![268
Substitute into Eq. (5)
Ans.
Ans.
Ans.
Ans.Vx = 1 = V2
2
x = 1
2
=
Pa
24EI
(4a2
– 3L2
)
v2 =
Pa
6EI
+ (3x2(x2 – L) + a2
)
v1 =
Px1
6EI
[x1
2
+ 3a(a – L)]
uA =
dv1
dx1
2
x1 = 0
=
Pa(a – L)
2EI
dv1
dx1
=
P
2EI
(x1
2
+ a2
- aL)
Ca =
Pa3
6
C1
C1 =
Pa2
2
–
PaL
2
Pa3
2
+ C1 = Pa3
-
PaL
2
EI v2 =
PL
4
x2
2
+ C3x3 + C4
EI
dv2
dx2
=
PL
2
x2 + C3
EI
d2
v2
dx2
= M2 =
PL
2
EIv1 =
Px1
2
6
+ C1x1 + C1
EI
dv1
dx2
=
Px1
2
2
+ C1
EI
d2
v1
dx1
2
= M1 = Px1
8–2. The bar is supported by a roller constraint at B, which
allows vertical displacement but resists axial load and
moment. If the bar is subjected to the loading shown,
determine the slope at A and the deflection at C. EI is
constant.
8–1. Continued
© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
A
C
B
L
2
L
2](https://image.slidesharecdn.com/chapter8-150430132416-conversion-gate02/85/structural-analysis-CE-engg-solved-ex-2-320.jpg)
![276
Support Reactions and Elastic Curve: As shown on FBD(a).
Moment Function: As shown on FBD(c) and (c).
Slope and Elastic Curve:
For ,
(1)
(2)
For ,
(3)
(4)
Boundary Conditions:
at , From Eq. [1],
at From Eq. [2],
Continuity Conditions:
At and From Eqs. [1] and [3],
At and . From Eqs. [2] and [4],
The Slope: Substituting into Eq. [1],
Ans.
The Elastic Curve: Substituting the values of C1, C2, C3, and C4
into Eqs. [2] and [4], respectively
Ans.
Ans.
Ans.vB = v2
2
x2 = 0
= -
41wa4
24EI
v2 =
w
24EI
( -x2
4
+ 28a3
x2 - 41a4
)
v1 =
wax1
12EI
(2x1
2
- 9ax1)
uC =
dv2
dx2
2
x1 = a
= -
wa3
EI
dv1
dx1
=
wax1
2EI
(x1 - 3a)
wa4
6
-
3wa4
4
= -
wa4
24
+
5wa4
6
+ C4 C4 = -
41wa4
8
v1 = v2x2 = a,x1 = a
wa3
2
-
3wa3
2
= - a -
wa3
6
+ C3b C3 =
7wa3
6
dv1
dx1
= -
dv2
dx2
x2 = a,x1 = a
C2 = 0x1 = 0v1 = 0
C1 = 0x1 = 0
dv1
dx1
= 0
EIv2 =
w
24
x2
4
+ C3x2 + C4
EI
dv2
dx2
= -
w
6
x2
3
+ C3
EI
d2
v2
dx2
2
= -
w
2
x2
2
M(x2) = -
w
2
x2
2
EIv1 =
wa
6
x1
3
-
3wa2
4
x1
2
+ C1x1 + C2
EI
dv1
dx1
=
wa
2
x1
2
-
3wa2
2
x1 + C1
EI
d2
v1
dx1
2
= wax1 -
3wa2
2
M(x1) = wax1 -
3wa2
2
EI
d2
v
dx2
= M(x)
*8–8. Determine the equations of the elastic curve using
the coordinates and , and specify the slope at C and
displacement at B. EI is constant.
x2x1
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
BA
a
x1
x3
x2
a
w
C](https://image.slidesharecdn.com/chapter8-150430132416-conversion-gate02/85/structural-analysis-CE-engg-solved-ex-10-320.jpg)
![277
Support Reactions and Elastic Curve: As shown on FBD(a).
Moment Function: As shown on FBD(b) and (c).
Slope and Elastic Curve:
For ,
(1)
(2)
For
(3)
(4)
Boundary Conditions:
at , From Eq. [1],
at , From Eq. [2],
Continuity Conditions:
At and From Eqs. [1] and [3],
At and , From Eqs.[2] and [4],
The Slope: Substituting the value of C3 into Eq. [3],
Ans.
The Elastic Curve: Substituting the values of C1, C2, C3, and C4
into Eqs. [2] and [4], respectively,
Ans.
Ans.
Ans.v3 =
w
24EI
( –x3
4
+ 8ax3
3
- 24a2
x3
2
+ 4a3
x3 - a4
)
vC = v1
2
x1
= a
= -
7wa4
12EI
v1 =
wax1
12EI
(2x1
2
- 9ax1)
uB =
dv3
dx3
2
x3 = 2a
= -
7wa3
6EI
dv3
dx3
=
w
2EI
(6ax3
2
- x3
3
- 12a2
x3 + a3
)
wa4
6
-
3wa4
4
=
wa4
3
-
wa4
24
- wa4
+
wa4
6
+ C4 C4 = -
wa4
24
v1 = v3x3 = a,x1 = a
C3 =
wa3
6
wa3
2
-
3wa3
2
= wa3
-
wa3
6
- 2wa3
+ C3
dv1
dx1
=
dv3
dx3
x3 = a,x1 = a
C2 = 0x1 = 0v1 = 0
C1 = 0x1 = 0
dv1
dx1
= 0
EIv3 =
wa
3
x3
3
-
w
24
x3
4
- wa2
x3
2
+ C3x3 + C4
EI
dv3
dx3
= wax3
2
-
w
6
x3
3
- 2wa2
x3 + C3
EI
d2
v3
dx3
2
= 2wax3 -
w
2
x3
2
- 2wa2
M(x3) = 2wax3 -
w
2
x2
3
- 2wa2
,
EIv1 =
wa
6
x1
3
-
3wa2
4
x1
2
+ C1x1 + C2
EI
dv1
dx1
=
wa
2
x1
2
-
3wa2
2
x1 + C1
EI
d2
v1
dx1
2
= wax1 -
3wa2
2
M(x1) = wax1 -
3wa2
2
EI
d2
v
dx2
= M(x)
8–9. Determine the equations of the elastic curve using
the coordinates and , and specify the slope at B and
deflection at C. EI is constant.
x3x1
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
BA
a
x1
x3
x2
a
w
C](https://image.slidesharecdn.com/chapter8-150430132416-conversion-gate02/85/structural-analysis-CE-engg-solved-ex-11-320.jpg)
![288
Using the diagram and the elastic curve shown in Fig. a and b,
respectively,Theorem 1 and 2 give
Then
+b
Ans.
Ans.= 3.86 mm T
=
54 kN # m3
EI
=
54(103
)N # m3
[200(109
) N>m2
][70(10–6
) m4
]
= 0.00386 m
¢C = ¢¿ - tC>A =
54 kN # m3
EI
- 0
=
24 kN # m2
EI
=
24(103
) N # m2
[200(109
) N>m2
][70(10–6
) m4
]
= 0.00171 rad
uC = uA + uC>A =
6 kN # m2
EI
+
18 kN # m2
EI
¢¿ =
9
6
tB>A =
9
6
a
36 kN # m3
EI
b =
54 kN # m3
EI
uA =
tB>A
LAB
=
36 kN # m3
>EI
6 m
=
6 kN # m2
EI
= 0
+ c
1
2
a -
12 kN # m
EI
b(3 m)d c
2
3
(3 m)d
tC>A = c
1
2
a
12 kN # m
EI
b(6 m)d(6 m) + c
1
2
a -
12 kN # m
EI
b(6 m)d c3 m +
1
3
(6 m)d
=
36 kN # m3
EI
tB>A = c
1
2
a
12 kN # m
EI
b(6 m)d(3 m) + c
1
2
a -
12 kN # m
EI
b(6 m)d c
1
3
(6 m)d
= -
18 kN # m
EI
=
18 kN # m
EI
uC>A =
1
2
a
12 kN # m
EI
b(6 m) +
1
2
a -
12 kN # m
EI
b(9 m)
M
EI
*8–20. Determine the slope and the displacement at the
end C of the beam. . Use
the moment-area theorems.
E = 200 GPa, I = 70(106
) mm4
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
BD
A C
3 m 3 m
8 kN
4 kN
3 m](https://image.slidesharecdn.com/chapter8-150430132416-conversion-gate02/85/structural-analysis-CE-engg-solved-ex-22-320.jpg)
![289
The real beam and conjugate beam are shown in Fig. a and b, respectively.
Referring to Fig. c
a
Referring to Fig. d,
a
Ans.
a
Ans.= 0.00386 m = 3.86 mm T
¢C = M¿C = -
54 kN # m3
EI
=
54 (103
) N # m3
[200(109
)N>m2
] [70(10-6
)m4
]
+ a
6 kN # m2
EI
b(3 m) = 0
M¿C + c
1
2
a
12 kN # m2
EI
b(3 m)d c
2
3
(3 m)d+ aMC = 0;
= 0.00171 rad
uC = V¿C = -
24 kN # m2
EI
=
24(103
) N # m2
[(200(109
) N>m2
)] [(70(10-6
) m4
]
-V¿C -
6 kN # m2
EI
-
1
2
a
12 kN # m
EI
b (3 m) = 0+ aFy = 0;
B¿y =
6 kN # m2
EI
- c
1
2
a
12 kN # m
EI
b(6 m)d c
2
3
(6 m)d = 0
B¿y(6 m) + c
1
2
a
12 kN # m
EI
b(6 m)d(3 m)+ aMA = 0;
8–21. Solve Prob. 8–20 using the conjugate-beam method.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
BD
A C
3 m 3 m
8 kN
4 kN
3 m](https://image.slidesharecdn.com/chapter8-150430132416-conversion-gate02/85/structural-analysis-CE-engg-solved-ex-23-320.jpg)
structural analysis CE engg. solved ex.
- 1. 267 For (1) (2) For (3) (4) Boundary conditions: at From Eq. (2) Due to symmetry: at From Eq. (3) Continuity conditions: at (5) at x1 = x2 = a dv1 dx1 = dv2 dx2 C1a # C4 = Pa3 2 - Pa3 L 2 Pa3 6 + C1a = Pa3 2 - Pa3 L 2 + C4 x1 = x2 = av1 = v2 C3 = PaL 2 0 = Pa L 2 + C3 x1 = L 2 dv1 dx1 = 0 C2 = 0 x = 0v1 = 0 EIv1 = Pax1 2 2 = C3x1 + C4 EI dv1 dx1 = Pax1 + C1 EI d2 v1 dx1 2 = Pa M1(x) = Pa EIv1 = Px1 2 6 C1x1 + C1 EI dv1 dx1 = Px1 2 2 + C1 EI d2 v1 dx1 2 = Px1 M1(x) = Px1 EI d2 v dx2 = M(x) *8–1. Determine the equations of the elastic curve for the beam using the and coordinates. Specify the slope at A and the maximum deflection. EI is constant. x2x1 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A B PP L x2 x1 a a
- 2. 268 Substitute into Eq. (5) Ans. Ans. Ans. Ans.Vx = 1 = V2 2 x = 1 2 = Pa 24EI (4a2 – 3L2 ) v2 = Pa 6EI + (3x2(x2 – L) + a2 ) v1 = Px1 6EI [x1 2 + 3a(a – L)] uA = dv1 dx1 2 x1 = 0 = Pa(a – L) 2EI dv1 dx1 = P 2EI (x1 2 + a2 - aL) Ca = Pa3 6 C1 C1 = Pa2 2 – PaL 2 Pa3 2 + C1 = Pa3 - PaL 2 EI v2 = PL 4 x2 2 + C3x3 + C4 EI dv2 dx2 = PL 2 x2 + C3 EI d2 v2 dx2 = M2 = PL 2 EIv1 = Px1 2 6 + C1x1 + C1 EI dv1 dx2 = Px1 2 2 + C1 EI d2 v1 dx1 2 = M1 = Px1 8–2. The bar is supported by a roller constraint at B, which allows vertical displacement but resists axial load and moment. If the bar is subjected to the loading shown, determine the slope at A and the deflection at C. EI is constant. 8–1. Continued © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. P A C B L 2 L 2
- 3. 269 Boundary conditions: At , At At At Ans. At Ans.vc = – PL3 6EI vc = Pa L 2 b 3 6EI – a 3 8 PL2 b a L 2 b + 0 x1 = L 2 dv1 dx1 = uA = - 3 8 PL2 EI x1 = 0 C4 = - 11 48 PL3 Pa L 2 b 2 2 + C1 = - Pa L 2 b 2 ; C1 = - 3 8 PL3 Pa L 2 b 2 6 + C1a L 2 b = PLa L 2 b 2 4 + C4 dv1 dx1 = - dv2 dx2 v1 = v2,x2 = L 2 ,x1 = L 2 , C3 = 00 + C3 = 0; dv2 dx2 = 0x2 = 0, C2 = 00 = 0 + 0 + C2; v1 = 0x1 = 0 EI v2 = PL 4 x2 2 + C3x2 + C4 EI dv2 dx2 PL 2 x2 + C3 EI d2 v2 dx2 = M2 = PL 2 EIv1 = Px1 2 6 + C1x1 + C2 EI dv2 dx1 = Px1 2 2 + C1 EI d2 v1 dx1 2 = M1 = Px1 8–2. Continued © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 8–3. Determine the deflection at B of the bar in Prob. 8–2. P A C B L 2 L 2
- 4. 270 Boundary conditions: At At At At , Ans.vB = - 11PL3 48EI x2 = 0 C4 = 11 48 PL3 Pa L 2 b 3 2 + C1 = - Pa L 2 b 2 ; C1 = - 3 8 PL2 Pa L 2 b 3 6 + C1a L 2 b = PLa L 2 b 2 4 + C4 dv1 dx1 = - dv2 dx2 v1 = v2,x2 = L 2 ,x1 = L 2 , C3 = 00 + C3 = 0; dv2 dx2 = 0x2 = 0, C2 = 00 = 0 + 0 + C2; v1 = 0x1 = 0, For (1)EI dv1 dx1 = - w 6 x1 3 + wa 2 x1 2 - wa2 2 x1 + C1 EI d2 v1 dx1 2 = - w 2 x1 2 + wax1 - wa2 2 M1(x) = - w 2 x1 2 + wax1 - wa2 2 EI d2 v dx2 = M(x) *8–4. Determine the equations of the elastic curve using the coordinates and , specify the slope and deflection at B. EI is constant. x2x1 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 8–3. Continued L A B a w x1 x2 x3 C
- 5. 271 (2) For (3) (4) Boundary conditions: At From Eq. (1), At From Eq. (2): Continuity conditions: At From Eqs. (1) and (3), From Eqs. (2) and (4), At The slope, from Eq. (3), Ans. The elastic curve: Ans. Ans. Ans.v1 = v2 2 x3 = L = wa3 24EI a - 4L + ab v2 = wa3 24EI a– 4x2 + ab v1 = w 24EI a– x1 4 + 4ax1 3 – 6a2 x1 2 b uB = dv2 dx2 = wa3 6EI C4 = wa4 24 - wa4 24 + wa4 6 - wa4 4 = - wa4 6 + C4; v1 = v2x2 = ax1 = a, C3 = - wa3 6 - wa3 6 + wa3 2 - wa3 2 = C3; dv1 dx1 = dv2 dx2 x2 = a;x1 = a, C2 = 0 v1 = 0x1 = 0, C1 = 0 dv1 dx1 = 0x1 = 0, EI v2 = C3x2 + C4 EI dv2 dx2 = C3 EI d2 v2 dx2 3 = 0M2(x) = 0; EIv1 = – w 24 x1 4 + wa 6 x1 3 – wa2 4 x1 2 + C1x1 + C2 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 8–4. Continued
- 6. 272 For (1) (2) For (3) (4) Boundary conditions: At From Eq. (1), At From Eq. (2), Continuity conditions: At At The slope Ans. The elastic curve: Ans. Ans. Ans.V2 = V3 2 x3 = 0 = wa3 24EI aa - 4Lb v3 = wa3 24EI a4x3 + a - 4Lb v1 = wx1 2 24EI a –x1 2 + 4ax1 - 6a2 b uB = dv3 dx3 2 x3 = 0 = wa3 6EI dv3 dx3 = wa3 6EI C4 = wa4 24 – wa3 L 6 - wa4 24 + wa4 6 - wa4 4 = wa3 6 (L - a) + C4; v1 = v2x3 = L - ax1 = a, C3 = + wa3 6 - wa3 6 + wa3 2 - -wa3 2 = -C3; dv1 dx1 = dv3 dx3 x3 = L – a;x1 = a, C2 = 00 = - 0 - 0 - 0 + 0 + C2; v1 = 0x1 = 0, C1 = 00 = - 0 + 0 - 0 + C1; dv1 dx1 = 0x1 = 0, EI v3 = C3x3 + C4 EI dv3 dx3 = C3 EI d2 v3 dx3 2 = 0M2(x) = 0; EIv1 = - w 24 x1 4 + wa 6 x1 3 - wa2 4 x1 2 + C1x1 + C2 EI dv1 dx1 = - w 6 x1 3 + wa 2 x1 2 - wa2 2 x1 + C1 EI d2 v1 dx1 2 = - w 2 x1 2 + wax1 - wa2 2 M1(x) = - w 2 x1 2 + wax1 - wa2 2 EI d2 v dx2 = M(x) 8–5. Determine the equations of the elastic curve using the coordinates and , and specify the slope and deflection at point B. EI is constant. x3x1 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. L A B a w x1 x2 x3 C
- 7. 273 Elastic curve and slope: For (1) (2) For (3) (4) Boundary conditions: at From Eq. (4): at From Eq. (4): at From Eq. (2): (5)0 = - wL4 24 + C1L + C2 x1 = Lv1 = 0 C3 = wL3 12 0 = -wL4 12 + C3L x2 = Lv2 = 0 C4 = 0 x2 = 0v2 = 0 EIv2 = -wLx3 2 412 + C3x3 + C4 EI dv2 dx2 = - wLx2 2 4 + C3 EI d2 v2 dx3 2 = - wLx2 2 M2(x) = -wLx2 2 EIv1 = - wx1 4 24 + C1x1 + C2 EI dv1 dx1 = -wx1 3 6 + C1 EI d1 v1 dx1 2 = - wx1 2 2 M1(x) = - wx1 2 2 EI d2 v dx2 = M(x) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 8–6. Determine the maximum deflection between the supports A and B. EI is constant. Use the method of integration. w A B L L x1 x2 C
- 8. 274 Continuity conditions: at From Eqs. (1) and (3) Substitute C1 into Eq. (5) (6) The negative sign indicates downward displacement (7) occurs when From Eq. (6) Substitute x2 into Eq. (7), Ans.(v2)max = wL4 1823EI x2 = L 23 L3 – 3Lx2 2 = 0 dv2 dx2 = 0(v2)max v2 = wL 12EI (L2 x2 - x2 3 ) (v1)max = – 7wL4 24EI (x1 = 0) v1 = w 24EI ( –x1 4 + 8L3 x1 – 7L4 ) uA = dv1 dx1 2 x1 = L = - dv2 dv3 2 x3 = L = wL3 6EI dv2 dx2 = w 12EI (L3 – 3Lx2 2 ) dv1 dx1 = w 6EI (2L3 – x1 3 ) C2 = 7wL4 24 C1 = wL3 3 - wL3 6 + C1 = - a - wL3 4 + wL3 12 b x1 = x2 = L dv1 dx1 = dv2 – dx2 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 8–6. Continued
- 9. 275 (1) (2) Boundary conditions: Due to symmetry, at From Eq. (1), At , From Eq. (2), From Eq. (1), Ans. From Eq. (2), Ans. Ans.vmax = v 2 x = L 3 = - woL4 120EI = woL4 120EI v = wox 960EIL (40L2 x2 - 16x4 - 25L4 ) uA = dv dx 2 x = 0 = - 5woL3 192EI = 5woL3 192EI dv dx = wo 192EIL (24L2 x2 - 16x4 - 5L4 ) C2 = 00 = 0 - 0 + 0 + C2; v = 0x = 0 0 = woL 8 a L2 4 b - wo 12L a L4 16 b + C1; C1 = - 5woL3 192 x = L 2 , dv dx = 0 EIv = woL 24 x3 - wo 60L x5 + C1x + C2 EI dv dx = woL 8 x2 - wo 12L x4 + C1 EI d2 v dx2 = woL 4 x - wo 3L x3 EI d2 v dx2 = M(x) 8–7. Determine the elastic curve for the simply supported beam using the x coordinate Also, determine the slope at A and the maximum deflection of the beam. EI is constant. 0 … x … L>2. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. L x A B w0
- 10. 276 Support Reactions and Elastic Curve: As shown on FBD(a). Moment Function: As shown on FBD(c) and (c). Slope and Elastic Curve: For , (1) (2) For , (3) (4) Boundary Conditions: at , From Eq. [1], at From Eq. [2], Continuity Conditions: At and From Eqs. [1] and [3], At and . From Eqs. [2] and [4], The Slope: Substituting into Eq. [1], Ans. The Elastic Curve: Substituting the values of C1, C2, C3, and C4 into Eqs. [2] and [4], respectively Ans. Ans. Ans.vB = v2 2 x2 = 0 = - 41wa4 24EI v2 = w 24EI ( -x2 4 + 28a3 x2 - 41a4 ) v1 = wax1 12EI (2x1 2 - 9ax1) uC = dv2 dx2 2 x1 = a = - wa3 EI dv1 dx1 = wax1 2EI (x1 - 3a) wa4 6 - 3wa4 4 = - wa4 24 + 5wa4 6 + C4 C4 = - 41wa4 8 v1 = v2x2 = a,x1 = a wa3 2 - 3wa3 2 = - a - wa3 6 + C3b C3 = 7wa3 6 dv1 dx1 = - dv2 dx2 x2 = a,x1 = a C2 = 0x1 = 0v1 = 0 C1 = 0x1 = 0 dv1 dx1 = 0 EIv2 = w 24 x2 4 + C3x2 + C4 EI dv2 dx2 = - w 6 x2 3 + C3 EI d2 v2 dx2 2 = - w 2 x2 2 M(x2) = - w 2 x2 2 EIv1 = wa 6 x1 3 - 3wa2 4 x1 2 + C1x1 + C2 EI dv1 dx1 = wa 2 x1 2 - 3wa2 2 x1 + C1 EI d2 v1 dx1 2 = wax1 - 3wa2 2 M(x1) = wax1 - 3wa2 2 EI d2 v dx2 = M(x) *8–8. Determine the equations of the elastic curve using the coordinates and , and specify the slope at C and displacement at B. EI is constant. x2x1 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. BA a x1 x3 x2 a w C
- 11. 277 Support Reactions and Elastic Curve: As shown on FBD(a). Moment Function: As shown on FBD(b) and (c). Slope and Elastic Curve: For , (1) (2) For (3) (4) Boundary Conditions: at , From Eq. [1], at , From Eq. [2], Continuity Conditions: At and From Eqs. [1] and [3], At and , From Eqs.[2] and [4], The Slope: Substituting the value of C3 into Eq. [3], Ans. The Elastic Curve: Substituting the values of C1, C2, C3, and C4 into Eqs. [2] and [4], respectively, Ans. Ans. Ans.v3 = w 24EI ( –x3 4 + 8ax3 3 - 24a2 x3 2 + 4a3 x3 - a4 ) vC = v1 2 x1 = a = - 7wa4 12EI v1 = wax1 12EI (2x1 2 - 9ax1) uB = dv3 dx3 2 x3 = 2a = - 7wa3 6EI dv3 dx3 = w 2EI (6ax3 2 - x3 3 - 12a2 x3 + a3 ) wa4 6 - 3wa4 4 = wa4 3 - wa4 24 - wa4 + wa4 6 + C4 C4 = - wa4 24 v1 = v3x3 = a,x1 = a C3 = wa3 6 wa3 2 - 3wa3 2 = wa3 - wa3 6 - 2wa3 + C3 dv1 dx1 = dv3 dx3 x3 = a,x1 = a C2 = 0x1 = 0v1 = 0 C1 = 0x1 = 0 dv1 dx1 = 0 EIv3 = wa 3 x3 3 - w 24 x3 4 - wa2 x3 2 + C3x3 + C4 EI dv3 dx3 = wax3 2 - w 6 x3 3 - 2wa2 x3 + C3 EI d2 v3 dx3 2 = 2wax3 - w 2 x3 2 - 2wa2 M(x3) = 2wax3 - w 2 x2 3 - 2wa2 , EIv1 = wa 6 x1 3 - 3wa2 4 x1 2 + C1x1 + C2 EI dv1 dx1 = wa 2 x1 2 - 3wa2 2 x1 + C1 EI d2 v1 dx1 2 = wax1 - 3wa2 2 M(x1) = wax1 - 3wa2 2 EI d2 v dx2 = M(x) 8–9. Determine the equations of the elastic curve using the coordinates and , and specify the slope at B and deflection at C. EI is constant. x3x1 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. BA a x1 x3 x2 a w C
- 12. 278 Using the diagram and the elastic curve shown in Fig. a and b, respectively, Theorem 1 and 2 give Ans. Ans.= 0.322 in T = 2700 (1728) k # in3 c29(103 ) k in2 d(500 in4 ) = 2700 k # ft3 EI ¢max = ¢C = ƒtB>A ƒ = c 1 2 a 90 k # ft EI b(6 ft)d c6 ft + 2 3 (6 ft)d = 270 k # ft2 EI = 270 (144) k # in2 c29(103 ) k in2 d (500 in4 ) = 0.00268 rad uB = ƒuB>A ƒ = 1 2 a 90 k # ft EI b(6 ft) M EI 8–10. Determine the slope at B and the maximum displacement of the beam. Use the moment-area theorems. Take E = 29(103 ) ksi, I = 500 in4 . © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6 ft 6 ft A B C 15 k
- 13. 279 The real beam and conjugate beam are shown in Fig. b and c, respectively. Referring to Fig. c, Ans. Referring to Fig. d, a Ans.= 2700 (123 ) k # in3 c29(103 ) k in2 d(500 in4 ) = 0.322 in T ¢ max = ¢C = MC¿ = - 2700 k # ft3 EI M¿C + c 1 2 a 90 k # ft EI b(6 ft)d c6 ft + 2 3 (6 ft)d = 0+ aMC = 0; = 270 (122 ) k # in2 c29(103 ) k in2 d (500 in4 ) = 0.00268 rad uB = V¿B = – 270 k # ft2 EI -V¿B - 1 2 a 90 k # ft EI b(6 ft) = 0+ c aFy = 0; 8–11. Solve Prob. 8–10 using the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6 ft 6 ft A B C 15 k
- 14. 280 Using the diagram and the elastic curve shown in Fig. a and b, respectively, Theorem 1 and 2 give Then, +b Ans. Ans.= 50625 k # ft3 EI T ¢C = ` tC>A ` – ¢¿ = 101250 k # ft3 EI - 50625 k # ft3 EI uC = –1125 k # ft2 EI + 5062.5 k # ft2 EI = 3937.5 k # ft2 EI uC = uA + uC>A uA = ƒtB>A ƒ LAB = 33750 k # ft3 >EI 30 ft = 1125 k # ft2 EI ¢¿ = 45 30 (tB>A) = 45 30 a 33750 k # ft3 EI b = 50625 k # ft3 EI = 101250 k # ft3 EI ƒtC>A ƒ = c 1 2 a 225 k # ft EI b(30 ft)d c15 ft + 1 3 (30 ft)d + c 1 2 a 225 k # ft EI b(15 ft)d c 2 3 (15 ft)d ƒtB>A ƒ = c 1 2 a 225 k # ft EI b(30 ft)d c 1 3 (30 ft)d = 33750 k # ft3 EI uC>A = 1 2 a - 225 k # ft EI b(45 ft) = - 5062.5 k # ft2 EI = 5062.5 k # ft2 EI M EI *8–12. Determine the slope and displacement at C. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A B C 15 ft 15 k 30 ft
- 15. 281 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a Referring to Fig. d, Ans. a Ans.¢C = M¿C = 50625 k # ft3 EI = 50625 k # ft3 EI T + aMC = 0; M¿C + c 1 2 a 225 k # ft EI b(15 ft)d(10 ft) + a 2250 k # ft2 EI b(15 ft) uC = V¿C = - 3937.5 k # ft2 EI = 3937.5 k # ft2 EI + c aFy = 0; -V¿C - 1 2 a 225 k # ft EI b(15 ft) - 2250 k # ft EI B¿y = 2250 k # ft2 EI + aMA = 0; B¿y(30 ft) - c 1 2 a 225 k # ft EI b(30 ft)d (20 ft) 8–13. Solve Prob. 8–12 using the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A B C 15 ft 15 k 30 ft
- 16. 282 Using the diagram and the elastic curve shown in Fig. a and b, respectively,Theorem 1 and 2 give Then Here, it is required that Choose the position root, Ans.a = 0.153 L 24a2 + 16La - 3L2 = 0 PL2 16EI - PaL 6EI = PL2 8EI - Pa2 2EI - PaL 2EI uB = uA>B uB = tD>B L = PL2 16EI - PaL 6EI = PL3 16EI - PaL2 6EI tD>B = c 1 2 a PL 4EI b(L)d a L 2 b + c 1 2 a - Pa EI b(L)d a L 3 b = PL2 8EI - Pa2 2EI - PaL 2EI uA>B = 1 2 a PL 4EI b(L) + 1 2 a - Pa EI b(a + L) M EI 8–14. Determine the value of a so that the slope at A is equal to zero. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A D P B C P a L__ 2 L__ 2
- 17. 283 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. d, a It is required that , Referring to Fig. c, Choose the position root, Ans.a = 0.153 L 24a2 + 16La - 3L2 = 0 c + aFy = 0; PL2 16EI - PaL 3EI - Pa2 2EI = 0 V'A = uA = 0 D¿y = PL2 16EI - PaL 3EI + aMB = 0; D'y(L) + c 1 2 a Pa EI b(L)d a 2 3 Lb - c 1 2 a PL 4EI b(L)d a 1 2 b = 0 8–15. Solve Prob. 8–14 using the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A D P B C P a L__ 2 L__ 2
- 18. 284 Using the diagram and the elastic curve shown in Fig. a and b, respectively,Theorem 2 gives It is required that Ans.a = L 3 PL3 96EI - PaL2 48EI = 1 2 c PL3 16EI - PaL2 6EI d tC>B = 1 2 tD>B = PL3 96EI - PaL2 48EI TC>B = c 1 2 a PL 4EI b a L 2 b d c 1 3 a L 2 b d + c 1 2 a– Pa 2EI b a L 2 b d c 1 3 a L 2 b d = PL3 16EI - PaL2 6EI tD>B = c 1 2 a PL 4EI b(L)d a L 2 b + c 1 2 a– Pa EI b(L)d a L 3 b M EI *8–16. Determine the value of a so that the displacement at C is equal to zero. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A D P B C P a L__ 2 L__ 2
- 19. 285 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a Here, it is required that . Referring to Fig. d, a Ans.a = L 3 L 96 - a 48 - L 32 + a 12 = 0 PL3 96EI - PaL2 48EI - PL3 32EI + PaL2 12EI = 0 –c PL2 16EI - PaL 6EI d a L 2 b = 0 –c 1 2 a Pa 2EI b a L 2 b d c 1 3 a L 2 b d + aMC = 0; c 1 2 a PL 4EI b a L 2 b d c 1 3 a L 2 b d M¿C = ¢C = 0 –B¿y = PL2 16EI - PaL 6EI + aMD = 0; c 1 2 a PL 4EI b(L)d a L 2 b - c 1 2 a Pa EI b(L)d a L 3 b - B¿y(L) = 0 8–17. Solve Prob. 8–16 using the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A D P B C P a L__ 2 L__ 2
- 20. 286 Using the diagram and the elastic curve shown in Fig. a and b, respectively, Theorem 1 and 2 give Then, Ans. Ans.¢C = tC>D - tB>D = 5Pa3 12EI - Pa3 6EI = Pa3 4EI c uC = uC>D = Pa2 4EI uC>D = 1 2 a Pa 2EI b(a) = Pa2 4EI tC>D = c 1 2 a Pa 2EI b(a)d aa + 2 3 ab = 5Pa3 12EI tB>D = c 1 2 a Pa 2EI b(a)d a 2 3 ab = Pa3 6EI M EI 8–18. Determine the slope and the displacement at C. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. a a a B A C P
- 21. 287 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a Referring to Fig. d Ans. a Ans.+ aMC = 0; M¿C - Pa2 4EI (a) = 0 ¢C = M¿C = Pa3 4EI c + c aFy = 0; Pa2 4EI - V¿C = 0 uC = V¿C = Pa2 4EI + aMA = 0; c 1 2 a Pa 2EI b(2a)d(a) - B¿y(2a) = 0 B¿y = Pa2 4EI 8–19. Solve Prob. 8–18 using the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. a a a B A C P
- 22. 288 Using the diagram and the elastic curve shown in Fig. a and b, respectively,Theorem 1 and 2 give Then +b Ans. Ans.= 3.86 mm T = 54 kN # m3 EI = 54(103 )N # m3 [200(109 ) N>m2 ][70(10–6 ) m4 ] = 0.00386 m ¢C = ¢¿ - tC>A = 54 kN # m3 EI - 0 = 24 kN # m2 EI = 24(103 ) N # m2 [200(109 ) N>m2 ][70(10–6 ) m4 ] = 0.00171 rad uC = uA + uC>A = 6 kN # m2 EI + 18 kN # m2 EI ¢¿ = 9 6 tB>A = 9 6 a 36 kN # m3 EI b = 54 kN # m3 EI uA = tB>A LAB = 36 kN # m3 >EI 6 m = 6 kN # m2 EI = 0 + c 1 2 a - 12 kN # m EI b(3 m)d c 2 3 (3 m)d tC>A = c 1 2 a 12 kN # m EI b(6 m)d(6 m) + c 1 2 a - 12 kN # m EI b(6 m)d c3 m + 1 3 (6 m)d = 36 kN # m3 EI tB>A = c 1 2 a 12 kN # m EI b(6 m)d(3 m) + c 1 2 a - 12 kN # m EI b(6 m)d c 1 3 (6 m)d = - 18 kN # m EI = 18 kN # m EI uC>A = 1 2 a 12 kN # m EI b(6 m) + 1 2 a - 12 kN # m EI b(9 m) M EI *8–20. Determine the slope and the displacement at the end C of the beam. . Use the moment-area theorems. E = 200 GPa, I = 70(106 ) mm4 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. BD A C 3 m 3 m 8 kN 4 kN 3 m
- 23. 289 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c a Referring to Fig. d, a Ans. a Ans.= 0.00386 m = 3.86 mm T ¢C = M¿C = - 54 kN # m3 EI = 54 (103 ) N # m3 [200(109 )N>m2 ] [70(10-6 )m4 ] + a 6 kN # m2 EI b(3 m) = 0 M¿C + c 1 2 a 12 kN # m2 EI b(3 m)d c 2 3 (3 m)d+ aMC = 0; = 0.00171 rad uC = V¿C = - 24 kN # m2 EI = 24(103 ) N # m2 [(200(109 ) N>m2 )] [(70(10-6 ) m4 ] -V¿C - 6 kN # m2 EI - 1 2 a 12 kN # m EI b (3 m) = 0+ aFy = 0; B¿y = 6 kN # m2 EI - c 1 2 a 12 kN # m EI b(6 m)d c 2 3 (6 m)d = 0 B¿y(6 m) + c 1 2 a 12 kN # m EI b(6 m)d(3 m)+ aMA = 0; 8–21. Solve Prob. 8–20 using the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. BD A C 3 m 3 m 8 kN 4 kN 3 m
- 24. 290 Using the diagram and the elastic curve shown in Fig. a and b, respectively. Theorem 2 gives It is required that Choose Ans.a = 0.152 L 56a2 - 48La + 6L2 = 0 7Pa3 6EI - Pa2 L EI + PaL2 8EI = 0 Pa 8EI (L2 – 4a2 ) + Pa3 3EI = 2c Pa 8EI - (L - 2a)2 d tD>C = 2 tB>C = - c Pa 8EI (L2 - 4a2 ) + Pa3 3EI d tD>C = a - Pa EI b a L - 2a 2 b aa + L - 2a 4 b + 1 2 a - Pa EI b(a)a 2 3 ab tB>C = a - Pa EI b a L - 2a 2 b a L - 2a 4 b = - Pa 8EI (L - 2a)2 M EI 8–22. At what distance a should the bearing supports at A and B be placed so that the displacement at the center of the shaft is equal to the deflection at its ends? The bearings exert only vertical reactions on the shaft. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A B a L PP a
- 25. 291 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a Referring to Fig. d, Referring to Fig. e, It is required that Choose Ans.a = 0.152 L 56a2 - 48La + 6L2 = 0 7Pa3 6EI - Pa2 L EI + PaL2 8EI = 0 Pa2 2EI (L – 2a) + Pa3 3EI = Pa 8EI (L – 2a)2 ƒ¢Dƒ = ¢C ¢C = M¿C = Pa 8EI (L – 2a)2 Pa 2EI (L - 2a)a L - 2a 2 b - Pa EI a L - 2a 2 b a L - 2a 4 b - M¿C = 0 ¢D = M¿D = - c Pa2 2EI (L - 2a) + Pa3 3EI d M¿D + Pa 2EI (L - 2a)(a) + c 1 2 a Pa EI b(a)d a 2 3 ab = 0 B¿y = Pa 2EI (L – 2a) B¿y (L - 2a) - c Pa EI (L – 2a)d a L – 2a 2 b = 0+ aMA = 0; 8–23. Solve Prob. 8–22 using the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A B a L PP a
- 26. 292 Using the diagram and elastic curve shown in Fig. a and b, respectively, Theorem 1 and 2 give Then, Ans. Ans.= 90 kN # m3 EI T ¢C = ƒ tC>D ƒ - ƒ tB>D ƒ = 103.5 kN # m3 EI - 13.5 kN # m2 EI uB = ƒ uB>D ƒ = 18 kN # m2 EI = 103.5 kN # m3 EI tC>D = c a - 12 kN # m EI b(1.5 m) d c 1 2 (1.5 m) + 3 md + c 1 2 a - 12 kN # m EI b(3 m)d c 2 3 (3 m)d tB>D = c a - 12 kN # m EI b(1.5 m) d c 1 2 (1.5 m) d = 13.5 kN # m3 EI uB>D = a - 12 kN # m EI b(1.5 m) = - 18 kN # m2 EI M EI *8–24. Determine the displacement at C and the slope at B. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A B C 3 m 1.5 m 1.5 m 4 kN4 kN 3 m
- 27. 293 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a Ans. Referring to Fig.d, a Ans.¢C = M¿C = - 90 kN # m3 EI = 90 kN # m3 EI T M¿C + a 18 kN # m2 EI b(3 m) + c 1 2 a 12 kN # m EI b(3 m)d c 2 3 (3 m)d = 0+ aMC = 0; B¿y = uB = 18 kN # m2 EI + aMA = 0; B¿y (3 m) - a 12 kN # m EI b(3 m)(1.5 m) = 0 8–25. Solve Prob. 8–24 using the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A B C 3 m 1.5 m 1.5 m 4 kN4 kN 3 m
- 28. 294 Using the diagram and elastic curve shown in Fig. a and b, respectively, Theorem 1 and 2 give Then Ans. Ans.AC = tB>C = 9Pa3 4EI T uB = uB>C = 7Pa2 4EI = 9Pa3 4EI tB>C = c 1 2 a Pa EI b(a)d a 2 3 ab + c Pa EI (a)d aa + 1 2 ab + c 1 2 a Pa 2EI b(a)d aa + 2 3 ab uB>C = 1 2 a Pa EI b(a) + a Pa EI b(a) + 1 2 a Pa 2EI b(a) = 7Pa2 4EI M EI 8–26. Determine the displacement at C and the slope at B. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A C B PP a a a a 2 P 2
- 29. 295 8–27. Determine the displacement at C and the slope at B. EI is constant. Use the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A C B PP a a a a 2 P 2 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a Ans. Referring to Fig. d, a Ans.¢C = M'C = - 9Pa3 4EI = 9Pa3 4EI T -M'C = 0 + c 1 2 a Pa 2EI b(a)d a a 3 b - 7Pa2 4EI (2a) + aMC = 0; c 1 2 a Pa 2EI b(a)d a 4 3 ab + c a Pa EI b(a)d a a 2 b uB = B¿y = 7Pa2 4EI -B¿y = (4a) = 0 + c a Pa EI b(2a) + 1 2 a Pa 2EI b(2a)d(2a) + aMA = 0; c 1 2 a Pa EI b(a)d a 2 3 ab + c 1 2 a Pa 2EI b(a)d a 10 3 ab
- 30. 296 Using the diagram and elastic curve shown in Fig. a and b, respectively, Theorem 2 gives It is required that Ans.F = P 4 Pa3 EI - 2Fa3 EI = 3 2 c Pa3 2EI - 2Fa3 3EI d tC>A = 3 2 tB>A = Pa3 2EI - 2Fa3 3EI + c 1 2 a - Fa EI b(a)d c 2 3 (a)d tC/A = c 1 2 a Pa 2EI b(2a)d(2a) + c 1 2 a - Fa EI b(2a)d c 1 3 (2a) + ad tB/A = c 1 2 a Pa 2EI b(2a)d(a) + c 1 2 a - Fa EI b(2a)d c 1 3 (2a)d = Pa3 2EI - 2Fa3 3EI M EI *8–28. Determine the force F at the end of the beam C so that the displacement at C is zero. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. a a a B D A C P F
- 31. 297 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a Here, it is required that Referring to Fig. d, a Ans.F = P 4 + aMC = 0; c 1 2 a Fa EI b(a)d c 2 3 (a)d - a Pa2 4EI - 2Fa2 3EI b(a) = 0 ¢C = M¿C = 0. B¿y = Pa2 4EI - 2Fa2 3EI + aMA = 0; c 1 2 a Pa 2EI b(2a)d(a) - c 1 2 a Fa EI b(2a)d c 2 3 (2a)d - B¿y (2a) = 0 8–29. Determine the force F at the end of the beam C so that the displacement at C is zero. EI is constant. Use the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. a a a B D A C P F
- 32. 298 Using the diagram and elastic curve shown in Fig. a and b, Theorem 1 and 2 give Then a Ans. Ans.¢C = ƒtC>A ƒ - ¢¿ = Pa3 EI - Pa3 4EI = 3Pa3 4EI T + uB = - Pa2 12EI + Pa2 2EI = 5Pa2 12EI uB = uA + uB>A ¢¿ = 3 2 ƒtB>A ƒ = 3 2 a Pa3 6EI b = Pa3 4EI uA = ƒtB>A ƒ LAB = Pa3 >6EI 2a = Pa2 12EI tC>A = c 1 2 a - Pa EI b(2a)d(a) = - Pa3 EI tB>A = c 1 2 a - Pa EI b(a)d c 1 3 (a)d = - Pa3 6EI uB>A = 1 2 a - Pa EI b(a) = - Pa2 2EI = Pa2 2EI M EI 8–30. Determine the slope at B and the displacement at C. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. a a a A C P P B
- 33. 299 The real beam and conjugate beam are shown in Fig. c and d, respectively. Referring to Fig. d, a Ans. Referring to Fig. c, a Ans.¢C = M¿C = - 3Pa3 4EI = 3Pa3 4EI T + aMC = 0; -M¿C - c 1 2 a Pa EI b(a)d a 2 3 ab - a 5Pa2 12EI b(a) = 0 uB = B¿y = 5Pa2 12EI + aMA = 0; c 1 2 a Pa EI b(a)d aa + 2 3 ab - B¿y (2a) = 0 8–31. Determine the slope at B and the displacement at C. EI is constant. Use the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. a a a A C P P B
- 34. 300 Using the diagram and the elastic curve shown in Fig. a and b, respectively, Theorem 1 and 2 give Then, Ans. Here Thus, a Ans.= 0.00802M0L2 EI T + c a 0.2887M0 EI b(0.2113L)d c 1 2 (0.2113L)d ¢max = ¢D = tB>D = c 1 2 a 0.2113M0 EI b(0.2113L)d c 1 3 (0.2113L)d + 0 = - M0L 24EI + M0 2EIL x2 x = L 212 = 0.2887L uD = uA + uD>A uD = 0. uA = ƒtB>A ƒ LAB = M0L2 >48EI L>2 = M0L 24EI tB>A = c 1 2 a M0 2EI b a L 2 b d c 1 3 a L 2 b d = M0L2 48EI uD>A = 1 2 a M0 EIL xb(x) = M0 2EIL x2 M EI *8–32. Determine the maximum displacement and the slope at A. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A M0 C B L__ 2 L__ 2
- 35. 301 8–33. Determine the maximum displacement at B and the slope at A. EI is constant. Use the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A M0 C B L__ 2 L__ 2 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c a Ans. Here it is required that . Referring to Fig. d, a Ans.= 0.00802M0L2 EI T ¢max = ¢D = M¿D = - 0.00802M0L2 EI - 1 2 a M0 EIL b a L 212 b a L 212 b c 1 3 a L 212 b d = 0 + aMD = 0; M¿D + a M0L 24EI b a L 212 b x = L 212 c aFy = 0; 1 2 a M0 EIL xb(x) - M0L 24EI = 0 uD = V¿D = 0 A¿y = uA = M0L 24EI + aMB = 0; A¿y(L) - c 1 2 a M0 2EI b a L 2 b d a L 3 b = 0
- 36. 302 Using the diagram and the elastic curve shown in Fig. a and b, respectively, Theorem 1 and 2 give Ans. Ans.= 25Pa3 6EI T + c 1 2 a Pa EI b(a)d a 2 3 ab = 0 ¢C = ƒtC>A ƒ = c 1 2 a Pa EI b(a)d aa + 2 3 ab + c a 2Pa EI b(a)d aa + a 2 b uC = ƒuC>A ƒ = 1 2 a Pa EI b(a) + a 2Pa EI b(a) + 1 2 a Pa EI b(a) = 3Pa2 EI M EI 8–34. Determine the slope and displacement at C. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. aa A C P B M0 ϭ Pa
- 37. 303 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, Ans. a Ans.¢C = M¿C = - 25Pa3 6EI = 25Pa3 6EI T + c 1 2 a Pa EI b(a)d aa + 2 3 ab = 0 + aMC = 0; M¿C + c 1 2 a Pa EI b(a)d a 2 3 ab + c a 2Pa EI b(a)d aa + a 2 b uC = V¿C = - 3Pa2 EI = 3Pa2 EI + c aFy = 0; -V¿C - 1 2 a Pa EI b(a) - a 2Pa EI b(a) - 1 2 a Pa EI b(a) = 0 8–35. Determine the slope and displacement at C. EI is constant. Use the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. aa A C P B M0 ϭ Pa
- 38. 304 Using the diagram and the elastic curve shown in Fig. a and b, respectively, Theorem 1 and 2 give Then Ans. Ans.¢C = 169 kN # m3 EI T ¢C + 56.25 kN # m3 EI = 1 2 a 112.5 kN # m3 EI + 337.5 kN # m3 EI b ¢C + tC>D = 1 2 (¢B + tB>D) uD = ¢B + tB>D LB>D = 112.5 kN # m3 >EI + 337.5 kN # m3 >EI 6 m = 75 kN # m2 EI tB>D = c 1 2 a 37.5 kN # m EI b(6 m)d(3 m) = 337.5 kN # m3 EI tC>D = c 1 2 a 37.5 kN # m EI b(3 m)d c 1 3 (3 m)d = 56.25 kN # m3 EI ¢B = ƒtB>A ƒ = c 1 2 a 37.5 kN # m EI b(3 m)d c 2 3 (3 m)d = 112.5 kN # m3 EI T M EI *8–36. Determine the displacement at C. Assume A is a fixed support, B is a pin, and D is a roller. EI is constant. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 3 m 3 m 25 kN 3 m A B D C
- 39. 305 8–37. Determine the displacement at C. Assume A is a fixed support, B is a pin, and D is a roller. EI is constant. Use the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 3 m 3 m 25 kN 3 m A B D C The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a Ans. Referring to Fig. d, a Ans.¢C = - 168.75 kN # m3 EI = 168.75 kN # m3 EI T - a 75 kN # m2 EI b(3 m) – M¿C = 0 + aMC = 0; c 1 2 a 37.5 kN # m EI b(3 m)d(1 m) uD = Dy¿ = 75 kN # m2 EI + c 1 2 a 37.5 kN # m EI b(3 m)d(2 m) - D¿y (6 m) = 0 + aMB = 0; c 1 2 a 37.5 kN # m EI b(6 m)d(3 m)
- 40. 306 Using the diagram and elastic curve shown in Fig. a and b, respectively, Theorem 1 and 2 give Then, +b Ans. Ans.= 10,368 k # ft3 EI T = 10368 k # ft3 EI + 3456 k # ft3 EI - 3456 k # ft3 EI ¢D = ƒtD>B ƒ + ¢¿ – ¢B uD = 144 k # ft2 EI + 864 k # ft2 EI = 1008 k # ft2 EI uD = uBR + uD>B uBR = ¢¿ LBD = 3456 k # ft3 >EI 24 ft = 144 k # ft2 EI ¢¿ = 2(¢B – ƒtC>B ƒ = 2a 3456 k # ft3 EI – 1728 k # ft3 EI b = 3456 k # ft3 EI tD>B = c 1 2 a - 72 k # ft EI b(24 ft)d(12 ft) = - 10368 k # ft3 EI tC>B = c 1 2 a - 72 k # ft EI b(12 ft)d c 1 3 (12 ft)d = - 1728 k # ft3 EI uD>B = 1 2 a - 72 k # ft EI b(24 ft) = - 864 k # ft2 EI = 864 k # ft2 EI ¢B = tB>A = c 1 2 a 72 k # ft EI b(12 ft)d c 2 3 (12 ft)d = 3456 k # ft3 EI c M EI 8–38. Determine the displacement at D and the slope at D. Assume A is a fixed support, B is a pin, and C is a roller. Use the moment-area theorems. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 ft 12 ft 12 ft A B C D 6 k
- 41. 307 The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a Referring to Fig. d, Ans. a Ans.M¿D = ¢D = - 10368 k # ft3 EI = 10,368 k # ft3 EI T + aMC = 0; M¿D + c 1 2 a 72 k # ft EI b(12 ft)d(8 ft) + a 576 k # ft2 EI b(12 ft) = 0 uD = V¿D = - 1008 k # ft2 EI = 1008 k # ft2 EI + c aFy = 0; -V¿D - 1 2 a 72 k # ft EI b(12 ft) - 576 k # ft2 EI = 0 C¿y = 576 k # ft2 EI = 0 + aMB = 0; C¿y(12 ft) – c 1 2 a 72 k # ft EI b(12 ft)d(16 ft) 8–39. Determine the displacement at D and the slope at D. Assume A is a fixed support, B is a pin, and C is a roller. Use the conjugate-beam method. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 ft 12 ft 12 ft A B C D 6 k