10.1 For the loading shown, use the doubleintegration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI that EI is is constant for each beam.
Fig. P10.1
Solution Integration of moment equation: d 2v EI 2 = M ( x) = − M 0 dx dv EI = − M 0 x + C 1 dx M 0 x 2 EI v = − + C1 x + C 2 2
(a) (b)
Boundary conditions: dv at x = 0 =0 dx v=0 at x=0 Evaluate constants: From Eq. (a), C 1 = 0. From Eq. (b), C 2 = 0 (a) Elastic curve equation:
x 2
0
EI v = −
∴
2
v=−
M 0 x2 2 EI
Ans.
(b) Deflection at the free end:
v B
=−
0
( L) 2
2 EI
= −
M 0 L2
Ans.
2 EI
(c) Slope at the free end:
dv dx B
= θ B = −
M 0 ( L) EI
= −
M0L EI
Ans.
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10.2 For the loading shown, use the doubleintegration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI that EI is is constant for each beam.
Fig. P10.2
Solution Integration of moment equation: d 2v wx 2 EI 2 = M ( x) = − dx 2
EI
dv dx
=−
EI v = −
wx3
+ C 1
6
wx 4 24
(a)
+ C1 x + C 2
(b)
Boundary conditions: dv at =0 = L dx v=0 at x=L Evaluate constants: Substitute x Substitute x = = L L and and dv/dx = dv/dx = 0 into Eq. (a) to determine C 1: 3 w( L) wL3 EI (0) = − + C1 ∴ C 1 = 6 6 Substitute x Substitute x = = L and v = 0 into Eq. (b) to determine C 2: L and w( L) 4 wL4 wL4 EI (0) = − + C1 ( L) + C2 = − + + C2 24 24 6 (a) Elastic curve equation:
EI v = −
wx 4 24
+
wL3 x 6
wL4
−
∴
8
v=−
w
∴ C 2 = −
⎡⎣ x4 − 4 L3 x + 3 L4 ⎤⎦ 24 EI
wL4 8 Ans.
(b) Deflection at the free end:
v A
=
w
⎡ − ( 0) 24 EI ⎣
4
+
4 L (0) − 3 L ⎤⎦ = − 3
4
3wL4 24 EI
4
= −
wL
8EI
Ans.
(c) Slope at the free end:
dv dx A
= θ A = −
w(0)3 6 EI
+
wL3 6 EI
=
wL3 6 EI
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.3 For the loading shown, use the doubleintegration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI that EI is is constant for each beam.
Fig. P10.3
Solution Integration of moment equation: d 2v w0 x3 EI 2 = M ( x) = − dx 6L 4 dv w0 x EI =− + C 1 dx 24 L w0 x5 EI v = − + C1 x + C 2 120 L
(a) (b)
Boundary conditions: dv at =0 = L dx v=0 at x=L Evaluate constants: Substitute x Substitute x = = L L and and dv/dx = dv/dx = 0 into Eq. (a) to determine C 1: 4 w0 ( L) w0 L3 EI (0) = − + C1 ∴ C 1 = 24 L 24 Substitute x Substitute x = = L L and and v = 0 into Eq. (b) to determine C 2: w0 ( L)5 w0 L5 w0 L3 EI (0) = − ( L) + C 2 + C1 ( L) + C2 = − + 120 L 120 L 24 ∴ C 2 =
w0 L4 120
w0 L4
−
24
=−
w0 L4 30
(a) Elastic curve equation: w0 x5 w0 L3 w0 L4 EI v = − x− + 120 L 24 30
∴v = −
w0
⎡⎣ x5 − 5 L4 x + 4 L5 ⎤⎦ 120L EI
Ans.
(b) Deflection at the free end:
v A
=−
w0
⎡ (0) 120 L EI ⎣
5
4
− 5L
(0) + 4 L ⎤⎦ 5
= −
w0 L4 30 EI
Ans.
(c) Slope at the free end:
dv dx A
= θ A = −
w0 (0) 4 2 4 L EI
+
w0 L3 24 EI
=
w0 L3 24 EI
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.4 For the beam and loading shown in Fig. P10.4, use the double-integration method to determine (a) the equation o the elastic curve for segment AB AB of the beam, (b) the deflection at B at B,, and (c) the slope at A at A.. Assume that EI that EI is is constant for the beam.
Fig. P10.4
Solution Integration of moment equation: d 2v P EI 2 = M ( x) = x dx 2 2 dv P x EI = + C 1 dx 4
P x3
EI v =
12
(a)
+ C1 x + C 2
(b)
Boundary conditions: v=0 at x=0
dv dx
=
0
at
L
x =
2
Evaluate constants: Substitute x Substitute x = = L/2 L/2 and and dv/dx = dv/dx = 0 into Eq. (a) to determine C 1: 2 P ( L / 2) PL2 EI (0) = + C1 ∴ C 1 = − 4 16 Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2: P (0)3 PL2 (0) EI (0) = − + C2 ∴ C 2 = 0 12 16 (a) Elastic curve equation:
P x3
EI v =
12
−
PL2 x
∴
16
v=−
Px
⎡⎣3L2 − 4 x 2 ⎤⎦ 48 EI
(0 ≤ x ≤
L 2
)
Ans.
(b) Deflection at B:
v B
P( L / 2) ⎡
=−
2 ⎛ L⎞ ⎤ ⎢3 L − 4 ⎜ ⎟ ⎥ = 48 EI ⎣⎢ ⎝ 2 ⎠ ⎦⎥ 2
−
PL3 48EI
Ans.
(c) Slope at A:
dv dx A
= θ A =
P(0) 2 4 EI
−
PL2 16 EI
= −
PL2 16 EI
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.5 For the beam and loading shown in Fig. P10.5, use the double-integration method to determine (a) the equation o the elastic curve for the beam, (b) the slope at A at A,, (c) the slope at B at B,, and (d) the deflection at midspan. Assume that EI that EI is is constant for the beam.
Fig. P10.5
Solution Beam FBD: Σ F y = Ay
+
By
=
0
∴ A y = − By Σ M A =
By L − M 0
∴ B y =
=0
0
and
L
Ay
M 0
=−
L
Moment equation: Σ M a − a =
M ( x) − Ay x − M 0
∴ M ( x) =
M 0
−
=
M ( x) +
M 0 L
x − M 0
=
0
M 0 x L
Integration of moment equation: d 2v M x EI 2 = M ( x) = M 0 − 0 dx L 2 dv M0x = M0x − + C 1 EI dx 2L M 0 x 2 M 0 x3 EI v = − + C1 x + C 2 2 6 L
(a) (b)
Boundary conditions: v=0 at x=0
v=0
at
x=L
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2: M 0 (0) 2 M 0 (0)3 EI (0) = − + C1 (0) + C2 2 6 L Substitute x Substitute x = = L L and and v = 0 into Eq. (b) to determine C 1:
∴ C 2 =
0
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EI (0) =
M 0 ( L) 2
−
2
L
0
∴ C 1 =
M 0 ( L)3 −
6 L M0L
=−
6 2 (a) Elastic curve equation: EI v =
M 0 x 2 2
−
M 0 x3 6 L
−
+ C1 ( L)
M0L 3
M 0 Lx
∴
3
v=−
M0 x
⎡⎣ x 2 − 3 Lx + 2 L2 ⎤⎦ 6 L EI
Ans.
(b) Slope at A:
dv dx A
= θ A = M 0 (0) −
M 0 (0) 2 2 L EI
−
M0L 3EI
= −
M0 L 3EI
Ans.
(c) Slope at B:
dv dx B
= θ B =
M 0 ( L) EI
−
M 0 ( L) 2 2 L EI
−
M0 L 3 EI
=
M0 6 EI
[6 L − 3L − 2 L] =
M0 L 6 EI
Ans.
(d) Deflection at midspan:
v x = L / 2
=−
⎤ ⎛ L⎞ 2 ⎢⎜ ⎟ − 3 L ⎜ ⎟ + 2 L ⎥ = 6 L EI ⎢⎣⎝ 2 ⎠ ⎝2⎠ ⎥⎦ 0
( L / 2) ⎡⎛ L ⎞
2
−
M 0 L2 16EI
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.6 For the beam and loading shown in Fig. P10.6, use the double-integration method to to determine determine (a) the the equation equation o the elastic curve for the beam, (b) the maximum deflection, and (c) the slope at A. A. Assume that EI is constant for the beam.
Fig. P10.6
Solution Moment equation: Σ M a − a =
M ( x) −
wL x 2
∴ M ( x) = −
+
wx 2
wx 2 2 +
=
0
wL x
2 2 Integration of moment equation: d 2v w x 2 w Lx + EI 2 = M ( x) = − dx 2 2 dv w x 3 w Lx 2 EI =− + + C 1 dx 6 4 wx 4 wL x 3 EI v = − + + C1 x + C 2 24 12
(a) (b)
Boundary conditions: v=0 at x=0
v=0
at
x=L
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2: w(0) 4 wL(0)3 EI (0) = − + + C1 (0) + C2 24 12 Substitute x Substitute x = = L L and and v = 0 into Eq. (b) to determine C 1: w( L) 4 wL( L) 3 EI (0) = − + + C1 ( L) 24 12 (a) Elastic curve equation:
EI v = −
wx 4 24
+
wL x 3 12
−
wL3 x
∴
24
v=−
∴ C 2 =
∴ C 1 =
0 w( L) 4 24 L
−
w( L) 4 12L
= −
wx
⎡⎣ x3 − 2 Lx2 + L3 ⎤⎦ 24 EI
wL3 24 Ans.
At x = = L/2 L/2:: (b) Maximum deflection: At x vmax
=−
w( L / 2) ⎡⎛ L ⎞
3
⎛ L⎞ ⎢⎜ ⎟ − 2 L ⎜ ⎟ 24 EI ⎢⎣⎝ 2 ⎠ ⎝2⎠
2
⎤ wL ⎡ L3 +L ⎥=− ⎢ 48EI ⎣ 8 ⎦⎥ 2
−
⎤ + L ⎥= 2 ⎦
L3
2
−
5 wL4 384EI
Ans.
(c) Slope at A:
dv dx A
= θ A = −
w(0)3 6
+
wL(0) 2 4
−
wL3 24
= −
wL3 24
Ans.
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10.7 For the beam and loading shown in Fig. P10.7, use the double-integration method to to determine determine (a) the the equation equation o the elastic curve for segment AB AB of the beam, (b) the deflection midway between the two supports, (c) the slope at A, A, and (d) the slope at B. B. Assume that EI is constant for the beam.
Fig. P10.7
Solution Beam FBD: Σ F y = Ay Σ M A =
+
By
By L −
∴ B y =
−
3 L
2 3 P
P = 0 P = 0 and
2
Ay
=−
P 2
Moment equation: Σ M a − a =
M ( x) +
P
x=0
2 Integration of moment equation: d 2v Px EI 2 = M ( x) = − dx 2 2 dv Px EI =− + C 1 dx 4 Px3 EI v = − + C1 x + C 2 12
∴ M ( x) = −
Px 2
(a) (b)
Boundary conditions: v=0 at x=0
v=0
at
x=L
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2: P (0)3 EI (0) = − + C1 (0) + C2 ∴ C 2 = 0 12 Substitute x Substitute x = = L L and and v = 0 into Eq. (b) to determine C 1: P ( L)3 PL2 EI (0) = − + C1 ( L) ∴ C 1 = 12 12
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(a) Elastic curve equation for segment AB of the beam:
Px3
EI v = −
12
+
PL2 x
Px
⎡⎣ L2 − x2 ⎤⎦ 12 EI
Ans.
⎤ PL ⎡ 3 L ⎤ PL3 ⎥= ⎢⎣ 4 ⎥⎦ = 32EI 2 4 EI ⎥⎦
Ans.
∴
12 12
v=
(b) Deflection at midspan:
v x = L / 2
=
P ( L / 2) ⎡
⎛ L⎞ ⎢L − ⎜ ⎟ 12 EI ⎢⎣ ⎝2⎠
2
2
(c) Slope at A:
dv dx A
= θ A = −
P(0) 2 4 EI
+
PL2 12 EI
=
PL2 12 EI
Ans.
(d) Slope at B:
dv dx B
= θ B = −
P( L) 2 4 EI
+
PL2 12 EI
= −
PL2 6 EI
Ans.
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10.8 For the beam and loading shown in Fig. P10.8, use the double-integration method to determine (a) the equation of the elastic curve for segment BC of the beam, (b) the deflection midway between B between B and C , and (c) the slope at C . Assume that EI that EI is is constant for the beam.
Fig. P10.8
Solution Beam FBD: Σ M B = PL + C y ( 4 L ) − P(5 L ) = 0 ∴ C y = Σ F y =
By
P
+
Cy
∴ B y =
P
−
2 P = 0
Moment equation: Σ M a − a = M ( x) − By x + P( L + x) = M ( x) − Px + P( L + x) = 0 ∴ M ( x) = − PL
Integration of moment equation: d 2v EI 2 = M ( x) = − PL dx dv EI = − PLx + C 1 dx PLx 2 EI v = − + C1 x + C 2 2
(a) (b)
Boundary conditions: v=0 at x=0
v=0
at
x = 4L
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2: PL(0) 2 EI (0) = − + C1 (0) + C2 ∴ C 2 = 0 2 Substitute x Substitute x = = 4 L and L and v = 0 into Eq. (b) to determine C 1: PL(4 L) 2 2 EI (0) = − + C1 ( 4 L) ∴ C1 = 2 PL 2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
(a) Elastic curve equation for segment BC of of the beam:
PLx 2
EI v = −
2
+
2 PL2 x
∴
v=
PLx 2 EI
[ 4 L − x]
Ans.
(b) Deflection at midspan:
v x = L / 2
=
PL(2 L) 2 EI
[ 4 L − (2 L)] =
2 PL3
Ans.
EI
(c) Slope at C :
dv dx
= θ C = − C
PL(4 L) EI
+
2 PL2 EI
= −
2 PL2 EI
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.9 For the beam and loading shown in Fig. P10.9, use the double-integration method to to determine determine (a) the the equation equation o the elastic curve for segment AB AB of the beam, (b) the deflection midway between A A and B, B, and (c) the slope at B. B. Assume that EI that EI is is constant for the beam.
Fig. P10.9
Solution Beam FBD: Σ M A = −
wL2 2
∴ B y = Σ F y =
−
⎛ 5L ⎞ ⎟ + By L = 0 ⎝ 4 ⎠
P⎜
wL
5 P
+
2 4 Ay + By − wL − P = 0
∴ A y =
wL 2
P
−
4
Moment equation: Σ M a − a =
M ( x) +
wx 2 2
∴ M ( x) = −
−
wx 2
Ay x = M ( x) + +
wL x
−
wx 2 2
−
w Lx 2
+
Px 4
=
0
Px
2 2 4 Integration of moment equation: d 2v wx 2 w L x Px + − EI 2 = M ( x) = − dx 2 2 4 3 2 2 dv wx w Lx Px EI =− + − + C 1 dx 6 4 8 wx 4 wLx3 Px3 EI v = − + − + C1 x + C 2 24 12 24
(a) (b)
Boundary conditions: v=0 at x=0
v=0
at
x=L
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2: w(0) 4 wL(0)3 P(0)3 EI (0) = − + − + C1 (0) + C2 24 12 24 Substitute x Substitute x = = L L and and v = 0 into Eq. (b) to determine C 1: w( L) 4 wL( L) 3 P( L) 3 EI (0) = − + − + C1 ( L) 24 12 24
∴ C 2 =
0
∴ C 1 = −
wL3 24
+
PL2 24
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(a) Elastic curve equation for segment AB of the beam: wx 4 wLx3 Px3 wL3 x PL2 x EI v = − + − − + 24 12 24 24 24 ∴
v=−
wx
Px
⎡⎣ x3 − 2 Lx 2 + L3 ⎤⎦ − ⎡⎣ x2 − L2 ⎤⎦ 24 EI 24EI
Ans.
(b) Deflection at midspan:
v x = L / 2
=−
w( L / 2) ⎡⎛ L ⎞
= −
3
⎛L⎞ ⎢⎜ ⎟ − 2 L ⎜ ⎟ 24 EI ⎢⎣⎝ 2 ⎠ ⎝2⎠ 5wL4
384 EI
+
2
⎤ P ( L / 2) ⎡ ⎛ L ⎞ 2 ⎤ 2 +L ⎥− ⎢⎜ ⎟ − L ⎥ 24EI ⎣⎢⎝ 2 ⎠ ⎦⎥ ⎦⎥ 3
PL3
Ans.
64EI
(c) Slope at B:
dv dx B
= θ B = −
w( L)3 6 EI
+
wL( L) 2 4 EI
−
P( L) 2 8 EI
−
wL3 24 EI
+
PL2 24EI
=
wL3 24EI
−
PL2 12EI
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.10 For the beam and loading shown in Fig. P10.10, use the double-integration method to to determine determine (a) the the equation equation o the elastic curve for segment AC of the beam, (b) the deflection at B at B,, and (c) the slope at A at A.. Assume that EI that EI is is constant for the beam.
Fig. P10.10
Solution Beam FBD: Σ M A = − w(3L) ∴ C y = Σ F y =
Ay
3 L
2 9 wL
0
4
+ Cy −
∴ A y =
+ C y ( 2 L) =
w(3 L) = 0
3wL −
9 wL 4
=
3wL 4
Moment equation: Σ M a − a =
3wL ⎛ x ⎞ ⎛ x⎞ M ( x) − Ay x + wx ⎜ ⎟ = M ( x) − x + wx ⎜ ⎟ = 0 4 ⎝2⎠ ⎝ 2⎠
∴ M ( x) = −
wx 2 2
+
3 wL x 4
Integration of moment equation: d 2v wx 2 3 wL x EI 2 = M ( x) = + dx 2 4 3 2 dv wx 3 w Lx EI =− + + C 1 dx 6 8 w x 4 3 wL x 3 EI v = − + + C1 x + C 2 24 24
(a) (b)
Boundary conditions: v=0 at x=0
v=0
at
x = 2L
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2: w(0) 4 3wL(0)3 EI (0) = − + + C1 (0) + C2 24 24 Substitute x Substitute x = = 2 L and L and v = 0 into Eq. (b) to determine C 1:
∴ C 2 =
0
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EI (0) = −
w(2 L )4 24
∴ C 1 =
+
8wL3 24
3wL(2 L )3 24
−
12 wL3 24
+ C1 ( 2 L)
=−
wL3 6
(a) Elastic curve equation for segment AC of of the beam: 4 3 3 wx 3 wL x wL x wx 3 ⎡⎣ x − 3Lx 2 + 4 L3 ⎤⎦ EI v = − + − =− 24 24 6 24 ∴
v=−
wx
⎡⎣ x3 − 3Lx 2 + 4 L3 ⎤⎦ 24 EI
(b) Deflection at B: w( L) ⎡⎣( L)3 − 3L( L) 2 v B = − 24 EI
Ans.
wL4
+ 4L ⎤ = − ⎦ 12EI 3
Ans.
(c) Slope at A:
dv dx A
= θ A = −
w(0)3 6 EI
+
3wL(0)2 8 EI
−
wL3 6 EI
= −
wL3 6EI
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.11 For the simply supported steel beam 6 4 [ E E = = 200 GPa; I GPa; I = = 129 × 10 mm ] shown in Fig. P10.11, use the double-integration method to determine the deflection at B. B. Assume L Assume L = = 4 m, P m, P = = 60 kN, and w = 40 kN/m.
Fig. P10.11
Solution Beam FBD:
⎛ L ⎞ ⎛L⎞ − P ⎜ ⎟ + C y ( L) = 0 ⎟ ⎝2⎠ ⎝2⎠
Σ M A = − wL ⎜
∴ C y = Σ F y =
wL
+
P
2 2 Ay + C y − w( L) − P = 0
∴ A y =
wL 2
Moment equation: wx 2 Σ M a − a = M ( x) + 2 ∴ M ( x) = −
P
+
2
Ay x = M ( x) +
−
wx 2 2
+
w Lx 2
+
wx 2 2
−
w Lx 2
−
Px 2
=
0
Px 2
Integration of moment equation: d 2v wx 2 w L x Px EI 2 = M ( x) = − + + dx 2 2 2 3 2 2 dv wx w Lx Px =− + + + C 1 EI dx 6 4 4 wx 4 wLx3 Px3 EI v = − + + + C1 x + C 2 24 12 12
(a) (b)
Boundary conditions: v=0 at x=0
dv dx
=
0
at
x =
L 2
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute x substitute x = = L/ 2 and dv/dx = dv/dx = 0 into Eq. (b) to determine C 1:
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EI (0) = −
w( L / 2)3 6 wL3
∴ C 1 =
−
+
w L ( L / 2) 2
wL3
4 −
PL2
+
=−
P( L / 2) 2 4 wL3
−
+ C 1
PL2
48 16 16 24 16 Elastic curve equation: wx 4 wLx3 Px3 wL3 x PL2 x EI v = − + + − − 24 12 12 24 16 ∴
v=−
wx
Px
⎡⎣ x3 − 2 Lx2 + L3 ⎤⎦ − ⎡⎣3 L2 − 4 x2 ⎤⎦ 24 EI 48EI
At x = = L/2 L/2:: Deflection at B: At x 4 3 5w L PL v B = − − 384 EI 48EI 6
4
Let E Let E = = 200 GPa, I GPa, I = = 129 × 10 mm , w = 40 kN/m, P kN/m, P = = 60 kN, and L and L = = 4 m. 4 5(40 N/mm)(4,000 mm) (60, 000 N)(4,000 mm)3 v B = − − 384(200, 00 000 N/ N/mm2 )(129 × 106 mm4 ) 48(200, 00 000 N/ N/mm2 )(129× 106 mm4 ) .1680 = −5.16 = −8.27
mm mm − 3.1008 mm
mm
Ans.
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10.12 For the cantilever steel beam [ E E = = 200 6 4 GPa; I = I = 129 × 10 mm ] shown in Fig. P10.12, use the double-integration method to determine the deflection at A at A.. Assume L Assume L = 2.5 m, P m, P = = 50 kN, and w = 30 kN/m.
Fig. P10.12
Solution Moment equation: Σ
a −a
= M ( x) +
wx 2 2
+
Px = 0
∴ M ( x) = −
wx 2 2
−
Px
Integration of moment equation: d 2v wx 2 − Px EI 2 = M ( x) = − dx 2 dv wx 3 Px 2 EI =− − + C 1 dx 6 2 wx 4 Px 3 EI v = − − + C1 x + C 2 24 6
(a) (b)
Boundary conditions: v=0 at x=L
dv dx
=
0
at
= L
Evaluate constants: Substitute x Substitute x = = L L and and dv/dx = dv/dx = 0 into Eq. (a) to determine C 1: 3 w( L) P ( L) 2 wL3 PL2 EI (0) = − − + C1 ∴ C 1 = + 6 2 6 2 Substitute x Substitute x = = L L and and v = 0 into Eq. (b) to determine C 2: w( L) 4 P( L) 3 wL3 PL2 wL4 EI (0) = − ( L) + ( L) + C2 = − − + 24 6 6 2 24 ∴ C 2 = −
wL4 8
−
+
wL4 6
−
PL3 6
+
PL3 2
+ C 2
PL3 3
Elastic curve equation: wx 4 wL3 x wL4 Px3 PL2 x PL3 + − − + − EI v = − 24 6 8 6 2 3 w P ⎡⎣ x 4 − 4 L3 x + 3 L4 ⎤⎦ − ⎡⎣ x3 − 3 L2 x + 2 L3 ⎤⎦ ∴v = − 24 EI 6 EI
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Deflection at A: w P 3wL4 PL3 4 3 4 3 2 3 ⎡ −(0) + 4 L (0) − 3L ⎤⎦ − ⎡(0) − 3 L (0) + 2 L ⎤⎦ = − v A = − 24 EI ⎣ 6 EI ⎣ 24 EI 3EI 6 4 Let E Let E = = 200 GPa, I GPa, I = = 129 × 10 mm , w = 30 kN/m, P kN/m, P = = 50 kN, and L and L = = 2.5 m. 4 3(30 N/mm)(2,500 mm) (50, 000 N)(2,500 mm)3 v A = − − 24(200, 00 000 N/ N/mm2 )(129 × 106 mm4 ) 3(200, 00 000 N/ N/mm2 )(129× 106 mm4 )
5.6777 77 = −5.67
mm − 10.0 10.093 937 7 mm mm
= −15.77 mm
Ans.
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10.13 For the cantilever steel beam [ E E = = 200 6 4 GPa; I = I = 129 × 10 mm ] shown in Fig. P10.13, use the double-integration method to determine the deflection at B at B.. Assume L Assume L = 3 m, M m, M 0 = 70 kN-m, and w = 15 kN/m.
Fig. P10.13
Solution Moment equation: Σ M a − a = − M ( x) −
∴ M ( x) = −
w( L − x) 2 2 w( L − x) 2 2
−
M 0
−
M 0
=
0
Integration of moment equation: d 2v w( L − x) 2 w 2 ⎡⎣ L − 2 Lx + x2 ⎤⎦ − M 0 EI 2 = M ( x) = − − M0 = − dx 2 2 2 2 3 dv w L x w Lx wx EI =− + − − M 0 x + C 1 dx 2 2 6 wL2 x 2 wLx3 wx4 M 0 x2 EI v = − + − − + C1 x + C 2 4 6 24 2
=−
wL2 2
+
wLx −
wx 2 2
−
M 0 (a) (b)
Boundary conditions: v=0 at x=0
dv dx
=
0
at
x = 0
Evaluate constants: Substitute x Substitute x = = 0 and dv/dx = dv/dx = 0 into Eq. (a) to determine C 1 = 0. Next, substitute substitute x x = = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Elastic curve equation: wL2 x 2 wLx3 EI v = − + 4 6 ∴v = −
w
⎡x 24 EI ⎣
4
−
w x4 24
−
M 0 x2 2
3
2
− 4 Lx + 6 L
x ⎤⎦ − 2
M 0 x2 2 EI
Deflection at B: w M 0 ( L) 2 4 3 2 2 ⎡( L) − 4 L( L) + 6 L ( L) ⎤⎦ − v B = − 24 EI ⎣ 2 EI 6
=−
wL4 8EI
−
M 0 L2 2EI
4
Let E Let E = = 200 GPa, I GPa, I = = 129 × 10 mm , w = 15 kN/m, M kN/m, M 0 = 70 kN-m, and L and L = = 3 m.
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v B
=−
(15 N/mm)(3,000 mm mm)4 8(200, 000 N/mm2 )(129 × 106 mm4 )
5.8866 66 = −5.88
−
(70 kN kN-m)(1,000 000 N/ N/kN) kN)(1,000 mm mm/m)(3,000 mm mm)2 2(200, 000 N/mm2 )(129× 106 mm4 )
mm − 12.2 12.209 093 3 mm mm
= −18.10 mm
Ans.
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10.14 For the cantilever steel beam [ E E = = 200 6 4 GPa; I = I = 129 × 10 mm ] shown in Fig. P10.14, use the double-integration method to determine the deflection at A at A.. Assume L Assume L = 2.5 m, P m, P = = 50 kN-m, and w0 = 90 kN/m.
Fig. P10.14
Solution Moment equation: Σ M a − a =
M ( x) +
⎛ x⎞ ( x) ⎜ ⎟ + Px = 0 2 L ⎝ 3⎠
w0 x
∴ M ( x) = −
w0 x3 6 L
−
Px
Integration of moment equation: d 2v w0 x3 EI 2 = M ( x) = − − Px dx 6L dv w0 x4 Px2 EI =− − + C 1 dx 24 L 2 w0 x5 Px3 − + C1 x + C 2 EI v = − 120 L 6
(a) (b)
Boundary conditions: v=0 at x=L
dv dx
=
0
at
= L
Evaluate constants: Substitute x Substitute x = = L L and and dv/dx = dv/dx = 0 into Eq. (a) to determine C 1: 4 w0 ( L) P( L) 2 w0 L3 PL2 EI (0) = − − + C1 ∴ C 1 = + 24 L 2 24 2 Substitute x Substitute x = = L L and and v = 0 into Eq. (b) to determine C 2: w0 ( L)5 P( L)3 w0 L3 PL2 w0 L4 w0 L4 EI (0) = − ( L) + ( L) + C2 = − − + + 120 L 6 24 2 120 24 ∴ C 2 = −
w0 L4 30
−
−
PL3 6
+
PL3 2
+
C 2
PL3 3
Elastic curve equation: w0 x5 Px3 w0 L3 PL2 w0 L4 PL3 EI v = − x+ x− − + − 120 L 6 24 2 30 3 w0 P ⎡⎣ x5 − 5 L4 x + 4 L5 ⎤⎦ − ⎡⎣ x3 − 3 L2 x + 2 L3 ⎤⎦ ∴v = − 120 L EI 6 EI Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
Deflection at A: 6 4 Let E Let E = = 200 GPa, I GPa, I = = 129 × 10 mm , w0 = 90 kN/m, P kN/m, P = = 50 kN, and L and L = = 2.5 m. w0 P ⎡⎣ (0)5 − 5 L4 (0) + 4 L5 ⎤⎦ − ⎡⎣(0)3 − 3L2 (0) + 2 L3 ⎤⎦ v A = − 120 L EI 6 EI =−
=−
w0 L4 30 EI
−
PL3 3EI
(90 N/mm)(2,500 mm)4 30(200, 00 000 N/ N/mm2 )(129 × 106 mm4 )
4.5422 22 = −4.54
−
(50, 000 N)(2,500 mm)3 3(200, 00 000 N/ N/mm2 )(129× 106 mm4 )
mm − 10.0 10.093 937 7 mm mm
= −14.64 mm
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.15 For the beam and loading shown in Fig. P10.15, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB beam AB,, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam.
Fig. P10.15
Solution Beam FBD:
w0 L ⎛ 2 L ⎞
Σ M A = − M A −
∴ M A = − Σ F y =
Ay
−
∴ A y =
⎟=0
⎜
2 ⎝ 3 ⎠ w0 L2
w0 L 2 w0 L
3 =
0
2
Moment equation: Σ M a − a =
⎛ x⎞ ( x) ⎜ ⎟ − Ay x 2 L ⎝3⎠
w0 x
M ( x) − M A +
= M ( x) +
w0 L2
∴ M ( x) = −
3
+
w0 x3 6 L
⎛ x⎞ w L ( x) ⎜ ⎟ − 0 ( x) = 0 2 L 2 ⎝3⎠
w0 x
+
w0 Lx 2
−
w0 L2 3
Integration of moment equation: d 2v w0 x3 w0 Lx w0 L2 EI 2 = M ( x) = − + − dx 6L 2 3 4 2 2 dv w0 x w0 Lx w0 L x EI =− + − + C 1 dx 24 L 4 3 w0 x5 w0 Lx3 w0 L2 x2 + − + C1 x + C 2 EI v = − 120 L 12 6
(a) (b)
Boundary conditions: v=0 at x=0
dv dx
=
0
at
x = 0
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute substitute x x = = 0 and dv/dx = dv/dx = 0 into Eq. (b) to determine C 1 = 0. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
(a) Elastic curve equation: w0 x5 w0 Lx3 w0 L2 x2 EI v = − + − 120 L 12 6 ∴
v=−
w0
⎡⎣ x5 − 10 L2 x3 + 20 L3 x2 ⎤⎦ 120 L EI
Ans.
(b) Deflection at the free end:
v B
=−
w0
⎡ ( L) 120 L EI ⎣
5
2
− 10 L
( L)
3
3
+ 20 L
( L) ⎤⎦ = 2
−
11w0 L5
Ans.
120L EI
(c) Slope at the free end:
dv dx B
= θ B = −
w0 ( L) 4 24 L
+
w0 L( L) 2 4
−
w0 L2 ( L) 3
=−
w0 L3 24
+
6 w0 L3 24
−
8 w0 L3 24
= −
w0 L3 8EI
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.16 For the beam and loading shown in Fig. P10.16, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB beam AB,, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam.
Fig. P10.16
Solution Beam FBD:
w0 L ⎛ L ⎞
Σ M A = − M A −
⎜ ⎟=0
2 ⎝3⎠ w0 L2
∴ M A = − Σ F y =
Ay
−
6
w0 L
∴ A y =
2 w0 L
=
0
2
Moment equation: Σ M a − a = − M ( x) −
M ( x) = − =−
=−
w0 6 L w0 6 L
2 L
3
=
0
−
x3 )
( L − x) 3 ( L3 − 3 L2 x + 3 Lx2
w0 L2 6
w0 ( L − x)3
+
w0 Lx 2
−
w0 x2 2
+
w0 x3 6 L
Integration of moment equation: d 2v w0 x3 w0 x2 w0 Lx w0 L2 EI 2 = M ( x) = − + − dx 6L 2 2 6 4 3 2 2 dv w0 x w0 x w0 Lx w0 L x EI = − + − + C 1 dx 24 L 6 4 6 w0 x5 w0 x 4 w0 Lx3 w0 L2 x2 EI v = − + − + C1 x + C 2 120 L 24 12 12
(a) (b)
Boundary conditions: v=0 at x=0
dv dx
=
0
at
x = 0
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Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute substitute x x = = 0 and dv/dx = dv/dx = 0 into Eq. (b) to determine C 1 = 0. (a) Elastic curve equation: w0 x5 w0 x 4 w0 Lx3 EI v = − + 120 L 24 12 ∴
v=
−
w0 L2 x2 12
w0
⎡⎣ x5 − 5Lx 4 + 10 L2 x3 − 10 L3 x2 ⎤⎦ 120 L EI
Ans.
(b) Deflection at the free end:
v B
=
w0
⎡( L) 120 L EI ⎣
5
4
2
− 5L ( L) + 10 L
( L)
3
3
− 10 L
( L) ⎤⎦ = − 2
4 w0 L5 120L EI
= −
w0 L4 30EI
Ans.
(c) Slope at the free end:
dv dx B
= θ B =
w0 ( L) 4 24 L EI
−
w0 ( L)3 6 EI
+
w0 L( L) 2 4 EI
−
w0 L2 ( L) 6 EI
= −
w0 L3 24EI
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.17 For the beam and loading shown in Fig. P10.17, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at B, B, (c) the deflection at the free end, and (d) the slope at the free end. Assume that EI that EI is is constant for the beam.
Fig. P10.17
Solution Beam FBD: Σ M A = − M A −
∴ M A = − Σ F y =
Ay
−
∴ A y =
wL ⎛ L
⎜ 2 ⎝2
+
L⎞
⎟=0
4⎠
3wL2
wL 2 wL
8 =
0
2
Consider beam segment AB (0 ≤ x ≤ L/2) Moment equation: Σ M a − a =
M ( x) − M A − Ay x = M ( x) +
∴ M ( x) = −
3wL2 8
+
3wL2 8
−
wL 2
x=0
wL x 2
Integration of moment equation: d 2v 3wL2 wL x EI 2 = M ( x) = − + dx 8 2 dv 3wL2 x wL x2 EI =− + + C 1 dx 8 4 3wL2 x 2 wL x3 EI v = − + + C1 x + C 2 16 12
(a) (b)
Boundary conditions: v=0 at x=0
dv dx
=
0
at
x = 0
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute substitute x x = = 0 and dv/dx = dv/dx = 0 into Eq. (b) to determine C 1 = 0. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
Elastic curve equation for beam segment AB: 3wL2 x 2 wL x3 EI v = − + 16 12 ∴
v=−
wL x 2 48 EI
[9 L − 4 x ]
Slope at B: Let x = = L/2 dv 3wL2 ( L / 2)
dx B
= θ B = −
8 EI
+
(0 ≤ x ≤ L / 2)
wL( L / 2) 2 4 EI
=−
wL3 8EI
Deflection at B: Let x = = L/2 w L( L / 2) 2 ⎡ 7 wL4 ⎛ L ⎞⎤ v B = − ⎢9 L − 4 ⎜ 2 ⎟ ⎥ = − 192EI 48 EI ⎣ ⎝ ⎠⎦
Consider beam segment BC ( L L/2 ≤ x ≤ L) Moment equation: Σ M b − b =
w⎛
M ( x) − M A − Ay x + 3wL2
= M ( x) +
∴ M ( x) = −
=−
−
8
w⎛
L⎞
2⎝
2⎠
⎜x−
wx 2 2
+
⎟
wL 2
2
wL x −
2
⎜x− 2⎟ ⎠
2⎝ x+
wL
+
L⎞
2
x−
w⎛
L⎞
⎜x− 2⎟ ⎠
2⎝
2
=
0
3wL2 8
wL2 2
Integration of moment equation: d 2v wx 2 wL2 EI 2 = M ( x) = − + wL x − dx 2 2 dv wx3 wL x2 wL2 x EI =− + − + C 3 dx 6 2 2 wx 4 wL x3 wL2 x2 EI v = − + − + C3 x + C 4 24 6 4
(c) (d)
Continuity conditions:
v=− dv dx
7 wL4 192 EI
=−
wL3 8 EI
at
x=
at
x =
L 2 L 2
Evaluate constants: Substitute the slope continuity condition into Eq. (c) for x for x = = L/ L/ 2 and solve for C 3: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
EI
dv dx
w( L / 2)3
=−
6
+
wL( L / 2) 2 2
−
wL2 ( L / 2) 2
+ C 3 = −
wL3 8
wL3
∴ C 3 =
48 Next, substitute the deflection continuity condition into Eq. (d) for x for x = = L/ L/ 2 and solve for C 4 4 3 2 2 3 w( L / 2) wL( L / 2) w L ( L / 2) wL 7 wL4 EI v = − ( L / 2) + C 4 = − + − + 24 6 4 48 192 ∴ C 4 = −
wL4 384
Elastic curve equation for beam segment BC : wx 4 wL x3 wL2 x2 wL3 x wL4 EI v = − + − − − 24 6 4 48 384 ∴
v=−
w
⎡⎣16 x 4 − 64 Lx3 + 96 L2 x2 − 8L3 x + L4 ⎤⎦ 384 EI
( L / 2 ≤ x ≤ L)
(a) Elastic curve equations for entire beam:
v=− v=−
wL x 2 48 EI
[9 L − 4 x ]
( 0 ≤ x ≤ L / 2)
Ans.
w
⎡⎣16 x 4 − 64 Lx3 + 96 L2 x2 − 8L3 x + L4 ⎤⎦ 384 EI
( L / 2 ≤ x ≤ L)
Ans.
(b) Deflection at B:
v B
=−
7 wL4
Ans.
192 EI
(c) Deflection at free end of cantilever:
w
⎡⎣16( L)4 − 64 L( L)3 + 96 L2 ( L)2 − 8 L3 ( L) + L4 ⎤⎦ = vC = − 384 EI (d) Slope at free end of cantilever: dv 8 w( L)3 24 wL( L)2 24 wL2 ( L) wL3 EI = − + − + dx 48 48 48 48 ∴
dv dx
= θ C = − C
7 wL3 48 EI
=−
−
41wL4 384EI
Ans.
7 wL3 48 Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
10.18 For the beam and loading shown in Fig. P10.18, use the double-integration method to determine (a) the equation o the elastic curve for the beam, and (b) the deflection at B at B.. Assume that EI that EI is is constant for the beam.
Fig. P10.18
Solution Beam FBD: Σ M A =
C y ( L) −
∴ C y = Σ F y =
Ay
wL ⎛ L ⎞
⎜ ⎟=0
2 ⎝4⎠
wL 8
+ C y −
∴ A y =
wL 2
0
=
3wL 8
Consider beam segment AB (0 ≤ x ≤ L/2) Moment equation: Σ M a − a =
M ( x) +
wx 2 2
∴ M ( x) = −
−
w x2 2
Ay x = M ( x) + +
wx 2 2
−
3 wL 8
x=0
3w L x 8
Integration of moment equation: d 2v w x 2 3 wL x EI 2 = M ( x) = − + dx 2 8 3 2 dv wx 3 wL x EI =− + + C 1 dx 6 16 w x 4 wL x 3 + + C1 x + C 2 EI v = − 24 16
(a) (b)
Boundary conditions: v=0 at x=0 Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Slope at B: Let x = = L/2 in Eq. (a).
EI
dv dx B
=
EIθ B
=−
w( L / 2)3 6
+
3wL( L / 2) 2 16
+ C1 = −
wL3 48
+
3 wL3 64
+ C1 =
5 wL3 192
+ C 1
(c)
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Deflection at B: Let x = = L/2 in Eq. (b).
EI v B
=−
w( L / 2) 4
wL( L / 2)3
+
24
16
+ C1 ( L / 2) = −
wL4 384
+
wL4 128
C1 L
+
2
=
wL4
+
192
C1 L 2
(d)
Consider beam segment BC ( L L/2 ≤ x ≤ L) Moment equation: Σ M b − b = − M ( x) + C y ( L −
∴ M ( x) =
wL 8
x) = − M ( x) +
( L − x) =
wL2 8
−
wL 8
( L − x) = 0
wL x 8
Integration of moment equation: d 2v wL x wL wL2 + EI 2 = M ( x) = − dx 8 8 2 2 dv wL x wL x EI =− + + C 3 dx 16 8 wL x3 wL2 x2 EI v = − + + C3 x + C 4 48 16
(e) (f)
Boundary conditions: v=0 at x=L Evaluate constants: Substitute x Substitute x = = L L and and v = 0 into Eq. (f) to find wL( L)3 wL2 ( L) 2 EI (0) = − + + C3 ( L) + C 4 48 16 ∴
C3 L + C 4
=−
wL4
24
(g)
Slope at B: Let x = = L/2 in Eq. (e).
EI
dv dx B
=
EI θ B
=−
w L ( L / 2) 2 16
+
wL2 ( L / 2) 8
+ C3 = −
wL3 64
+
wL3 16
+ C3 =
3 wL3 64
+ C 3
(h)
Deflection at B: Let x = = L/2 in Eq. (f).
EI v B
=−
wL ( L / 2 ) 3 48
+
wL2 ( L / 2) 2 16
+ C3 ( L / 2) + C4 =
5 wL4 384
+
C3 L 2
+
C 4
Continuity conditions: Since the slope at B at B must must be the same for both beam segments, equate Eqs. (c) and an d (h): 3 3 5w L 3 wL + C1 = + C 3 192 64
(i)
(j)
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Further, the deflection at B at B must must be the same for both segments; therefore, equate Eqs. (d) and (i): wL4 C1 L 5 wL4 C3 L + = + + C 4 192 2 384 2
(k)
Evaluate constants: Solve Eqs. (g), (j), and (k) simultaneously to determine the values of constants C 1, C 3, and C 4: 9wL3 17 wL3 wL4 C1 = − C3 = − C 4 = 384 384 384 (a) Elastic curve equation for beam segment AB: w x 4 wL x3 9 wL3 x EI v = − + − 24 16 384 ∴
v=−
wx
⎡⎣16 x3 − 24 L x 2 + 9 L3 ⎤⎦ 384 EI
(0 ≤ x ≤ L / 2)
Ans.
( L / 2 ≤ x ≤ L)
Ans.
(a) Elastic curve equation for beam segment BC : wL x3 wL2 x2 17 wL3 x wL4 EI v = − + − + 48 16 384 384 ∴
v=−
wL
⎡⎣8 x3 − 24 2 4 Lx2 + 17 L2 x − L3 ⎤⎦ 384 EI
(b) Deflection at B:
EI v B
=
wL4 192
−
9 wL4 768
=−
5wL4 768
∴
vB
=−
5wL4 768 EI
Ans.
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10.19 For the beam and loading shown in Fig. P10.19, use the double-integration method to to determine determine (a) the the equation equation o the elastic curve for the entire beam, (b) the deflection at C , and (c) the slope at B at B.. Assume that EI that EI is is constant for the beam.
Fig. P10.19
Solution Beam FBD: Σ M A =
⎛ ⎝
By (3 L) − wL ⎜ 3 L +
∴ B y = Σ F y =
Ay
+
L ⎞
⎟=0
2⎠
7 wL 6 By
∴ A y = −
−
wL = 0
wL 6
Consider beam segment AB (0 ≤ x ≤ 3 L) Moment equation: Σ M a − a =
⎛ wL ⎞ ⎟x=0 ⎝ 6 ⎠
M ( x) − Ay x = M ( x) + ⎜
∴ M ( x) = −
wL x 6
Integration of moment equation: d 2v wL x EI 2 = M ( x) = − dx 6 dv wL x 2 EI =− + C 1 dx 12 wL x3 EI v = − + C1 x + C 2 36 Boundary conditions: v=0 at x=0
and
(a) (b)
v=0
at
x = 3L
Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute substitute x x = = 3 L and L and v = 0 into Eq. (b) and solve for C 1: wL(3 L)3 9 wL3 wL3 EI (0) = − + C1 (3 L) ∴ C 1 = = 36 36 4
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Slope at B: Let x = = 3 L in Eq. (a).
EI
dv dx B
=
EI θ B
=−
wL(3 L)2 12
wL3
+
4
= −
wL3 2
(c)
Consider beam segment BC (3 L ≤ x ≤ 4 L) Moment equation: Σ M b − b = − M ( x) − ∴ M ( x) = −
w
(4 L − x) 2
2 w(4 L − x)2
=
0
2
Integration of moment equation: d 2v w(4 L − x) 2 EI 2 = M ( x) = − dx 2 3 dv w(4 L − x) EI = + C 3 6 dx w(4 L − x)4 EI v = − + C3 x + C 4 24
(e) (f)
Boundary conditions: v=0 at x = 3L Substitute x Substitute x = = 3 L and L and v = 0 into Eq. (f) to find w(4 L − 3L)4 wL4 EI (0) = − + C3 (3 L) + C4 = − 24 24 ∴ C4 =
wL4 24
+
C3 (3 L) + C 4
− (3L)C 3
(g)
Slope at B: Let x = = 3 L in Eq. (e).
EI
dv dx B
=
EI θ B
=
w(4 L − 3 L)3 6
+ C3 =
wL3 6
+ C 3
Continuity conditions: Since the slope at B at B must must be the same for both beam segments, equate Eqs. (c) and an d (h): wL3 wL3 2 wL3 C 3 = − − = + C3 2 6 3 Backsubstitute this result into Eq. (g) to determine C 4:
C4
=
wL4 24
− (3L )C3 =
(h)
(i)
⎛ −2 wL3 ⎞ 49 wL4 − (3 L) ⎜ ⎟= 24 3 24 ⎝ ⎠
wL4
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(a) Elastic curve equation for beam segment AB: wL x3 9 wL3 x + EI v = − 36 36 ∴
v=−
wL x
⎡⎣ x 2 − 9 L2 ⎤⎦ 36 EI
( 0 ≤ x ≤ 3 L)
Ans.
(a) Elastic curve equation for beam segment BC : w(4 L − x )4 2 wL3 x 49 wL4 EI v = − − + 24 3 24 ∴
v=−
w
⎡⎣(4 L − x)4 + 16 L3 x − 49 L4 ⎤⎦ 24 EI
(b) Deflection at C : w ⎡⎣(4 L − 4 L) 4 vC = − 24 EI ∴
vC = −
3
+ 16 L
(3 L ≤ x ≤ 4 L)
(4 L) − 49 49 L ⎤⎦ = − 4
w
⎡ 64 L 24EI ⎣
4
−
Ans.
49 L ⎤⎦ = − 4
5wL4
15wL4 24EI Ans.
8 EI
(c) Slope at B: Let x = = 3 L in Eq. (a).
EI
dv dx B
=
EI θ B
=−
wL3 2
∴
dv dx
= θ B = − B
wL3 2 EI
Ans.
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10.20 For the beam and loading shown in Fig. P10.20, use the double-integration method to determine (a) the equation o the elastic curve for the beam, (b) the location of the maximum deflection, and (c) the maximum beam deflection. Assume that EI that EI is is constant for the beam.
Fig. P10.20
Solution Beam FBD: Σ M A =
By L −
∴ B y = Σ F y =
Ay
+
∴ A y =
w0 L ⎛ 2 L ⎞
⎟=0
⎜
2 ⎝ 3 ⎠ w0 L 3
By
−
w0 L 2
=
0
w0 L 6
Moment equation: Σ M a − a =
M ( x) +
= M ( x) +
∴ M ( x) = −
w0 x 2 ⎛ x ⎞ ⎜ ⎟ − Ay x 2 L ⎝ 3 ⎠ w0 x 2 ⎛ x ⎞ w0 L x ⎜ ⎟− 6 2 L ⎝ 3 ⎠
w0 x3 6 L
+
=
0
w0 L x 6
Integration of moment equation: d 2v w0 x3 w0 L x + EI 2 = M ( x) = − dx 6L 6 dv w0 x4 w0 L x2 EI =− + + C 1 24 L 12 dx w0 x5 w0 L x3 EI v = − + + C1 x + C 2 120 L 36
(a) (b)
Boundary conditions: v=0 at x=0
v=0
at
x=L
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Evaluate constants: Substitute x Substitute x = = 0 and v = 0 into Eq. (b) to determine C 2 = 0. Next, substitute substitute x x = = L L and and v = 0 into Eq. (b) and solve for C 1: 7 w0 L3 w0 ( L)5 w0 L( L) 3 EI (0) = − + + C1 ( L) ∴ C 1 = − 120 L 36 360 (a) Elastic curve equation:
EI v = −
w0 x5 120 L
+
w0 L x3 36
−
7 w0 L3 x
∴
360
v=−
w0 x
⎡⎣3 x4 − 10 L2 x2 + 7 L4 ⎤⎦ 360L EI
Ans.
(b) Location of maximum deflection: The maximum deflection occurs where the beam slope is zero. Therefore, set the beam slope equation [Eq. (a)] equal to zero: dv w0 x4 w0 L x2 7 w0 L3 EI = − + − =0 dx 24 L 12 360 Multiply by −360 L/w0 to obtain: 15 x 4 − 30 L2 x2 + 7 L4 = 0 Solve this equation numerically to obtain:
= 0.51932962236 L = 0.51933L (c) Maximum beam deflection: w (0.51933L ) ⎡⎣3(0.51933L )4 vmax = − 0 360 L EI =−
w0 (0.51933) 360 EI
⎡⎣ 4.52118 L ⎤⎦ = − 4
Ans.
2
− 10 L
(0.51933L )2
(0.0065222)w0 L4 EI
+
7L4 ⎤⎦
= − 0.00652
w0 L4 EI
Ans.
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