10.21 For the beam and loading shown in Fig. P10.21, integrate the load distribution to determine (a) the equation of the elastic curve for the beam, and (b) the maximum deflection for the beam. Assume that EI is constant for the beam.
Fig. P10.21
Solution Integrate the load distribution: d 4v wx EI 4 = − 0 dx L 3 d v w0 x2 EI 3 = − + C 1 dx 2L d 2v w0 x3 EI 2 = − + C1 x + C 2 dx 6L dv w0 x 4 C1 x2 =− + + C2 x + C 3 EI dx 24 L 2 w0 x5 C1 x3 C2 x2 EI v = − + + + C3 x + C 4 120 L 6 2 Boundary conditions and evaluate constants: d 3v at x = 0, V = EI 3 = 0 dx d 2v at x = 0, M = EI 2 = 0 dx dv w0 ( L) 4 at x = L, =0 − + C3 = 0 dx 24 L w0 ( L)5 w0 L3 ( L) at x = L, v = 0 − + + C4 = 0 120 L 24
∴ C 1 = 0 ∴ C 2 = 0 ∴ C 3 =
w0 L3
24 w0 L4 ∴ C 4 = − 30
(a) Elastic curve equation:
EI v = −
w0 x5 120 L
+
w0 L3 x 24
−
w0 L4
∴v=−
30
w0
⎡⎣ x5 − 5 L4 x + 4 L5 ⎤⎦ 120L EI
Ans.
(b) Maximum deflection:
vmax = −
w0
w0 L4
⎡ (0) − 5 L (0) + 4 L ⎤⎦ = − 120 LEI L EI ⎣ 30EI 5
4
5
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.
10.22 For the beam and loading shown in Fig. P10.22, integrate the load distribution to determine (a) the equation of the elastic curve for the beam, and (b) the deflection midway between the supports. Assume that EI that EI is is constant for the beam.
Fig. P10.22
Solution Integrate the load distribution: d 4v wx EI 4 = − 0 dx L 3 d v w0 x2 + C 1 EI 3 = − dx 2L d 2v w0 x3 EI 2 = − + C1 x + C 2 dx 6L dv w0 x 4 C1 x2 EI =− + + C2 x + C 3 dx 24 L 2 w0 x5 C1 x3 C2 x2 EI v = − + + + C3 x + C 4 120 L 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx d 2v w0 ( L)3 at x = L, M = EI 2 = 0 − + C1 ( L) = 0 dx 6L at x = 0, v = 0
at x = L, v = 0
−
w0 ( L)5
+
120 L
w0 L x3 36
+ C3 x = 0
∴ C 2 = 0 ∴ C 1 =
w0 L
6 ∴ C 4 = 0
∴ C 3 = −
7 w0 L3 360
(a) Elastic curve equation:
EI v = −
w0 x5 120 L
+
w0 L x3 36
−
7 w0 L3 x
∴ v=−
360
w0
⎡⎣3 x5 − 10 L2 x3 + 7 L4 x⎤⎦ 360L EI
Ans.
(b) Maximum deflection:
vmax = −
w0
5w0 L4
⎡3( L / 2) − 10 L ( L / 2) + 7 L ( L / 2) ⎤⎦ = − 360 LEI L EI ⎣ 768EI 5
2
3
4
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.
10.23 For the beam and loading shown in Fig. P10.23, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection at the left end of the beam, and (c) the support reactions R reactions R B and M B. Assume that EI that EI is is constant for the beam.
Fig. P10.23
Solution Integrate the load distribution: d 4v w0 x3 EI 4 = − 3 dx L d 3v w0 x4 EI 3 = − + C 1 dx 4 L3 d 2v w0 x5 EI 2 = − + C1 x + C 2 dx 20 L3 dv w0 x6 C1 x2 EI =− + + C2 x + C 3 dx 120 L3 2 w0 x 7 C1 x3 C2 x2 EI v = − + + + C3 x + C 4 840 L3 6 2 Boundary conditions and evaluate constants: d 3v at x = 0, V = EI 3 = 0 dx d 2v at x = 0, M = EI 2 = 0 dx dv w0 ( L) 6 at x = L, =0 − + C3 = 0 dx 120 L3 w0 ( L) 7 w0 L3 ( L) at x = L, v = 0 − + + C4 = 0 840 L3 120 (a) Elastic curve equation:
EI v = −
w0 x 7 3
840 L
+
7 w0 L3 x 840
−
6 w0 L4
∴ v=−
840
∴ C 1 = 0 ∴ C 2 = 0 ∴ C 3 =
w0 L3
120 6 w0 L4 ∴ C 4 = − 840
w0
⎡⎣ x7 − 7 L6 x + 6 L7 ⎤⎦ 840L EI 3
Ans.
(b) Deflection at left end of beam:
vmax = −
w0
6 w0 L7
6w0 L4
⎡(0) − 7 L (0) + 6 L ⎤⎦ = − = − 840 L3 EI ⎣ 840L3 EI 140EI 7
6
7
Ans.
(c) Support reactions R B and M B:
V B = EI
d 3v dx3 x = L
M B = EI
=−
d 2v dx 2 x = L
w0 ( L) 4
=−
4 L3
=−
w0 ( L)5 20 L3
w0 L
=−
4 w0 L2 20
∴ RB = ∴ M B =
w0 L 4
↑
w0 L2 20
Ans.
(cw)
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.
10.24 For the beam and loading shown in Fig. P10.24, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, and (c) the support reactions R A and R and R B. Assume that EI is is constant for the beam.
Fig. P10.24
Solution Integrate the load distribution: d 4v w0 x3 EI 4 = − 3 dx L 3 d v w0 x4 EI 3 = − + C 1 dx 4 L3 d 2v w0 x5 EI 2 = − + C1 x + C 2 dx 20 L3 dv w0 x6 C1 x2 EI =− + + C2 x + C 3 dx 120 L3 2 w0 x 7 C1 x3 C2 x2 EI v = − + + + C3 x + C 4 840 L3 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx d 2v w0 ( L)5 at x = L, M = EI 2 = 0 − + C1 ( L) = 0 dx 20 L3 at x = 0, v = 0
at x = L, v = 0
−
w0 ( L) 7 840 L3
+
w0 L( L) 3 120
∴ C 2 = 0 ∴ C 1 =
w0 L
20 ∴ C 4 = 0
+ C3 ( L) = 0
∴ C 3 = −
6 w0 L3 840
(a) Elastic curve equation:
EI v = −
w0 x 7 3
840 L
+
w0 Lx3 120
−
6 w0 L3 x
∴ v=−
840
w0
⎡⎣ x7 − 7 L6 x3 + 6 L6 x ⎤⎦ 840L EI 3
Ans.
(b) Deflection midway between the supports:
v x = L / 2
3 ⎡⎛ L ⎞7 ⎤ 13w0 L4 6⎛ L⎞ 6⎛ L⎞ =− ⎢⎜ ⎟ − 7 L ⎜ ⎟ + 6 L ⎜ ⎟ ⎥ = − 840 L3 EI ⎣⎢⎝ 2 ⎠ 5120EI ⎝ 2⎠ ⎝ 2 ⎠ ⎦⎥
w0
Ans.
(c) Support reactions R A and R B:
V A = EI V B = EI
d 3v dx3 x = 0 d 3v dx3 x = L
=− =−
w0 (0) 4 4 L3 w0 ( L) 4 4 L3
+
w0 L
+
w0 L
20 20
=
w0 L 20
=−
4 w0 L 20
∴ RA =
w0 L
∴ RB =
w0 L
20 5
↑
Ans.
↑
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.
10.25 For the beam and loading shown in Fig. P10.25, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection at the left end of the beam, and (c) the support reactions R reactions R B and M B. Assume that EI that EI is is constant for the beam.
Fig. P10.25
Solution Integrate the load distribution: π x d 4v EI 4 = − w0 cos dx 2L 3 π x d v 2 w0 L EI 3 = − sin + C 1 π dx 2L π x d 2 v 4 w0 L2 EI 2 = cos + C1 x + C 2 2 π dx 2L π x dv 8w0 L3 C1 x2 EI sin = + + C2 x + C 3 3 π dx 2L 2 π x 16w0 L4 C1 x3 C2 x2 EI v = − cos + + + C3 x + C 4 4 π 2 L 6 2 Boundary conditions and evaluate constants: d 3v at x = 0, V = EI 3 = 0 dx π (0) d 2v 4 w0 L2 at x = 0, M = EI 2 = 0 c o s + C2 = 0 2 π dx 2L π ( L) dv 8 w0 L3 4 w0 L2 ( L) at x = L, sin =0 − + C3 = 0 3 2 π π dx 2L
at x = L, v = 0
− ∴ C 4 =
2 w0 L4 π
3
16w0 L4 π
4
cos
π ( L)
2 L
−
4 w0 L2 ( L)2 2π 2
−
∴ C 1 = 0 ∴ C 2 = − ∴ C 3 = −
4 w0 L3 ( L) π
3
4 w0 L2 π
2
4 w0 L3 π
3
(2 − π )
(2 − π ) + C 4 = 0
(4 − π )
(a) Elastic curve equation: π x 16w0 L4 4 w0 L2 x2 4w0 L3 2 w0 L4 EI v = − cos (2 − π ) + (4 − π ) − − 4 3 3 π π π 2 L 2π 2
∴v=−
π x ⎡ ⎤ 4 2 2 2 3 4 π L x + 8π L x( 2 − π ) − 4π L ( 4 − π ) 3 2 L c o s 4 + ⎥⎦ 2π EI ⎢⎣ 2L
w0 4
Ans.
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(b) Deflection at left end of beam: π (0) w ⎡ ⎤ v A = − 40 ⎢32 L4 cos + 4π 2 L2 (0)2 + 8π L3 (0)(2 − π ) − 4π L4 (4 − π ) ⎥ 2π EI ⎣ 2L ⎦
=−
w0
w0 L4
⎡32 L − 4π L (4 − π )⎤⎦ = − 4 [ 32 − 4π (4 − π )] 2π 4 EI ⎣ 2π EI
= −0.1089
4
4
w0 L4
Ans.
EI
(c) Support reactions R B and M B:
V B = EI
d 3v dx3 x = L
M B = EI
=−
d 2v dx 2 x = L
=
2 w0 L π
4 w0 L2 π
2
sin
π ( L)
cos
2L π ( L)
2L
=− −
2 w0 L
∴ RB =
π
4 w0 L2 π
2
=−
4 w0 L2 π
2
2 w0 L
∴ M B =
π
↑
4 w0 L2 π
2
Ans.
(cw)
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.
10.26 For the beam and loading shown in Fig. P10.26, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, (c) the slope at the left end of the beam, and (d) the support reactions R reactions R A and R and R B. Assume that EI that EI is is constant for the beam.
Fig. P10.26
Solution Integrate the load distribution: π x d 4v EI 4 = − w0 sin dx L 3 π x d v wL EI 3 = 0 cos + C 1 π dx L π x d 2 v w0 L2 EI 2 = sin + C1 x + C 2 2 π dx L 2 π x C1 x dv w0 L3 EI = − 3 cos + + C2 x + C 3 π dx L 2 π x w0 L4 C1 x3 C2 x2 + + + C3 x + C 4 EI v = − 4 sin π L 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx π ( L) d 2v w0 L2 at x = L, M = EI 2 = 0 s i n + C1 ( L) = 0 2 π dx L at x = 0, v = 0
at x = L, v = 0
w0 L4
−
π
4
sin
π ( L)
L
∴ C 2 = 0 ∴ C 1 = 0 ∴ C 4 = 0
+ C3 ( L) = 0
∴ C 3 = 0
(a) Elastic curve equation:
EI v = −
w0 L4 π
4
sin
πx
∴ v=−
L
w0 L4 π
4
EI
sin
π x
Ans.
L
(b) Deflection midway between the supports:
v x = L / 2 = −
w0 L4 π
4
EI
sin
π ( L / 2)
L
= −
w0 L4 π
4
Ans.
EI
(c) Slope at the left end of the beam:
EI
dv dx A
= EI θ A = −
w0 L3 π
3
cos
π (0)
L
=−
w0 L3 π
3
∴ θ A = −
w0 L3 π
3
EI
Ans.
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(d) Support reactions R A and R B:
V A = EI V B = EI
d 3v dx
3
=
w0 L
=
w0 L
x = 0
d 3v dx3 x = L
π
π
cos cos
π (0)
L π ( L)
L
=
w0 L π
=−
w0 L π
∴ RA =
w0 L
∴ RB =
w0 L
π
π
↑
Ans.
↑
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.
10.27 For the beam and loading shown in Fig. P10.27, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, (c) the slope at the left end o the beam, and (d) the support reactions R reactions R A and R and R B. Assume that EI that EI is is constant for the beam.
Fig. P10.27
Solution Integrate the load distribution: π x d 4v EI 4 = − w0 sin dx 2L 3 π x d v 2 w0 L EI 3 = cos + C 1 π dx 2L π x d 2 v 4 w0 L2 EI 2 = sin + C1 x + C 2 2 π dx 2L π x dv 8 w0 L3 C1 x2 EI cos =− + + C2 x + C 3 3 π dx 2L 2 π x 16w0 L4 C1 x3 C2 x2 EI v = − sin + + + C3 x + C 4 4 π 2 L 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx π ( L) d 2v 4 w0 L2 at x = L, M = EI 2 = 0 sin + C1 ( L) = 0 2 π dx 2L at x = 0, v = 0
at x = L, v = 0
−
16w0 L4 π
4
sin
π ( L)
2 L
−
4 w0 L( L)3 6π
2
+ C3 ( L) = 0
∴ C 2 = 0 ∴ C 1 = −
4 w0 L π
2
∴ C 4 = 0 ∴ C 3 =
2 w0 L3 3π
4
(24 + π 2 )
(a) Elastic curve equation: π x 16w0 L4 4 w0 L x3 2 w0 L3 x EI v = − sin (24 + π 2 ) − + 4 2 4 π 2 L 6π 3π
∴v=−
π x 2 w0 ⎡ 4 2 3 2 3 ⎤ π L x − ( 24 + π ) L x 2 4 L s i n + 4 ⎥ 3π EI ⎢⎣ 2L ⎦
Ans.
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(b) Deflection midway between the supports:
v x = L / 2 = − =− =−
2w0 ⎡
4
⎢ 24 L sin
π ( L / 2)
3π 4 EI ⎢⎣ 2 w0 L4 ⎡
2L
π
π
2
⎢ 24sin 4 + 8 −
+π
3π 4 EI
3 ⎛ L⎞ ⎛ L ⎞⎤ L⎜ ⎟ − (24 + π 2 ) L3 ⎜ ⎟ ⎥ ⎝2⎠ ⎝ 2 ⎠ ⎥⎦
(24 + π 2 ) ⎤
3π 4 EI ⎣ 2 w0 L4
2
⎥ ⎦
2
(1.2694611)
= −0.0086882
w0 L4 EI
= − 0.00869
w0 L4
Ans.
EI
(c) Slope at the left end of the beam: 2 π (0) 2 w0 L(0) dv 8 w0 L3 2 w0 L3 EI cos (24 + π 2 ) = EI θ A = − − + 3 2 4 π π dx A 2L 3π
=−
8w0 L3 π
3
+
2w0 L3 3π 4
∴ θ A = − 0.0262
2 ⎤ ⎡ 8 16 − − = −0.026209w0 L3 3 4 2 ⎥ π 3π ⎦ ⎣π
(24 + π 2 ) = − w0 L3 ⎢
w0 L3
Ans.
EI
(d) Support reactions R A and R B:
V A = EI
d 3v dx
∴ R A = V B = EI
=
3
∴ R B =
π
x = 0
2 w0 L π
2
d 3v dx
2 w0 L
2 w0 L
x = L
4 w0 L π
2
π (0)
2L
−
4 w0 L π
2
=
2 w0 L π
2
(π − 2)
(π − 2) ↑
=
3
cos
↑
π
cos
Ans. π ( L)
2L
−
4 w0 L π
2
=−
4 w0 L π
2
Ans.
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10.28 For the beam and loading shown in Fig. P10.28, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection at the left end of the beam, and (c) the support reactions R reactions R B and M B. Assume that EI that EI is is constant for the beam.
Fig. P10.28
Solution Integrate the load distribution: d 4v π x EI 4 = − w0 sin dx 2L d 3v 2 w0 L π x EI 3 = cos + C 1 dx π 2L d 2 v 4 w0 L2 π x EI 2 = sin + C1 x + C 2 2 dx π 2L dv 8 w0 L3 π x C1 x2 EI cos =− + + C2 x + C 3 3 dx π 2L 2 16w0 L4 π x C1 x3 C2 x2 EI v = − sin + + + C3 x + C 4 4 2 L 6 2 π Boundary conditions and evaluate constants: d 3v 2 w0 L π (0) at x = 0, V = EI 3 = 0 cos + C1 = 0 dx π 2L d 2v 4 w0 L2 π (0) 2 w0 L(0) at x = 0, M = EI 2 = 0 s i n − + C2 = 0 2 dx π 2L π dv 8 w0 L3 π ( L) 2 w0 L ( L) 2 at x = L, cos =0 − + + C3 = 0 3 dx 2L 2π π π ( L) 16w0 L4 2 w0 L( L)3 w0 L3 ( L) at x = L, v = 0 sin − − − + C 4 = 0 4 π π 2 L 6π
∴ C 4 =
2 w0 L4 3π
4
∴ C 1 = −
2 w0 L π
∴ C 2 = 0 ∴ C 3 = −
w0 L3 π
(24 − π 3 )
(a) Elastic curve equation: π x 16w0 L4 2 w0 L x3 w0 L3 x 2 w0 L4 EI v = − sin (24 − π 3 ) − − + 4 4 π π 2 L 6π 3π
∴v=−
π x ⎡ 4 3 3 3 3 3 4⎤ π L x − 3π L x − 2( 24 − π ) L 4 8 L s i n + ⎥ 3π EI ⎢⎣ 2L ⎦
w0 4
Ans.
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(b) Deflection at left end of beam: w ⎡ π (0) ⎤ v A = − 40 ⎢ 48 L4 sin + π 3 L(0)3 − 3π 3 L3 (0) − 2(24 − π 3 ) L4 ⎥ 3π EI ⎣ 2L ⎦
=−
w0
w0 L4
⎡ −2(24 − π ) L ⎤⎦ = −0.0479509 3π 4 EI ⎣ EI
= −0.04795
3
4
w0 L4
Ans.
EI
(c) Support reactions R B and M B:
V B = EI
d 3v dx3 x = L
M B = EI
=
d 2v dx 2 x = L
2 w0 L π
=
∴ M B = −
cos
4 w0 L2 π
2
2 w0 L2 π
2
π ( L)
2L
sin
−
π ( L)
2L
(π − 2) =
2 w0 L π
−
=−
2 w0 L
2 w0 L( L) π
2 w0 L2 π
2
∴ RB =
π
=
(π − 2)
4 w0 L2 π
2
−
(cw)
2 w0 L π
↑
Ans.
2 w0 L2 π
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.
