11.1 A beam is loaded and supported as shown in Fig. P11.1. Use the doubleintegration method to determine the magnitude of the moment M 0 required to make the slope at the left end of the beam zero.
Fig. P11.1
Solution Moment equation:
⎛ x ⎞ Σ M a − a = M ( x) + wx ⎜ ⎟ − M 0 = 0 ⎝2⎠ ∴ M ( x) = M 0 −
wx 2 2
Integration: d 2v wx 2 EI 2 = M ( x) = M 0 − dx 2 3 dv wx EI = M0x − + C 1 dx 6 M 0 x 2 wx 4 EI v = − + C1 x + C 2 2 24 Boundary conditions and evaluate constants: dv w( L)3 at x = L, =0 M 0 ( L) − + C 1 = 0 dx 6
∴ C1 =
wL3 6
− M0L
Beam slope equation: dv wx3 wL3 EI = M0x − + − M0L dx 6 6 Constraint: At x At x = = 0, the slope of the beam is to be zero; z ero; therefore, 3 3 dv w(0) wL EI = M 0 (0) − + − M0L = 0 dx A 6 6
∴ M 0 =
wL2 6
Ans.
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11.2 When moment M moment M 0 is applied to the left end of the cantilever beam shown in Fig. P11.2, the slope of the beam is zero. Use the double-integration method to determine the magnitude of the moment M moment M 0.
Fig. P11.2
Solution Moment equation: Σ M a − a = M ( x) − Px + M 0 = 0
∴ M ( x) = Px − M 0
Integration: d 2v EI 2 = M ( x) = Px − M 0 dx dv P x 2 EI = − M 0 x + C 1 dx 2 Px3 M 0 x 2 EI v = − + C1 x + C 2 6 2 Boundary conditions and evaluate constants: dv P( L) 2 at x = L, =0 − M 0 ( L) + C 1 = 0 dx 2
PL2 ∴ C1 = − + M0L 2 Beam slope equation: dv P x 2 PL2 EI = − M0x − + M0L dx 2 2 Constraint: At x At x = = 0, the slope of the beam is to be zero; z ero; therefore,
EI
dv dx A
=
P(0) 2 2
∴ M 0 =
− M 0 ( 0) − PL 2
PL2 2
+ M0L = 0 Ans.
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11.3 When the load P load P is is applied to the right end of the cantilever beam shown in Fig. P11.3, the deflection at the right end of the beam is zero. Use the double-integration method to determine the magnitude of the load P load P .
Fig. P11.3
Solution Moment equation:
⎛ L − x ⎞ Σ M a − a = − w( L − x ) ⎜ ⎟ + P( L − x) − M ( x) = 0 2 ⎝ ⎠ w ∴ ( x) = − ( L − x) 2 + P( L − x) 2
Integration: d 2v w EI 2 = M ( x) = − ( L − x) 2 + P( L − x) dx 2 dv w P EI = ( L − x )3 − ( L − x) 2 + C 1 dx 6 2 w P EI v = − ( L − x) 4 + ( L − x) 3 + C1 x + C 2 24 6 Boundary conditions and evaluate constants: dv w P at x = 0, =0 ( L − 0)3 − ( L − 0)2 + C 1 = 0 dx 6 2
∴ C 1 = −
wL3
+
6
at x = 0, v = 0
∴ C 2 =
PL2 2 w
−
24 wL4
−
24
( L − 0)4 +
P ( L − 0)3 + C1 (0) + C 2 = 0 6
PL3 6
Beam elastic curve equation: w P wL x3 PL2 x wL4 PL3 4 3 EI v = − ( L − x) + ( L − x) − + + − 24 6 6 2 24 6
=−
w 24
( L − x) − 4
wL x 3 6
+
wL4 24
+
P 6
( L − x) + 3
PL2 x 2
−
PL3 6
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Constraint: At x At x = = L L,, the deflection of the beam is to be zero; therefore, w wL( L)3 wL4 P PL2 ( L) PL3 4 3 EI v B = − ( L − L) − + + ( L − L) + − =0 24 6 24 24 6 2 6 which simplifies to wL4 wL4 PL3 PL3 wL4 PL3 EI v B = − + + − =− + =0 6 24 2 6 8 3 Therefore, the magnitude of P of P is is
P = =
3wL 8
Ans.
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11.4 A beam is loaded and supported as shown in Fig. P11.4. Use the doubleintegration method to determine the reactions at supports A supports A and and B B..
Fig. P11.4
Solution Beam FBD: Σ F y = RA + RB = 0
Σ M A = − M A + RB L − M 0 = 0
Moment equation: Σ a − a = − M ( x) − M 0 + RB ( L − x) = 0
∴ M ( x) = RB ( L − x) − M 0
Integration: d 2v EI 2 = M ( x) = R B ( L − x) − M 0 dx dv R EI = − B ( L − x) 2 − M 0 x + C 1 dx 2 R B M 0 x2 3 EI v = ( L − x) − + C1 x + C 2 6 2 Boundary conditions and evaluate constants:
at x = 0,
dv dx
=0
−
R B 2
at x = 0, v = 0
R B
at x = L, v = 0
R B
R B L3
6
=
6
( L − 0) − M 0 (0) + C1 = 0
( L − 0) 0) − 3
( L − L) − 3
M 0 L2
3
M 0 (0) 2
+ C1 (0) + C2 = 0
2 M 0 ( L) 2
∴ R B =
2
∴ C 1 =
2
+
RB L2
2 3M 0
2
=
2 L
3M 0
( L) −
RB L3 6
RB L2
∴ C 2 = −
2 RB L3 6
=0
↑
2L
Backsubstitute into equilibrium equations:
Σ F y = RA + RB = 0
RA = − RB = −
Σ M A = − M A + RB L − M 0 = 0 ∴ M A =
M 0 2
=
M 0 2
(cw)
3 M 0
∴ RA =
2 L
3M 0
↓
2L
Ans.
⎛ 3 M 0 ⎞ ⎟ L − M 0 = 0 ⎝ 2 L ⎠
M A = RB L − M 0 = ⎜
Ans.
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11.5 A beam is loaded and supported as shown in Fig. P11.5. (a) Use the double-integration method to determine the reactions at supports A supports A and and B B.. (b) Draw the shear-force and bendingmoment diagrams for the beam.
Fig. P11.5
Solution Beam FBD: Σ F y = RA + RB − wL = 0
⎛ L ⎞ Σ M A = M B + RB L − wL ⎜ ⎟ = 0 ⎝2⎠ Moment equation:
Σ
a −a
⎛ x ⎞ = M ( x) + wx ⎜ ⎟ − RA x = 0 ⎝2⎠
∴ M ( x) = −
wx 2
+ RA x
2
Integration: d 2v wx 2 EI 2 = M ( x) = − + R A x dx 2 dv wx3 R A x2 =− + + C 1 EI dx 6 2 wx 4 R A x3 EI v = − + + C1 x + C 2 24 6 Boundary conditions and evaluate constants: w(0)4 R A (0)3 at x = 0, v = 0 − + + C1 (0) + C2 = 0 24 6
at x = L,
dv dx
=0
−
w( L)3
+
R A ( L) 2
6
at x = L, v = 0
−
w( L) 4
+
2
R A ( L) 3
24
6
∴ C 2 = 0
+ C1 = 0
∴ C 1 =
+ C1 ( L) = 0
∴ C 1 =
wL3
−
RA L2
6
2
wL3
RA L2
−
24
6
(a)
(b)
Solve Eqs. (a) and (b) simultaneously to find: C1 = −
wL3
R A =
and
48
3w L
=
3 wL
8
↑
Ans.
8
Backsubstitute into equilibrium equations:
Σ F y = RA + RB − wL = 0
RB = wL − RA = wL −
⎛ L ⎞ Σ M A = M B + RB L − wL ⎜ ⎟ = 0 ⎝2⎠ ∴ M B = −
wL2 8
=
wL2 8
(cw)
MB =
3wL 8 wL2 2
=
5wL
∴ RB =
8
− RB L =
wL2 2
−
5 wL2 8
=−
5 wL 8
↑
Ans.
wL2 8 Ans.
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11.6 A beam is loaded and supported as shown in Fig. P11.6. Use the doubleintegration method to determine the reactions at supports A supports A and and B B..
Fig. P11.6
Solution Beam FBD:
Σ F y = RA + RB −
w0 L 2
Σ M A = M B + RB L −
=0
w0 L ⎛ 2 L ⎞
⎟= 0
⎜
2 ⎝ 3 ⎠
Moment equation:
Σ M a − a = M ( x) +
w0 x 2 ⎛ x ⎞
⎜ ⎟ − RA x = 0
2 L ⎝ 3 ⎠
∴ M ( x) = −
w0 x3 6 L
+ R A x
Integration: d 2v w0 x3 EI 2 = M ( x) = − + R A x dx 6L dv w0 x4 R A x2 EI =− + + C 1 dx 24 L 2 w0 x5 R A x3 EI v = − + + C1 x + C 2 120 L 6 Boundary conditions and evaluate constants: w0 (0)5 R A (0)3 at x = 0, v = 0 − + + C1 (0) + C2 = 0 120 L 6
at x = L,
dv dx
=0
at x = L, v = 0
− −
w0 ( L) 4
+
R A ( L) 2
24 L
2
w0 ( L)5
R A ( L) 3
120 L
+
6
+ C1 = 0 + C1 ( L) = 0
∴ C 2 = 0 ∴ C 1 = ∴ C 1 =
w0 L3
−
RA L2
24
2
w0 L3
RA L2
120
−
6
(a)
(b)
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Solve Eqs. (a) and (b) simultaneously to find: C1 = −
w0 L3
R A =
and
120
w0 L
=
10
w0 L
↑
Ans.
10
Backsubstitute into equilibrium equations:
Σ F y = RA + RB −
w0 L 2
Σ M A = M B + RB L − ∴ M B = −
=0
RB =
w0 L
w0 L ⎛ 2 L ⎞
⎜
⎟= 0
2 ⎝ 3 ⎠
w0 L2 15
=
w0 L2 15
(cw)
2
− RA =
w0 L
−
w0 L
2
MB =
10 w0 L2 3
=
4 w0 L
∴ RB =
10
− RB L =
w0 L2 3
−
2 w0 L2 5
=−
2 w0 L 5
↑
Ans.
w0 L2 15 Ans.
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11.7 A beam is loaded and supported as shown in Fig. P11.7. Use the fourth-order integration method to determine the reaction at roller support B support B..
Fig. P11.7
Solution Integrate the load distribution: d 4v w0 x 2 EI 4 = − 2 dx L 3 d v w0 x3 EI 3 = − + C 1 dx 3 L2 d 2v w0 x 4 EI 2 = − + C1 x + C 2 dx 12 L2 dv w0 x5 C1 x2 =− + + C2 x + C 3 EI dx 60 L2 2 w0 x 6 C1 x3 C2 x2 EI v = − + + + C3 x + C 4 360 L2 6 2 Boundary conditions and evaluate constants: w0 (0)6 C1 (0)3 C 2 (0) 2 at x = 0, v = 0 − + + + C3 (0) + C4 = 0 360 L2 6 2 dv w0 (0)5 C1 (0) 2 at x = 0, =0 − + + C 2 (0) + C3 = 0 dx 60 L2 2
at x = L, v = 0
−
at x = L, M = EI
d 2v dx 2
=0
−
w0 ( L) 6 360 L2 w0 ( L) 4 12 L2
+
C1 ( L) 3
+
C2 ( L) 2
6
2
=0
+ C1 ( L) + C2 = 0
∴ C 4 = 0 ∴ C 3 = 0 ∴ C1 L + 3C 2 = ∴ C1 L + C 2 =
w0 L2
(a)
60
w0 L2 12
(b)
Solve Eqs. (a) and (b) simultaneously to obtain: 2C2 = C1L =
w0 L2
−
w0 L2
60
12 12
w0 L2
w0 L2
12
+
30
=−
4 w0 L2
∴ C 2 = −
60
=
7 w0 L2
∴ C 1 =
60
w0 L2 30
7 w0 L 60
Roller reaction at B:
V B = EI
d 3v dx
=−
3 x = L
w0 ( L)3 2
3L
+
7 w0 L 60
=−
20 w0 L 60
+
7 w0 L 60
=−
13w0 L 60
∴ RB =
13 w0 L 60
↑
Ans.
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11.8 A beam is loaded and supported as shown in Fig. P11.8. Use the fourth-order integration method to determine the reaction at roller support A support A..
Fig. P11.8
Solution Integrate the load distribution: d 4v w0 x 2 EI 4 = − 2 dx L 3 d v w0 x3 EI 3 = − + C 1 dx 3 L2 d 2v w0 x 4 EI 2 = − + C1 x + C 2 dx 12 L2 dv w0 x5 C1 x2 =− + + C2 x + C 3 EI dx 60 L2 2 w0 x 6 C1 x3 C2 x2 EI v = − + + + C3 x + C 4 360 L2 6 2 Boundary conditions and evaluate constants: w0 (0)6 C1 (0)3 C 2 (0) 2 at x = 0, v = 0 − + + + C3 (0) + C4 = 0 360 L2 6 2 d 2v w0 (0) 4 at x = 0, M = EI 2 = 0 − + C1 (0) + C2 = 0 dx 12 L2
at x = L,
dv dx
=0
−
at x = L, v = 0
−
w0 ( L)5 60 L2 w0 ( L) 6 360 L2
+
C1 ( L) 2 2
+
C1 ( L) 3 6
∴ C 4 = 0 ∴ C 2 = 0
+ C3 = 0
∴ C1 L + 2C 3 =
+ C3 ( L) = 0
∴ C1 L + 6C 3 =
2
2
w0 L3 30 w0 L3 60
(a) (b)
Solve Eqs. (a) and (b) simultaneously to obtain: w0 L3
−4C3 = C1L = 2
w0 L3
−
30
60
w0 L3
w0 L3
30
+
120
=
w0 L3
∴ C 3 = −
60
=
5w0 L3
∴ C 1 =
120
w0 L3 240
5w0 L 120
Roller reaction at A:
V A = EI
d 3v dx
=−
3 x = 0
w0 (0)3 2
3L
+
5 w0 L 120
=
5 w0 L 120
∴ RA =
5 w0 L 120
↑
Ans.
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11.9 A beam is loaded and supported as shown in Fig. P11.9. Use the fourth-order integration method to determine the reaction at roller support A support A..
Fig. P11.9
Solution Integrate the load distribution: d 4v π x 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 dx 2L π π x dv 8 w0 L3 C1 x2 =− + + C2 x + C 3 EI cos 3 dx 2L 2 π 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: π (0) 16w0 L4 C1 (0)3 C2 (0) 2 at x = 0, v = 0 − sin + + + C3 (0) + C4 = 0 ∴ C 4 = 0 4 2 L 6 2 π π (0) d 2v 4 w0 L2 at x = 0, M = EI 2 = 0 s i n + C1 (0) + C2 = 0 ∴ C 2 = 0 2 dx 2L π π ( L) dv 8 w0 L3 C1 ( L) 2 at x = L, =0 − cos + + C3 = 0 ∴ C1 L2 + 2C 3 = 0 3 π dx 2L 2
at x = L, v = 0
−
16w0 L4 π
4
sin
( L)
π
+
C1 ( L)3
2 L
6
+ C3 ( L) = 0
∴ C1 L + 6C 3 = 2
96 w0 L3 4 π
(a)
(b)
Solve Eqs. (a) and (b) simultaneously to obtain:
−4C3 = −
96w0 L3 π
∴ C 3 =
4
⎛ 24w0 L3 ⎞ C1L = −2 ⎜ ⎟ 4 ⎝ π ⎠
24w0 L3 4 π
∴ C 1 = −
2
48w0 L 4 π
Roller reaction at A:
V A = EI
d 3v dx
=
3 x = 0
2 w0 L π
cos
(0)
π
2L
−
48 w0 L π
4
∴ RA =
2 w0 L π
−
48 w0 L π
4
Ans.
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11.10 A beam is loaded and supported as shown in Fig. P11.10. Use the fourth-order integration method to determine the reactions at supports A supports A and and B B..
Fig. P11.10
Solution Integrate the load distribution: π x d 4v EI 4 = w0 cos dx 2L 3 d v 2 w0 L π x EI 3 = sin + C 1 π dx 2L d 2v 4 w0 L2 π x EI 2 = − cos + C1 x + C 2 2 π dx 2L π x dv 8 w0 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 2 L 6 2 π Boundary conditions and evaluate constants: dv 8 w0 L3 C1 (0) 2 π (0) at x = 0, =0 − sin + + C2 (0) + C3 = 0 3 π dx 2L 2 π (0) 16w0 L4 C1 (0)3 C2 (0) 2 at x = 0, v = 0 cos + + + C4 = 0 4 π 2 L 6 2
dv
at x = L,
dx
=0
−
8 w0 L3 π
3
16 w0 L4
at x = L, v = 0
π
4
sin cos
(L)
π
+
C1 ( L) 2
2L ( L)
π
2
+
C1 (L)3
2 L
∴ C 3 = 0 ∴ C 4 = −
+ C2 ( L) = 0 +
C2 ( L) 2
6
−
2
16 w0 L4 4 π
∴ C1 L + 2C 2 =
16 w0 L4 π
4
=0
∴ C1 L + 3C 2 =
16 w0 L2 3 π
96 w0 L2 4 π
(a) (b)
Solve Eqs. (a) and (b) simultaneously to obtain:
−C2 =
16w0 L2 π
−C1 L = −
−
3
96w0 L2 π
48w0 L2 π
3
+
∴ C 2 =
4
192w0 L2 π
4
∴ C 1 =
16w0 L2 4 π
48w0 L 4 π
[6 − π ]
[π − 4]
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.
Reactions at supports A and B
V A = EI
d 3v dx
2 w0 L
=
3
sin
π
x = 0
V B = EI
dx
=
3
2 w0 L π
x = L
∴ R B =
M A = EI
d 2v dx 2 x = 0
=− =−
2
4 w0 L2 π
2
∴ M A =
M B = EI
d 2v dx 2 x = L
=−
4 w0 L2 π
=
2
48w0 L2 π
( L)
4
∴ M B =
+
cos
+
π
48w0 L 4 π
[π − 4]
4
Ans.
[π − 4] =
2 w0 L 4 π
⎡⎣π 3 + 24π − 96⎤⎦
⎡⎣96 − 24π + π 3 ⎤⎦ ↓ (0)
π
+
48 4 8w0 L(0)
2L
π
16 w0 L2 4 π
4w0 L2 4 π
cos
[π − 4] =
48 w0 L
2L
4 π
4
[π − 4] ↓
π
2 w0 L
4 w0 L2 π
π
4 π
sin
48 w0 L
+
2L
48w0 L
∴ R A = d 3v
( 0)
π
[π − 4] +
( L)
+
48 w0 L( L)
2L
[π − 4] + 32w0 L2
16 w0 L2 4 π
[ 6 − π ]
[6 − π ]
⎡⎣ 24 − 4π − π 2 ⎤⎦
π
4 π
4
Ans.
π
16w0 L2 π
4
4
(cw)
Ans.
[π − 4] +
[6 − π ] =
[π − 3] (ccw)
16 w0 L2
16w0 L2 π
4
4 π
[ 6 − π ]
[3π − 12 + 6 − π ] =
16 w0 L2 π
4
[ 2π − 6]
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.
11.11 A beam is loaded and supported as shown in Fig. P11.11. Use the fourth-order integration method to determine the reactions at supports A supports A and and B B..
Fig. P11.11
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 dx L π π x d 2v w0 L2 + C1 x + C 2 EI 2 = − 2 sin dx L π π x dv w0 L3 C1 x2 EI = 3 cos + + C2 x + C 3 π dx L 2 w0 L4 C1 x3 C2 x2 π x EI v = − 4 sin + + + C3 x + C 4 π L 6 2 Boundary conditions and evaluate constants: π (0) dv w0 L3 C1 (0) 2 w0 L3 at x = 0, =0 cos + + C2 (0) + C3 = 0 ∴ C 3 = − 3 3 dx L 2 π π 4 3 2 π (0) wL C ( 0) C (0) at x = 0, v = 0 − 0 4 sin + 1 + 2 + C3 (0) + C4 = 0 ∴ C 4 = 0 π L 6 2
dv
at x = L,
dx
w0 L3
=0
π
at x = L, v = 0
−
3
cos
w0 L4 π
4
( L)
π
sin
+
C1 ( L) 2
L
+ C2 ( L) −
2 ( L)
π
L
+
C1 ( L) 3
+
C2 ( L) 2
6
2
w0 L3 π
−
3
=0
w0 L3 π
3
( L) = 0
∴ C1 L + 2C 2 = ∴ C1 L + 3C 2 =
4 w0 L2 3 π
6 w0 L2 3 π
(a) (b)
Solve Eqs. (a) and (b) simultaneously to obtain: C2 =
6 w0 L2
C1 L =
π
3
−
π
4 w0 L2 π
4 w0 L2
3
3
⎛ 2 w0 L2 ⎞ − 2⎜ ⎟ 3 ⎝ π ⎠
∴ C 2 =
2 w0 L2 3 π
∴ C 1 = 0
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Reactions at supports A and B
V A = EI
d 3v dx
=−
3
w0 L
∴ R A =
V B = EI
dx3 x = L
=−
w0 L
w0 L
M A = EI
d 2v dx 2 x = 0
M B = EI
dx 2 x = L
=−
2
w0 L2 π
2
Ans.
=
w0 L
L
π
↓
Ans.
π
( 0)
π
sin
∴ M A =
d 2v
↓
( L)
w0 L
w0 L2 π
π
π
π
=−
w0 L
π
cos
∴ R B =
=−
L
π
x = 0
d 3v
(0)
π
cos
2w0 L2 3 π
sin
∴ M B =
+
L
π
( L)
π
L
2 w0 L2 3 π
2 w0 L2 3
3 π
(cw)
+
2 w0 L2 π
=
2 w0 L2
3
(ccw)
Ans.
=
2 w0 L2 3 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.
11.12 A beam is loaded and supported as shown in Fig. P11.12. (a) Use the double-integration method to determine the reactions at supports A supports A and and C . (b) Draw the shear-force and bendingmoment diagrams for the beam. (c) Determine the deflection in the middle of the span.
Fig. P11.12
Solution Beam FBD: from symmetry,
R A = RC =
P 2
and M A = M C
Moment equation:
Σ
a −a
= M ( x) − M A −
P 2
x=0
∴ M ( x) =
Px 2
+ M A
Integration: d 2v Px + M A EI 2 = M ( x) = dx 2 dv P x 2 EI = + M A x + C 1 dx 4 P x3 M A x 2 EI v = + + C1 x + C 2 12 2 Boundary conditions and evaluate constants: dv P (0) 2 =0 + M A (0) + C1 = 0 at x = 0, dx 4 P (0)3 M A (0) 2 at x = 0, v = 0 + + C2 = 0 12 2
∴ C 1 = 0 ∴ C 2 = 0
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(a) Beam reaction forces:
R A = RC =
P
Ans.
2
(a) Beam reaction moments: L dv at x = , =0 2 dx
P ( L / 2) 2 4
⎛ L⎞ ⎟=0 2 ⎝ ⎠
+ M A ⎜
PL P L P L M A = − = (c (ccw) 8 8
M C =
PL 8
(cw)
Ans.
Elastic curve equation:
EI v =
P x3
+
M A x 2
12
∴v = −
=
2 Px 2
P x3
−
PLx2
12
16
=−
Px2 48
[3L − 4 x ]
[3L − 4 x ]
48 EI
(c) Midspan deflection:
P( L / 2) 2 PL3 v B = − [ 3 L − 4( L / 2) ] = − 48 EI 192EI
Ans.
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11.13 A beam is loaded and supported as shown in Fig. P11.13. (a) Use the double-integration method to determine the reactions at supports A supports A and and B B.. (b) Draw the shear-force and bendingmoment diagrams for the beam. (c) Determine the deflection in the middle of the span.
Fig. P11.13
Solution Beam FBD: from symmetry,
R A = RB =
wL 2
and M A = M B
Moment equation:
⎛ x ⎞ ⎛ wL ⎞ Σ M a − a = M ( x) − M A + wx ⎜ ⎟ − ⎜ ⎟x= 0 ⎝2⎠ ⎝ 2 ⎠ ∴ M ( x) = −
wx2 2
+
wL x 2
+ M A
Integration: d 2v w x 2 wL x EI 2 = M ( x) = − + + M A dx 2 2 dv w x 3 wL x 2 EI =− + + M A x + C 1 dx 6 4 w x 4 wL x3 M A x2 EI v = − + + + C1 x + C 2 24 12 2 Boundary conditions and evaluate constants: dv w(0)3 wL(0) 2 at x = 0, =0 − + + M A (0) + C1 = 0 dx 6 4 w(0)4 wL(0)3 M A (0) 2 at x = 0, v = 0 − + + + C2 = 0 24 12 2
∴ C 1 = 0 ∴ C 2 = 0
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(a) Beam reaction forces:
R A = RB =
wL
Ans.
2
(a) Beam reaction moments: L dv at x = , =0 2 dx
M A = −
wL2
=
12
wL2 12
−
w( L / 2)3
+
6 (c (ccw)
wL( L / 2) 2 4
⎛ L⎞ ⎟= 0 ⎝2⎠
+ M A ⎜ M B = −
wL2 12
=
wL2 12
(cw)
Ans.
Elastic curve equation: w x 4 wL x3 M A x2 w x4 wL x3 wL2 x2 w x2 2 w x2 2 ⎡⎣ x − 2 Lx + L ⎤⎦ = − + + =− + − =− EI v = − ( x − L)2 24 12 2 24 12 24 24 24
∴v = −
wx 2 24 EI
( x − L) 2
(c) Midspan deflection:
v x = L / 2
w( L / 2) 2 ⎡⎛ L ⎞
2
wL4 ⎤ =− ⎢⎜ ⎟ − L) ⎥ = − 384EI 24 EI ⎣⎝ 2 ⎠ ⎦
Ans.
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11.14 A beam is loaded and supported as shown in Fig. P11.14. (a) Use the double-integration method to determine the reactions at supports A supports A and and C . (b) Determine the deflection in the middle of the span.
Fig. P11.14
Solution Beam FBD: from symmetry,
R A = RC =
w0 L 2
and M A = M C
Moment equation:
Σ M a − a
w0 x 2 ⎛ x ⎞ ⎛ w0 L ⎞ = M ( x) − M A + ⎜ ⎟−⎜ ⎟x= 0 2 L ⎝ 3 ⎠ ⎝ 2 ⎠
∴ M ( x) = −
w0 x3 6 L
+
w0 L x 2
+ M A
Integration: d 2v w0 x3 w0 L x EI 2 = M ( x) = − + + M A dx 6L 2 dv w0 x4 w0 L x2 EI =− + + M A x + C 1 dx 24 L 4 w0 x5 w0 L x3 M A x2 EI v = − + + + C1 x + C 2 120 L 12 2 Boundary conditions and evaluate constants: dv w0 (0) 4 w0 L(0) 2 at x = 0, =0 − + + M A (0) + C1 = 0 dx 24 L 4 w0 (0)5 w0 L(0)3 M A (0) 2 − + + + C2 = 0 at x = 0, v = 0 120 L 12 2
∴ C 1 = 0 ∴ C 2 = 0
(a) Beam reaction forces:
R A = RC =
w0 L 2
Ans.
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(a) Beam reaction moments: dv at x = L, =0 dx
M A = −
5w0 L2 24
=
−
w0 ( L) 4
+
w0 L( L) 2
24 L
5w0 L2 24
4
+ M A ( L) = 0 M C = −
(c (ccw)
5w0 L2 24
=
5 w0 L2 24
(cw)
Ans.
Elastic curve equation: w0 x5 w0 L x3 M A x2 w0 x5 w0 L x3 5 w0 L2 x2 EI v = − + + =− + − 120 L 12 2 120L 12 48
=−
2 w0 x5
+
20 w0 L2 x3
240 L
∴v = −
w0 x
240 L
−
25w0 L3 x2
=−
240L
w0 x2
⎡⎣ 2 240L
3
− 20 L2 x + 25L3 ⎤⎦
2
⎡⎣ 2 x3 − 20 L2 x + 25L3 ⎤⎦ 240 L EI
(c) Midspan deflection:
v B = −
w0 ( L) 2
7 w0 L4
⎡ 2( L) − 20 L ( L) + 25 L ⎤⎦ = − 240 L EI ⎣ 240 EI 3
2
3
Ans.
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11.15 A beam is loaded and supported as shown in Fig. P11.15. (a) Use the double-integration method to determine the reactions at supports A supports A and and C . (b) Draw the shear-force and bendingmoment diagrams for the beam. (c) Determine the deflection in the middle of the span.
Fig. P11.15
Solution Beam FBD: Σ F y = RA + RC − P = 0
⎛ L ⎞ Σ M A = M C + RC L − P ⎜ ⎟ = 0 ⎝2⎠ Moment equation: Σ M a − a = M ( x) − RA x = 0
L ⎞ ⎛ ⎜0 ≤ x ≤ 2 ⎟ ⎝ ⎠
∴ M ( x) = R A x
⎛ ⎝
Σ M b − b = M ( x) − RA x + P ⎜ x − ∴
( x) = R A x − P x +
L ⎞
⎟=0
2⎠
⎛L ⎜ 2 ≤ x≤ ⎝
PL 2
Integration: For beam segment AB: d 2v EI 2 = M ( x) = R A x dx dv R A x2 EI = + C 1 dx 2 R A x3 EI v = + C1 x + C 2 6
⎞ ⎠
L⎟
For beam segment BC : d 2v PL EI 2 = M ( x) = R A x − P x + dx 2 2 2 dv R A x Px PL x EI = − + + C 3 dx 2 2 2 R A x3 P x3 PL x2 EI v = − + + C3 x + C 4 6 6 4
Boundary conditions and evaluate constants: R A (0)3 at x = 0, v = 0 + C1 (0) + C2 = 0 6
at x = L,
dv dx
=0
R A ( L) 2 2
−
P( L) 2 2
+
PL( L) 2
∴ C 2 = 0 + C3 = 0
∴ C 3 = −
RA L2 2
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R A ( L)3
at x = L, v = 0
−
P( L)3
6
+
PL( L) 2
6
−
RA L2
4
2
( L) + C4 = 0
∴ C 4 =
RA L3 3
−
PL3 12
Slope continuity condition at x = = L/2: L dv dv at x = , = 2 dx AB dx BC
R A ( L / 2) 2 2
+ C 1 =
RA ( L / 2) 2
P( L / 2) 2
−
2
+
PL( L / 2)
2
∴ C 1 =
PL2
−
−
2
RA L2 2
R A L2
8
2
Deflection continuity condition at x = = L/2: L at x = , v B AB = vB BC 2
R A x3
PL2 x
RA L2 x
RA x3
P x3
PL x2
RA L2 x
RA L3
PL3
+ − = − + − + − 6 8 2 6 6 4 2 3 12 eliminate terms and rearrange: R A L3 P x3 PL x2 PL2 x PL3 = − + + 3 6 4 8 12 Substitute x Substitute x = = L L/2 /2 to obtain: R A L3 P( L / 2)3 PL( L / 2) 2 PL2 ( L / 2) PL3 5 PL3 = − + + = 3 6 4 8 12 48 ∴ R A =
5 P 16
(a) Beam reaction forces:
R A =
5 P
RC =
16
11P
Ans.
16
(a) Beam reaction moment:
PL 11PL 3 PL ⎛ L ⎞ M C = P ⎜ ⎟ − RC L = − =− 2 16 16 ⎝2⎠
M C = −
3 PL
=
3 PL
16
16
(cw)
Ans.
Elastic curve equation for beam segment AB: R A x3 PL2 x RA L2 x 5 Px3 PL2 x 5 PL2 x 5 Px3 3 PL2 x EI v = + − = + − = − 6 8 2 96 8 32 96 96 P x ⎡⎣5 x 2 − 3L2 ⎤⎦ ∴v = 96 EI (c) Midspan deflection:
⎤ P( L / 2) ⎡ ⎛ L ⎞ 7 PL3 2 v B = ⎢5 ⎜ ⎟ − 3L ⎥ = − 96 EI ⎢⎣ ⎝ 2 ⎠ 768EI ⎥⎦ 2
Ans.
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11.16 A beam is loaded and supported as shown in Fig. P11.16. (a) Use the double-integration method to determine the reactions at supports A supports A and and C . (b) Draw the shear-force and bendingmoment diagrams for the beam.
Fig. P11.16
Solution Beam FBD: Σ F y = RA + RC = 0
∴ RC = − RA
Σ M A = − M A + RC L − M 0 = 0
Moment equation: Σ M a − a = M ( x) − RA x − M A = 0
∴ M ( x) = R A x + M A
L ⎞ ⎛ ⎜0 ≤ x ≤ 2 ⎟ ⎝ ⎠
Σ M b − b = M ( x) − RA x − M A − M 0 = 0 ∴
( x) = R A x + M A + M 0
Integration: For beam segment AB: d 2v EI 2 = M ( x) = R A x + M A dx dv R A x2 EI = + M A x + C 1 dx 2 R A x3 M A x 2 EI v = + + C1 x + C 2 6 2
⎛ L ⎜ 2 ≤ x≤ ⎝
⎞ ⎠
L⎟
For beam segment BC : d 2v EI 2 = M ( x) = R A x + M A + M 0 dx dv R A x2 EI = + M A x + M 0 x + C 3 dx 2 R A x3 M A x2 M 0 x2 EI v = + + + C3 x + C 4 6 2 2
Boundary conditions and evaluate constants for segment AB: R A (0)3 M A (0) 2 at x = 0, v = 0 + + C1 (0) + C2 = 0 6 2
∴ C 2 = 0
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at x = 0,
dv dx
R A (0) 2
=0
2
+ M A (0) + C1 = 0
∴ C 1 = 0
Slope continuity condition at x = = L/2: L dv dv at x = , = 2 dx AB dx BC
R A x 2
+
x =
A
2
RA x2
+ M A x + M 0 x + C 3
2
∴ C 3 = −
M 0 L 2
Deflection continuity condition at x = = L/2: L at x = , v B AB = vB BC 2
R A x3
+
M A x2
6
=
2
RA x3
+
M A x2
6
+
M 0 x2
2
∴ C 4 =
−
M0 L x
2
2
+ C 4
M 0 L2 8
Boundary condition for segment BC :
R A ( L)3
at x = L, v = 0
6
+
M A ( L) 2 2
+
M 0 ( L) 2
−
M0 L
2
2
( L) +
M 0 L2 8
=0
∴ R A L + 3M A = −
3 M 0 4
Also, the beam moment equilibrium equation can be written as: R A L + M A = − M 0 (a) Beam Reactions: Solve these two equations simultaneously to obtain:
M A =
M 0 8
=
M0 8
(cw)
RA = −
9M 0 8 L
=
9M 0 8L
↓
RC =
9M 0 8L
↑
Ans.
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11.17 A beam is loaded and supported as shown in Fig. P11.17. (a) Use the double-integration method to determine the reactions at supports A supports A and and C . (b) Draw the shear-force and bendingmoment diagrams for the beam.
Fig. P11.17
Solution Beam FBD:
Σ F y = RA + RC −
wL 2
Σ M A = M C + RC L −
=0 wL ⎛ L ⎞
⎜ ⎟=0
2 ⎝4⎠
Moment equation:
⎛ x ⎞ Σ M a − a = M ( x) + wx ⎜ ⎟ − RA x = 0 ⎝2⎠ ∴ M ( x) = −
wx 2
L⎞ ⎛ ⎜0 ≤ x ≤ 2 ⎟ ⎝ ⎠
L⎞ x − ⎟ − RA x = 0 ⎜ 2 ⎝ 4⎠ wL ⎛ L⎞ ⎛L ( x) = − ⎜ x − ⎟ + R A x ⎜ ≤ x ≤ 2 ⎝ 4⎠ ⎝2
Σ M b − b = M ( x) + ∴
2
+ R A x
wL ⎛
Integration: For beam segment AB: d 2v w x2 EI 2 = M ( x) = − + R A x dx 2 dv w x3 R A x2 EI =− + + C 1 dx 6 2 w x 4 R A x3 EI v = − + + C1 x + C 2 24 6
⎞ ⎠
L⎟
For beam segment BC : d 2v wL ⎛ L⎞ EI 2 = M ( x) = − x − ⎜ ⎟ + R A x dx 2 ⎝ 4⎠
EI
dv
=−
dx
EI v = −
wL ⎛
L⎞
2
⎜x− 4⎟ + ⎠
R A x2
4 ⎝
wL ⎛
L⎞
3
⎜x− 4⎟ + ⎠
12 ⎝
Boundary conditions and evaluate constants for segment AB: w(0)4 R A (0)3 at x = 0, v = 0 − + + C1 (0) + C2 = 0 24 6
2
R A x3 6
+ C 3
+ C3 x + C 4
∴ C 2 = 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.
Slope continuity condition at x = = L/2: Equate the slope expressions for the two beam segments:
−
w x3
wL ⎛
R A x 2
L⎞
2
RA x2
+ + C1 = − ⎜x− 4⎟ + 6 2 4 ⎝ ⎠ Set x Set x = = L L/2 /2 and solve for the constant C 1: C1 = C3 +
w x3
−
6
∴ C1 = C 3 +
wL ⎛
L⎞
2
+ C 3
2
w( L / 2)3
⎜ x − 4 ⎟ = C3 + ⎠
4 ⎝
−
6
wL ⎛ L
⎜
4 ⎝2
−
L⎞
2
wL3
wL w L3
⎟ = C3 + 48 − 64
4⎠
wL3 192
Deflection continuity condition at x = = L/2: Equate the deflection expressions for the two beam segments:
−
w x4
wL ⎛
R A x3
L⎞
3
RA x3
+ + C1 x = − ⎜x− 4⎟ + 24 6 12 ⎝ ⎠ Set x Set x = = L L/2 /2 and solve for the constant C 4: −
w( L / 2) 4 24
6
+ C3 x + C4
3
⎡ wL3 ⎤ ⎛ L ⎞ wL ⎛ L L ⎞ ⎛ L⎞ + ⎢C3 + =− − ⎟ + C3 ⎜ ⎟ + C 4 ⎥ ⎜ ⎟ ⎜ 192 ⎦ ⎝ 2 ⎠ 12 ⎝ 2 4 ⎠ ⎝ 2⎠ ⎣
wL4
4 wL4 ⎛ L ⎞ wL ⎛ L⎞ − + C3 ⎜ ⎟ + =− + C3 ⎜ ⎟ + C 4 384 768 ⎝ 2 ⎠ 384 ⎝ 2⎠
∴ C 4 =
wL4 768
Boundary conditions and evaluate constants for segment BC :
at x = L,
dv dx
=0
−
wL ⎛
L⎞
2
⎜L− 4⎟ + ⎠
R A ( L) 2
4 ⎝
2
+ C3 = 0
∴ C 3 =
9 wL3 64
−
RA L2 2
at x at x = = L L,, v = 0 3
wL ⎛ L⎞ R A ( L)3 wL4 − ⎜ L − 4 ⎟ + 6 + C3 ( L) + 768 = 0 12 ⎝ ⎠
−
27 wL4
+
768
−
⎡ 9 wL3 RA L2 ⎤ wL4 +⎢ − ⎥ ( L) + 768 = 0 6 6 4 2 ⎣ ⎦
R A L3
27 wL4 768
+
R A L3 6
+
9 wL4
−
RA L3
64
+
wL4
2
768
R A L3
3 RA L3
6
−
=0 =
26 wL4
6
768
R A L3
82 wL4
3
=
768
−
108wL4 768
∴ R A =
41wL 128
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Solve for C 3: C 3 =
9 wL3
−
41wL3
64
= −
5wL3
256
256
wL3
11wL3
and for C 1: C 1 = −
5wL3 256
+
= −
192
768
(a) Beam force reactions:
R A =
41wL
RC =
wL
Ans.
128 2
− RA =
wL
−
41wL
2
=
23wL
128
∴ RC =
128
23wL
↑
Ans.
128
Beam moment reaction:
M C =
wL2 8
− RC L =
wL2 8
−
23wL2 128
=−
7 wL2 128
∴ M C = −
7 wL2 128
=
7 wL2 128
(cw)
Ans.
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