 = RA
M = P x − P hx − ai − MA
P 2 P
2
EIv ′ =
x − hx − ai − MA x + C1
2
2
MA 2
P 3 P
3
x − hx − ai −
x + C1 x + C2
EIv =
6
6
2
(b)
Solution of Eqs. (a) and (b) is
C1 = −78.75 kN · m3
RA = 13.71 kN ◭
Boundary conditions:
1. v ′ |x=0 = 0
7.12
C1 = 0
2. v|x=0 = 0
C2 = 0
2
P 3a
3a
P a 2
′
3. v |x=3a/2 = 0
=0
−MA
−
2
2
2 2
2
MA = MB =
0
M = −MA + RA x − 120 hx − 2i kN · m
RA 2
EIv ′ = −MA x +
x − 120 hx − 2i + C1 kN · m2
2
MA 2 RA 3
2
x +
x − 60 hx − 2i + C1 x
EIv = −
2
6
+C2 kN · m3
2P a
◭
3
(b)
δ mid = −v|x=3a/2
" 2 #
3
P a 3 P a 3a
1 P 3a
−
−
= −
EI 6
2
6 2
3
2
Boundary conditions:
1. v|x=0 = 0
C2 = 0
′
vx=0
=0
C1 = 0
2.
5P a3
=
↓ ◭
24EI
3. v ′ |x=5 m = 0
− MA (5) +
−5MA +
7.11
4. v|x=5 m = 0
RA 2
(5) − 120(3) = 0
2
25
RA − 360 = 0
2
(a)
MA 2 RA 3
(5) +
(5) − 60(3)2 = 0
2
6
125
25
RA − 540 = 0
(b)
− MA +
2
6
−
Solution of Eqs. (a) and (b) is
M = RA x − 24 hx − 3i − 30 hx − 6i kN · m
x2
2
2
− 12 hx − 3i − 15 hx − 6i + C1 kN · m2
EIv ′ = RA
2
x3
3
3
− 4 hx − 3i − 5 hx − 6i
EIv = RA
6
+C1 x + C2 kN · m3
MA = 14. 40 kN · m ◭
RA = 34.56 kN ◭
From FBD of beam:
ΣFy = 0
ΣMB = 0
RA − RB = 0
MB + MA + 120 − 5RA = 0
Substituting for MA and RA , we get
RB = 34.6 kN ◭
MB = 38.4 kN · m ◭
168
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
7.13
1. v|x=0 = 0
C2 = 0
2. v ′ |x=0 = 0
C1 = 0
3. v ′ |x=6 m = 0
(a) Due to symmetry MA = MB , v ′ |x=0 = 0 and
RA = RB = 400 N ◭
4. v|x=6 m = 0
For the right half of the beam:
100
2
hx − 4i N · m
2
50
3
EIv ′ = MC x −
hx − 4i + C1 N · m2
3
MC 2 50
4
EIv =
x −
hx − 4i + C1 x + C2 N · m3
2
12
M = MC −
′
2. v |x=8 m = 0
MA = 21.33 kN · m ◭
3. v|x=8 m = 0
ΣMB = 0
C1 = 0
RA + RB − 24 = 0
MA − MB − 6RA + 24(4) = 0
Substituting for MA and RA , we get
50
MC (8) − (4)3 = 0
3
RB = 6.22 kN ◭
400
N·m
3
7.15
A
C2 = −32 000 N · m3
RA
MB = 10.65 kN · m ◭
30 kN
y
400/3 2 50 4
(8) − (4) + C2 = 0
2
12
C
B
3m
x
6m
3m
RB
RC
From the FBD of the beam:
From FBD of right half of beam:
ΣMB = 0
400
− 800 = 0
MB +
3
RA = 17.78 kN ◭
From FBD of beam:
ΣFy = 0
MC =
MA 2 RA 3
(6) +
(6) − 4(4)3 = 0
2
6
−18MA + 36RA − 256 = 0
(b)
−
Solving Eqs. (a) and (b):
Boundary conditions:
1. v ′ |x=0 = 0
RA 2
(6) − 12(4)2 = 0
2
−6MA + 18RA − 192 = 0
(a)
− MA (6) +
ΣMC = 0
ΣFy = 0
MB + MC − 400(2) = 0
2000
MB = MA =
N·m ◭
3
+ 12RA + 6RB − 30(9) = 0
+ ↑ RA + RB + RC − 30 = 0
(a)
(b)
M = RA x − 30 hx − 3i + RB hx − 6i
hx − 6i2
x2
2
− 15 hx − 3i + RB
+ C1
2
2
3
hx − 6i
x3
3
− 5 hx − 3i + RB
+ C1 x + C2
EIv = RA
6
6
Boundary conditions:
1. v|x=0 = 0
C2 = 0
2. v|x=6 m = 0
EIv ′ = RA
(b)
EIv|x=0 = C2 = −32 000 N · m3 = 32 000 N · m3 ↓ ◭
7.14
0 = 36RA − 135 + 6C1 = 0
(c)
0 = 288RA − 3645 + 36RB + 12C1 = 0
(d)
3. v|x=12′ = 0
M = −MA + RA x − 24 hx − 2i kN · m
RA 2
2
EIv ′ = −MA x +
x − 12 hx − 2i + C1 kN · m2
2
MA 2 RA 3
3
x +
x − 4 hx − 2i + C1 x + C2 kN · m3
EIv = −
2
6
Solution of Eqs. (a)-(d) is C1 = −50.6 kN·m2 and
RA = 12.19 kN ◭
RB = 20.6 kN ◭
RC = −2.81 kN ◭
169
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
7.17
7.16
M = −MA + RA x + RB hx − 6i − (120x)
x
N·m
2
EI = 200 × 103 (30 × 106 ) = 6 × 1012 N · mm2
RA 2 RB
2
x +
hx − 6i − 20x3
2
2
+C1 N · m2
MA 2 RA 3 RB
3
x +
x +
hx − 6i − 5x4
EIv = −
2
6
6
+C1 x + C2 N · m3
EIv ′ = −MA x +
M = −MA − RA x + RB hx − 4000i N · mm
RA 2 RB
2
EIv ′ = −MA x −
x +
hx − 4000i + C1 N · mm2
2
2
MA 2 RA 3 RB
3
x −
x +
hx − 4000i + C1 x +
EIv = −
2
6
6
C2 N · mm3
Boundary conditions:
Boundary conditions:
1. v ′ |x=0 = 0
C1 = 0
1. v ′ |x=0 = 0
C1 = 0
2. v|x=0 = 0
C2 = 0
2. v|x=0 = 0
C2 = 0
3. v|x=6 m = 0
−
MA 2 RA 3
(6) +
(6) − 5(6)4 = 0
2
6
−18MA + 36RA − 6480 = 0
3. EIv|x=4000 mm = −EIδ 0
−
(a)
4. v|x=12 m = 0
−
From FBD of beam:
MA
RA
RB 3
(12)2 +
(12)3 +
(6) − 5(12)4 = 0
2
6
6
−72MA + 288RA + 36RB − 103 680 = 0
MA
RA
(4000)2 −
(4000)3 = − 6 × 1012 (12)
2
6
3MA + 4000RA = 27 × 106
(a)
ΣMB = 0
ΣFy = 0
(b)
MA + 4000RA − (10000)(4000) = 0 (b)
− RA + RB − 1000 = 0
(c)
Solution of Eqs. (a) and (b) is
From FBD of beam:
MA = −6500000 N · mm ◭
ΣMC = 0
−MA + 12RA + 6RB − (120 × 12)(6) = 0
(c)
ΣFy = 0
(d)
RA + RB + RC − 120 × 12 = 0
RA = 11625 N ◭
From Eq. (c)
−11625 + RB − 10000 = 0
RB = 21625 N ◭
Solution of Eqs. (a)-(c) is
MA = 309 N · m ◭
RA = 334 N ◭
RB = 823 N ◭
7.18
From Eq. (d)
RC = 283 N ◭
π EIθ 0 = (72 × 109 )(126 × 10−6 ) 0.75 ×
180
3
2
= 118.75 × 10 N · m = 118.75 kN · m2
M = RA x − 200 hx − 2i kN · m
RA 2
2
EIv ′ =
x − 100 hx − 2i + C1 kN · m2
2
RA 3 100
3
x −
hx − 2i + C1 x + C2 kN · m3
EIv =
6
3
170
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Boundary conditions:
1. v|x=0 = 0
7.20
C2 = 0
RA 2
(4) −100(2)2 +C1 = 118.75
2
8RA + C1 − 518. 75 = 0
(a)
2. EIv ′ |x=4m = EIθ0
RA 3 100 3
(4) −
(2) + C1 (4) = 0
6
3
16RA + 6C1 − 400 = 0
3. v|x=4 m = 0
(b)
Solution of Eqs. (a) and (b) yields
RA = 84.77 kN ◭
1
(15RA )(15) = 112.5RA
2
1
A2 = (−13 500)(15) = −67 500 N · m2
3
1
A3 = (−6750)(15) = −25 310 N · m2
4
From FBD of beam:
ΣMB = 0
ΣFy = 0
A1 =
MB + 4RA − 200(2) = 0
RA + RB − 200 = 0
Substituting for RA , we get
MB = 60.9 kN · m ◭
7.19
RB = 115.2 kN ◭
EItA/B = area of M -diagram|B
A · x̄/A = 0
2
3
0 = (112.5RA)
× 15 − (67 500)
× 15
3
4
4
−(25 310)
× 15
5
RA = 945 N ◭
18 kN
A
3m
3m
B
C
tA/C
RB
3RB
M(kN . m)
0
−54
−108
7.21
(a)
C
EItB/C = area of M -diagram|B · x̄/B = 0
1
1
0 = (3RB × 3)(2) − (54 × 3)(1.5) − (54 × 3)(2)
2
2
RB = 45 kN ◭
(b)
C
EItA/C = area of M -diagram|A · x̄/A
1
1
= (3RB × 3)(5) − (108 × 6)(4)
2
2
1
1
= (3 × 45 × 3)(5) − (108 × 6)(4)
2
2
= −283.5 kN · m3
EItA/B = area of M -diagram|B
A · x̄/A = 0
1
8 1
4 1
2
(1280 × 8) − (1280 × 4) − (8MB )
×8 =0
2
3 3
4 2
3
For W150 × 37.1: I = 22.2 × 106 mm4 = 22.2 × 10−6 m4
283.5 × 103
= 0.0639 m
(200 × 109 ) (22.2 × 10−6 )
= 69.9 mm ◭
MB = 560 N · m ◭
δ A = tA/C =
171
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
From FBD of beam:
7.22
ΣMC = 0
ΣFy = 0
− 15RA + 9RB − 36 450 = 0
− RA + RB + RC − 8100 = 0
(b)
(c)
Solution of Eqs. (a)-(c) is
RA = 910 N ↓ ◭
RC = 3440 N ↑ ◭
EIθA/B = area of M -diagram|B
A
1 PL L
= MB L −
=0
×
2
2
2
PL
MB =
◭
8
RB = 5570 N ↑ ◭
7.24
From FBD of beam:
ΣFy = 0
RA − P = 0
ΣMA = 0
MA + MB −
yielding
RA = P ◭
MA =
PL
=0
2
3P L
◭
8
EItB/A = area of M -diagram|B
A · x̄/B
10 1 5000
10
1
× 10
= (10RA × 10) −
2
3
4
3
5
1
= (500RA − 25 000) N · m3
3
EItC/A = area of M -diagram|C
A · x̄/C
1
20 1 40 000
20
= (20RA × 20) −
× 20
2
3
4
3
5
1
10
+ (10RB × 10)
2
3
1
= (4000RA − 800 000 + 500RB )
3
7.23
EItB/A = area of M -diagram|B
A · x̄/B
6
1
= − (6RA × 6) = −36RA N · m3
2
3
EItC/A = area of M -diagram|C
A · x̄/C
1
15 1
9
= − (15RA × 15) + (9RB × 9)
2
3
2
3
9
1
− (36 450 × 9)
3
4
= −562.5RA + 121.5RB − 246.0 × 103 N · m3
From geometry of deflection diagram:
tC/A = 2tB/A
2
1
(4000RA − 800 000 + 500RB ) = (500RA − 25 000)
3
3
3000RA + 500RB − 750 000 = 0
(a)
From FBD of beam:
ΣMC = 0
From geometry of deflection digram:
tC/A
tB/A
=
6
15
36RA
−562.5RA + 121.5RB − 246.0 × 103
−
=
6
15
31.5RA = 8.10RB − 16 400
ΣFy = 0
20
=0
20RA + 10RB − 2000
3
RA + RB + RC − 2000 = 0
(b)
(c)
Solution of Eqs. (a)-(c) is
RA = 42 N ◭
RB = 1250 N ◭
RC = 708 N ◭
(a)
172
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
7.25
7.26
(a) Utilize symmetry and consider right half of beam only.
EItB/A =
EIθ B/C = area of M -diagram|C
B
L 1 w0 a2
= MC × −
×a =0
2
3
2
w0 a3
MC =
3L
area of M -diagram|B
A · x̄/B
8
8
1
(−8RA × 8) + (2000 × 8)
2
3
2
= −85.33RA + 64 000
=
EItC/A = area of M -diagram|C
A · x̄/C
16 1
8
1
= (−16RA × 16) + (8RB × 8)
2
3
2
3
4
16
+(2000 × 16) − (2000 × 4)
2
2
= −682.67RA + 85.55RB + 240 000
From FBD of right half of beam:
a
− MB − MC = 0
2
w0 a2
w0 a3
w0 a2
− MC =
−
MB = MA =
2
2
3L
w0 a2
(3L − 2a) ◭
=
6L
ΣMB = 0
From geometry of deflection diagram:
tC/A = 2tB/A
(b)
−682.67RA + 85.33RB + 240 000 = 2 (−85.33RA + 64 000)
−512RA + 85.33RB + 112 000 = 0
(a)
EIδ C = EItC/B = area of M -diagram|B
C · x̄/C
2
L L 1 w0 a
L a
−
×a
−
= MC ×
2 4
3
2
2
4
3 2
3
3
w0 a L
w0 a
L a
w0 a
=
=−
−
−
(L − a)
3L 8
6
2
4
24
w0 a3
=
(L − a) ↓ ◭
24
From FBD of beam:
ΣMC = 0
ΣFy = 0
16RA − 8RB = 0
− RA + RB − RC = 0
(b)
(c)
Solution of Eqs. (a)-(c) is
RA = 328 N ◭
RB = 656 N ◭
(w0 a)
RC = 328 N ◭
173
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
7.27
7.29
For aluminum rod:
Pal = σ w A = 120 × 106 (40 × 10−6 ) = 4800 N = 4.8 kN
Pal L
4800(5)
=
= 8.571 × 10−3 m
EA
(70 × 109 )(40 × 10−6 )
For steel beam:
δB =
EIθ A/B = area of M -diagram|B
A = 0
1
1
0 = (9RA × 9) − 9MA + (4050 × 3)
2
3
1
− (22 050 × 7)
3
0 = 40.5RA − 9MA − 47 400
EItB/A =
EIδ B = −area of M -diagram|C
B · x̄/B
1
4 1
4
= − (9.6 × 2) + (2P × 2) + (2P × 2) (1.0)
2
3 2
3
= −12.8 + 6.667P kN · m3
(a)
= 12.8 × 103 + 6.667P N · m3
−12.8 × 103 + 6.667P
δB =
(200 × 109 )(50 × 10−6 )
area of M -diagram|B
A · x̄/B = 0
9
9 1
3
1
(9RA × 9) − (9MA ) + (4050 × 3)
2
3
2 3
4
7
1
− (22 050 × 7)
3
4
0 = 125.5RA − 40.5MA − 87 000
(b)
0 =
= −1.28 × 10−3 + (666.7 × 10−9 )P
8.571 × 10−3 = −1.28 × 10−3 + (666.7 × 10−9 )P
P = 14 780 N = 14.78 kN ◭
Solving Eqs. (a) and (b) yields
RA = 2080 N ◭
MA = 4090 N · m ◭
7.30
7.28
EIδ A = −EItA/B = −area of M -diagram|B
A · x̄/A
1
1
4
= − (3RA × 3) (2) + (150 × 2) 1 +
2
2
3
Using segment AB:
= −9RA + 350 kN · m3 = −9RA + 350 × 103 N · m3
For spring
δA =
EIδ B = −EItB/A = −area of M -diagram|B
A · x̄/B
1
= − (6RB × 3) (2) + (6000 × 3) (1.5)
2
1
+ (18000 × 3)(2)
2
= −36RB + 81000 N.m3
RA
k
Thus
RA
−9RA + 350 × 103
=
EI
k
RA
−9RA + 350 × 103
=
(200 × 109 )(60 × 10−6 )
660 × 103
(−36RB + 81000) (1000)3
= 5 mm
(200 × 103 )(12.5 × 106 )
RB = 1902.8 N ◭
δB =
RA = 12 880 N = 12.88 kN ◭
174
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
30 kN
7.31
B
A
3m
RA
3m
20.63 kN
C
6m
RC
From the FBD of the beam:
Choose MA and MB as the redundant reactions.
Constraints are θA = θB = 0.
ΣMC = 0
EIθA =
MA L MB L
w0 a2
(a − 2L)2 −
−
24L
6
3
60(9)2
MA (12) MB (12)
2
0 =
[9 − 2(12)] −
−
24(12)
6
3
0 = 3797 − 2MA − 4MB
12RA + 6(20.63) − 30(9) = 0
RA = 12.185 kN ◭
ΣFy = 0
+ ↑ 12.185 + 20.63 + RC − 30 = 0
2
w0 a
MA L MB L
(2L2 − a2 ) −
−
24L
3
6
MA (12) MB (12)
60(9)2 2
2
2(12) − 9 −
−
0 =
24(12)
3
6
0 = 3493 − 4MA − 2MB
+
RC = −2.82 kN ◭
(a)
7.33
(b)
Choose MA and RB as redundant reactions. Constraints
are θA = δ B = 0.
EIθB =
Solution of Eqs. (a) and (b) is
MA = 531.5 N · m ◭
w0 L3
RB L2
MA L
−
−
24
16
3
MA (12)
120(12)3 RB (12)2
−
−
0 =
24
16
3
0 = 8640 − 9RB − 4MA
EIθA =
MB = 683.5 N · m ◭
From FBD of beam:
ΣMB = 0
531.5 − 683.5 + 540(4.5) − 12RA = 0
RA = 189.8 N ◭
5w0 L4
RB L3
MA L2
−
−
384
48
16
MA (12)2
5(120)(12)4 RB (12)3
−
−
0 =
384
48
16
0 = 32 400 − 36RB − 9MA
ΣFy = 0
RA + RB − 540 = 0
RB = 540 − RA = 540 − 189.8 = 350 N ◭
7.32
↓ EIδ B =
30 kN
3m
(b)
Solution of Eqs. (a) and (b) is
B
A
(a)
C
MA = 309 N · m ◭
6m
3m
RB
RB = 823 N ◭
From FBD of beam:
Let RB be the redundant reaction. The constraint is
δ B = 0.
RB L3
Pb L
3
2
2
3
(x − a) + (L − b )x − x −
=0
↓ EIδ B =
6L b
48
30(9) 12
RB (12)3
0=
(6 − 3)3 + (122 − 92 )(6) − 63 −
6(12) 9
48
0 = 742.5 − 36RB
RB = 20.63 kN ◭
ΣMA = 0
(823 − 1440) (6) + 12RC + 309 = 0
RC = 283 N ◭
ΣFy = 0
823 − 1440 + RA + RC = 0
RA = 617 − RC = 617 − 283 = 334 N ◭
175
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
7.34
7.36
Choose MB as the redundant reaction. Constraint is
θB = 0.75◦ , or
π EIθB = (72 × 109 )(126 × 10−6 ) 0.75 ×
180
3
2
= 118.75 × 10 N · m = 118.75 kN · m2
Choose MA as the redundant reaction. Constraint is
θ B = 0.
EIθB =
0 =
P L2
MB L
−
16
3
200(4)2 MB (4)
−
MB = 60.9 kN · m ◭
118.75 =
16
3
EIθB =
P (L/2)2
− MB L
2
ΣFy = 0
ΣMA = 0
MA =
− 60.9 − 4RA + 200(2) = 0
RA = 84.8 kN ◭
ΣFy = 0
RA + RB − 200 = 0
PL
◭
8
RA − P = 0
RA = P ◭
PL
L
MA +
−P =0
8
2
3P L
◭
8
7.37
RB = 200 − RA = 200 − 84.8 = 115.2 kN ◭
7.35
MB =
From FBD of beam:
From FBD of beam:
ΣMB = 0
P a2
− MB L
2
180 N/m
Choose RB as the redundant reaction. Constraint is
δ B = 0.
120 N/m
A
RA
B
15 m
↓ EIδ B =
Let RA be the redundant reaction. The constraint is
δ A = 0.
w0 a2
(−L + x)(a2 − 4Lx + 2x2 )
24L
RB bx 2
(L − x2 − b2 )
6L
900(9)2
(−15 + 9) 92 − 4(15)(9) + 2(9)2
0 =
24(15)
RB (9)(6)
152 − 62 − 92
−
6(15)
RB = 5570 N ↑ ◭
−
w0 L4
w1 L4
RA L3
−
−
=0
3
8
30
w0 L w1 L
RA
−
−
0 =
3
8
30
RA
120(15) 180(15)
0 =
−
−
3
8
30
RA
− 315
RA = 945 N ◭
0 =
3
↑ EIδ B =
From FBD of beam:
ΣMC = 0
15RA − 5570(9) + 8100(4.5) = 0
RA = 910 N ↓ ◭
ΣFy = 0
− RA + 5570 − 8100 + RC = 0
RC = 910 − 5570 + 8100 = 3440 N ↑ ◭
176
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
7.38
EI = 200 × 109 (20.0 × 10−6 ) = 4.0 × 106 N · m2
Choose RB and RC as the redundant reactions.
Constraints are δ B = δ C = 0.
12 kN . m
6m
δC'
δB'
δB''
↓ EIδ B =
6m
4.0 × 10
x
δC''
6
RB L3
MA L2
−
16
48
20 × 103 (6)2
RB (6)3
(4.25 × 10 ) =
−
16
48
RB = 6220 N = 6.22 kN ◭
−3
RB
δB'''
7.40
δC'''
RC
M0 x2B
12(12)2
=
= 864 kN · m3
2
2
12(6)2
M0 x2C
=
= 216 kN · m3
EIδ ′C =
2
2
RB (12)2
RB x2B
(3aB − xB ) =
(3 × 12 − 12)
EIδ ′′B =
6
6
= 576RB kN · m3
RB x2C
RB (6)2
EIδ ′′C =
(3aB − xC ) =
(3 × 12 − 6)
6
6
= 180RB kN · m3
RC a2C
RC (6)2
EIδ ′′′
=
(3x
−
a
)
=
(3 × 12 − 6)
B
C
B
6
6
= 180RC kN · m3
RC (6)2
RC a2C
(3x
−
a
)
=
(3 × 6 − 6)
EIδ ′′′
=
C
C
C
6
6
= 72 kN · m3
EIδ ′B =
Use superposition with RB as the redundant reaction. The
constraint is δ AB = δ BC at contact point, or
3
RB L3
w0 L4
+
3
8
3 (220) (6)
3w0 L
=
= 247.5 N
RB =
16
16
RB (L)
3
= −
From FBD of beam AB: ΣMA = 0
MA − (247.5) (6) = 0
MA = 1485 N · m ◭
From FBD of beam BC: ΣMC = 0
6
− 247.5 (6) − MC = 0
20 (6)
2
MC = 2475 N · m ◭
EIδ B = EI(δ ′B + δ ′′B + δ ′′′
B) = 0
864 − 576RB − 180RC = 0
EIδ C = EI(δ ′C + δ ′′C + δ ′′′
C) = 0
(a)
216 − 180RB − 72RC = 0
(b)
7.41
The solution of Eqs. (a) and (b) is RB = 2.57 kN and
RC = −3.43 kN. Hence
RB = 2.57 kN ↑◭
25(150)3
= 7.03 × 106 mm4
12
50(250)3
= 65.1 × 106 mm4
ICD =
12
IAB =
Both beams are simply supported, loaded at the midspan.
Letting R be the contact force, the compatibility equation
is δ AB = δ CD at midspan, or
RC = 3.43 kN ↓◭
(5000 − R) L3AB
RL3CD
=
48EIAB
48EICD
(5000 − R)(8 × 1000)3
R(10 × 1000)3
=
7.03 × 106
65.1 × 106
R = 4129 N ◭
7.39
Use superposition with RB as the redundant reaction.
Constraint is δ B = 4.25 mm.
177
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
7.42
7.44
Let the contact force R be the redundant reaction.
Constraint is δ A = δ C + 4 mm.
Choose RB and RC as the redundant reactions.
Constraints are δ B = δ C = 0. Note that due to symmetry
RB = RC = R.
Displacement at B due to w0 :
↓ δA =
(2 − R) L3
(2 − R)(0.180)3
=
3EIAB
3(200 × 109 )(3 × 10−12 )
w0 x 3
(LAB − 2LAB x2 + x3 )
24
22
w0 L (3L)3 − 2(3L)L2 + L3 =
w0 L4 ↓
=
24
24
EIδ ′B =
= (6.480 − 3.240R) × 10−3 m = 6.480 − 3.240R mm
RL3
R(0.180)3
↓ δC =
=
3EICD
3(200 × 109 )(5 × 10−12 )
= 1944 × 10−3 R m = 1.944R mm
Displacement at B due to RB and RC :
EIδ ′′B =
X P bx
(L2 − x2 − b2 )
6LAB AB
R(2L)(L) =
(3L)2 − L2 − (2L)2
6(3L)
R(L)(L) +
(3L)2 − L2 − L2
6(3L)
8
7
5
=
RL3 + RL3 = RL3 ↑
18
18
6
6.480 − 3.240R = 1.944R + 4
R = 0.478 N ◭
7.43
Let MA and MB be the redundant reactions. Constraints
are θA = θB = 0.
δ B = δ ′B − δ ′′B = 0
MA LAB
MB LAB
w0 a2
(a − 2LAB )2 −
−
EIθA =
24LAB
3
6
w0 L2
M
(2L)
M
(2L)
A
B
2
0 =
[L − 2(2L)] −
−
24(2L)
3
6
2MA L MB L
3w0 L3
−
−
(a)
0 =
16
3
3
RB = RC = R =
P
5
22
w0 L4 − RL3 = 0
24
6
11
w0 L ◭
10
Fy = 0 yields
RA = RD =
3
11
2
w0 L − w0 L = w0 L ◭
2
10
5
MA LAB
MB LAB
w0 a2
(2L2AB − a2 ) −
−
24LAB
6
3
2 w0 L
MA (2L) MB (2L)
0 =
2(2L)2 − L2 −
−
24(2L)
6
3
3
MA L 2MB L
7w0 L
−
−
(b)
0 =
48
3
3
7.45
Solution of Eqs. (a) and (b) is
Use symmetry to consider right half of beam only. Choose
RC as the redundant reaction. Constraint is δ C = 0.
EIθ B =
MA =
11w0 L2
◭
48
MB =
5w0 L2
◭
48
RC L3
w0 x2
6L2BD − 4LBD x + x2 −
24
3
3
R
L
w0 L2 C
6(L + b)2 − 4(L + b)L + L2 −
0 =
24
3
w0
2
2
RC =
3L + 8Lb + 6b
8L
↓ EIδ C =
178
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
From statics:
Boundary conditions:
2
w0 b
2
w0 (L + b)2
− RC L
MB =
2
w0
w0 (L + b)2
3L2 + 8Lb + 6b2
−
=
2
8
w0 2
=
(L − 2b2 )
8
MC =
C2 = 0
2. v ′ |x=L = 0
RA L −
w0 L2
+ C1 = 0
4
w0 L2
C1 = −RA L +
4
3. v|x=L =
w0 L3
w0 L2
RA L2
L=0
−
+ −RA L +
0
2
12
4
Hence the requirement MC = MB is
w0 b2
w0 2
=
(L − 2b2 )
2
8
1. v|x=0 = 0
L
b= √ ◭
6
RA =
w0 L
◭
3
7.48
7.46
Use moment-area method.
Use integration
A
4m
B
RA
9.6 kN
1.6 m 2.4 m
C
RB
8RA
M (kN .m)
w0 x2
2
Boundary conditions:
M = MA −
′
1. v |x=0 = 0
C1 = 0
2. v ′ |x=L = 0
MA L −
EIv ′ = MA x −
w0 x3
+ C1
6
−23.04
4RB
M (kN . m)
w0 L3
=0
6
MA =
4RA
0
w0 L2
◭
6
0
tA/C = area of M -diagram|C
A · x̄/A = 0
1
2
×8
0 = (8RA × 8)
2
3
1
2
− (23.04 × 2.4) 5.6 + × 2.4
2
3
2
1
+ (4RB × 4) 4 + × 4
2
3
0 = 170.67RA + 53.33RB − 199.07
7.47
(a)
tB/C = area of M -diagram|C
B · x̄/B = 0
1
2
1
0 = (4RA × 4)
× 4 + (4RA × 4)
×4
2
3
2
2
1
− (23.04 × 2.4) 1.6 + × 2.4
2
3
1
2
+ (4RB × 4)
×4
2
3
0 = 53.33RA + 21.33RB − 88.47
(b)
Use integration.
w0 x2
x
M = RA x −
L
2
w
x
L
M
0
RA −
=
v ′′ =
EI
EI
2
0
2
L
w0 x
v′ =
RA x −
+ C1
EI0
4
2
L
RA x
w0 x3
v =
−
+ C1 x + C2
EI0
2
12
I = I0
Solution of Eqs. (a) and (b) is:
RA = −0.593 kN ◭
RB = 5. 63 kN ◭
179
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
7.51
7.49
4 N/mm
A
1.5 m
F
6m
B B
Utilize symmetry to analyse only the left half of the beam.
Use superposition with MB as the redundant reaction.
The constraint is θB = 0.
F
For the beam:
I =
δ ′B =
=
=
P ab
MB L
(2L − b) −
6L
3
MB (5)
24(3)(2)
[2(5) − 2] −
0 =
6(5)
3
MB = 23.0 kN · m ◭
EIθ B =
bh3
60(30)3
=
= 135000 mm4
12 12
w0 L4AB
F L3AB
1
−
EI
8
3
1
4(1500)4 F (1500)3
−
(E)(135000)
8
3
1
(2.53 × 1012 − 1.125 × 109 F )
(E)(135000)
7.52
For the wire:
π(3.75)2
πd2
=
= 11.04 mm2
4
4
FL
F (6000)
F
δ ′′B =
=
= 543.5
EA
(E)(11.04)
E
A =
Utilize symmetry to analyse only the left half of the beam.
Use superposition with RA as the redundant reaction. The
constraint is δ A = 0.
δ ′B = δ ′′B
(2.53 × 1012 − 1.125 × 109 F )
= 543.5F
135000
F = 2111 N ◭
RA a3
P x2
(3L − x) −
6
3
RA a3
P a2
[3(a + b) − a] −
0 =
6
3
3b
↑ ◭
RA = RC = P 1 +
2a
↓ EIδ A =
7.50
From FBD of beam:
3b
1+
− 2P − RB = 0
2a
ΣFy = 0
2P
RB = 3P
b
↓ ◭
a
Use superposition with RB as the redundant reaction.
Constraint is δ B = 0.
7.53
RB L3BC
MA x2
−
2
3
4000(4)2 RB (4)3
−
RB = 1500 N ↑ ◭
0 =
2
3
↓ EIδ B =
Use superposition with the contact force R between the
beams as the redundant reaction. At B the displacements
of the beams must be equal.
For beam ABC (simply supported, L = 12 m, b = 4 m,
x = 8 m):
From FBD diagram of beam:
ΣMC = 0
ΣFy = 0
4000 − 1500(4) + MC = 0
1500 − RC = 0
MC = 2000 N · m ◭
EIδ ′B =
RC = 1500 N ↓ ◭
R(4)(8)
P bx 2
122 − 82 − 42
(L − x2 − b2 ) =
6L
6(12)
= 28.44R N · m3 ↓
180
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
For beam AC:
For beam BD (cantilever, L = 8 m):
RL3AB
w0 x2
6L2 − 4Lx + x2 −
24
3
i
3000(4)2 h
2
=
6 (6) − 4 (6 × 4) + (4)2
24
R(4)3
−
3
= 272.0 × 103 − 21.33R N ↓
(1400 − R) (8)3
P L3
=
3
3
= 238 900 − 170.67R N · m3 ↓
EIδ ′′B =
EIδ ′′B =
δ ′B = δ ′′B
28.44R = 238 900 − 170.67R
R = 1200 N
δ ′B = δ ′′B
21.33R = 272.0 × 103 − 21.33R
R = 6375 N = 63.8 kN ◭
7.56
In ABC : Mmax = MB = (400)(8) = 3200 N · m ◭
In BC : Mmax = MD = 200(8) = 1600 N · m ◭
7.54
Due to M0 alone:
M0 L2
L
M0 LAB L
M0 (L/2) L
=
=
↓
δ 1 = θ1 =
2
6EI
2
6EI
2
24EI
Use superposition with RB as the redundant reaction.
Constraint is δ B = δ 0 .
Due to spring force P alone:
RB L3
5w0 L4
−
↓ EIδ B =
384
48
(P L/2)(L/2)
P L2
MB LAB
=
=
3EI
3EI
12EI
P (L/2)3
P L3
P L3BC
′
=
=
δ2 =
3EI
3EI 24EI
P L2 L
P L3
P L3
L
+
=
↑
δ 2 = θ2 + δ ′2 =
2
12EI 2
24EI
12EI
θ2 =
If reactions are equal, then RB = w0 L/3. Therefore,
5w0 L4
7w0 L4
w0 L L3
EIδ 0 =
−
δ0 =
◭
384
3
48
1152EI
7.55
P = kδ total ↓ = k(δ 1 − δ 2 )
k
M0 L2
P L3
kL3 M0
P =
−
−P
=
EI
24
12
12EI 2L
−1
M0
12EI
P =
1+
Q.E.D.
2L
kL3
Use superposition with the contact force R as the
redundant reaction. Constraint is: displacements of the
beams at B must be equal.
For beam AB:
EIδ ′B =
R(4)3
RL3AB
=
= 21.33R N ↓
3
3
181
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
C7.1 MathCad worksheet for
beam (b)
C7.1
Given:
L := 10 · m
w(x) := (1000 · N · m−1 )
x
L
Computations:
C1 :=
Using Table 6.3, the end rotations of a simply supported
beam carrying the concentrated load P are
θA =
P ab
(2L − a)
6LEI
θB =
C2 :=
P ab
(2L − b)
6LEI
1
·
L2
1
·
L2
θ ′B =
1
6LEI
0
Z L
0
wx(L − x)(L + x)dx
6EI
MA + 2MB =
1
L2
Z L
0
Z L
0
:= Find(MA , MB )
MA = 3.333 × 103 N · m
MB = 5 × 103 N · m
L := 9 · m
(0 · N · m−1 ) if x < 2 · m
w(x) := (0 · N · m−1 ) if x < 6 · m
(900 · N · m−1 ) otherwise
3EI
The compatibility conditions θ ′′A = θ ′A and θ′′B = θ′B yield
the simultaneous equations
1
2MA + MB = 2
L
Given:
MA L MB L
θ′′B =
+
6EI
MA
MB
C7.1 MathCad worksheet for
beam (c)
The end rotations due to the end moments are (see
Table 6.3):
3EI
w(x) · x · (L − x) · (L + x)dx
2 · MA + MB = C1
MA + 2 · MB = C2
wx(L − x)(2L − x)dx
MA L MB L
+
θ′′A =
0
Given:
Integrating the result over the length of the beam (adding
the contributions of all the load elements), we get
1
6LEI
Z L
w(x) · x · (L − x) · (2 · L − x)dx
MA := 1 · N · m
Trial values used in the solution
MB := 1 · N · m
(wdx)x(L − x)
(2L − x)
dθ ′A =
6LEI
(wdx)x(L − x)
[2L − (L − x)]
dθ ′B =
6LEI
θ′A =
0
Solve simultaneous equations for the end moments:
If we set P = wdx, a = x and b = L − x, we obtain the end
rotations dθ ′A and dθ ′B caused by the distributed load
element wdx:
Z L
Z L
Computations:
wx(L − x)(2L − x)dx
Z L
1
·
w(x) · x · (L − x) · (2 · L − x)dx
L2 Z 0
L
1
w(x) · x · (L − x) · (L + x)dx
C2 := 2 ·
L
0
C1 :=
wx(L − x)(L + x)dx
Solve simultaneous equations for the end moments:
MA := 1 · N · m
Trial values used in the solution
MB := 1 · N · m
182
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Given
Computations:
n
X
1
3
Pi · (ai ) · (3 · L − ai )
·
· w0 · L +
8
2 · L3 i=1
n
X
1
M (X) := − · w0 · (L − x)2 −
Pi · (ai − x)
2
i=1
·Φ(ai − x) . . . + RB · (L − x)
2 · MA + MB = C1
RB :=
MA + 2 · MB = C2
MA
MB
:= Find(MA , MB )
MA = 4.089 × 103 N · m
MB = 3.378 × 103 N · m
Φ(z) is the Heaviside step function:
Φ(z) = 0 if z < 0 and Φ(z) = 1 otherwise
C7.2
x := 0, 0.005 · L..L
Plotting range and interval
5·104
M (N· m)
2.5·104
0
–2.5·104
–5·104
The displacements of end B shown above where obtained
form Table 6.2. The compatibility condition δ B = 0 is
4
n
X
Pi a2
–7.5·104
–1·105
3
RB L
w0 L
i
+
(3L − ai ) −
=0
8EI
6EI
3EI
i=1
which yields
0
1
2
3
4
5
x (m)
6
7
8
C7.3
n
3
1 X
RB = w0 L +
Pi a2i (3L − ai )
8
2L3 i=1
The expression for the bending moment is (using bracket
functions):
n
X
1
Pi hai − xi + RB (L − x)
M = − w0 (L − x)2 −
2
i=1
Let δ ij be the displacement at i due to a unit load acting
at j. The expression for δ ij can be obtained from Table 6.3
by substituting P = 1, x = ai , a = aj and b = bj . The
result is

b j ai 2


(L − a2i − b2j )
if ai ≤ aj

6L
EIδ ij = b
L
j


(ai − aj )3 + (L2 − b2j )ai − a3i if ai > aj

6L bj
C7.2 MathCad worksheet
Let δ i be the displacement at i due to the uniform loading
on intensity w0 . With x = ai Table 6.3 yields
w0 ai 3
(L − 2La2i + a3i )
EIδ i =
24
Using superposition, the condition of zero displacement at
i is
3
X
δ ij Rj − δ i = 0
Given:
L := 8 · m w0 :=5 · 103· N · m−1
 
10
2
 12 
4
3

 
n := 4
P := 
 −8  · 10 · N a :=  5  · m
15
6
j=1
183
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Applying this condition to all three middle supports, gives
us the following three simultaneous equations for the
support reactions Rj :
3
X
δ ij Rj = δ i
Computations:
i := 1, 2..3
bi := L − ai
Let ui,j = EIδ ij and vi = EIδ i ,
b j · ai 2
2
2
if ai ≤ aj
L − (ai ) − (bj )
6·L



L
ui,j :=
3
·
(a
−
a
)
·
·
·
i
j

 bj  bj
·
 otherwise

2
6·L
2
3
+ L − (bj ) · ai − (ai )
w · ai 3
vi :=
L − 2 · L · (ai )2 + (ai )3
24
i = 1, 2, 3
j=1
C7.3 Mathcad worksheet for
Part (a)
Given:


5
a :=  9  · m
12
Solve the simultaneous equations uR = v:


4.439 × 105
R := u−1 · v
R = 4.492 × 105  N
4.439 × 105
w := 100000 · N · m−1
L := 16 · m
Computations:
i := 1, 2..3
bi := L − ai
j := 1, 2..3
C7.4
Let ui,j = EIδ ij and vi = EIδ i ,
b j · ai 2
L − (ai )2 − (bj )2 if ai ≤ aj
6·L



L
ui,j :=
3

 bj  bj · (ai − aj ) · · ·
·
 otherwise

2
6·L
2
3
+ L − (bj ) · ai − (ai )
w · ai 3
L − 2 · L · (ai )2 + (ai )3
vi :=
24
Consider the right half OB of the beam. Letting MO be
the bending moment at O (due to symmetry VO = 0), we
obtain from the FBD
Solve the simultaneous equations uR = v:


5.474 × 105
R := u−1 · v
R = 2.722 × 105  N
4.177 × 105
M = MO −
B
M
diagram
θB/O = area of
EI
O
Z L
w0 x2
1
MO −
dx
0 =
EI
2Z
0 Z
L
L 2
1
w0
x
0 = MO
dx −
dx
2
0 I
R L 0 2I −1
w0 0 x I dx
MO =
RL
2
I −1 dx
Given:

3.54
a :=  8  · m
12.46
L := 16 · m
w0 x2
2
Due to symmetry, θO = 0. Thus the first area-moment
theorem yields
C7.3 Mathcad worksheet for
Part (b)

j := 1, 2..3
0
w := 100000 · N · m−1
184
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
C7.4 Mathcad worksheet for Part
(a)
C7.4 Mathcad worksheet for
Part (b)
Given:
Given:
h1 := 450 mm
h2 := 1080 mm
L := 9 m
b := 240 mm
w := 3600 N/m
x 2 0
h(x) := h1 + (h2 − h1 ) ·
L
h1 := 660 mm
h2 := 660 mm
L := 9 m
b := 240 mm
w := 3600 N/m
x 2 0
h(x) := h1 + (h2 − h1 ) ·
L
Computations:
I(x) :=
b · h(x)3
12
Z L
w0
·
M0 :=
2
0
Computations:
I(x) :=
x2 · I(x)−1 dx
Z L
b · h(x)3
12
Z L
w0
·
M0 :=
2
Bending moment at O
I(x)−1 dx
0
0
x2 · I(x)−1 dx
Z L
Bending moment at O
I(x)−1 dx
0
w0 · x2
6 · M (x)
σ(x) :=
2
b · h(x)2
x := 0, 0.01 · L..L
Plotting range and interval
w0 · x2
6 · M (x)
σ(x) :=
2
b · h(x)2
x := 0, 0.01 · L..L
Plotting range and interval
M (x) := M0 −
M (x) := M0 −
3
3
2
1
Stress (MPa)
Stress (MPa)
2.5
0
2
1.5
–1
1
–2
0.5
–3
0
1
2
3
4
5
x (m)
6
7
8
9
0
1
2
3
4
5
x (m)
6
7
8
9
185
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Given
C7.5
R 2
M C · a2
RB · a3
− C·a (3L − a) −
=0
3
6
2
w0 · L4
RB · a2
RC · L3
MC · L2
−
· (3 · L − a) −
−
=0
30
6
3
2
RB · a2
RC · L2
w0 · L3
−
−
− MC · L = 0
24
2
2


RB


 RC  := Find (RB , RC , MC )
MC
vB −
Choose RB , RC and MC as the redundant reactions. Using
superposition and Table 6.2 we get the following
displacements and rotations at the redundant supports:
w0 a2
(10L3 − 10L2a + 5La2 − a3 )
120L
RB a3
RC a2
M C a2
−
−
(3L − a) −
3
6
2
RB a2
RC L3
MC L2
w0 L4
−
(3L − a) −
−
EIδ C =
30
6
3
2
RB a2
RC L2
w0 L3
−
−
− MC L
EIθC =
24
2
2
The compatibility conditions
EIδ B =
M (x) : = MC + RC · (L − x) + RB · (a − x) · Φ(a − x)
w0 · (L − x)3
−
6·L
x := 0, 0.01 · L..L
1000
EIδ B = EIδ C = EIθ C = 0
M (N·m)
yield three simultaneous equations in the redundant
reactions. After solving these equations, the bending
moment anywhere in the beam can be evaluated from
0
–1000
3
w0 (L − x)
6L
(note the bracket function associated with RB ).
M = MC + RC (L − x) + RB ha − xi −
–2000
0
1
2
x (m)
3
4
C7.5 MathCad worksheet for
part (b)
Given:
L := 4 · m
C7.5 MathCad worksheet for
part (a)
M (N·m)
a := 2 · m
w0 := 6 · 103 · N · m−1
Computations:
EIδ B
due to
w0 · a
· (10 · L3 − 10 · L2 · a + 5 · L · a2 − a3 ) applied
vB :=
120 · L
load
0
–1000
–2000
2
RB := 1 · N RC := 1 · N MC := 1 · N · m
w0 := 6 · 103 · N · m−1
1000
Given:
L := 4 · m
a := 1.55 · m
0
1
2
x (m)
3
4
Trial
values
used
in the
solution
186
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
C7.6
C7.6 MathCad worksheet for
Part (a)
Given:
L := 40 m
b := 10 m
P1 := 200 × 103
P2 := 300 × 103 N
Computations:
L
3
x
·
· L2 − x2
if x ≤
12
4
2
v(x) := L−x
3
·
otherwise
· L2 − (L − x)2
12
4
Table 6.3 contains the following formula for the
displacement caused by a load P :
EIδ =
P bx 2
(L − x2 − b2 ) if x ≤ a
6L
We need the midspan displacement v due to a unit load
acting at an arbitrary distance x from A. This can be
obtained from above formula by substituting P = 1, a = x,
b = L − x and x = L/2. The result is
"
#
2
L
(L − x)(L/2)
2
2
L −
− (L − x)
EIv(x) =
6L
2
L−x 3 2
L − (L − x)2
if x ≥ L/2
=
12
4
RB :=
Φ(z) is the Heaviside step function: Φ(z) = 1 if z > 0
Φ(z) = 0 otherwise
x := −b, −b + 0.01 · L..L Plotting range and interval
If the load is on the left half of the span, we switch a and
b; that is, we substitute a = L − x and b = x, yielding
x 3 2
2
if x ≤ L/2
L −x
EIv(x) =
12 4
×105
5
4.5
4
3.5
Reaction (N)
Considering RB as a redundant reaction, the compatibility
condition is that the displacement of B must be zero due
to P1 , P2 and RB acting simultaneously:
P1 v(x) + P2 v(x + b) − RB v(L/2) = 0
Thus
RB =
P1 · v(x) · Φ(x) + P2 · v(x + b) · Φ(L − x − b)
L
v
2
3
2.5
2
1.5
1
0.5
P1 v(x) + P2 v(x + b)
v(L/2)
0
–10
Note that if x < 0, P1 is not on the span; consequently, the
term containing P1 must be set to zero. For the same
reason, the term containing P2 must be set to zero if
x + b > L.
–5
0
5
10
15 20
x (m)
25
30
35
40
187
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
C7.6 MathCad worksheet for
Part (b)
Given:
L := 40 m
b := 20 m
P1 := 200 × 103
P2 := 300 × 103 N
×105
4
3.5
Reaction (N)
3
2.5
2
1.5
1
0.5
0
–20
–10
0
10
x (m)
20
30
40
188
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Chapter 8
8.1
8.5
Spherical vessel, Do = 2000 mm, Di = 1800 mm.,
p = 6 MPa.
Cylinder : r = 2 − 0.025 = 1.975 m
r=
Di
= 900 mm
2
σ=
t=
pr
1.5(1.975)
=
= 59.25 MPa ◭
2t
2(0.025)
pr
= 2σ ℓ = 118.5 MPa ◭
σc =
t
√
√
Spherical cap: rsph = 2r = 2 (1.975) = 2.793 m
σℓ =
Do − Di
= 100 mm
2
pr
6(900)
=
= 27 MPa ◭
2t
2(100)
σ=
8.2
8.6
Spherical vessel, r = 0.5 m = 500 mm, t = 5.6 mm,
σ w = 30 MPa.
Maximum allowable pressure is given by
σw =
pr
2t
p=
2tσ w
2(5.6)(30)
=
= 0.672 MPa ◭
r
500
r=
360
− 3.8 = 176.2 mm
2
1.5(2.793)
prsph
=
= 83.8 MPa ◭
2t
2(0.025)
Spherical balloon, t = 0.2 mm, p = 1500 Pa,
σ ult = 10 MPa, N = 1.2.
Setting σ = σ ult /N , we get
σ ult
pr
=
N
2t
2(0.2 × 10−3 ) 10 × 106
2tσ ult
=
= 2.222 m
r =
Np
1.2(1500)
D = 2r = 4.44 m ◭
8.3
8.7
1.2(176.2)
pr
=
= 27.8 MPa ◭
2t
2(3.8)
σ c = 2σ ℓ = 55.6 MPa ◭
225
− 15 = 97.5 mm
2
The highest tensile stress in the tank is the circumferential
stress in the cylinder:
σℓ =
r=
pr
18(97.5)
=
= 117 MPa
t
15
The factor of safety is
340
σ ult
=
= 2.9 ◭
N=
σc
117
σc =
8.4
From the test results, the allowable stresses are
1 80000
2000
1 Pc
=
=
MPa
N bt
2 180t
9t
1 160000
4000
1 Pℓ
=
=
MPa
σℓ =
N bt
2 (180t)
9t
σc =
8.8
2σ w t
2(54)(30)
=
=◭
r
10 × 1000
54
σw
(1 − 0.3) = 1.89 × 10−4
(1 − ν) =
(b) ε =
E
200 × 103
∆r = εr = (1.89 × 10−4 )(10 × 1000) = 1.89 mm
(a)
In a cylindrical vessel σ c = 2σ ℓ , which means that σ c
governs.
2000
t
pr
tσ c
9t
σc =
r=
=
r = 249 m
t
p
0.9
D = 2r = 494 mm ◭
p=
If A is the surface area of the container, the change in the
volume is approximately
∆V = A ∆r = (4πr2 )∆r = 4π(10000)2(1.89)
= 2375 × 106 mm3 ◭
189
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.9
8.11
Cylindrical vessel, r = 400 mm, t = 8 mm, p = 1.2 MPa,
E = 200 GPa, ν = 0.3.
1.2 × 106 (0.4)
pr
=
= 30.0 × 106 Pa
σℓ =
2t
2(8 × 10−3 )
Cylindrical tube attached to rigid walls, r = 45 mm,
L = 180 mm, t = 3 mm, p = 3 MPa, E = 84 GPa, ν = 1/3.
σc =
σ c = 2σ ℓ = 60.0 × 106 Pa
3(45)
pr
=
= 45 MPa ◭
t
3
Rigid walls prevent longitudinal strain. Thus
σ ℓ − νσ c
=0
E
1
σ ℓ = νσ c = (45) = 15 MPa ◭
3
εℓ =
σ c − νσ ℓ
[60.0 − 0.3(30.0)] × 106
εc =
=
E
200 × 109
−6
= 255.0 × 10
∆r = εc r = 255.0 × 10−6 (400) = 0.102 mm ◭
8.12
8.10
Cylinder with hemispherical ends, Do = 400 mm, L = 600
mm, t = 18 mm, p = 3.6 MPa, E = 200 GPa, ν = 0.3.
The inner radius is
Pipe: Do = 450 mm, t = 10 mm, p = 3.5 MPa.
r=
450
Do
−t=
− 10 = 215 mm
2
2
r=
Do
400
−t=
− 18 = 182 mm
2
2
Cylindrical part:
Bolts: d = 40 mm, σ w = 80 MPa, σ init = 55 MPa, n =
number of bolts.
(a)
σℓ =
pr
(3.6 × 106 )(0.182)
=
= 18.20 × 106 Pa
2t
2(0.018)
σ c = 2σ ℓ = 36.40 × 106 Pa
[18.20 − 0.3(36.40)] × 106
σ ℓ − νσ c
(0.6)
L=
E
200 × 109
= 21.84 × 10−6 m = 0.02184 mm
∆L = εℓ L =
Pp = πr2 p = π(0.215)2 (3.5 × 106 ) = 508.3 × 103 N
Hemispherical cap:
π(0.04)2
πd2
(σ w − σ init ) = n
(80 − 55) × 106
4
4
31.42 × 103 n N
Pb = n
=
Pp = Pb
508.3 × 103 = 31.42 × 103 n
σ = σ ℓ = 18.20 × 106 Pa
(1 − ν)σ
(1 − 0.3)(18.20 × 106 )
(0.182)
r=
E
200 × 109
= 11.59 × 10−6 m = 0.01159 mm
n = 16.17
∆r = εr =
Use 17 bolts ◭
Overall change of length is
(b)
σc =
(3.5 × 106 )(0.215)
pr
=
t
0.01
= 75.3 × 106 Pa = 73.5 MPa ◭
∆L + 2∆r = 0.02184 + 2 (0.01159) = 0.0450 mm ◭
190
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
At section m-n:
8.13
(0.06P )(0.014)
M ctop
P
P
−
−
=
−6
A
I
130 × 10
7800 × 10−12
= −100 000P Pa
P
(0.06P )(0.011)
M cbot
P
σ bot =
+
+
=
A
I
130 × 10−6
7800 × 10−12
= 92 308P Pa
σ top =
Cylinder of elliptic cross section, a > b.
Maximum σ c occurs at section m-n. Letting L be the
length of the vessel, we have from the FBD:
Compression at the top:
ΣF = 0
2 (σ c )max tL − paL = 0
pa
◭
(σ c )max =
2t
−100 000P = −30 × 106
gives P = 300 N
Tension at the bottom:
Minimum σ c can be found in a similar manner. The result
is
pb
◭
(σ c )min =
2t
92 308P = 15 × 106 + gives P = 162.5 N ◭
8.17
8.14
The cross-sectional properties are
A = 0.0052 = 25 × 10−6 m2
0.0053
S =
= 20.83 × 10−9 m3
6
S =
At section m-n we have M = 250e N·m.
= 580.1 mm3
i
h
2
2
A = π R2 − r2 = π (11.25) − (11.25 − 1.88)
P
M
+
S
A
250
250e
+
150 × 106 =
20.83 × 10−9 25 × 10−6
e = 11.66 × 10−3 m = 11.66 mm ◭
σ max =
= 121.8 mm2
M
100
24000
P
+
=−
+
A
S
121.8
580.1
= 40.6 MPa (T) ◭
M
100
24000
P
=−
−
σB = − −
A
S
121.8
580.1
= −42.2 MPa (C) ◭
σA = −
8.15
σ max = −
4
4
π R4 − r 4
(11.25) − (11.25 − 1.88)
=π
4R
4 (11.25)
P
Mc
P
(P e)(h/2)
P
+
=− +
= − 2 (h − 6e)
A
I
bh
bh3 /12
bh
σ max = 0
e=
8.18
h
◭
6
8.16
0.06(0.18)2
bh2
=
= 324.0 × 10−6 m3
6
6
A = bh = 0.06(0.18) = 10.8 × 10−3 m2
S =
191
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Maximum normal stress (compression) occurs at A.
P = 24 kN
M = 24(0.18) − 18(0.54) = −5.40 kN · m
P
My
P
(P e) y
+
=
+
A
I
A
I
P (44.3)(222.3)
P
= 8.2669 × 10−5 P MPa
+
=
16950
416 × 106
|σ|max =
(−5.40)
M
24
P
−
−
=
σA =
−3
A
S
10.8 × 10
324.0 × 10−6
2
= 18 890 kN/m = 18.89 MPa (T) ◭
P
(−5.40)
M
24
σB =
+
+
=
A
S
10.8 × 10−3 324.0 × 10−6
2
= −14 440 kN/m = 14.44 MPa (C) ◭
σ w = |σ|max
P = 1088.7 kN ◭
For W360 × 101 without the plate
P = σ w A = 90(12900) = 1161 kN
which is larger than the allowable value with plate.
8.19
The properties of the cross section are
8.21
A = π(R2 − r2 ) = π(602 − 502 ) = 3455.8 mm2
π
π
S =
(R4 − r4 ) =
[604 − 504 ] = 87833.7 mm3
4R
4(60)
P
M
P
M
+
σb = −
+
S
A
S
A
20 − 100
σa + σb
A=
(3455.8)
∴P =
2
2
= 138232 N = 138.232 kN ◭
20 + 100
σa − σb
S=
(87833.7)
M =
2
2
= 5270022 N · mm = 5270.022 kN · mm ◭
σa =
8.20
90 = 8.26698 × 10−5 P
15 mm
ΣMB = 0
0.1 (P cos 40◦ ) + 1.0 (P sin 40◦ ) − 3RA = 0
RA = 0.2398P
ΣFy = 0
P sin 40◦ − RA − By = 0
By = P sin 40◦ − 0.2398P = 0.4030P
ΣFx = 0
Bx − P cos 40◦ = 0
Bx = 0.7660P
At section just to the left of C (no axial force):
M = 2RA = 2(0.2398P ) = 0.4796P N · m
M
0.4796P
M
= 2 =
= 719.4P N/m2
σ max =
S
bh /6
0.1(0.2)2 /6
270 mm
e
NA
356 mm
At section just to the right of C (axial force = Bx ):
P
y-
M = 1.0By = 0.4030P N · m
Bx
M
0.7660P
0.4030P
σ max =
+
=
+
A
S
(0.1 × 0.2) 0.1(0.2)2 /6
.A
For: A = 12900 mm2 , I¯ = 301 × 106 mm4
For the section:
= 642.8P N/m2 (smaller than at left of C)
A = 12900 + 4050 = 16950 mm2
12900(356/2) + 4050(356 + 7.5)
ΣAi ȳi
=
ȳ =
ΣAi
16950
= 222.3 mm
I = Σ I¯i + Ai (ȳi − ȳ)2
2
356
6
= 301 × 10 + 12900
− 222.3
2
270(15)3
2
+
+ 4050 (356 + 15 − 222.3)
12
= (326.4 + 89.6) × 106 = 416 × 106 mm4
356
= 44.3 mm
e = 222.3 −
2
σ w = σ max
10 × 106 = 719.4P
P = 13 900 N = 13.90 kN ◭
192
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.24
8.22
Px = P cos 17.5◦ = 0. 9537P
Py = P sin 17.5◦ = 0.3007P
M = 1.0Py − 0.2Px = 0.3007P − 0.2 (0. 9537P )
2
3
= 0.1100P N · m
3
For W130 × 28.1: A = 3590 mm , S = 167 × 10 mm .
From FBD of entire member:
3
4
RB + 0.3
RB − 1.0(45) = 0
ΣMA = 0
0.9
5
5
RB = 50.0 kN
Px
0.1100P (0.1)
M cA
0. 9537P
−
−
=
A
I
8000 × 10−6
50 × 10−6
= −100.8P Pa
Px
0.1100P (0.2)
M cB
0. 9537P
σB =
+
+
=
−6
A
I
8000 × 10
50 × 10−6
= 559.2P Pa
σA =
From FBD of segment to the right of section m-n:
With P = 100 × 103 N we get
ΣMC = 0
M + 0.2(50) − 0.325(36) = 0
M = 1.70 kN · m
ΣFaxial = 0
σ A = −10.08 × 106 Pa = 10.08 MPa (C) ◭
σ B = 55.9 × 106 Pa = 55.9 MPa (T) ◭
P = 27 kN (C)
P
1.70
M
27
±
±
=−
A
S
3590 × 10−6
167 × 10−6
= −7521 ± 10 180 kPa = −7.52 ± 10.18 MPa
σ max,min = −
σ max = 2.66 MPa ◭
8.25
σ min = −17.70 MPa ◭
From solution of Prob. 8.24:
σ A = 100.8P Pa (C)
σ B = 559.2P Pa (T)
Because the given working stress in tension (8 MPa) is
smaller than in compression (12 MPa), the tensile stress at
B governs. Therefore,
8.23
8 × 106 = 559.2P
8.26
P
B
2m
Ay
RB
60o
For W200 × 41.7: S = 398 × 10−6 m3 ,
A = 5.32 × 10−3 m2
From the FBD of the beam:
Stress at B governs. Letting |σ B | = σ w , we have
96 = 4.5162 × 10
40 kN/m
Ax A
M cA
P
630P (180)
P
= 4.694 × 10−5 P
+
=
+
A
I
99000 3078 × 106
P
M cB
P
630P (270)
σB =
−
=
−
A
I
99000 3078 × 106
= −4.5162 × 10−5 P
σA =
−5
P = 14 310 N = 14.31 kN ◭
ΣMA = 0
P = 2126 kN ◭
+
RB = 80.0 kN
ΣFx = 0
+→
◦
2RB cos 60◦ − (40 × 2)(1.0) = 0
Ax − RB sin 60◦ = 0
Ax = 80 sin 60 = 69.28 kN
193
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Maximum compressive stress occurs in the top flange at
midspan (where M is maximized).
Maximum stress (compression) occurs at the top of the
cross section at B where
2
1
M = P N·m
Paxial = − P
2
3
Mmax
Ax
+
S
A
40(2)2
w0 L2
=
= 20.0 kN · m
Mmax =
8
8
69.28
20.0
+
∴ (σ c )max =
398 × 10−6
5.32 × 10−3
2
= 63.3 × 103 kN/m = 63.3 MPa ◭
(σ c )max =
Paxial
M
P/2
2P/3
−
=−
−
A
S
0.1 × 0.4 (0.1 × 0.42 )/6
= −262.5P
σ =
Setting |σ| = σ w , we get
262.5P = 120 × 106
8.27
P = 457 × 103 N = 457 kN ◭
8.29
From FBD of frame:
ΣMA = 0
3(40) + 5(72) − 8Ey = 0
Ey = 60 kN
ΣMA = 0
From FBD of member BD:
ΣMB = 0
1(40) + 3(72) − 4Dy = 0
2(64) − 4(60) + 2D = 0
Dy = 64 kN
Dx = 56 kN
σ =
M
120 × 103
100 × 103
P
−
=−
−
A
S
0.1 × 0.3
0.1 × 0.32 /6
= −70.7 × 106 Pa = 70.7 MPa (C) ◭
Mmax = 64 kN·m occurs at the 72-kN load.
σ max =
NC = 60 kN
Maximum compressive stress occurs at the section just
below B where
5
P = −120 kN
M = (60) = 100 kN · m
3
From FBD of member CDE:
ΣMC = 0
2(150) − 5NC = 0
P
Mmax
56 × 103
64 × 103
+
=
+
A
S
0.1 × 0.4 (0.1 × 0.42 )/6
8.30
= 25.4 × 106 Pa = 25.4 MPa (T) ◭
8.28
ΣMA = 0
ΣFx = 0
ΣFy = 0
4
T
5
ΣFx′ = 0:
5
2P − 3
=0
T = P
6
1
3
Ax = P
Ax − T = 0
5
2
4
2
Ay − P + T = 0
Ay = P
5
3
σ dA + (60 dA cos 30◦ ) cos 30◦ − (40 dA sin 30◦ ) sin 30◦ = 0
ΣFy′ = 0:
σ = −35 MPa ◭
τ dA − (60 dA cos 30◦ ) sin 30◦ − (40 dA sin 30◦ ) cos 30◦ = 0
τ = 43.3 MPa ◭
194
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
σx + σ y
σ x − σy
−
cos 2θ − τ xy sin 2θ
2
2
= 5 − 35 cos 60◦ − 0 = −12.5 MPa ◭
8.31
σy′ =
σx − σ y
sin 2θ + τ xy cos 2θ
2
◦
= −35 sin 60 + 0 = −30.3 MPa ◭
τ x′ y ′ = −
ΣFx′ = 0:
3
4
σ dA − 1280(0.8 dA) − 2400(0.6 dA) = 0
5
5
σ = 1683 N/m2 ◭
ΣFy′ = 0:
4
3
τ dA − 2400(0.6 dA) + 1280(0.8 dA) = 0
5
5
8.34
2
τ = 538 N/m ◭
σx = σy = 0
8.32
τ xy = 8 MPa
θ = −30◦
σx + σ y
σ x − σy
+
cos 2θ + τ xy sin 2θ
2
2
◦
= 0 + 0 + 8 sin(−60 ) = −6.93 MPa ◭
σ x′ =
x'
τ
dA
25o dA cos25o
y'
dA cos25o
σ
σx + σ y
σ x − σy
−
cos 2θ − τ xy sin 2θ
2
2
◦
= 0 + 0 − 8 sin(−60 ) = 6.93 MPa ◭
σy′ =
12 MPa
12 MPa
6 MPa
ΣFx′ = 0:
◦
σx − σy
sin 2θ + τ xy cos 2θ
2
= 0 + 8 cos(−60◦ ) = 4.0 MPa ◭
τ x′ y ′ = −
◦
σ dA + (6 dA cos 25 ) cos 25
−(12 dA cos 25◦ ) sin 25◦ − (12 dA sin 25◦ ) cos 25◦ = 0
σ + 6 cos2 25◦ − 24 cos 25◦ sin 25◦ = 0
σ = 4.26 MPa ◭
ΣFy′ = 0:
τ dA − (6 dA cos 25◦ ) sin 25◦
−(12 dA cos 25◦ ) cos 25◦ + (12 dA sin 25◦ ) sin 25◦ = 0
τ − 6 cos 25◦ sin 25◦ − 12(cos2 25◦ − sin2 25◦ ) = 0
8.35
τ = 10.01 MPa ◭
σ x = 60 MPa
σ y = 80 MPa
θ = −35◦
8.33
σ x = 40 MPa
σ y = −30 MPa
τ xy = 0
θ = 30
σx + σ y
σ x − σy
= 5 MPa
= 35 MPa
2
2
σ x + σy
= 70 MPa
2
sin 2θ = −0.9397
◦
τ xy = −100 MPa
σx − σ y
= −10 MPa
2
cos 2θ = 0.3420
σx − σy
σx + σy
+
cos 2θ + τ xy sin 2θ
2
2
= 70 + (−10) (0.3420) + (−100)(−0.9397)
σ x′ =
σx − σy
σx + σy
+
cos 2θ + τ xy sin 2θ
2
2
◦
= 5 + 35 cos 60 + 0 = 22.5 MPa ◭
σ x′ =
= 160.6 MPa ◭
195
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
σ x + σy
σx − σ y
−
cos 2θ − τ xy sin 2θ
2
2
= 70 − (−10)(0.3420) − (−100)(−0.9397)
σ x − σy
σx + σ y
+
cos 2θ + τ xy sin 2θ
2
2
= 3.5 + 2.0(−0.1736) + 11.3(−0.9848)
σ y′ =
σ x′ =
= −20.6 MPa ◭
= −7.98 MPa ◭
σx − σ y
sin 2θ + τ xy cos 2θ
2
= −(−10)(−0.9397) + (−100)(0.3420)
= −43.6 MPa ◭
τ x′ y ′ = −
43.6 MPa
σx + σ y
σ x − σy
−
cos 2θ − τ xy sin 2θ
2
2
= 3.5 − 2.0(−0.1736) − 11.3(−0.9848)
= 14.98 MPa ◭
σy′ =
20.6 MPa
160.6 MPa
σx − σ y
sin 2θ + τ xy cos 2θ
2
= −2(−0.9848) + 11.3(−0.1736)
= 0.01 MPa ◭
τ x′ y ′ = −
x
35o
x'
7.98 MPa
0.01 MPa
14.98 MPa
8.36
x
σ x = σ y = −8 MPa
τ xy = −10 MPa
σ x + σy
= −8 MPa
2
sin 2θ = 0.8660
θ = 60◦
50o
σx − σy
=0
2
cos 2θ = −0.5
x'
8.38
σx + σy
σ x − σy
+
cos 2θ + τ xy sin 2θ
2
2
= −8 + 0 + (−10)(0.8660) = −16.66 MPa ◭
σ x′ =
σx − σy
σx + σy
−
cos 2θ − τ xy sin 2θ
2
2
= −8 − 0 − (−10)(0.8660) = 0.66 MPa ◭
σ y′ =
σx = 0
σ y = 60 MPa
s
2
σ x − σy
R =
2
= 42.43 MPa
σx − σy
sin 2θ + τ xy cos 2θ
2
= 0 + (−10)(−0.5) = 5.0 MPa ◭
τ x′ y ′ = −
+ τ 2xy =
τ xy = 30 MPa
s
0 − 60
2
2
+ 302
σ x + σy
0 + 60
±R=
± 42.43
2
2
σ 1 = 72.4 MPa ◭
σ 2 = −12.4 MPa ◭
σ 1,2 =
30
τ xy
=
= 0.7070
R
42.43
σ x − σy
0 − 60
cos 2θ1 =
=
= −0.7070
2R
2(42.43)
2θ1 = 135◦
θ1 = 67.5◦ ◭
sin 2θ1 =
8.37
σ x = 5.5 MPa
θ = −50
σ y = 1.5 MPa
τ xy = 11.3 MPa
◦
σ x + σy
= 3.5 MPa
2
sin 2θ = −0.9848
σx − σy
= 2.0 MPa
2
cos 2θ = −0.1736
196
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.41
8.39
σ x = −4 MPa
σ y = 6 MPa
s
σ x − σy
R =
2
= 9.434 MPa
2
+ τ 2xy =
s
σ x = 30 MPa
τ xy = 8 MPa
−4 − 6
2
2
s
σ x − σy
R =
2
= 91.24 MPa
+ 82
σy = 0
2
τ xy = 90 MPa
+ τ 2xy =
s
30 − 0
2
2
+ 902
σ x + σy
30 + 0
±R=
± 91.24
2
2
σ 1 = 106.2 MPa ◭
σ 2 = −76.2 MPa ◭
σx + σy
−4 + 6
±R=
± 9.434
2
2
σ 1 = 10.43 MPa ◭
σ 2 = −8.43 MPa ◭
σ 1,2 =
σ 1,2 =
τ xy
90
=
= 0.9864
R
91.24
σ x − σy
30 − 0
cos 2θ1 =
=
= 0.1644
2R
2(91.24)
2θ1 = 80.5◦
θ1 = 40.3◦ ◭◭
8
τ xy
=
= 0.8480
sin 2θ1 =
R
9.434
σx − σy
−4 − 6
cos 2θ1 =
=
= −0.5300
2R
2(9.434)
2θ1 = 122.0◦
θ 1 = 61.0◦ ◭
sin 2θ1 =
8.42
8.40
σ x = −10 MPa
σ x = σ y = 6 MPa
R=
s
σx − σ y
2
2
+ τ 2xy =
σy = 0
σx − σ y
= −5 MPa
2
τ xy = −12 MPa
p
02 + (−12)2 = 12 MPa
σ̄ =
τ xy = −10 MPa
σx + σy
= −5 MPa
2
s
2
q
σx − σ y
2
+ τ xy2 = (−5) + (−10)2
2
= 11.18 MPa ◭
τ max =
6+6
σx + σy
±R =
± 12 = 6 ± 12 MPa
2
2
σ 1 = 18 MPa ◭
σ y = −6 MPa ◭
σ 1,2 =
−5
σx − σy
=−
= −0.50
2τ xy
−10
2θ = −26.57◦
θ = −13.28◦ ◭
tan 2θ = −
τ xy
σx − σy
= −1.0
cos 2θ1 =
=0
R
2R
2θ1 = −90◦
θ1 = −45◦ ◭
sin 2θ1 =
σx − σ y
sin 2θ + τ xy cos 2θ
2
= −(−5) sin(−26.57◦) + (−10) cos(−26.57◦)
τ x′ y ′ = −
6 MPa
= −11.18 MPa
45o
18 MPa
197
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.43
8.45
σ x = −30 MPa
σ y = 30 MPa
τ xy = −40 MPa
σx − σ y
σx + σy
= −30 MPa
σ̄ =
=0
2
2
s
2
q
σx − σ y
τ max =
+ τ xy2 = (−30)2 + (−40)2
2
= 50 MPa ◭
σ x = 4 MPa
σ y = −6 MPa
σx − σ y
= 5 MPa
2
τ xy = 3 MPa
σx + σ y
σ̄ =
= −1 MPa
2
s
2
p
σx − σy
+ τ xy2 = 52 + 32 = 5.83 MPa ◭
2
−5.0
σx − σy
=−
= −1.6667
tan 2θ = −
2τ xy
3
2θ = −59.04◦
θ = −29.5◦ ◭
τ max =
σx − σy
−30
=−
= −0.75
2τ xy
−40
2θ = −36.87◦
θ = −18.43◦ ◭
tan 2θ = −
σx − σy
sin 2θ + τ xy cos 2θ
2
= −5.0 sin(−59.04◦) + 3 cos(−59.04◦) = +5.83 MPa
τ x′ y ′ = −
σx − σy
sin 2θ + τ xy cos 2θ
2
= −(−30) sin(−36.87◦) + (−40) cos(−36.87◦)
τ x′ y ′ = −
= −50 MPa
8.46
8.44
R=
σ x = 70 MPa
σ y = 50 MPa
s
2
σx − σ y
+ τ 2xy =
τ max =
2
= 31.6 MPa ◭
τ xy = 30 MPa
s
70 − 50
2
2
p
102 + 102 = 14.14 MPa ◭
σ̄ = 0 ◭
+ 302
σx − σ y
70 − 50
=−
= −0.3333
2τ xy
2(30)
2θ = −18.435◦
θ = −9.22◦ ◭
tan 2θ = −
8.47
R = 10 MPa ◭
σx − σy
sin 2θ + τ xy cos 2θ
2
70 − 50
sin(−18.435◦) + 30 cos(−18.435◦)
= −
2
= 31.6 MPa
τ x′ y ′ = −
σ̄ = −10 MPa ◭
60 MPa
31.6 MPa
9.22o
60 MPa
198
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.48
8.52
R=0 ◭
8.49
σ̄ = −p ◭
σ x′ = −8 − 10 cos 30◦ = −16.66 MPa ◭
σ y′ = −8 + 10 cos 30◦ = 0.66 MPa ◭
t
60 MPa
x
y
x
20 MPa
−60
30
30 MPa
20
s
y
Stresses in MPa
p
402 + 302 = 50 MPa ◭ σ̄ = −20 MPa ◭
8.53
t
y'
10
50
60
R=
R
τ x′ y′ = 10 sin 30◦ = 5.0 MPa ◭
30
−20
8.50
20 60°
y
−40
x
s
80
52.0
x
120°
σ x′ = −8 sin 60◦ = −6.93 MPa ◭
σ y′ = 8 sin 60◦ = 6.93 MPa ◭
τ x′ y′ = 8 cos 60◦ = 4.0 MPa ◭
60°
x'
Stresses in MPa
x'
σ x′ = 20 − 60 cos 60◦ = −10 MPa ◭
σ y′ = 20 + 60 cos 60◦ = 50 MPa ◭
τ x′ y′ = 60 sin 60◦ = 52.0 MPa ◭
8.54
x'
t
8.51
4.33
5.88
x
50°
2 α
−4
0.33
R
y'
8
2
2
s
x'
25°
y
y'
p
62 + 22 = 6.325 MPa
2
α = tan−1 = 18.43◦
6
σ x′ = 2 − 6.325 cos(50◦ + 18.43◦ ) = −0.33 MPa ◭
R =
p
R =
602 + 402 = 72.11 MPa
40
= 56.31◦
a = 90◦ − tan−1
60
σ y′ = 2 + 6.325 cos(50◦ + 18.43◦ ) = 4.33 MPa ◭
τ x′ y′ = 6.325 sin(50◦ + 18.43◦) = 5.88 MPa ◭
σ x′ = 72.11 cos 56.31◦ = 40.0 MPa ◭
σ y′ = −σ x′ = −40.0 MPa ◭
τ x′ y′ = 72.11 sin 56.31◦ = 60.0 MPa ◭
199
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.55
8.57
p
352 + 302 = 46.10 MPa
30
= 29.40◦
α = 180◦ − 110◦ − tan−1
35
R =
(a)
p
√
22 + 62 = 40 MPa
6
θ1 = 35.8◦
2θ1 = tan−1 = 71.57◦
2
R =
σ x′ = 25 + 46.10 cos 29.40◦ = 65.2 MPa ◭
σ y′ = 25 − 46.10 cos 29.40◦ = −15.2 MPa ◭
τ x′ y′ = 46.10 sin 29.40◦) = 22.6 MPa ◭
√
σ 1 = −6 + 40 = 0.32 MPa ◭
√
σ 2 = −6 − 40 = −12.33 MPa ◭
(b)
8.56
τ max = R =
√
40 = 6.32 MPa ◭
8.58
p
252 + 802 = 83.82 MPa
80
= 47.35◦
α = 120◦ − tan−1
25
R =
(a)
σ x′ = −25 − 83.82 cos 47.35◦ = −81.8 MPa ◭
σ y′ = −25 + 83.82 cos 47.35◦ = 31.8 MPa ◭
p
√
62 + 62 = 72 MPa
6
θ 1 = 22.5◦ ◭
2θ1 = tan−1 = 45◦
6
R =
τ x′ y′ = 83.82 sin 47.35◦ ) = 61.7 MPa ◭
√
σ 1 = −2 + 72 = 6.49 MPa ◭
√
σ 2 = −2 − 72 = −10.49 MPa ◭
(b)
τ max = R =
√
72 = 8.49 MPa ◭
200
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.59
(a)
115.6
t
y
90
−60
p
402 + 302 = 50 MPa
30
= 36.87◦
θ 2 = 18.4◦ ◭
2θ2 = tan−1
40
R =
31.7°
R tmax
30
2
85.6
−15
1
2q1
90
s
σ 1 = 40 + 50 = 90 MPa ◭
σ 2 = 40 − 50 = −10 MPa ◭
15
76.7°
(b)
15
τ max = R = 50 MPa ◭
x
100.6
Stresses in MPa
8.62
(a)
p
R =
452 + 902 = 100.6 MPa
90
= 63.4◦
θ1 = 31.7◦ ◭
2θ1 = tan−1
45
σ 1 = −15 + 100.6 = 85.6 MPa ◭
(a) Transform the shear stress due to the second loading to
the xy-axes:
σ 2 = −15 − 100.6 = −115.6 MPa ◭
(b)
τ max = R = 100.6 MPa ◭
8.60
σ x = 4 sin 60◦ = 3.464 MPa
◦
τ xy = 4 cos 60 = 2 MPa
σ y = −σ x = −3.464 MPa
Superimpose the stresses acting on the xy-planes caused
by the two loadings:
σ x = 0 + 3.464 = 3.464 MPa
σ y = 0 − 3.464 = −3.464 MPa
(a)
p
√
R =
62 + 42 = 52 MPa
4
θ2 = 73.2◦ ◭
2θ2 = 180◦ − tan−1 = 146.31◦
6
√
σ 1 = −12 + 52 = −4.79 MPa ◭
√
σ 2 = −12 − 52 = −19.21 MPa ◭
(b)
τ max = R =
τ xy = 3 + 2 = 5 MPa
√
52 = 7.21 MPa ◭
8.61
p
3.4642 + 52 = 6.083 MPa
5
= 55.29◦
θ 1 = 27.6◦ ◭
2θ1 = tan−1
3.464
σ 1 = R = 6.08 MPa ◭
σ 2 = −R = −6.08 MPa ◭
R =
201
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.63
8.64
(a) Transform the shear stress due to the second loading to
the xy-axes:
Stress state (b):
20
t
x
x'
90°
y'
−20
y'
x'
−10
45°
s
10
20
y
y
x
σ x = 4 MPa
σ y = −4 MPa
τ xy = 0
Stresses in MPa
Superimposing stress states (a) and (b):
Superimpose the stresses acting on the xy-planes caused
by the two loadings:
σ x = 0 + 4 = 4 MPa ◭
10
30
20
σ y = 0 − 4 = −4 MPa ◭
y
t
x
τ xy = 3 + 0 = 3 MPa ◭
10
x
(b)
R
20
2
30
−10
10
45°
1
38.3
20
y
22.5°
x
18.3
Stresses in MPa
R =
p
R =
42 + 32 = 5 MPa
3
θ1 = 18.4◦ ◭
2θ1 = tan−1 = 36.87◦
4
σ 1 = R = 5 MPa ◭
σ 2 = −R = −5 MPa ◭
p
202 + 202 = 28.3 MPa
σ 1 = 10 + 28.3 = 38.3 MPa ◭
σ 2 = 10 − 28.3 = −18.3 MPa ◭
θ2 = 22.5◦ ◭
202
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.65
8.68
(a) τ max = 20 MPa ◭
8.69
◦
Plot point x .
Knowing that R = τ max = 10 MPa, construct diameter of
circle.
There are two ways to draw the circle; hence there are two
solutions.
p
AB = 102 − 82 = 6 MPa
(b) τ abs = 40 MPa ◭
t
xy-plane
−6
tmax = tabs
10
s
Stresses
in MPa
(a) τ max = 8 MPa ◭
(b) τ abs = 8 MPa ◭
σ y = 30 ± 2AB = 30 ± 2(6) MPa
σ y = 18 MPa or 42 MPa ◭
8.70
8.66
(a) τ max =
p
(75 − 30)2 + 302 = 54.1 MPa ◭
(b) τ abs = τ max = 54.1 MPa ◭
12 + σ x
12 − σ x
σ aa =
+
cos 60◦ = 4 MPa
2
2
σ x = −20 MPa ◭
8.71
8.67
(a) τ max = 1.5 MPa ◭
p
(a) τ max = (3.6 − 2.4)2 + 1.22 = 1.697 MPa ◭
2.4 + 1.697
2.4 + τ max
=
= 2.05 MPa ◭
(b) τ abs =
2
2
(b) τ abs = 2.5 MPa ◭
203
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.72
8.76
p
(a) τ max = (80 − 55)2 + 152 = 29.2 MPa ◭
55 + τ max
55 + 29.2
(b) τ abs =
=
= 42.1 MPa ◭
2
2
8.73
t
xy-plane
tmax
24
−18
−32
−25
R
s
Because σ w > 2τ w and |σ max | = 2τ max , shear stress
governs.
1P
τ max = τ w
= τw
2A
24
x
(a)
R=
20 + 15
= 17.5 MPa ◭
2
8.77
y
−50
τ abs =
p
242 + 72 = 25 MPa
τ max = R = 25 MPa ◭
P = 2τ w A = 2 20 × 106
π (0.06)2
4
= 113.1 × 10 N = 113.1 kN ◭
(b) Where are the other two Mohr’s circles that should
appear in the figure? One coincides with the circle shown,
the other one is a point at the origin.
Therefore,
τ abs = τ max = 25 MPa ◭
3
8.78
8.74
τ abs = 10 MPa ◭
σx =
8.75
t
zx-plane
tabs
−10
Let the x′ -axis be perpedicular to the plane of the joint.
From Mohr’s circle:
yz-plane
xy-plane
10
20
P
200 × 103
=
= 40 × 106 Pa = 40 MPa
A
0.05 × 0.1
s
σ x′ = 20 − 20 cos 80◦ = 16.53 MPa ◭
τ x′ y′ = 20 sin 80◦ = 19.70 MPa ◭
Stresses in MPa
20 − (−10)
τ abs =
= 15 MPa ◭
2
204
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.81
8.79
P
600 × 103
=−
= −0.0530×103 MPa = −53 MPa
A
π(120)2 /4
Assuming that shear stress govens:
For thin-walled tube:
σx = −
2
D
πD3 t
=
2
4
π(1.5 × 1000)3(8)
=
= 21.2 × 109 mm4
4
J = AR2 = (πDt)
τ max = τ w
R = 60 MPa
Assuming that normal stress governs:
53
+ R = 78
2
R = 51.5 MPa ◭ use
s
2
53
= 44.2 MPa
τ xy = 51.52 −
2
(3000 × 1000)(.75 × 1000)
TR
= 0.1061 GPa
=
J
21.2 × 109
= 106.1 MPa
|σ|max = σ w
τ xy =
From Mohr’s circle:
πd3 rxy
π(120)3 (44.2)
=
= 14996.7 kN · mm
16
16
2π(250)(14996.7) × (10−3 )
2πN T
=
= 392.6 kw
P=
60
60
σ x′ = 106.1 sin 40◦ = 68.2 MPa ◭
T =
τ x′ y′ = 106.1 cos 40◦ = 81.3 MPa ◭
8.80
8.82
π 2
π
(D − d2 ) = (902 − 752 ) = 1943.86 mm2
4
4
108000
P
= −55.6 MPa
σl = − = −
A
1943.86 θ
0.8π/180
τ cl = G r = (84 × 103 )
(45) = 10.6 MPa
L
5000
A =
P
600 × 103
=−
= −76.39 × 106 Pa
A
π(0.1)2 /4
= −76.39 MPa
θ
1.5π/180
dθ
(0.05)
τ xy = G r = G r = (80 × 109 )
dx
L
8
σx = −
= 13.09 × 106 Pa = 13.09 MPa
s
σl
±
σ 1,2 =
2
= −
2
76.39
τ max =
+ 13.092 = 40.4 MPa ◭
2
76.39
|σ|max =
+ 40.4 = 78.6 MPa ◭
2
r σ 2
55.6
−
2
l
2
s
+ τ 2cl
55.6
2
2
+ 10.62
|σ|max = |σ 2 | = 57.6 MPa ◭
s
r 2
σl 2
55.6
2
τ max =
+ τ cl =
+ 10.62
2
2
= 29.8 MPa ◭
205
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.85
8.83
T =
2.5 × 103
P
=
= 198.94 N · m
2πf
2π(2)
tcl
c
sl
l
32(5500 × 103 )
32M
=
= 76.8 MPa
πd3
π(90)3
Assuming that shear stress governs:
σy =
50 × 103
P
=
−
= −70.36 × 106 Pa
πd2 /4
π(0.03)2 /4
= −70.36 MPa
16(198.94)
16T
=
= 37.53 × 106 Pa = 37.53 MPa
τ cl =
3
πd
π(0.03)3
σl = −
τ max = τ w
R = 112 MPa
Assuming that normal stress governs:
76.8
+ R = 108
2
R = 69.6 MPa ◭ (use)
s
2
76.8
τ xy = 69.62 −
= 58 MPa
2
σ max = σ w
σl
σ 1,2 =
±
2
= −
r σ 2
l
70.36
±
2
2
s
+ τ 2cl
−
70.36
2
|σ|max = |σ 2 | = 86.6 MPa ◭
8.84
2
+ 37.532
π(90)3 (58)
πd3 τ xy
=
= 8302 kN · mm
16
16
T =
8.86
z
M
x
V
T
y
At cross section A:
At the critical point B:
M = (16 + 8.4)(0.24) = 5.856 kN · m
32M
4P
+ 2
πd3
πd
32(125 × 103 )(0.02) 4(125 × 103 )
+
=
π(0.08)3
π(0.08)2
σy =
T = (16 − 8.4)(0.3) = 2.28 kN · m
Stresses acting on the element shown:
= 74.60 × 106 Pa = 74.60 MPa
M
5.865
59.74
σz =
kPa
=
=
S
πd3 /32
d3
16(2.28)
11.612
16T
=
=
kPa
τ zx =
3
3
πd
πd
d3
r σz 2
+ τ 2zx
τ max =
2
1p
80 × 103 = 3 59.742 + 11.6122
d
d = 0.0913 m = 91.3 mm ◭
Assuming that shear stress governs:
τ max = τ w
R = 80 MPa
Assuming that normal stress governs:
74.60
+ R = 100
2
R = 62.70 MPa ⊳ use
σ max = σ w
τ xy =
T =
s
62.702 −
74.60
2
2
= 50.40 MPa
π(0.08)3 (50.40 × 106 )
πd3 τ xy
=
= 5070 N · m ◭
16
16
206
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.87
8.89
τ xy =
At the critical point B:
σx =
32M
32(1.2 × 103 )
=
= 97.78 × 106 Pa = 97.78 MPa
πd3
π(0.05)3
At point A:
Assuming that shear stress governs:
τ max = τ w
σx =
R = 80 MPa ◭ (use)
4P
32M
4(160 × 103 ) 32(6 × 103 )
+
=
+
πd2
πd3
π(0.1)2
π(0.1)3
= 81.49 × 106 Pa = 81.49 MPa
Assuming that normal stress governs:
97.78
σ max = σ w
+ R = 140
2
R = 91.11 MPa
s
2
97.78
= 63.32 MPa
τ xy =
802 −
2
T =
16T
16(9 × 103 )
=
= 45.84 × 106 Pa = 45.84 MPa
πd3
π(0.1)3
s
2
81.49
τ max =
+ 45.842 = 61.3 MPa ◭
2
81.49
σT =
+ 61.3 = 102.0 MPa ◭
2
π(0.05)3 (63.32 × 106 )
πd3 τ xy
=
= 1554 N · m ◭
16
16
8.88
At point B:
σx =
4P
32M
4(160 × 103 ) 32(6 × 103 )
−
=
−
πd2
πd3
π(0.1)2
π(0.1)3
= −40.74 × 106 Pa = −40.74 MPa
At the critical point C:
4(2 × 103 )(1.2)
9600
4M
=
=
Pa
πr3
πr3
πr3
3
2(2 × 10 )(1.5)
6000
2T
=
=
Pa
τ xy =
3
3
πr
πr
πr3
p
(9600/2)2 + 60002
7684
τ max =
=
Pa
3
πr
πr3
Assuming that shear stress governs:
σx =
τ max =
σC =
s
40.74
2
+ 45.842 = 50.2 MPa
40.74
+ 50.2 = 70.6 MPa ◭
2
7684
= 60 × 106
r = 0.0344 m
πr3
Assuming that normal stress governs:
τ max = τ w
9600/2 + 7684
= 120 × 106
πr3
r = 0.0321 m = 32.1 mm ◭
σ max = σ w
207
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
π
π
(D4 − d4 ) =
(4804 − 4504 ) = 1.1857 × 109 mm4
32
32
1.1857 × 109
J
=
= 0.5929 × 109 mm4
I =
2
2
480
D
−t=
− 15 = 225 mm (inner radius)
r =
2
2
8.90
J =
At the critical point B:
pr
5(225)
=
= 75 MPa
t
15
5(225) (30 × 106 )(480/2)
pr M (D/2)
+
=
+
σl =
2t
I
2(15)
0.5929 × 109
= 49.6 MPa
T (D/2)
(120 × 106)(480/2)
τ lc =
= 24.3 MPa
=
J
1.1857 × 109
σc =
At the section containing A and B:
V = 2.5 kN
M = 2.5(0.8) = 2.0 kN · m
T = 2.5(0.6) = 1.5 kN · m
Considering the cross section as a difference of the outer
160-mm and the inner 150-mm squares:
From Mohr’s circles:
s
2
75 − 49.6
+ 24.32 = 27.4 MPa
R =
2
75 + 49.6
σ max =
+ 27.4 = 89.7 MPa ◭
2
σ max
89.7
τ abs =
=
= 44.85 MPa
2
2
0.164 − 0.154
I =
= 12.426 × 10−6 m4
12
Q = (0.16 × 0.08)(0.04) − (0.15 × 0.075)(0.0375)
= 90.125 × 10−6 m3
At point A:
(2.0 × 103 )(0.08)
Mc
= 12.876 × 106 Pa
=
I
12.426 × 10−6
T
1.5 × 103
τ =
=
= 6.244 × 106 Pa
2A0 t
2(0.155)2 (0.005)
(A0 is the area enlosed by the median line)
s
r 2
σ 2
12.876
+ τ2 =
+ 6.2442 × 106
τ max =
2
2
σ =
8.92
= 8.97 × 106 Pa = 8.97 MPa ◭
At point B:
T
VQ
+
2A0 t I(2t)
(2.5 × 103 )(90.125 × 10−6 )
= 6.244 × 106 +
(12.426 × 10−6 )(2 × 0.005)
σ max = τ =
σc =
pr
2t
pr
t
σt =
pr
t
τ abs =
From Mohr’s circles:
σ max =
= 8.06 × 106 Pa = 8.06 MPa ◭
pr
2t
Because σ w > 2τ w , shear stress governs.
2r
pr
= τw
= 30
2t
2(15)
r = 450 mm d = 2r = 900 mm (inner dia.) ◭
τ abs = τ w
8.91
208
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.93
Assume that normal stress governs:
π
π
A = (D2 − d2 ) = (6002 − 5702 ) = 27567.5 mm2
4
4
600
D
−t=
− 15 = 285 mm (inner radius)
r =
2
2
43.5 + 21.75
σ w = σ 1 100 =
+
2
τ ℓc = 66.5 MPa
pr
1.5(285)
=
= 28.5 MPa
t
15
1.5(285) 120 × 103
pr P
+
=
+
= 18.6 MPa
σℓ =
2t
A
2(15)
27567.5
28.5
σ max = 28.5 MPa ◭
τ abs =
= 14.25 MPa
2
s
43.5 − 21.75
2
2
2
+ rℓc
σc =
Assume that shear stress governs:
s
2
43.5 − 21.75
2
+ rℓc
τ w = τ max 50 =
2
τ ℓc = 48.8 MPa ⇐= (use)
48.8(439.7 × 106 )
τ ℓc J
=
= 119.2 kN · m ◭
T =
r
180
8.94
8.96
t
c
A = 27567.5 mm2
tcl
c
b
l
a
r = 285 mm (inner radius)
sl
0.9(285)
pr
=
= 17.1 MPa
t
15
pr P
0.9(285) (−300 × 103 )
σℓ =
+
=
+
= −2.33 MPa
2t
A
2(15)
27567.5
17.1 + 2.33
σ max = 17.1 MPa ◭
τ max =
= 9.7 MPa
2
R
weld
16 174
60°
s
10 020
From the solution to Prob. 8.93:
l
σc =
P
200
=
= 16 174 Pa
2
A
π(0.5 − 0.4842 )/4
16(30)(0.5)
16T D
=
= 10 020 Pa
τ cl =
π(D4 − d4 )
π(0.54 − 0.4844 )
σl =
s
2
16 174
+ 10 0202 = 12 876 Pa
2
10 020
α = sin−1
= 51.10◦
12 876
β = 60◦ − 51.10◦ = 8.90·
16 174
+ 12 876 cos 8.90◦
σ weld =
2
= 20 800 Pa = 20.8 kPa ◭
R =
8.95
J = Ar̄2 = (2πr̄t)r̄2 = 2πr̄3 t = 2π(180)3 (12)
= 439.7 × 106 mm4
pri
3(180 − 12/2)
σc =
+
= 43.5 MPa
t
12
σc
= 21.75 MPa
σℓ =
2
209
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.97
M2
A
T
8.99
M
t
M1
s
The highest stresses occur at point A located at the
support. The stress resultants at the support are
M1 = 120(0.2) = 24 N · m
M2 = 180(0.2) = 36 N · m
p
242 + 362 = 43.27 N · m
M =
T = 120(0.15) = 18 N · m
The corresponding stresses at point A are
At the critical section, which is just to the left of C:
p
12002 + 3002 = 1236.9 N · m
M =
32(43.27)
32M
=
= 130.59 × 106 Pa = 130.59 MPa
3
πd
π(0.015)3
16(18)
16T
=
= 27.16 × 106 Pa = 27.16 MPa
τ=
πd3
π(0.015)3
s
r 2
σ
130.59
σ 2
130.59
2
σ1 = +
+τ =
+ 27.162
+
2
2
2
2
= 136.0 MPa ◭
σ=
T = 300 N · m
32(1236.9)
32M
=
= 100.79 × 106 Pa
3
πd
π(0.05)3
= 100.79 MPa
16T
16(300)
τ xy =
=
= 12.22 × 106 Pa = 12.22 MPa
πd3
π(0.05)3
σx =
8.98
r σ
x
2 =
+ rxy
s
100.79
2
2
+ 12.222
2
= 51.9 MPa ◭
100.79
σx
+ τ max =
+ 51.9 = 102.3 MPa ◭
σ max =
2
2
τ max =
8.100
At the critical section, which is just to the left of C:
M = 1500 N · m
T = 300 N · m
32M
16T
σx −
τ xy =
3
πs
πd3
τ max =
r σ
x
2 =
+ rxy
16 p 2
M + T2
πd3
2
p
16
15002 + 3002
80 × 106 =
πd3
d = 0.0460 m = 46.0 mm ◭
σx =
32M
πd3
r σ
τ xy =
16T
πd3
16 p 2
M + T2
2
πd3
p
16
σx
[M + M 2 + T 2 ]
+ τ max =
σ max =
3
2
πd
τ max =
x
+ τ 2xy =
210
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Assuming that stresses at section B govern:
8.102
16 p
1762 + 542
d = 4.5 mm
πd3
p
16 25 =
d = 4.2 mm
176 + 1762 + 542
3
πd
10 =
bh3
20(120)3
=
= 2.88 × 106 mm4 = 2.88 × 10−6 m4
12
12
Q = A′ ȳ ′ = (20 × 40)(40) = 32 × 103 mm3 = 32 × 10−6 m3
Assuming that stresses at section D govern:
I=
16 p
1202 + 1802
d = 4.8 mm ◭
πd3
p
16
2 + 1802
120
+
120
d = 4.1 mm
25 =
πd3
10 =
MyB
40000
P
(30000 × 0.3)(0.02)
+
+
=−
I
A
2.88 × 10−6
0.02 × 0.12
= −45.83 × 106 Pa
VQ
30000(32 × 10−6 )
τ xy =
=
= 16.67 × 106 Pa
Ib
(2.88 × 10−6 )(0.02)
σx = −
8.101
From Mohr’s circle:
bh3
20(120)3
=
= 2.88 × 106 mm4 = 2.88 × 10−6 m4
12
12
Q = A′ ȳ ′ = (20 × 40)(40) = 32 × 103 mm3 = 32 × 10−6 m3
I=
R =
s
45.83
2
2
2α = 60◦ − sin−1
40000
(30000 × 0.25)(0.02)
M yA P
+
+
=
−6
I
A
2.88 × 10
0.02 × 0.12
= 68.75 × 106 Pa
30000(32 × 10−6 )
VQ
=
= 16.67 × 106 Pa
τ xy =
Ib
(2.88 × 10−6 )(0.02)
σx =
+ 16.672 = 28.34 MPa ◭
16.67
= 23.97◦
28.34
45.83
+ 28.34 cos 23.97◦ = 2.98 MPa ◭
2
τ x′ y′ = 28.34 sin 23.97◦ = 11.51 MPa ◭
σ y′ = −
From Mohr’s circle:
s
2
68.75
τ max = R =
+ 16.672 = 38.20 MPa ◭
2
68.75
σ1 =
+ 38.20 = 72.6 MPa ◭
2
68.75
σ2 =
− 38.20 = −3.8 MPa ◭
2
16.67
2θ1 = sin−1
2θ1 = 25.87◦
θ 1 = 12.9◦ ◭
38.20
211
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
From Mohr’s circle:
s
2
27.90
R =
+ 6.012 = 15.190 MPa
2
6.01
2θ1 = sin−1
= 23.31◦
θ1 = 11.7◦ ◭
15.190
8.103
P
4(180)
4P
=
= 15.9 MPa
=
2
A
πd
π(120)2
16T
16(6 × 106 )
τ xy =
=
= 17.7 MPa
3
πd
π(120)3
σx =
27.90
+ 15.190 = 29.1 MPa ◭
2
27.90
σ2 =
− 15.190 = −1.24 MPa ◭
2
σ1 =
From Mohr’s circle:
s
2
15.9
+ 17.72 = 19.4 MPa
R =
2
17.7
2α = 70 − sin−1
= 4.16◦
19.4
15.9
− 19.4 cos 4.16◦ = −11.4 MPa ◭
σ ′y =
2
τ x′ y′ = 19.4 sin 4.16◦ = 1.4 MPa
8.105
See solution of Prob. 8.104 for Mohr’s circle.
τ max = R = 15.19 MPa ◭
27.90
σ̄ =
= 13.95 MPa
2
8.104
8.106
200(300)3 180(260)3
−
= 186.36 × 106 mm4
12
12
= 186.36 × 10−6 m4
I =
Q = (200 × 20)(140) = 560 × 103 mm3 = 560 × 10−6 m3
From FBD of end-plate, the net axial force on cylinder is
P =2
3
M yA
(40 × 10 )(0.13)
= 27.90 × 106 Pa
=
I
186.36 × 10−6
(40 × 103 )(560 × 10−6 )
VQ
=
= 6.01 × 106 Pa
=
Ib
(186.36 × 10−6 )(0.02)
σx =
τ x′ y ′
π(300)2
− 4(40 × 103 ) = −18628.3 N
4
18628.3
P
=−
= −2.57 MPa
A
π(3152 − 3002 )/4
2(150)
pr
=
= 40 MPa
σc =
t
7.5
Note that σ ℓ and σ c have opposite signs. Therefore,
σℓ =
τ max =
σ c − σℓ
40 − (−2.57)
=
= 21.3 MPa ◭
2
2
212
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
into the expression for σ̄, we obtain
8.107
1 E
1 E
[(1 + ν)(εx + νεy )] =
(εx + εy )
2 1 − ν2
21−ν
E
= ε̄
Q.E.D.
1−ν
σ̄ =
Working stresses are: (σ T )w = 40 MPa, (σ C )w = 28 MPa,
and τ w = 22 MPa. From the above FBD of an end-plate,
the net axial force is
8.109
2
P =
π (300)
p − 160 000 = 70685.8p − 140 000
4
P
70685.8p − 16 0000
=
= 9.756p − 19.3 MPa
A
π (3152 − 3002 ) /4
p (150)
pr
=
= 20p
σc =
t
7.5
σt =
Equation (f) of Art. 8.9 is
β = εy sin θ cos θ − εx sin θ cos θ − γ xy sin2 θ
Assume longitudinal stress governs:
When p = 0, σ l = 19.3 MPa (compression)< (σ C )w .
Therefore, σ l will not cause the cylinder to fail.
With εx = ε2 , εy = ε1 and γ xy = 0, this becomes
β = (ε1 − ε2 ) sin θ cos θ =
Assume circumferential stress governs:
σ c = 20p = 40
ε1 − ε2
sin 2θ
2
From Mohr’s circle:
p = 2 MPa
γ x′ y ′
ε1 − ε2
=
sin 2θ = β
2
2
Assume shear stress governs:
σ c − σl
20p − (9.756p − 19.3)
= 22
2
2
p = 6.2 MPa ◭
Q.E.D.
τ max =
8.110
8.108
The radii of the stress and strain circles are
γ
Rσ = τ max
Rε = max
2
From Mohr’s circle for strain:
s
2
−500 − 260
Rε =
+ 3602 × 10−6 = 523.5 × 10−6
2
−500 + 260
ε̄ =
× 10−6 = −120 × 10−6
2
But according to Hooke’s law
τ max = Gγ max =
E
γ
2(1 + v) max
Therefore,
Rσ =
E
E
γ
=
Rε
2(1 + v) max 1 + v
200 × 109
E
= 523.5 × 10−6
1+ν
1 + 0.3
= 80.54 × 106 Pa = 80.54 MPa
E
200 × 109
σ̄ = ε̄
= (−120 × 10−6 )
1−ν
1 − 0.3
= −34.29 × 10−6 Pa = −34.29 MPa
Q.E.D.
Rσ = Rε
The centers of the stress and strain circles are located at
σ̄ =
1
(σ x + σ y )
2
ε̄ =
1
(εx + εy )
2
Substituting Hooke’s law
E
(εx + vεy )
σx =
1 − v2
σ 1 = σ̄ + Rσ = −34.29 + 80.54 = 46.25 MPa ◭
σ 2 = σ̄ − Rσ = −34.29 − 80.54 = −114.83 MPa ◭
E
σy =
(εy + vεx )
1 − v2
213
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.111
8.113
From Mohr’s circle for strain:
s
2
800 + 400
Rε =
+ 3002 × 10−6 = 670.8 × 10−6
2
800 − 400
ε̄ =
× 10−6 = 200 × 10−6
2
From Mohr’s circle for strain:
s
2
800 + 200
Rε =
+ 4002 × 10−6 = 640.3 × 10−6
2
200 − 800
ε̄ =
× 10−6 = −300 × 10−6
2
400
= 38.66◦
2θ = sin−1
640.3
200 × 103
E
= 670.8 × 10−6
= 103.2 MPa
1+ν
1 + 0.3
E
200 × 103
σ̄ = ε̄
= (200 × 10−6 )
= 57.1 MPa
1−ν
1 − 0.3
σ 1 = σ̄ + Rσ = 57.1 + 103.2 = 160.3 MPa ◭
Rσ = Rε
200 × 109
E
= (640.3 × 10−6 )
1+v
1 + 0.3
6
= 98.51 × 10 Pa = 98.51 MPa
E
200 × 109
σ̄ = ε̄
= (−300 × 10−6 )
1−ν
1 − 0.3
6
= −85.71 × 10 Pa = −85.71MPa
Rσ = Rε
σ 2 = σ̄ − Rσ = 57.1 − 103.2 = −46.1 MPa ◭
8.112
g /2
t
x
y'
x
270
2q
Re
84
−620
e
From Mohr’s circle for stress:
68
.2
37.49°
90°
σ x′ = −85.71 − 98.51 cos 1.34◦ = −184.2 MPa ◭
τ x′ y′ = 98.51 sin 1.34◦ = 2.3 MPa ◭
s
52.51°
ey
Units: 10−6
y
x'
Units: MPa
76.6
s
8.114
2
620 + 84
Rε =
+ 2702 × 10−6 = 443.6 × 10−6
2
84 − 620
ε̄ =
× 10−6 = −268 × 10−6
2
270
= 37.49◦
2θ = sin−1
443.6
By inspection, we have for the 45◦ rosette θb = 45◦ and
εx = εα
200 × 103
E
= 443.6 × 0−6
= 68.2 MPa
Rσ = Rε
1+ν
1 + 0.3
E
200 × 103
σ̄ = ε̄
= (−268 × 10−6 )
= 76.6 MPa
1−ν
1 − 0.3
σ x′ = −76.6 − 68.2 cos 52.51◦ = −118.1 MPa ◭
εy = εc
Q.E.D.
Equation (b) of Art. 8.10 is
γ xy
εx + εy
εx − εy
+
cos 2θb +
sin 2θb
2
2
2
γ xy
εa − εc
εa + εc
+
cos 90◦ +
sin 90◦
=
2
2
2
γ xy
εa + εc
+
=
2
2
εa + εc
= εb −
Q.E.D.
2
εb =
τ x′ y′ = 76.6 sin 52.51◦ = 60.8 MPa ◭
60.8 MPa 118.1 MPa
γ xy
2
45°
214
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.115
8.117
By inspection, we have for the 60◦ rosette θb = 60θ ,
θc = 120◦ and
εa = εx
Q.E.D.
Equation (b) of Art. 8.10 is:
State of strain:
γ xy
εx + εy
εx − εy
+
cos 2θb +
sin 2θb
2
2
2
γ xy
εa − εy
εa + εy
+
cos 120◦ +
sin 120◦
=
2
2
2
√
Substituting cos 120◦ = −1/2 and sin 120◦ = 3/2, we get
εx = εa = 160 × 10−6
2εb + 2εc − εa
2(−220) + 2(360) − 160
εy =
=
× 10−6
3
3
= 40 × 10−6
γ xy
εb − εc
−220 − 360
√
√
=
=
× 10−6 = −334.9 × 10−6
2
3
3
εb =
εb =
√
1
(εa + 3εy + 3γ xy )
4
(a)
From Mohr’s circle for strain:
s
2
160 − 40
Rε =
+ (−334.9)2 × 10−6 = 340.2 × 10−6
2
40 + 160
ε̄ =
× 10−6 = 100
2
Equation (c) of Art. 8.10 yields
γ xy
εx + εy
εx − εy
+
cos 2θc +
sin 2θc
2
2
2
γ xy
εa − εy
εa + εy
+
cos 240◦ +
sin 240◦
=
2
2
2
√
With cos 240◦ = −1/2 and sin 240◦ = − 3/2, this becomes
εc =
εc =
√
1
εa + 3εy − 3γ xy
4
70 × 103
E
= 340.2 × 10−6
= 18.3 MPa
1+ν
1 + 0.3
70 × 103
E
= (100 × 10−6 )
= 5.4 MPa
σ̄ = ε̄
1−ν
1 + 0.3
σ 1 = σ̄ + Rσ = 5.4 + 18.3 = 23.7 MPa ◭
σ 2 = σ̄ − Rσ = 5.4 − 18.3 = −12.9 MPa ◭
Rσ = Rε
(b)
Adding Eqs. (a) and (b), we get
εb + εc =
1
(εa + 3εy )
2
εy =
2εb + 2εc − εa
Q.E.D.
3
τ max = Rσ = 18.3 MPa ◭
Subtracting Eq. (b) from Eq. (a) yields
√
γ xy
εb − εc
3
Q.E.D.
γ
= √
εb − εc =
2 xy
2
3
8.118
8.116
Equations (8.19) are
εx = εa
State of strain:
εy = εc
εa + εc
γ xy = εb −
2
εx = εa = 340 × 10−6
2(−550) + 2(−180) − 340
2εb + 2εc − εa
=
× 10−6
εy =
3
3
= −600 × 10−6
γ xy
εb − εc
−550 − (−180)
√
= √
=
× 10−6 = −213.6 × 10−6
2
3
3
If gages a and c are aligned with the principal directions,
then γ xy = 0 and the third equation yields
εb =
εa + εc
◭
2
215
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
From Mohr’s circle for strain:
s
2
340 + 600
Rε =
+ 213.62 × 10−6 = 516.2 × 10−6
2
340 − 600
ε̄ =
× 10−6 = −130 × 10−6
2
σ 1 = σ̄ + Rσ = 61.43 + 83.25 = 144.7 MPa ◭
σ 2 = σ̄ − Rσ = 61.43 − 83.25 = −21.8 MPa ◭
70 × 103
E
= 516.2 × 10−6
= 27.8 MPa
1+ν
1 + 0.3
E
70 × 103
σ̄ = ε̄
= (−130 × 10−6 )
= −13 MPa
1−ν
1 − 0.3
σ 1 = σ̄ + Rσ = −13 + 27.8 = 14.8 MPa ◭
σ 2 = σ̄ − Rσ = −13 − 27.8 = −40.8 MPa ◭
Rσ = Rε
8.120
τ max = Rσ = 27.8 MPa ◭
8.119
εx = εa = 300 × 10−6
εy = εc = 100 × 10−6
γ xy
εa + εc
300 + 100
× 106
= εb −
= 600 −
2
2
2
= 400 × 10−6
From Mohr’s circle for strain:
s
2
300 − 100
+ 4002 × 10−6 = 412.3 × 10−6
Rε =
2
300 + 100
× 10−6 = 200 × 10−6
ε̄ =
2
400
2θ1 = sin−1
= 75.97◦
θ1 = 38.0◦
◭
412.3
State of strain:
εx = εa = 550 × 10−6
εy = εc = −120 × 10−6
γ xy
εa + εc
550 + (−120)
× 10−6
= εb −
= −210 −
2
2
2
= −425 × 10−6
From Mohr’s circle for strain:
s
2
550 + 120
Rε =
+ 4252 × 10−6 = 541.1 × 10−6
2
550 − 120
ε̄ =
× 10−6 = 215 × 10−6
2
425
= 51.76◦
2θ1 = sin−1
541.1
200 × 109
E
= (412.3 × 10−6 )
1+ν
1 + 0.3
6
= 63.43 × 10 Pa = 63.43MPa
200 × 109
E
= (200 × 10−6 )
σ̄ = ε̄
1−ν
1 − 0.3
= 57.14 × 106 Pa = 57.14 MPa
Rσ = Rε
From Mohr’s circle for stress:
200 × 109
E
= 541.1 × 10−6
Rσ = Rε
1+ν
1 + 0.3
= 83.25 × 10−6 Pa = 83.25 MPa
E
200 × 109
σ̄ = ε̄
= (215 × 10−6 )
1−ν
1 − 0.3
= 61.43 × 10−6 Pa = 61.43 MPa
σ 1 = σ̄ + Rσ = 57.14 + 63.43 = 120.6 MPa ◭
σ 2 = σ̄ − Rσ = 57.14 − 63.43 = −6.3 MPa ◭
216
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.121
8.123
εx = εa = 300 × 10−6
2εb + 2εc − εa
2(−400) + 2(100) − 300
εy =
=
× 10−6
3
3
= −300 × 10−6
γ xy
−400 − 100
εb − εc
√
√
=
× 10−6 = −288.7 × 10−6
=
2
3
3
Components of the load are
Px = 4000 cos 15◦ = 3864 N
Py = 4000 sin 15◦ = 1035 N
From FBD of beam:
ΣMB = 0
From Mohr’s circle for strain:
p
Rε =
3002 + 288.72 × 10−6 = 416.4 × 10−6
288.7
= 43.89◦
θ1 = 21.9◦ ◭
2θ1 = sin−1
416.4
ΣFx = 0
ΣFy = 0
6RA + 0.375 (3864) − 4 (1035) = 0
3864 − Bx = 0
By + RA − 1035 = 0
which yield
RA = 448.5 N
E
200 × 109
Rσ = Rε
= (416.4 × 10−6 )
1+ν
1 + 0.3
6
= 64.06 × 10 Pa = 64.06 MPa
Bx = 3864 N By = 586.5 N
On section m-n:
P = 3864 N (C)
From Mohr’s circle for stress:
M = 586.5 (3) = 1760 N · m
6 1760 × 103
3864
6M
P
−
(σ T )max =
−
=
2
2
2
bh
bh
90 (270)
90 (270)
= 1.6 MPa ◭
6 1760 × 103
3864
P
6M
+
+
(σ C )max =
=
2
2
bh2
bh
90 (270)
90 (270)
= 1.61 MPa ◭
σ 1 = σ̄ + Rσ = 0 + 64.06 = 64.1 MPa ◭
σ 2 = σ̄ − Rσ = 0 − 64.06 = −64.1 MPa ◭
8.122
8.124
Maximum tensile stress, which occurs at B, is:
σB = −
(P e)
M
P
P
P
+
= − 2 + 3 = 3 (−r + 4e)
A
S
πr
πr /4
πr
From Mohr’s circle drawn for the limiting case of σ 1 = 0:
p
τ xy = 102 − 52 = 8.66 MPa ◭
Setting σ B = 0 yields
e=
r
4
◭
217
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.125
8.127
p
502 + 602 = 78.10 MPa
60
= 50.19◦
θ 1 = 25.1◦ ◭
2θ1 = tan−1
50
σ 1 = −30 + 78.10 = 48.1 MPa ◭
R =
p
122 + 42 = 12.649 MPa
12
= 71.57◦
α = tan−1
4
β = 100◦ − 71.57◦ = 28.43◦
R =
σ 2 = −30 − 78.10 = −108.1 MPa ◭
σ x′ = −4 + 12.649 cos 28.43◦ = 7.12 MPa ◭
8.126
σ y′ = −4 − 12.649 cos 28.43◦ = −15.12 MPa ◭
τ x′ y′ = 12.649 sin 28.43◦ = 6.02 MPa ◭
8.128
Use Mohr’s circle to transform the second state of stress to
xy-coordinates:
σ x = −35 − 65 cos 60◦ = −67.5 MPa
σ y = −35 + 65 cos 60◦ = −2.5 MPa
τ xy = 65 sin 60◦ = 56.3 MPa
(a)
Superimposing the two states of stress, we get
σ x = −27.5 MPa ◭
p
σ x = −20 − 602 − 302 = −72.0 MPa ◭
q
p
σ y = −σ̄ + R2 − τ 2xy = −20 + 602 − 302
σ y = 57.5 MPa ◭
τ xy = 56.3 MPa ◭
= 32.0 MPa ◭
(b)
2θ1 = 180◦ − sin−1
30
= 150.0◦
60
θ 1 = 75.0◦ ◭
218
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.129
8.131
The strain rosette is equivalent to the 60◦ rosette shown.
500
= 22.6◦
θ = 11.3◦ ◭
2θ1 = tan−1
1200
1600 + (−800)
ε̄ =
× 10−6 = 400 × 10−6
2
s
2
1600 + 800
+ 5002 × 10−6 = 1300 × 10−6
Rℓ =
2
εx = εa = 600 × 10−6
2εb + 2εc − εa
εy =
3
2(−110) + 2(200) − 600
× 10−6 = −140 × 10−6
=
3
γ xy
−110 − 200
εb − εc
√
√
=
× 10−6 = −179.0 × 10−6
=
2
3
3
70 × 103
E
ε̄ =
(400 × 10−6 ) = 38.9 MPa
1−ν
1 − 0.28
E
70 × 103
Rσ =
Rε =
(1300 × 10−6 )
1+ν
1 + 0.28
σ 1 = σ̄ + Rσ = 38.9 + 71 = 109.9 MPa
σ 2 = σ̄ − Rσ = 38.9 − 71 = −32.1 MPa
σ̄ =
From Mohr’s circle:
600 − 140
× 10−6 = 230 × 10−6
2
s
2
600 + 140
+ 179.02 × 10−6 = 411.0 × 10−6
R =
2
ε̄ =
ε1 = ε̄ + R = (230 + 411.0) × 10−6 = 641 × 10−6 ◭
8.130
ε2 = ε̄ − R = (230 − 411.0) × 10−6 = −181 × 10−6 ◭
179.0
2θ1 = tan−1
= 25.8◦
θ1 = 12.9◦ ◭
(600 + 140)/2
8.132
540 + 180
× 10−6 = 360 × 10−6
ε̄ =
2
R = ε1 − ε̄ = (660 − 360) × 10−6 = 300 × 10−6
s
2
|γ xy |
540 − 180
2
× 10−6 = 240 × 10−6
=
300 −
2
2
|γ xy | = 480 × 10−6 ◭
ε2 = ε̄ − R = (360 − 300) × 10
−6
= 60 × 10
−6
From Table 5.4, the section modulus of the vessel is
S =
π
π
(R4 − r4 ) =
(7504 − 7304 )
4R
4(750)
= 33.95 × 106 mm4 = 33.95 × 10−6 m4
The smallest longitudinal stress occurs at the top of the
vessel. Setting this stress to zero, we get
◭
pr M
−
=0
2t
S
(2 × 106 )(0.730)(33.95 × 10−6 )
prS
=
∴M =
2t
2(0.02)
= 1239 N · m ◭
(σ ℓ )min =
219
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.133
8.134
A = (240 × 60) = 14400 mm2
bh3
60(240)
I =
=
= 69.12 × 106 mm4
12
12
4 600 × 103
P
4P
σ = − =− 2 =−
= −8.488 × 106 Pa
2
A
πd
π (0.3)
16T
16T
τ =
=
3 = 188.63T
πd3
π (0.3)
Move the 40 kN load to point C and resolve into
components. At section AB:
P = 34.64 kN
r σ
σ 2
σ1 = +
+ τ2
2
2
Solving for τ gives:
r
σ 2 σ 2 p
σ1 −
τ=
−
= σ 1 (σ 1 − σ)
2
2
V = 20 kN
M = 20(360) + 34.64(300) = 17592 kN · mm
At point A:
P
Mc
34.64 × 103
17592 × 103 (120)
+
=
+
A
I
14400
69.12 × 106
= 32.9 MPa ◭
σ1 =
188.63T =
σ2 = 0 ◭
p
1.8 [1.8 − (−8.488)] × 106
T = 22 810 N · m = 22.8 kN · m ◭
At point B:
34.64 × 103
P
=
= 2.4 MPa
A
14400
3V
3(20 × 103 )
τ =
=
= 2.1 MPa
2A
2(14400)
r
σ σ
σ1
+ τ2
±
=
σ2
2
2
s
2
2.4
2.4
=
+ 2.12
±
2
2
= 1.2 ± 2.42
σ 1 = 3.62 MPa ◭
σ 2 = −1.22 MPa ◭
σ =
8.135
Maximum stress resultants occur at the built-in support.
They are M = T = P R. The corresponding stresses acting
on the shaded element are
σz =
32P R
32M
=
πd3
πd3
τ yz =
16T
16P R
=
πd3
πd3
√
16P R p 2
σ z 2
16 5P R
2
2
τ max =
+ τ yz =
2 +1 =
2
πd3
πd3
3
3
π(0.035)
πd
√ τ max = √
(100 × 106 )
P =
16 5R
16 5(0.75)
= 502 N ◭
r
220
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
8.136
8.137
l
sl
tcl
A
tcl
l
tcl
tcl
B
c
sl
c
At the cross section containing A and B the internal forces
are
M = 20 × 103 0.175 = 3500 N · m
π D2 − d2
π 452 − 332
A =
=
= 735 mm2
4
4
π D4 − d4
π 454 − 334
I =
=
= 143 × 103 mm4
64
64
J = 2I = 286 × 103 mm4
T = (20 × 103 )(0.035) = 700 N · m
V = 20 000 N
(a)
(a) Critical point at E:
At point A:
P = 480 N M = 480 (300) = 144000 N · mm
P
Mc
480 144000 (45/2)
σ= +
= 23.3 MPa (C)
=
+
A
I
735
143 × 103
σ
τ max = = 11.65 MPa ◭
2
3500
M
=
= 103.94 × 106 Pa
S
π(0.035)3 /4
= 103.94 MPa
2T
2(700)
τ cl =
=
= 10.394 × 106 Pa
πr3
π(0.035)3
= 10.394rMPa
σ 2
σl
l
σ 1,2 =
+ τ 2cl
±
2
2
s
2
103.94
103.94
+ 10.3942
±
=
2
2
σ 1 = 105.0 MPa ◭
σ 2 = −1.029 MPa ◭
σl =
(b) Critical point is at the top (or bottom) of section A:
V = 480 N M = 480 (600) = 288000 N · mm
T = 480 (300) = 144000 N · mm
Mc
288000 (45/2)
σ =
= 45.3 MPa
=
I
143 × 103
144000 (45/2)
Tr
= 11.3 MPa
=
τ =
J
286 × 103
s
r 2
σ 2
45.3
2
τ max =
+ 11.32
+τ =
2
2
= 25.3 MPa ◭
(b) At Point B:
τ cl =
Q =
πr2
2
4
r
3π
2T
VQ
+
πr3
I(2r)
=
2 3
2
r = (0.035)3
3
3
8.138
= 18.58 × 10−6 m3
πr4
π
π
I(2r) =
(2r) = r5 = (0.035)5 = 82.50 × 10−9 m5
4
2
2
(20 000)(18.58 × 10−6 )
82.50 × 10−9
6
= 14.898 × 10 Pa = 14.898 MPa
σ 1,2 = ±τ cl
∴ τ cl = 10.394 × 106 +
∴ σ 1 = 14.898 MPa ◭
Critical section is just left of C, where
σ 2 = −14.898 MPa
M = 198000 N · mm
T = 108000 N · mm
221
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
At the most highly stressed point
σ1
σ2
Mτ
4M
Mc
= 4 = 3
I
πr /4
πr
Tr
Tr
2T
τ =
= 4 = 3
J
πr /2
πr
r p
2
σ
σ
2
σ max =
+ r2 = 3 (M + M 2 + T 2 )
+
2
2
πr
p
2
(198000 + 1980002 + 1080002)
18 =
πr3
r = 24.7 mm ◭
σ =
s
2
σℓ − σc
+ τ 2ℓc
2
s
2
11.0 − 22.0
11.0 + 22.0
+ 0.6082
±
=
2
2
= 16.5 ± 5.53 MPa
σℓ + σc
±
2
=
σ 1 = 22.0 MPa ◭
σ 2 = 10.97 MPa ◭
C8.1 MathCad Worksheet for
Part (a)
Given:
8.139
P := 20000 N
h := 90 mm
At the section of interest:
V = 550 N
d := 60 mm
tw := 11.25 mm
a := 48 mm
t := 15 mm
b := 30 mm
M = 550(0.96) = 528 N · m
For a semicircle of diameter d:
2 d3
2d
πd
′ ′
=
Q = A ȳ =
8
3π
12
Computations:
Determine X-sectional properties of section from
properties of component rectangles:
For the tube:
i := 1, 2..3
D3 − d3
0.123 − 0.113
Q =
=
= 33.08 × 10−6 m3
12
12
π(0.124 − 0.114 )
π(D4 − d4 )
=
= 2.992 × 10−6 m4
I =
64
64
A1 := a · t
A2 := b · t
t
t
y1 :=
y2 := h −
2
2
X
Ai
Atot :=
i
!
X
1
yC :=
Ai · yi
·
Atot
i
(a) At point A:
M c pr
(2 × 106 )(0.11/2)
528(0.12/2)
σℓ =
+
+
=
I
2t
2.992 × 10−6
2(0.005)
= 21.6 × 106 Pa
(2 × 106 )(0.11/2)
pr
=
= 22.0 × 106 Pa
σc =
t
0.005
σ 1 = σ c = 22.0 MPa ◭
σ 2 = σ ℓ = 21.6 MPa ◭
I :=
A3 := (h − 2 · t) · tw
h
y3 :=
2
X-sectional area
Location of neutral axis
from left edge of section
a · t3
b · t3
tw · (h − 2 · t)3
+
+
···
12
12
12
X
Ai · (yi − yC )2
Moment of inertia about NA
+
i
(b) At point B:
Compute the stresses:
pr (2 × 106 )(0.11/2)
+
= 11.0 × 106 Pa
2t
2(0.005)
pr
(2 × 106 )(0.11/2)
σc =
=
= 22.0 × 106 Pa
t
0.005
VQ
550(33.08 × 10−6 )
τ ℓc =
=
= 0.608 × 106 Pa
Ib
(2.992 × 10−6 )(2 × 0.005)
P · (d + yC) · yC
P
σ t = 54.2 MPa
+
I
Atot
P · (d + yC) · (h − yC)
P
σ c :=
σ c = −82.9 MPa
+
I
Atot
σ t :=
σℓ =
222
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
C8.1 MathCad Worksheet for
Part (b)
C8.2 MathCad Worksheet for Part
(b)
Given:
Given:
P := 20000 N
h := 90 mm
tw := 11.25 mm
a := 49.5 mm
MPa := 106 · Pa
σ x := 80 MPa σ y := 80 · MPa τ xy := −30 · MPa
d := 60 mm
t := 15 mm
b := 20.7 mm
Computations:
..
.
Computations:
θ := 0, 2.5 · deg.. 180 · deg
..
.
Plotting range and interval
150
Compute the stresses:
P
P · (d + yC) · yC
+
I
Atot
P
P · (d + yC) · (h − yC)
+
σ c :=
I
Atot
σ t :=
Stress (MPa)
100
σ t = 55.5 MPa
s
50
σ c = −78.6 MPa
t
0
–50
C8.2 MathCad Worksheet for
Part (a)
0
30
60
90
120
Theta (deg)
1
· atan2(σ x − σ y , 2 · τ xy )
2
σ 1 := σ(θ1 )
θ2 := θ1 + 90 · deg
σ 2 := σ(θ2 )
θ1 :=
Given:
MPa := 106 · Pa
σ x := 60 MPa σ y := −30 · MPa τ xy := 80 · MPa
150
180
θ1 = −45 deg
σ 1 = 110 MPa
θ2 = 45 deg
σ 2 = 50 MPa
Computations:
C8.3 MathCad Worksheet for
Part (a)
σx + σy
σ x + σy
+
· cos(2 · θ) + τ xy · sin(2 · θ)
2
2
−σ x + σ y
+ · sin(2 · θ) + τ xy · cos(2 · θ)
τ (θ) :=
2
σ(θ) :=
θ := 0, 2.5 · deg..180 · deg
Given:
Plotting range and interval
θa := 0 · deg θb := 60 · deg θc := 120 · deg
εa := 300 · 10−6 εb := −400 · 10−6 εc := 100 · 10−6
150
E := 200 · 109 · Pa ν := 0.3
Stress (MPa)
100
s
50
0
t
–50
–100
0
30
60
90
120
Theta (deg)
150
180
223
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Computations:
Set up and solve simultaneous equations–see Eqs. (a)-(c),
p. 334
C8.4 MathCad Worksheet for
Part (a)
εx := 1
Given:
εy := 1
γ xy := 1 Trial values used in solution
Given:
P := 10000 N
e := 0.75 m
γ xy
εx + εy
εx − εy
+
· cos(2 · θ a ) +
· sin(2 · θ a ) = εa
2
2
2
γ
εx + εy
εx − εy
xy
+
· cos(2 · θb ) +
· sin(2 · θb ) = εb
2
2
2
γ xy
εx − εy
εx + εy
+
· cos(2 · θc ) +
· sin(2 · θc ) = εc
2
2
2


εx
 εy  := Find(εx , εy , γ xy )
γ xy
E
Rσ :=
·
1+ν
s
By trial the smallest S for which the σ < σ w is
S := 636 mm
Computations:
b(h) := 0.5 · S − h
σ av :=
εx − εy
2
2
+
εx + εy
E
·
1−ν
2
γ
xy
2
2
Max. Stress (MPa)
0.13
0.12
0.11
0.10
0.09
0.08
60
εa := 100 · 10
εb := 300 · 10
9
E := 200 · 10 · Pa
ν := 0.3
θ c := 120 · deg
−6
2 #
h
2
0.14
Given:
θ b := 75 · deg
t · h3
+ b(h) · t ·
I(h) := 2 ·
12
P · L · (0.5 · h)
P ·e
τ (h) :=
I(h)
2 · b(h) · h · t
s
2
σ(h)
σ(h)
+
+ τ (h)2
σ max (h) :=
2
2
h := 0.1 · S, 0.11 · S..0.4 · S
C8.3 MathCad Worksheet for
Part (b)
−6
"
σ(h) :=
σ 1 := σ av + Rσ
σ 2 := σ av − Rσ
7
σ 1 = 6.405 × 10 Pa
σ 2 = −6.405 × 107 Pa
1
· atan2(εx − εy , γ xy )
θ 2 := θ1 + 90 · deg
θ1 :=
2
θ1 = −21.949deg
θ2 = 68.051 deg
θa := 30 · deg
t := 3.75 mm
L := 0.75 m
80 100 120 140 160 180 200 220 240 260
h (mm)
h := 0.25 · S
εc := −200 · 10−6
Trial value used in the
solution for optimal h
Computations:
..
.
σ 1 := σ av + Rσ
7
Given
d
σ max (h) = 0
h := Find(h)
dh
h = 187.9 mm
b(h) = 130 mm
σ 2 := σ av − Rσ
σ 1 = 4.43 × 10 Pa
σ 2 = −7.287 × 107 Pa
1
θ1 :=
· atan2(εx − εy , γ xy )
θ 2 := θ1 + 90 · deg
2
θ1 = 63.401 deg
θ2 = 153.401 deg
224
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Critical sections are A and B, where
C8.4 MathCad Worksheet for
Part (b)
MA (d/2)
θ
6Edb
sin
=
I
L2
2
Gd
T (d/2)
=
θ
τ =
J
2L
s
2
p
θ
6Eb
d
2
2
+ (Gθ)2
τ max =
sin
(σ/2) + r =
2L
L
2
σ =
Given:
P := 10000 N
t := 3.75 mm
e := 0.5 m
L := 1 m
Analysis of arm OA
By trial the smallest S for which the σ < σ w is S := 636
mm
Computations:
θ
=0
2
θ
θ
θ
GJ
24EIb
C = T + VA b cos =
b cos
sin
θ+
2
L
L3
2
2
2
GJ
12EIb
=
sin θ
θ+
L
L3
ΣMO = 0
0.16
0.15
Max. Stress (MPa)
0.14
0.13
C − T − VA b cos
0.12
0.11
0.10
0.09
0.08
60
C8.5 MathCad Worksheet
80 100 120 140 160 180 200 220 240 260
h (mm)
Given:
h := 0.25 · S Trial value used in the solution for optimal h
Given
L := 1800 mm
E := 200 GPa
d
σ max (h) = 0
h := Find(h)
dh
h = 213 mm
b(h) = 115.6 mm
b := 180 mm
d := 7.5 mm
ν := 0.28
τ yp := 300 MPa
Computations:
C8.5
G :=
E
2 · (1 + ν)
I :=
π · d4
64
J := 2 · I
s
2
d
θ
6·E ·b
τ max (θ) :=
+ (G · θ)2
·
· sin
2·L
L
2
θ := 0, 2 · deg..180 · deg
Plotting range and interval
Analysis of bar AB
600
Max. Shear Stress (GPa)
GJ
θ
T =
l
Using Table 6.3:
VA L2
MA L
−
=0
2EI
EI
MA L2
θ
VA L3
−
= 2b sin
δA = ∆
3EI
2EI
2
24EIb
θ
θ
12EIb
VA =
sin
sin
MA =
L3
2
L2
2
θA = 0
500
400
300
200
100
0
0
20
40
60
80 100 120 140 160 180
Theta (deg)
225
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Computations:
12 · E · I · b2
G·J
· sin(θ)
·θ+
C(θ) :=
L
L3
θ := 90 · deg
Trial value of θ at yielding
Given τ max (θ) = τ yp
θyp := Find(θ)
π · D2
· (γ w ) + π · D · t · (γ s )
4
π · D3 · t
D2 · t
I :=
Q(θ) :=
· sin(θ)
8
2
w0 · x2
w0 · L
w0 · L
·x−
V (x) :=
− w0 · x
M (x) :=
2
2
2
p·D
p · D M (x) · D · cos(θ)
+
σ y (x, θ) :=
σ x (x, θ) :=
2·t
2·I
4·t
V (x) · Q(θ)
τ xy (x, θ) :=
τ max (x, θ)
s2· I · t
2
σ x (x, θ) − σ y (x, θ)
τ max (x, θ) :=
+ τ xy (xθ)2
2
w0 :=
θ yp = 113.61 deg Values of θ and C at yielding
C(θ yp ) = 2.8631 × 104 N · mm
C8.6
σ x (x, θ) + σ y (x, θ)
+ τ max (x, θ)
2
σ x (x, θ) + σ y (x, θ)
− τ max (x, θ)
σ 2 (x, θ) :=
2


0.5 · |σ 1 (x, θ)|
τ abs (x, θ) := max  0.5 · |σ 2 (x, θ)| 
τ max (x, θ)
L
θ := 0, 10 · deg.. 360 · deg
x :=
2
σ 1 (x, θ) :=
From the FBD of shell segment in Fig. (a):
V =
w0 L
− w0 x
2
M=
w0 L
w0 x2
x−
2
2
The intensity of the distributed load is
w0 =
πD2
γ + πDtγ s
4 w
The first moment of cross-sectional area shown in Fig. (b),
taken about the NA, is
D2 t
R sin θ
=
sin θ
θ
2
Shear Stress (Pa)
Q = Aȳ = (2Rθt)
2·107
The state of stress at a point is given by
pD M (D/2) cos θ
+
2t
I
VQ
τ xy =
2It
σx =
σy =
pD
4t
1.5·107
tabs
1·107
tmax
5·106
0
0
60
120
180
240
300
360
Theta (deg)
C8.6 MathCad Worksheet
θ := 0, 10 · deg.. 360 · deg
x := 0
1.4·107
Shear Stress (Pa)
Given:
D := 2 · m L := 10 · m t := 10 · mm
γ s := 7850 · 9.81 · N · m−3
p := 250 · 103 · Pa
γ w := 1000 · 9.81 · N · m−3
First printing incorrectly lists p = 250.106 Pa
tabs
1.2·107
1·107
tmax
8·106
6·106
0
60
120
180
240
Theta (deg)
300
360
226
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Chapter 9
Without reinforcement:
9.1
(see solution of Prob. 9.1) M0 = 15.36 kN · m
∆M = M − M0 = 23.23 − 15.36 = 7.87 kN · m ◭
9.3
1440(256)3 (1440 − 200)(240)3
−
12
12
= 584.8 × 106 mm4 = 584.8 × 106 m4
I =
From the solution to Prob. 9.1: I = 584.8 × 10−6 m4
Assuming that wood governs:
Mmax = 16(2) − 16(0.5) = 24.0 kN · m
M (0.12)
M cwd
8 × 106 =
I
584.8 × 106
M = 39.0 × 103 N · m = 39.0 kN · m
σ wd =
Mmax cwd
(24.0 × 103 )(0.12)
=
I
584.8 × 10−6
6
= 4.92 × 10 Pa = 4.92 MPa ◭
(σ wd )max =
Assuming that steel governs:
M cst
M (0.128)
120 × 106 = 15
I
584.8 × 106
M = 36.55 × 103 N · m = 36.55 kN · m ⊳ use
σ st = n
(24.0 × 103 )(0.128)
Mmax cst
= 15
I
584.8 × 10−6
6
= 78.8 × 10 Pa = 78.8 MPa ◭
(σ st )max = n
Without reinforcement:
σ wd =
M0 cwd
I0
8 × 106 =
M0 (0.12)
0.2(0.24)3/12
9.4
M0 = 15.36 × 103 N · m = 15.36 kN · m
∆M = M − M0 = 36.55 − 15.36 = 21.19 kN · m ◭
Mmax =
9.2
P
(9 × 1000) = 4500P N · mm
2
P
A ȳ
(60 × 100)(55) + (800 × 5)(2.5)
P i i =
= 34 mm
Ai
(60 × 100) + (800 × 5)
X
I¯i + Ai (ȳi − ȳ)2
I =
ȳ =
60(100)3
+ (60 × 100)(55 − 34)2
12
800(53)
+ (800 × 5)(2.5 − 34)2
+
12
= 1162.3 × 104 mm4
=
480(256)3 (480 − 200)(240)3
−
12
12
= 348.5 × 106 mm4 = 348.5 × 106 m4
I =
Assuming that wood governs:
Assuming that wood governs:
M (0.12)
M cwd
8 × 106 =
σ wd =
I
348.5 × 106
M = 23.23 × 103 N · m = 23.23 kN · m ⊳ use
Mmax cwd
I
P = 261.9 N
σ wd =
7.2 =
(4500P )(105 − 34)
1162.3 × 104
Assuming that steel governs:
Assuming that aluminum governs
Mmax cst
I
P = 410.2 N
M cal
M (0.128)
80 × 106 = 5
I
348.5 × 106
M = 43.56 × 103 N · m = 43.56 kN · m
σ st = n
σ al = n
108 = 20
(4500P )(34)
1162.3 × 104
227
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Assuming that wood governs:
9.5
σ wd =
M cwd
I
10 × 106 =
(40 × 103 )(0.125)
(0.19531 + 5.073b) × 10−3
b = 60.1 × 10−3 m = 60.1 mm ◭
Assuming that steel governs:
M cst
M cwd
σ st = 20
σ wd =
I
I
σ st
20cst 10800
20ȳ
=
=
σ wd
cwd
7200
105 − ȳ
ȳ = 45 mm
σ st = n
M cst
I
120 × 106 = 15
(40 × 103 )(0.135)
(0.19531 + 5.073b) × 10−3
b = 94.6 × 10−3 m = 94.6 mm ◭
9.8
P
Ai ȳi
ȳ = P
Ai
(60 × 100)(55) + (20b × 5)(2.5)
45 =
(60 × 100) + (20b × 5)
b = 14.12 mm ◭
0.15(0.2)3
5b(0.006)3
2
I =
+2
+ (5b × 0.006)(0.103)
12
12
= (100 + 636.7b) × 10−6 m4
9.6
Assuming that wood governs:
σ wd =
M cwd
I
10 × 106 =
(14 × 103 )(0.1)
(100 + 636.7b) × 10−6
b = 62.8 × 10−3 m = 62.8 mm
The properties of the equivalent cross section were
calculated in the solution of Prob. 9.4:
Assuming that steel governs:
σ st = n
ȳ = 34 mm I = 1162.3 × 104 mm4
Mmax = 3000(10) − 3000(5) = 15000 N · m
M cst
I
80 × 106 = 5
(14 × 103 )(0.106)
(100 + 636.7b) × 10−6
b = −11.39 × 10−3 m = −11.39 mm
9.9
(15000 × 103 )(105 − 34)
Mmax cwd
=
I
1162.3 × 104
= 91.6 MPa ◭
σ wd =
For C250×30: I = 32.8 × 106 mm4
For equivalent section made of wood:
(15000 × 103 )(34)
Mmax cst
= 20
σ st = n
I
1162.3 × 104
= 877.6 MPa ◭
200(254)3
+ 2(20)(32.8 × 106 )
12
= 1.5851 × 109 mm4 = 1585.1 × 10−6 m4
I = Iwd + 2nIC =
Assuming that wood governs:
9.7
M (0.254/2)
Mc
8 × 106 =
I
1585.1 × 10−6
M = 99.8 × 103 N · m = 99.8 kN · m
σ wd =
Assuming that steel governs:
I =
0.15(0.25)3
15b(0.01)3
+2
+ (15b × 0.01)(0.13)2
12
12
= (0.19531 + 5.073b) × 10−3 m4
Mc
M (0.254/2)
120 × 106 = 20
I
1585.1 × 10−6
M = 74.9 × 103 N · m = 74.9 kN · m ◭
σ st = n
228
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Assuming that steel governs:
9.10
C of channel
.
15.4
M cst
M (259/2 + 8)
108 = 3
I
361.9 × 106
6
M = 94.8 × 10 N · mm ⊳ use
σ st = n
69.6
200
NA
Without reinforcement:
M0
M0
σ al =
90 =
S0
1090 × 103
254
For C250×30 channel: A = 3790 mm2 , I¯ = 1.17 × 106 mm4
For one channel about NA:
M0 = 98.1 × 106 N · mm
Increase in strength due to reinforcement:
M − M0
(94.8 − 98.1) × 106
×100% =
×100% = −3.36% ◭
M0
98.1 × 106
Ich = I¯ + Ad2 = 1.17 × 106 + 3790(100 + 15.4)2
= 51.64 × 106 mm4
(b) Increase in flexural rigidity:
361.9 × 106 − 142 × 106
Eal I − Eal I0
× 100% =
× 100%
Eal I0
142 × 106
= 154.9% ◭
For equivalent section made of wood:
254(200)3
+ 2(20)(51.64 × 106 )
12
= 2235 × 106 mm4 = 2235 × 10−6 m4
I = Iwd + 2nIch =
Assuming that wood governs:
9.12
M (0.1)
M cwd
8 × 106 =
σ wd =
I
2235 × 10−6
3
M = 178.8 × 10 N · m = 178.8 kN · m
73
Eal
=
= 23.55
Evi
3.1
For the transformed (vinyl) cross section:
n=
Assuming that steel governs:
π(R4 − r4 )
πr4
+n
4
4
π(50)4
π(534 − 504 )
=
+ 23.55
4
4
= 35.25 × 106 mm4 = 35.25 × 10−6 m4
I = Ivi + nIal =
M cst
M (0.1 + 0.0696)
120 × 106 = 20
I
2235 × 10−6
3
M = 79.1 × 10 N · m = 79.1 kN · m ◭
σ st = n
The maximum stresses are
Mr
6000(0.05)
= 8.51 × 106 Pa
=
I
35.25 × 10−6
= 8.51 MPa ◭
6000(0.053)
MR
= 212 × 106 Pa
= 23.55
σ al = n
I
35.25 × 10−6
= 212 MPa ◭
9.11
σ vi =
For W250 × 89
I0 = 142 × 106 mm4
S0 = 1090 × 103 mm3
For equivalent section made of aluminum:
771(8)3
6
2
I = 142 × 10 + 2
+ (771 × 8)(133.5)
12
9.13
13.0
Esp
=
= 4.194
Evi
3.1
For the transformed (vinyl) section
n=
= 361.9 × 106 mm4
πr4
π(R4 − r4 )
+n
4
4
π(534 − 504 )
π(50)4
+ 4.194
=
4
4
= 10.312 × 106 mm4 = 10.312 × 10−6 m4
(a) Assuming that aluminum governs
I = Ivi + nIsp =
M (259/2)
M cal
90 =
I
361.9 × 106
M = 251.5 × 106 N · mm
σ al =
229
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
Assume that the stress in vinyl governs:
9.16
Mr
M (0.05)
(σ w )sp =
40 × 106 =
I
10.312 × 10−6
M = 8250 N · m
C of channel
15.4
NA
.
200
Assume that the stress in spruce governs:
M (0.053)
MR
30 × 106 = 4.194
I
10.312 × 10−6
M = 1392 N · m ◭
(σ w )sp = n
For equivalent section made of wood (see solution of Prob.
9.10): I = 2235 × 10−6 m4
For a channel:
Q = nAy = 20(3790)(100 + 15.4)
= 8.747 × 106 mm3 = 8.747 × 10−3 m3
9.14
From Eq. (5.10a) the shear force in a bolt is (note that
there are two rows of bolts)
(40 × 103 )(8.747 × 10−3 )(0.3)
V Qe
=
= 23 480 N
2I
2(2235 × 10−6 )
23 480
F
=
= 74.7 × 106 Pa = 74.7 MPa ◭
τ=
πd2 /4
π(0.02)2 /4
F −
P
Ai ȳi
(60 × 100)(55) + (900 × 5)(2.5)
= 32.5 mm
=
ȳ = P
(60 × 100) + (900 × 5)
Ai
X
I¯i + Ai (ȳi − ȳ)2
I=
60(100)3
+ (60 × 100)(55 − 32.5)2
12
900(53)
+ (900 × 5)(2.5 − 32.5)2
+
12
= 1209.7 × 104 mm4
VQ
V (60 × 72.5)(72.5/2)
τ=
0.5 =
Ib
(1209.7 × 104 )(60)
V = 2301.5 N ◭
=
9.17
For equivalent section made of wood:
1920(8)3
200(240)3
+2
+ (1920 × 8)(124)2
I =
12
12
= 702.9 × 106 mm4 = 702.9 × 10−6 m4
E = 10 × 109 Pa
9.15
From Table 6.3:
Because Q about the neutral axis is zero for each channel,
there is no tendency for the channels to slide relative to
the wood core. Hence there is no shear force in the bolts,
so that
τ =0 ◭
δ max =
5w0 L4
5(20 × 103 )(4)4
=
384EI
384(10 × 109 )(702.9 × 10−6 )
= 9.48 × 10−3 m = 9.48 mm ◭
230
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
For the transformed (wood) section:
9.18
I = Iwd +nIst = [341.3+15(131.60)]×104 = 2315×104 mm4
First moment of one-half of the transformed cross section:
Qwd = (40 × 80)(20) = 64000 mm3
3.75
= 15703 mm3
Qst = 375 40 +
2
Q = Qwd + nQst = 64000 + 15(15703) = 299545 mm3
Use equivalent section made of wood.
P
Ai ȳi
ȳ = P
Ai
(600 × 10)(8.5) + (30 × 60)(50) + (200 × 20)(10)
=
(600 × 10) + (30 × 60) + (200 × 20)
= 54.24 mm
X
I =
[I¯i + Ai (ȳi − ȳ)2 ]
Maximum shear stress in wood:
τ max =
600(10)3
+ (600 × 10)(85 − 54.24)2
12
30(60)3
+ (30 × 60)(50 − 54.24)2
+
12
200(20)3
+ (200 × 20)(10 × 54.24)2
+
12
= 1426.1 × 104 mm4
V (299545)
2315 × 104 (80)
V = 18548 N ◭
For the equivalent section made of wood (see solution of
Prob. 9.4):
I = 1162.3×104 mm4 E =
200 × 103
Est
=
= 10×103MPa
n
20
From Table 6.3:
δ max =
Between wood and steel:
V Qst
V (600 × 10)(85 − 54.24)
=
Ib
1426.1 × 104 (30)
= 4.31 × 10−4 V MPa ◭
P L3
1500(18 × 1000)3
=
= 78.4 mm ◭
48EI
48(200 × 103 )(1162.3 × 104 )
9.21
2
2nAst h 2nAst
h
+
−
=0
d
bd d
bd
2nAst
2(6)(1500)
=
= 0.160
(a) bd
250(450)
2
h
h
h
+ 0.160 − 0.160 = 0
= 0.3279
d
d
d
h = 0.3279(450) = 147.6 mm ◭
Between aluminum and wood:
τ=
3=
9.20
=
τ=
VQ
Ib
V Qal
V (200 × 20)(54.24 − 10)
=
Ib
1426.1 × 104 (30)
= 4.14 × 10−4 V MPa
2(10)(1500)
2nAst
=
= 0.2667
bd
250(450)
2
(b) h
h
+ 0.2667 − 0.2667 = 0
d
d
h = 0.400(450) = 180.0 mm ◭
9.19
80(80)3
= 341.3 × 104 mm4
Iwd =
12
"
2 #
100(3.75)3
3.75
Ist = 2
+ 375 40 +
12
2
h
= 0.400
d
= 131.6 × 104 mm4
231
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
we get
2
h
h
+ 0.166 67 − 0.166 67 = 0
d
d
h
= 0.3333
d
By substituting σ st = (σ st )yp into Eq, (9.7), we obtain the
bending moment that results in yielding of the
reinforcement:
h
M = 1−
Ast (σ st )yp d
3d
0.3333
= 1−
(125)(360)(150)
3
9.22
(a) Equation (9.5) for finding the distance h between the
neutral axis and the top of the beam is
2
2nAst h 2nAst
h
+
−
=0
d
bd d
bd
Substituting
2(8)(125)
2nAst
=
= 0.13889
bd
(80)(180)
we get
= 6 × 106 N · mm ◭
2
h
h
+ 0.138 89 − 0.138 89 = 0
d
d
(b) Equation (9.7) also yields the maximum stress in the
concrete. After some rearrangement we get
1 2h
h
M = bd
1−
(σ co )max
2
d
3d
0.3333
1
2
6
(σ co )max
6 × 10 = (80)(150) (0.3333) 1 −
2
3
(σ co )max = 22.5 MPa ◭
which yields
h
= 0.3096
d
By substituting σ st = (σ st )yp into Eq, (9.7), we obtain the
bending moment that results in yielding of the
reinforcement:
h
Ast (σ st )yp d
M = 1−
3d
0.3096
(125)(360)(180)
= 1−
3
9.24
= 7.264 × 106 N · mm ◭
2
2nAst h 2nAst
h
+
−
=0
d
bd d
bd
2(8)(1500)
2nAst
=
= 0.100
bd
400(600)
2
h
h
+ 0.100 − 0.100 = 0
d
d
h = 0.2702(600) = 162.12 mm
(b) Equation (9.7) also yields the maximum stress in the
concrete. After some rearrangement we get
1
h
h
M = bd2
1−
(σ co )max
2
d
3d
0.3096
1
(σ co )max
7.264 × 106 = (80)(180)2 (0.3096) 1 −
2
3
(σ co )max = 20.2 MPa ◭
h
1
(σ co )max
M = bh d −
2
3
1
0.16212
3
60 × 10 = (0.4)(0.16212) 0.6 −
(σ ∞ )max
2
3
9.23
(a)
h
= 0.2702
d
Equation (9.5) is
(σ ∞ )max = 3.39 × 106 106 Pa = 3.39 MPa ◭
2
2nAst h 2nAst
h
+
−
=0
d
bd d
bd
M =
Substituting
60 × 103 =
2(8)(125)
2nAst
=
= 0.166 67
bd
(80)(150)
h
d−
3
Ast σ st
0.16212
(1500 × 10−6 )σ st
0.6 −
3
σ st = 73.3 × 106 Pa = 73.3 MPa ◭
232
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
9.25
9.27
2
2nAst h 2nAst
h
+
−
=0
d
bd d
bd
2nAst
2(10)(6000)
=
= 0.320
bd
500(750)
2
h
h
h
+ 0.320 − 0.320 = 0
= 0.4279
d
d
d
h = 0.4279(750) = 320.9 mm
1
h
M = bh d −
(σ co )max
2
3
1
0.3209
3
270 × 10 = (0.5)(0.3209) 0.75 −
(σ co )max
2
3
2
2nAst h 2nAst
h
−
=0
d
bd d
bd
2nAst
2(10)(200)
=
= 0.2222
bd
(100)(180)
2
h
h
+ 0.2222 − 0.2222 = 0
d
d
h = 0.3732(180) = 67.18 mm
h
= 0.3732
d
Assuming that concrete governs:
1
h
Mmax = bh d −
(σ co )w
2
3
67.18
1
(10.8)
= (100)(67.18) 180 −
2
3
(σ co )max = 5.23 × 106 Pa = 5.23 MPa ◭
h
M = d−
Ast σ st
3
0.3209
3
(6000 × 10−6 )σ st
270 × 10 = 0.75 −
3
= 5.718 × 106 N · mm = 5.718 × 103 N · m
Assuming that steel governs:
h
Ast (σ st )w
Mmax = d −
3
67.18
= 180 −
(200)(120)
3
σ st = 70.0 × 106 Pa = 70.0 MPa ◭
9.26
= 3.783 × 106 N · mm = 3.783 × 103 N · m ◭ use
2
2nAst h 2nAst
h
+
−
=0
d
bd d
bd
2(8)(1400)
2nAst
=
= 0.16593
bd
300(450)
2
h
h
+ 0.16593 − 0.16593 = 0
d
d
h = 0.3327(450) = 149.72 mm
Total load on the beam is
(100 × 200)
6
W = (w0 + γ co A) L = w0 + (2400 × 9.8)
10002
= (6w0 + 2822.4) N
h
= 0.3327
d
Assuming that concrete governs:
1
h
M = bh d −
(σ co )w
2
3
0.14972
1
(12 × 106 )
= (0.3)(0.14972) 0.45 −
2
3
WL
8
(6w0 + 2822.4)(6)
3
3.783 × 10 =
8
Mmax =
w0 = 370.3 N/m ◭
= 107.8 × 103 N · m = 107.8 kN · m
Assuming that steel governs:
h
M = d−
Ast (σ st )w
3
0.14972
= 0.45 −
(1400 × 10−6 )(140 × 106 )
3
= 78.4 × 103 N · m = 78.4 kN · m ◭
233
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
9.28
Qst = Qco
h
1
(σ co )max
M = bh d −
2
3
h
1
6
(20)
20 × 10 = (120)h 180 −
2
3
26400ȳ = (500 × 150)(575 − ȳ) + 250
0 = ȳ 2 − 1811.2ȳ + 595000
ȳ = 431.1 mm
h = 118.7 mm
500(218.9)3 250(68.9)3
−
3
3
9
4
−3
4
= 6.627 × 10 mm = 6.627 × 10 m
I = 26400(431.1)2 +
2
h
2nAst h 2nAst
+
−
=0
d
bd d
bd
2
2n h
h
+
− 1 Ast = 0
d
bd d
2
2(8)
118.7
118.7
+
− 1 Ast = 0
180
120(180)
180
Ast = 1723.8 mm
M =
h
d−
3
(500 − ȳ)2
2
2
Assuming that concrete governs:
M cco
I
M (0.2189)
6
12 × 10 =
6.627 × 10−3
(σ co )w =
◭
M = 363 × 103 N · m
Assuming that steel governs:
M cst
I
M (0.4311)
6
M = 269 × 103 N · m
140 × 10 = 8
6.627 × 10−3
M = 269 kN · m ◭
Ast σ st
118.7
(1723.8)
20 × 106 = 180 −
3
σ st = 82.6 MPa ◭
(σ st )w = n
9.31
9.29
Qtens = Qcomp
Qst = Qco
(575 − ȳ)2
2
2
0 = ȳ − 1256.7ȳ + 343960
ȳ = 402.8 mm
(180 − ȳ)2
3600ȳ = (300 × 40)(200 − ȳ) + 120
2
2
0 = ȳ − 620ȳ + 72400
ȳ = 156.1 mm
3
180(23.9)3
300(63.9)
−
I = 3600(156.1)2 +
3
3
4
4
= 11300 × 10 mm
M cco
(200 × 106 )(63.9)
σ co =
= 113 MPa ◭
=
I
11300 × 104
M cst
8(200 × 106 )(156.1)
σ st = n
= 2210.3 MPa ◭
=
I
11300 × 104
12000ȳ = 4000(500 − ȳ) + 300
300(172.2)3
3
= 2.495 × 109 mm4 = 2.495 × 10−3 m4
I = 12000(402.8)2 + 4000(97.2)2 +
Assuming that concrete governs:
M (0.1722)
M cco
12 × 106 =
I
2.495 × 10−3
M = 173.9 × 103 × 103 N · m = 173.9 kN · m
(σ co )w =
9.30
Assuming that steel governs:
M cst
M (0.4028)
140 × 106 = 10
I
2.495 × 10−3
3
3
M = 86.7 × 10 × 10 N · m = 86.7 kN · m ◭
(σ st )w = n
234
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
9.32
9.34
Qtens = Qcomp
Qtens = Qcomp
(210 − ȳ)2
2
0 = ȳ 2 − 526.7ȳ + 5370
ȳ = 138.24 mm
120(71.76)3
I = 3200(138.24)2 + 3200(41.76)2 +
3
= 8152 × 104 mm4
(20 × 106 )(71.76)
M cco
= 17.6 MPa ◭
=
(σ co )max =
I
8152 × 104
M ctens
(20 × 106 )(138.24)
(σ st )tens = n
=8
I
8152 × 104
= 271.3 MPa ◭
(20 × 106 )(41.76)
M ccomp
=8
(σ st )comp = n
I
8152 × 104
= 82 MPa ◭
300ȳ 2
4000(575 − ȳ) = 12000(ȳ − 75) +
2
0 = ȳ 2 + 106.67ȳ − 21.333
3200ȳ = 3200(180 − ȳ) + 120
ȳ = 102.16 mm
300(102.16)3
3
= 1009.8 × 106 mm4 = 1009.8 × 10−6 m4
I = 4000(472.84)2 + 12000(27.16)2 +
Assuming that concrete governs:
M (0.10216)
M cco
12 × 106 =
I
1009.8 × 10−6
M = 118.6 × 103 N · m = 118.6 kN · m
(σ co )w =
Assuming that steel governs:
M cst
M (0.47284)
140 × 106 = 10
I
1009.8 × 10−6
3
M = 29.9 × 10 N · m = 29.9 kN · m ◭
(σ st )w = n
9.35
h=
9.33
Ast (σ st )yp
250(400)
=
= 41.3 mm
0.72b(σco )ult
0.72(120)(28)
Mnom = (d − 0.425h) T = (d − 0.425h) Ast (σ st )yp
= (240 − 0.425 × 41.3)(250)(400)
= 22.2 × 106 N · mm
Qtens = Qcomp
The strain in the reinforcement at failure is
(210 − ȳ)2
3200ȳ = 1600(180 − ȳ) + 120
2
2
0 = ȳ − 500ȳ + 48900
ȳ = 133.38 mm
120(76.62)3
I = 3200(133.38)2 + 1600(46.62)2 +
3
= 7840 × 104 mm4
Ccco
(20 × 106 )(76.62)
(σ co )max =
=
= 19.6 MPa ◭
I
(7840 × 104 )
d−h
240 − 41.3
= 0.003
h
41.3
= 0.0144
εst = 0.003
Because εst > 0.005, we have φ = 0.9, and the ultimate
moment becomes
Mult = 0.9Mnom = 0.9 22.2 × 106
= 2.0 × 107 N · mm ◭
M ctens
(20 × 106 )(133.38)
=8
I
7840 × 104
= 272 MPa ◭
(20 × 106 )(46.62)
M ccomp
=8
(σ st )comp = n
I
7840 × 104
= 95 MPa ◭
(σ st )tens = n
235
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
9.36
9.38
h=
580(400)
Ast (σ st )yp
=
= 95.9 mm
0.72b(σco )ult
0.72(120)(28)
Substituting εst = 0.010 into Eq. (9.8), we get
0.010 = 0.003
Mnom = (d − 0.425h) T = (d − 0.425h) Ast (σ st )yp
= (240 − 0.425 × 95.9)(580)(400)
2 × 106 /0.9
Mnom
=
C =
d − 0.425h
d(1 − 0.425(0.2308))
2.464 × 106
N
=
d
The strain in the reinforcement at failure is
d−h
240 − 95.9
= 0.003
h
95.9
= 0.004507
εst = 0.003
which upon substitution into Eq. (9.9) results in
From Eq. (9.13b):
2.464 × 106
= 0.72(0.5d)(0.2308d)(28)
d
d = 101.9 mm ◭
φ = 0.483 + 83.3(0.004 151) = 0.8288
= 38.3 × 107 N · mm ◭
h = 0.2308d
From Eq. (9.12) we obtain
= 46.2 × 106 N · mm
Mult = 0.8288Mnom = 0.8288 46.2 × 106
d−h
h
Therefore,
b = 0.5(101.9) = 50.95 mm ◭
The equation T = C is
2.464 × 106
d
2.464 × 106
Ast (400) =
101.9
9.37
Ast (σ st )yp =
(1.5 × 10−3 )(400 × 106 )
Ast (σ st )yp
=
h =
0.72b(σco )ult
0.72(0.2)(28 × 106 )
= 0.148 81 m
Ast = 60.5 mm2 ◭
9.39
Mnom = (d − 0.425h)Ast (σ st )yp
= (0.36 − 0.425 × 0.148 81)(1.5 × 10−3 )(400 × 106 )
With εst = 0.008, Eq. (9.8) becomes
3
= 178.05 × 10 N · m
500 − h
h
h = 136.36 mm = 0.136 36 m
0.008 = 0.003
d−h
0.36 − 0.148 81
= 0.003
h
0.148 81
= 0.004 257
εst = 0.003
C = 0.72bh(σco )ult = 0.72(0.3)(0.136 36)(28 × 106 )
= 824.7 × 103 N
φ = 0.483 + 83.3εst = 0.483 + 83.3(0.004 257)
= 0.8376
Mult = φMnom = 0.8376(178.05 × 103 )
T = C = Ast (σ st )yp
824.7 × 103 = Ast (400 × 106 )
−3
2
Ast = 2.062 × 10 m = 2.062 × 103 mm2 ◭
= 149.1 × 103 N · m = 149.1 kN · m ◭
Mult = φ(d − 0.425h)C
= 0.9(0.5 − 0.425 × 0.136 36)(824.7 × 103 )
= 328 × 103 N · m = 328 kN · m ◭
236
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
C9.1 MathCad worksheet for Part
(a)
9.40
With εst = 0.005, Eq. (9.8) becomes
Given:
420 − h
h
h = 157.50 mm = 0.157 50 m
0.005 = 0.003
b := 114 mm
L := 6 m
n := 5
C = 0.72bh(σco )ult = 0.72b(0.157 50)(28 × 106 )
= 3.175 × 106 b
h := 300 mm
w0 := 600 N/m
t := 15 mm
Computations:
w0 · L2
w0 · L
V max :=
8
2
(n − 1) · b · (h − 2 · t)3
(n · b) · h3
−
I :=
12
12
h−t
Q := (n · b) · t ·
2
h
Mmax ·
−t
2
σ 1 :=
σ 1 = 0.68 MPa
I
h
Mmax ·
2
σ 2 = 3.79 MPa
σ 2 := n ·
I
Vmax · Q
τ :=
τ = 0.036 MPa
I ·b
Acost := b · (h + 14 · t)
Acost = 58140mm2
Mmax :=
Mult = φ(d − 0.425h)C
400 × 103 = 0.9(0.42 − 0.425 × 0.157 50)(3.175 × 106 b)
b = 0.396 5 m = 397 mm ◭
C = T
3.175 × 106 b = Ast (σ st )yp
3.175 × 106 (0.396 5) = Ast (400 × 106 )
Ast = 3.15 × 10−3 m2 = 3.15 × 103 mm2 ◭
C9.1
C9.1 MathCad worksheet for Part
(b)
Given:
Equivalent section made of material 1
3
b := 88.2 mm
L := 6 m
n := 5
3
(nb)h
(n − 1)b(h − 2t)
−
12
12
Q = first moment of shaded area about NA
h−t
= (nb)t
2
I =
h := 300 mm
w0 := 600 N/m
t := 22.32 mm
Computations:
w0 · L
w0 · L2
V max :=
8
2
(n − 1) · b · (h − 2 · t)3
(n · b) · h3
−
I :=
12
12
h−t
Q := (n · b) · t ·
2
h
−t
Mmax ·
2
σ 1 = 0.69 MPa
σ 1 :=
I
h
Mmax ·
2
σ 2 := n ·
σ 2 = 4.03 MPa
I
Vmax · Q
τ :=
τ = 0.056 MPa
I ·b
Acost := b · (h + 14 · t)
Acost = 54021mm2
Mmax :=
The cost of beam is proportional to
Acost = A1 + 8A2 = b(h − 2t) + 8(2bt) = b(h + 14t)
237
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i
i
i
i
i
C9.2
C9.2 MathCad worksheet for
Part (c)
Given:
At := 2080 · mm2
b := 267 · mm
n := 8
Computations:
Find location of NA:
Equivalent section made of concrete
h := 0.5 · d
C9.2 MathCad worksheet for
Part (b)
Ac := 0 · mm2
d := 325 · mm
M := 90.103 · N · m
1
· b · h2 + (n − 1) · Ac · (h − a) − n · At · (d + a − h) = 0
2
h := Find(h)
h = 162.66 mm
a := 75 · mm
I :=
Computations:
Find location of NA:
h := 0.5 · d
b · h3
+ (n − 1) · Ac · (h − a)2 + n · At · (d + a − h)2
3
h
I
M · (d + a − h)
σ st tens := n ·
I
M · (h − a)
σ st comp := n ·
I
Trial value used in the solution
σ co := M ·
Given
1
· b · h2 + (n − 1) · Ac · (h − a) − n · At · (d + a − h) = 0
2
h := Find(h)
h = 162.63 mm
I :=
Trial value used in the solution
Given
Given:
At := 1859 · mm2
b := 267 · mm
n := 8
Ac := 680 · mm2
d := 325 · mm
a := 75 · mm
M := 100 · 103 · N · m
σ co = 1.1987 × 107 Pa
σ st tens = 1.3993 × 108 Pa
σ st comp = 5.168 × 107 Pa
b · h3
+ (n − 1) · Ac · (h − a)2 + n · At · (d + a − h)2
3
h
I
M · (d + a − h)
σ st tens := n ·
I
M · (h − a)
σ st comp := n ·
I
σ co := M ·
σ co = 1.199 × 107 Pa
σ st tens = 1.4 × 108 Pa
σ st comp = 5.1685 × 107 Pa
238
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
i
i
i
i