CHAPTER 9
PROBLEM 9.1
For the loading shown, determine (a) thee equation of the elastic
curve for the cantilever beam AB, (b) the deflection
d
at the free end,
(c) the slope at the free end.
SOLUTION
ΣM J = 0: − M − P( L − x) = 0
M = − P ( L − x)
EI
d2y
= − P( L − x) = − PL + Px
dx 2
EI
dy
1
= − PLx + Px 2 + C1
dx
2
dy


 x = 0, dx = 0  :


0 = −0 + 0 + C1
C1 = 0
1
1
EIy = − PLx 2 + Px3 + C1x + C2
2
6
[ x = 0, y = 0] :
(a)
0 = −0 + 0 + 0 + C2
C2 = 0
y =−
Elastic curve:
Px 2
(3L − x) 
6EI
dy
Px
=−
(2L − x)
2EI
dx
(b)
y at x = L :
(c)
dy
at x = L :
dx
yB = −
dy
dx
=−
B
PL2
PL3
(3L − L) = −
6EI
3EI
PL
PL2
(2L − L) = −
2EI
2EI
yB =
θB =
PL3
↓ 
3EI
PL2
2EI

PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.2
For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the
slope at the free end.
SOLUTION
 MK = 0 : − M0 + M = 0
M = M0
d2y
= M = M0
dx
dy
EI
= M 0 x + C1
dx
EI
dy


 x = L, dx = 0  :


0 = M 0 L + C1
EIy =
[ x = L, y = 0]
(a)
0=
C1 = − M 0 L
1
M 0 x 2 + C1x + C2
2
1
M 0 L2 − M 0 L2 + C2
2
Elastic curve:
C2 =
1
M 0 L2
2
y=
M0 2
( x − 2Lx + L2 ) 
2EI
y =
(b)
y at x = 0 :
(c)
dy
at x = 0 :
dx
yA =
M0
( L − 0)2
2EI
M0
( L − x)2 
2EI
yA =
M 0 L2
↑ 
2EI
dy
M
M
M L
= − 0 ( L − x) = − 0 ( L − 0) = − 0
dx
EI
EI
EI
θA =
w0 L
EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.3
For the loading shown, determine (a) the equ
uation of the elastic curve
for the cantilever beam AB, (b) the deflectioon at the free end, (c) the
slope at the free end.
SOLUTION
ΣM J = 0: (wx)
x
+M =0
2
1
M = − wx 2
2
d2y
1
= M = − wx 2
2
dx 2
dy
1
EI
= − wx3 + C1
dx
6
EI
dy
1 3
1 3


 x = L, dx = 0  : 0 = − 6 wL + C1 C1 = 6 wL


EI
dy
1
1
= − wx3 + wL3
dx
6
6
EIy = −
1
1
wx 4 + wL3 x + C2
24
6
[ x = L, y = 0] 0 = −
1
1
wL4 + wL4 + C2 = 0
24
6
1
3
 1
−  wL4 = − wL4
C2 = 
24
 24 6 
(a)
Elastic curve:
(b)
y at x = 0:
(c)
dy
at x = 0:
dx
y =−
yA = −
dy
dx
=
A
3wL4
wL4
=−
24EI
8EI
wL3
6EI
w
( x 4 − 4L3 x + 3L4 ) 
24
2 EI
yA =
θA =
wL4
↓ 
8EI
wL3
6EI

PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PRO
OBLEM 9.4
For thhe loading shown, determine (a) the equation of the elastic curve
for thhe cantilever beam AB, (b) the deflection at the freee end, (c) the
slopee at the free end.
SOLUTION
 Fy = 0:
1
wL = 0
2
1
RA = w0 L
2
RA =
 M A = 0:
2L wL
⋅
=0
3
2
1
= − w0 L2
3
− MA =
MA
ΣM J = 0:
w x2 x
1
1
w0 L2 − w0 Lx + 0 ⋅ + M = 0
3
2
2L 3
w x3
1
1
M = − w0 L2 + w0 Lx − 0
3
2
6L
EI
d2y
w0 x3
1
1
2
w
L
w
Lx
=
−
+
−
0
0
3
2
6L
dx 2
dy
w x4
1
1
= − w0 L2 x + w0 Lx 2 − 0 + C1
dx
3
4
24L
dy

= 0,
= 0 :
0 = −0 + 0 − 0 + C1
dx

EI

x

C1 = 0
1
1
w x5
w0 Lx3 − 0 + C2
EIy = − w0 L2 x 2 +
6
12
120L
[ x = 0, y = 0] :
0 = − 0 + 0 − 0 + 0 + C2
(a)
Elastic curve:
y=−
C2 = 0
w0  1 3 2 1 4
1 5
Lx +
x  
 Lx −
EIL  6
12
120 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.4 (Continued)
(b)
y at x = L :
yB = −
w0 L4  1
1
1 
11 w0 L4
−
+
=
−


120 EI
EI  6 12 120 
yB =
(c)
dy
at x = L :
dx
dy
dx
=−
B
11 w0 L4
↓ 
120 EI
w0 L3  1 1
1 
1 w0 L4
−
+
=
−


EI  3 4 24 
8 EI
θB =
1 w0 L3
8 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLE
EM 9.5
For the caantilever beam and loading shown, determine (a) thee equation of
the elastic curve for portion AB of the beam, (b) the deflection
n at B, (c) the
slope at B..
[ x = 0, y = 0]
dy


 x = 0, dx = 0 


SOLUTION
Using ABC as a free body,
ΣFy = 0: R A − wL = 0 R A = wL
 L  wL
ΣM A = 0: − M A − ( wL)   +
=0
6
2
1
M A = − wL2
3
Using AJ as a free body, (Portion AB only)
x
ΣM J = 0: M + ( wx)   − RA x − M A = 0
2
1 2
M = − wx + RA x + M A
2
1 2
1
= − wx + wLx − wL2
2
3
2
1
1
d2y
= − wx 2 + wLx − wL2
2
2
3
dx
1 3 1
1
dy
= − wx + wLx 2 − wLx + C1
EI
6
2
3
dx
dy


 x = 0, dx = 0  : − 0 + 0 − 0 + C1 = 0 ∴ C1 = 0


EI
EIy = −
1
1
1
wx 4 + wLx3 − wLx 2 + C2
24
6
6
[ x = 0, y = 0] : − 0 + 0 − 0 + C2 = 0 ∴ C2 = 0
(a)
y=
Elastic curve over AB:
w
(− x 4 + 4 Lx 3 − 4 L2 x 2 ) 
24 EI
dy
w
=
(− x 3 + 2 Lx 2 − L2 x )
dx 6 EI

(b)
y at x = L :
yB = −
(c)
dy
at x = L :
dx
dy
dx
wL
L4
24EI
E
=0
yB =
wL4

24 EI
θB = 0 
B
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.6
For the cantilever beam and loading shown, deteermine (a) the equation of
the elastic curve for portion AB of the beam, (b) the
t deflection at B, (c) the
slope at B.
SOLUTION
FBD ABC:
Using ABC as a free body,
ΣFy = 0: RA + 2wa −
2
wa = 0
3
4
4
RA = − wa = wa ↓
3
3
2

ΣM A = 0: −M A + (2wa)(a) −  wa  (3a) = 0
3

MA = 0
Using AJ as a free body,
FBD AJ:
4

 x
ΣM J = 0: M +  wa  ( x) − (wx)   = 0
3

2
1
4
M = wx 2 − wax
2
3
d2y
1
4
= wx 2 − wax
2
3
dx 2
dyy
1
2
EI
= wx3 − wax 2 + C1
dxx
6
3
EI
dy


 x = 0, dx = 0  : 0 = 0 − 0 + C1 ∴ C1 = 0


EIy =
1
2
wx 4 − wax3 + C2
24
9
[ x = 0, y = 0]: 0 = 0 − 0 + C2 ∴ C2 = 0
(a)
Elastic curve over AB:


(b)
y at x = 2a :
(c)
dy
at x = 2a :
dx
y =
w
(3x 4 − 16ax3 ) 
72 EI
dy
w 3
=
( x − 4ax 2 )
dx
6EI
yB = −
10wa 4
9EI
d 
4wa3
 dy
=
−
 
3EI
d B
 dx
yB =
θB =
10wa 4
↓ 
9 EI
4wa3
3EI

PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.7
For the beam and loading shown, determine (a) the equation of the
elastic curve for portion BC of the beam, (b) the deflection at midspan,
(c) the slope at B.
x=−
L x = 0


2  y = 0
 x = L
 y = 0


SOLUTION
Using ABC as a free body,
 wL  3L 
L
ΣM C = 0: 
  − RB L + ( wL)   = 0
5
2

 
2
4
RB = wL
5
For portion BC only, (0 < x < L)
wL  L
x
 4
 + x  − wLx + ( wx) + M = 0
5 2
2
 5
3
1
1
M = wLx − wx 2 − wL2
5
2
10
ΣM J = 0:
d2y
3
1
1
= wLx − wx 2 −
wL2
5
2
10
dx 2
dy
3
1
1
=
EI
wLx 2 − wx3 −
wL2 x + C1
dx 10
6
10
1
1
1
EIy =
wLx3 −
wx 4 −
wL2 x 2 + C1x + C2
10
24
20
[ x = 0, y = 0]: 0 = 0 − 0 − 0 + 0 + C2 C2 = 0
EI
1
1 
1
−  wL4 + C1 L + 0
[ x = L, y = 0]: 0 =  −
 10 24 20 
(a)
Elastic curve:
y=
C1 = −
w 1 3 1 4 1 2 2
1 3 
Lx −
x −
Lx −
L x

24
20
120
EI  10

1
wL3
120

dy w  3 2 1 3 1 2
1 3
=
Lx − x − L x −
L

dx EI  10
6
10
120 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.7 (Continued)
(b)
y@ x =
L
:
2
yM =
=
w
EI
 1  L 3 1  L 4 1 2  L 2
1 3  L  
L   −
L
 L  −   −

20  2  120  2  
10  2  24  2 
wL4  1
1
1
1 
13wL4
−
−
 −
=−
EI  80 384 80 240 
1920 EI
yM =
(c)
dy
@x = 0:
dx
dy
dx
=
B
w
EI
13wL4

1920 EI
1 3
wL3

0
−
0
−
0
−
L
=
−

120 
120 EI

θB =
wL3
120 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.8
For the beeam and loading shown, determine (a) the equation of the elastic
curve for portion AB of the beam, (b) the deflection at midspan, (c) the
slope at B.
SOLUTION
Reactions:
1
1  1
 1 

ΣM B = 0: − RA L +  w0 L   L  −  w0 L  L  = 0
2
3  4
 6 

1
RA = w0 L
8
Boundary conditions: [ x = 0, y = 0] [ x = L, y = 0]
(0 ≤ x < L)
For portion AB only,
ΣM J = 0: −
M =
d2y 1
= w0 Lxx −
8
dx 2
dy
1
EI
w0 Lx
L 2
=
dx 16
1
EIy =
w0 Lx
L 3
48
EI
1
1w
w0 Lx +  0
8
2 L
1
1 w0 3
w0 Lx −
x
8
6 L
1 w0 3
x
6 L
1 w0 4
−
x + C1
24 L
1 w0 5
−
x + C1x + C2
120 L
[ x = 0, y = 0]: 0 = 0 − 0 + 0 + C2
[ x = L, y = 0] : 0 =
(a)
C2 = 0
1
1
1
w0 L4 −
w0 L4 + C1L C1 = − w0 L3
80
48
120
y =
Elastic curve:
 x
x  ( x)   + M = 0
 3
w0  1 2 3
1 5
1 4 
Lx −
x −
L x 

EIL  48
120
80

dy
w  1
1 4
1 4
= 0  L2 x 2 −
x −
L 
dx
EIL  16
24
80 
L
:
2
(b)
y at x =
(c)
dy
at x = L :
dx
yL 2 =
dy
dx
=
B
w0
EIL
 L5
L5
L5 
15w0 L4
−
−

 = −
3840EI
 384 3840 160 
w0  L4 L4 L4 
2w0 L3
−
−

 = +
EIL  16 24 80 
240EI
yL 2 =
θB =
w0 L4
↓ 
256EI
w0 L3
120EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.9
Knowing that beam AB is a W130 × 23.8 rolled shape and that
P = 50 kN, L = 1.25 m, and E = 200 GPa, determine (a) the slope
at A, (b) the deflection at C.
[ x = 0, y = 0]
[ x = L, y = 0]
L dy


 x = 2 , dx = 0 


SOLUTION
Use symmetry boundary condition at C.
By symmetry,
1
P
2
RA = RB =
0≤ x≤
Using free body AJ,
L
2
ΣM J = 0: M − RA x = 0
M = RA x =
1
Px
2
d2y
1
= Px
2
dx 2
1
dy
EI
= Px 2 + C1
4
dx
1
EIy =
Px3 + C1x + C2
12
EI
[ x = 0, y = 0]: 0 = 0 + 0 + C2 ∴ C2 = 0
2
L dy
1 L
1


2
 x = 2 , dx = 0 : 0 = 4 P  2  + C1 C1 = − 16 PL


 
PL
Elastic curve:
y =
(4 x3 − 3L2 x)
48EI
dy
PL
=
(4 x 2 − L2 )
dx 16EI
dy
dx
Slope at x = 0:
Deflection at x =
L
:
2
=−
A
PL3
16 EI
yC = −
PL3
48EI
θA =
PL2
16EI
yC =
PL3
48EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.9 (Continued)
Data:
I = 8.80 × 106 mm 4 = 8.80 × 10−6 m 4
P = 50 × 103 N,
E = 200 × 109 Pa
EI = 1.76 × 106 N ⋅ m 2
(a)
θA =
(50 × 10 )(1.25)
(16)(1.76 × 106 )
(b)
yC =
(50 × 103 )(1.25)3
= 1.156 × 10 −3 m
(48)(1.76 × 106 )
3
2
L = 1.25 m
θ A = 2.77 × 10−3 rad

yC = 1.156 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.10
Knowing that beam AB is an S200 × 27.4 rolled shape and that
w0 = 60 kN/m, L = 2.7 m, and E = 200 GPa, determine (a) the
slope at A, (b) the deflection at C.
[ x = 0, y = 0] [ x = L, y = 0]
L dy


 x = 2 , dx = 0 


SOLUTION
Use symmetry boundary conditions at C.
Using free body ACB and symmetry,
RA = RB =
0< x<
For
1
w0 L
4
L
2
w=
2w0 x
L
2w x
dV
= −w = − 0
dx
L
w x2
w 1
dM

= V = − 0 + RA = 0  L2 − x 2 
dx
L
L 4

M=
M = 0 at
But
EI
0=
EIy =
[ x = 0, y = 0]:
hence CM = 0
x = 0;
d2y
w 1
1 
= 0  L2 x − x3 
2
L 4
3 
dx
EI

L dy

 x = 2, dx = 0 :


w0  1 2
1 
L x − x 3  + CM
L  4
3 
dy
w 1
1 4
= 0  L2 x 2 −
x + C1
dx
L 8
12 
w0  1 4
1 4
L −
L + C1 = 0
L  32
192 
C1 = −
5
w0 L3
192
w0  1 2 3
1 5
5
Lx −
x −
w0 L3 x + C2
L  24
120  192
0 = 0 − 0 + 0 + C2
C2 = 0
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.10 (Continued)
y=
Elastic curve:
1 5
5 4 
w0  1 2 3
x −
L x
 Lx −
60
192
EIL  24

1 4
5 4
dy
w 1
= 0  L2 x 2 −
x −
L 
12
192 
dx
EIL  8
w0 = 60 kN/m,
Data:
E = 200 GPa,
I = 23.9 × 106 mm 4
EI = (200 × 109 )(23.9 × 10−6 ) = 4.78 × 106 Nm 2
L = 2.7 m
(a)
Slope at x = 0:
  5 

dy
60000
(2.7) 4  = −6.434 × 103
=
− 

6
dx (4.78 × 10 )(2.7)   192 

θ A = 6.43 × 10−3 rad
(b)

Deflection at x = 1.35 m:
y=
 1 

60000
1
5
2
3
5
(2.7) 4 (1.35)  = −0.0056 × 10−3 m
  (2.7) (1.35) − (1.35) −
6
60
192
(4.78 × 10 )(2.7)  24 

= 5.6 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.11
(a) Determine the location and magnitude of the maximum deflection of
beam AB. (b) Assuming that beam AB is a W360 × 64, L = 3.5 m, and
E = 200 GPa, calculate the maximum allowable value of the applied
moment M0 if the maximum deflection is not to exceed 1 mm.
SOLUTION
Using entire beam as a free body,
ΣM B = 0: M 0 − RA L = 0 RA =
M0
L
Using portion AJ ,
[ x = 0, y = 0]
ΣM J = 0: M 0 −
[ x = L, y = 0]
M =
M0
x+M =0
L
M0
( x − L)
L
d2y
M
= 0 ( x − L)
L
dx 2
dy
M 1

= 0  x 2 − Lx  + C1
EI
dx
L 2

EI
EIy =
[ x = 0, y = 0]
0 = 0 − 0 + 0 + C2
[ x = L, y = 0]
0=
y=
(a)
M0  1 3 1 2 
 x − Lx  + C1x + C2
L 6
2

M0  1 3 1 3 
L − L  + C1L + 0
L  6
2 
M0  1 3 1 2 1 2 
 x − Lx + L x 
EIL  6
2
3

To find location maximum deflection, set
1 2
1
xm − Lxm + L2 = 0
2
3
C2 = 0
1
M 0L
3
dy
M 1
1 
= 0  x 2 − Lx + L2 
dx
EIL  2
3 
dy
= 0.
dx
xm = L −
 1  1  
L2 − (4)   L2  = 1 −
 2  3  
= 0.42265L
ym =
C1 =

M 0 L2  1 
1
1
3
2
  (0.42265) −   (0.42265) +   (0.42265) 
EI  6 
2
3

1
L
3 
xm = 0.423L 
ym = 0.06415
M 0 L2

EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.11 (Continued)
Solving for M 0 ,
(b)
Data:
M0 =
EIym
0.06415L2
E = 200 × 109 Pa, I = 178 × 106 mm 4 = 178 × 10−6 m 4
L = 3.5 m
M0 =
ym = 1 mm = 10−3 m
(200 × 109 )(178 × 10 −6 )(10 −3 )
= 45.3 × 103 N ⋅ m
(0.06415)(3.5) 2
M 0 = 45.3 kN ⋅ m 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.12
[ x = 0, y = 0]
[ x = L, y = 0]
For the beam and loading shown, (a) express the magnitude and location
of the maximum deflection in terms of w0, L, E, and I. (b) Calculate the
value of the maximum deflection, assuming that beam AB is a W460 × 74
rolled shape and that w0 = 60 kN/m, L = 6 m, and E = 200 GPa.
SOLUTION
Using entire beam as a free body,
1
 L 
ΣM B = 0: − RA L +  w0 L   = 0
2

 3 
1
RA = w0 L
6
Using AJ as a free body,
 1 w0 x 2   x 
1
w0 Lx + 
   + M = 0
6
 2 L  3 
1
1 w0 3
M = w0 Lx −
x
6
6 L
ΣM J = 0: −
d2y
1
= w0 Lx −
6
dx 2
1
dy
EI
=
w0 Lx 2
dx 12
1
EIy =
w0 Lx3
36
EI
Elastic curve:
[ x = 0, y = 0]:
0 = 0 − 0 + 0 + C2
[ x = L, y = 0]:
0=
y =
w0
EI
1 w0 3
x
6 L
1 w0 4
−
x + C1
24 L
1 w0 5
−
x + C1x + C2
120 L
∴ C2 = 0
1
1
7w L3
w0 L4 −
w0 L4 + C1L + 0 ∴ C1 = − 0
36
120
360
 1
1 x5
7 3 
3
L x
−
 Lx −
120 L
360
 36

1 x4
7 3 
dy
w  1
= 0  Lx 2 −
−
L
24 L
360 
dx
EI 12
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.12 (Continued)
To find location of maximum deflection,
dy
= 0.
dx
set
15xm4 − 30L2 xm2 + 7 L4 = 0
xm2 =
30L2 − 900L4 − 420L4
30

8  2
2
xm2 = 1 −
 L = 0.2697 L
15 

ym =
w0
EI
xm = 0.5193L 
 1

1 (0.5193L5 )
7 3
3
−
−
L
(0.5193
L
)
L (0.5193L) 

120
L
360
 36

= −0.00652
w0 L4
EI
or 0.00652
w0 L4
EI
Data:
w0 = 60 kN/m = 60 × 103 N/m
For
W460 × 74, I = 333 × 106 mm 4 = 333 × 10−6 m 4
ym =

L=6m
(0.00652)(60 × 103 )(6) 4
= 7.61 × 10 −3 m
(200 × 109 )(333 × 10 −6 )
ym = 7.61 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.13
For the beam and loading shown, dettermine the deflection at
point C. Use E = 200 GPa.
[ x = 0, y = 0]
[ x = L, y = 0]
[ x = a, y = y]
dy
dy 

 x = a, dx = dx 


SOLUTION
b=L−a
Pb
,
Reactions:
RA =
L
Bending moments:
Let
RB =
Pa
L
0 < x < a:
a < x < L:
0< x<a
Pb
x
L
P
M = [bx − L( x − a)]
L
a< x< L
M =
2
d y
P
= (bx)
2
L
dx
dy
P1

EI
=  bx 2  + C1
dx
L2

EI
EIy =
P1 3
bx  + C1x + C2
L  6

EI
(1)
dy
P 1
1

=  bx 2 − L( x − a)2  + C3
dx
L 2
2

EI
EIy =
(2)
[ x = 0, y = 0]
d2y
P
= [bx − L( x − a)]
L
dx 2
Eq. (2):
(3)
P 1 3 1

bx − L( x − a)3  + C3 x + C4
L  6
6

0 = 0 + 0 + C2
(4)
∴ C2 = 0

dy dy 
P 1 2 
P1 2

 x = a, dx = dx  Eqs. (1) and (3): L  2 ba  + C1 = L  2 ba + 0 + C3 ∴ C3 = C1






P 1 3 
[ x = a, y = y] Eqs. (2) and (4):
 ba  + C1a + C2
L 6

=

P1 3
 ba + 0  + C1a + C4 ∴ C4 = C2 = 0
L6

PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.13 (Continued)
[ x = L, y = 0] Eq. (4):
C1 = C3 =
P 1 3 1

bL − L(L − a)3  + C3L = 0

L 6
6

P 1
1
1
 P1

( L − a)3 − bL2  =  b3 − bL2 

L 6
6
6
 L6

Make x = a in Eq. (2).
yC =
Data:
3
3
2
P 1 3 1 3
1
 P(ba + b a − L ab)
ba + b a − bL2a  =

6
6
6 EIL
EIL  6

P = 150 kN, E = 200 GPa
L = 4.5 m, a = 1.5 m, b = 3 m
I = 122 × 10−6 m 4 , EI = 24.4 × 106 Nm 2
yC =
150 × 103
[(3)(1.5)3 + (3)3 (1.5) − (4.5)2 (1.5)(3)]
6
(24.4 × 10 )(4.5)
= −9.22 × 10−3 m
yC = 9.2 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.14
For the beam and loading shown, knowing that a = 2 m,
w = 50 kN/m, and E = 200 GPa, deterrmine (a) the slope at
support A, (b) the deflection at point C.
SOLUTION
Using ACB as a free body and notinng that L = 3a,
a
 M A = 0: RB L − (wa)   = 0
2
[ x = 0, y = 0]
[ x = L, y = 0]
RB = (wa)
[ x = a, y = y ]
a 1
= wa
2L 6
dy 
dy

 x = a, dx = dx 


 Fy = 0:
0≤ x≤a
 MK = 0 :
 x
M − RA x + (wx)   = 0
 2
d2y
1
= RA x − wx 2
2
dx 2
dy 1
1
EI
= RA x 2 − wxx3 + C1
dx 2
6
1
1
EIy = RA x3 −
w 4 + C1 x + C2
wx
6
24
0 = 0 − 0 + 0 + C2 C2 = 0
1
1
RA x 3 −
wxx 4 + C1x
6
24

1
dy 1
3
2
EI
= RA x − wx + C1
6
dx 2
EIy =
−M + RB ( L − x) = 0
1 2
wx
2
EI
[ x = 0, y = 0]
5
wa
6
a≤x≤L
MJ = 0 :
M = RA x −
RA =
RA + RB − wa = 0
M = RB ( L − x)
d2y
= RB ( L − x)
dx 2
dy
1
EI
= − RB ( L − x) 2 + C3
2
dx
1
EIy = RB ( L − x)3 + C3 x + C4
6
EI
[ x = L, y = 0]
0 = 0 + C3L + C4
C4 = − C3 L
1
RB (L − x)2 − C3 ( L − x) 
6
dy
1
EI
= − RB (L − x)2 + C3
dx
2
EIy =
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.14 (Continued)
dy dy 
1
1
1

RAa 2 − wa3 + C1 = − RB (2a)2 + C3
 x = a, dx = dx 
2
6
2


1
1
1
7
C3 = C1 + RAa 2 − wa3 + RB (2a)2 = C1 + wa3
2
6
2
12
1
1
1
7
[ x = a, y = y ] 6 RAa3 − 24 wa 4 + C1a = 6 RB (2a)3 −  C1 + 12 wa3  (2a)


1
1
1
7
25
3C1a = − RAa3 +
wa 4 + RB (2a)3 −
wa 2 (2a) = −
wa 4
6
24
6
12
24
25
C1 = − wa3
72
5
1
25
For 0 ≤ x ≤ a, EIy =
wax3 −
wx 4 −
wa3 x
36
24
72
dy
5
1
25
EI
=
wax 2 − wx3 −
wa3
dx 12
6
72
Data:
w = 50 × 103 N/m, a = 2 m, E = 200 × 109 Pa
I = 84.9 × 106 mm 4 = 84.9 × 10−6 m 4 , EI = 16.98 × 106 N ⋅ m 2
(a)
Slope at x = 0:
16.98 × 106
dy
dx
dy
dx
(b)
=0 − 0 −
A
25
(50 × 103 )(2)3
72
= θ A = −8.18 × 10−3
θ A = 8.18 × 10−3 rad

A
Deflection at x = 2 m:
5
1
25 4
1
wa 4 −
wa 4 −
wa = − wa 4
36
24
72
4
1
16.98 × 106 yC = − (50 × 103 )(2)4 yC = − 11.78 × 10−3 m
4
EIyC =
yC = 11.78 mm ↓ 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.15
For the beam and loading shown, determine the deflection at
point C. Use E = 200 GPa.
SOLUTION
Reactions:
RA = M 0/L ↑,
RB = M 0 /L ↓
 M J = 0:
0 < x < a:
M0
x+M =0
L
M
M= 0x
L
−
[ x = 0, y = 0]
[ x = L, y = 0]
[ x = a, y = y ]
M K = 0:
a < x < L:
dy
dy 

 x = a, dx = dx 


M0
x + M0 + M = 0
L
M
M = 0 ( x − L)
L
−
0< x<a
a< x<L
d 2y M0
=
x
L
dx 2
dy M 0  1 2 
EI
=
 x  + C1
dx
L 2 
EI
EIy =
[ x = 0,
M0  1 3 
x + C1x + C2
L  6 
y = 0] Eq. (2):
d 2 y M0
=
( x − L)
L
dx 2
dy M 0  1 2

EI
=
 x − Lx  + C3
dx
L 2

EI
(1)
(2)
0 = 0 + 0 + C2
EIy =
M0  1 3 1 2 
x − Lx  + C3 x + C4
L  6
2

(3)
(4)
C2 = 0
dy dy 
M0  1 2 
M0  1 2


 x = a, dx = dx  Eqs. (1) and (3): L  2 a  + C1 = L  2 a − La  + C3






C3 = C1 + M 0a
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.15 (Continued)
[ x = a,
y = y ] Eqs. (2) and (4):
[ x = L,
y = 0] Eq. (4):
M0
L
C1 =
Elastic curve for 0 < x < a :
M0  1 3 
M0  1 3 1 2 
 a  + C1a =
 a − La  + ( C1 + M 0a)a + C4
L 6 
L 6
2

1
C4 = − M 0 a 2
2
1
 1 3 1 3
2
 L − L  + (C1 + M 0a) L − M 0a = 0
6
2
2


M0  1 2 1 2

 L + a − aL 
L 3
2

y=
M0 1 3  1 2 1 2
 
 x +  L + a − aL  x 
3
2
EIL  6

 
M0  1 3 1 2
1
M 2
1


a + L a + a3 − a 2 L  = 0  a3 + L2a − La 2 

EIL  6
3
2
3
 EIL  3

Make x = a.
yC =
Data:
E = 200 × 109 Pa, I = 34.4 × 106 mm 4 = 34.4 × 10−6 m 4 , M 0 = 60 × 103 N ⋅ m
a = 1.2 m, L = 4.8 m
yC =
(60 × 103 ) (2)(1.2)3 / 3 + (4.8) 2 (1.2) / 3 − (4.8)(1.2) 2 
(200 × 109 )(34.4 × 10−6 )(4.8)
= 6.28 × 10−3 m
yC = 6.28 mm ↑ 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.16
Knowing that beam AE is an S200 × 27.4 rolled shape and that
P = 17.5 kN, L = 2.5 m, a = 0.8 m and E = 200 GPa, determine
(a) the equation of the elastic curve for portion BD, (b) the deflection at
the center C of the beam.
SOLUTION
Consider portion ABC only. Apply symmetry about C.
RA = RE = P
Reactions:
dy
dy  
L dy


=
= 0
, x = ,
Boundary conditions: [ x = 0, y = 0], [ x = a, y = y],  x = a,

dx
dx  
2 dx


a< x< L−a
0< x<a
2
EI
d y
= M = Px
dx 2
EI
dy
1
= Px 2 + C1
dx
2
EIy =
d2y
= M = Pa
dx 2
dy
EI
= Pax + C3
dx
1
EIy = Pax 2 + C3 x + C4
2
EI
(1)
1 3
Px + C1x + C2
6
(2)
L dy
1


 x = 2 , dx = 0  → C3 = − 2 PaL


[ x = 0, y = 0] → C2 = 0
L dy
dy 

 x = 2 , dx = dx 


1 2
1
Pa + C1 = Pa 2 − PaL
2
2
L


x = 2 , y = y



1 3 1 2
1
1
1
Pa +  Pa − PaL  a = Pa3 − Pa 2 L + C4
6
2
2
2
 2

C1 =
C4 =
(a)
1 2 1
Pa − PaL
2
2
1 3
Pa
6
Elastic curve for portion BD:
y =
1 1

Pax 2 + C3 x + C4 
EI  2

y=
P 1 2 1
1 
ax − aLx + a3  

EI  2
2
6 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.16 (Continued)
For deflection at C,
x=
L
.
2
yC =
P
EI
set
=−
1 2 1 2 1 3
 aL − aL + a 
4
6 
8
Pa  1 2 1 2 
 L − a 
EI  8
6 
I = 23.9 × 106 mm 4 = 23.9 × 10−6 m 4 ,
Data:
E = 200 × 109 Pa
P = 17.5 × 103 N,
L = 2.5 m, a = 0.8 m
(b)
yC = −
 2.52 0.82 
(17.5 × 103 )(0.8)
−3
−
 = −1.976 × 10 m
9
6 
6 
(200 × 10 )(23.9 × 10 )  8
yC = 1.976 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.17
For the beam and loading shown, determine (a) the equation of the
elastic curve, (b) the deflection at the free end.
SOLUTION
Boundary conditions are shown at right.
[ x = 0, y = 0]
[ x = L, V = 0]
dy
[ x = 0,
= 0]
dx
[ x = L, M = 0]
2

dV
x x 
= − w = − w0 1 − 4   + 3   
dx
 L   L  


2 x 2 x3 
V = − w0  x −
+ 2  + CV
L
L 

[ x = L, V = 0]: 0 = − w0 [ L − 2 L + L] + CV = 0
CV = 0

dM
2 x 2 x3 
= V = − w0  x −
+ 2
dx
L
L 

 x 2 2 x3
x4 
M = − w0  −
+ 2  + CM
3L 4 L 
2
2
1 
1
[ x = L, M = 0]: 0 = −w0  L2 − L2 + L2  + CM
2
3
4 

EI
CM =
1
w0 L2
12
1
d2 y
2 x3 1 x 4 1 2 
= M = − w0  x 2 −
+
− L 
dx
3 L 4 L2 12 
2
EI
1
dy
1 x4
1 x5 1 2 
= − w0  x3 −
+
− L x  + C1
dx
6 L 20 L2 12
6

[ x = 0,
dy
= 0]
dx
C1 = 0
1
1 x5
1 x6 1 2 2 
EIy = − w0  x 4 −
+
−
L x  + C2
30 L 120 L2 24
 24

[ x = 0, y = 0]
C2 = 0
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.17 (Continued)
(a)
Elastic curve:
(b)
Deflection at x = L :
y=−
yB = −
w0  1 2 4 1 5
1 6 1 4 2
L x − Lx +
x −
Lx  
2 
30
120
24
EIL  24

w0  1 6 1 6
w0 L4
1 6 1 6
L
L
L
L
−
+
−
=
−


30
120
24 
40 EI
EIL2  24
yB =
w0 L4
↓ 
40 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.18
For the beam and loading shown, determine (a) the
equation of the elastic curve, (b) the slope at end A, (c) the
deflection at the midpoint of the span.
[ x = 0, M = 0]
[ x = L, M = 0]
[ x = 0, y = 0]
[ x = L, y = 0]
SOLUTION
Boundary conditions at A and B are noted.
w0
(4Lx − 4 x 2 )
L2
w
= −w = 20 (4 x 2 − 4Lx)
L
w 4

= V = 20  x3 − 2Lx 2  + C1
L 3

w=
dV
dx
dM
dx
M =
[ x = 0, M = 0]
0 = 0 + 0 + 0 + C2
[ x = L, M = 0]
0=
EI
w0  1 4 2 4 
 L − L  + C1L + 0
3 
L2  3
C1 =
1
w0 L
3
dy
w  1
1
1

= 20  x5 − Lx 4 + L3 x 2  + C3
dx
6
6
L  15

EIy =
w0  1 6
1
1 3 3
x −
Lx5 +
L x  + C3 x + C4
2 
30
18
L  90

[ x = 0, y = 0]
0 = 0 + 0 + 0 + 0 + C4
[ x = L, y = 0]
0=
Elastic curve:
C2 = 0
d2y
w 1
2
1

= M = 20  x 4 − Lx3 + L3 x 
2
3
3
dx
L 3

EI
(a)
w0  1 4 2 3 
 x − Lx  + C1x + C2
3
L2  3

C4 = 0
w0  1 6
1 6
1 6
L −
L +
L  + C3 L + 0
2 
30
18 
L  90
y=
C3 = −
1
w0 L3
30
w0  1 6
1 5 1 3 3
1 5 
x −
Lx +
Lx −
L x 
2
30
18
30
EIL  90

dy
w 1
1
1
1 5
= 0 2  x5 − Lx 4 + L3 x 2 −
L
dx
6
6
30 
EIL  15
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.18 (Continued)
(b)
Set x = 0 in
Slope at end A:
dy
.
dx
dy
dx
=−
A
1 w0 L3
30 EI
θA =
(c)
Deflection at midpoint:
yC =
=
Set x =
1 w0 L3
30 EI

L
in y.
2
6
5
3
w0 L4  1  1 
1 1
1  1  
 1  1 
−
+
  
  
  −
 
EI  90  2 
18  2 
30  2  
 30  2 
w0 L4  1
1
1
1
61 w0 L4
−
+
−
=
−


EI  5760 960 144 60 
5760 EI
yC =
61 w0 L4
↓ 
5760 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.19
For the beam and loading shown, dettermine the reaction at the
roller support.
[ x = 0, y = 0]
[ x = L, y = 0]
dy


 x = L, dx = 0


SOLUTION
Reactions are statically indeterm
minate.
Boundary conditions are shown above.
Using free body AJ,
ΣM J = 0: M 0 − RA x + M = 0
M = RA x − M 0
d2y
= RA x − M 0
dx 2
1
dy
= RA x 2 − M 0 x + C1
EI
2
dx
EI
dy
1


2
 x = L, dx = 0  0 = 2 RA L − M 0 L + C1


C1 = M 0 L −
EIy =
1
RA L2
2
1
1
RA x3 − M 0 x3 + C1x + C2
6
2
[ x = 0, y = 0]
C2 = 0
[ x = L, y = 0] 0 =
1
1
1


RA L3 − M 0 L2 +  M 0 L − RA L2  L + 0
6
2
2


RA =
3 M0
↑ 
2 L
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.20
For the beam and loading shown, determine the reeaction at the
roller support.
[ x = 0, y = 0]
[ x = L, y = 0]
dy


 x = 0, dx = 0 


SOLUTION
Reactions are statically indeterminate.
Boundary conditions are shown above.
Using free body KB,
L − x
M K = 0: RB (L − x) − w( L − x) 
−M =0
 2 
M = RB (L − x) −
1
w( L − x) 2
2
d2y
1
= RB ( L − x) − w( L − x) 2
2
2
dx
1
1
dy
= − RB ( L − x) 2 + w( L − x)3 + C1
EI
dx
6
2
EI
dy
1
1 3


2
 x = 0, dx = 0  : 0 = − 2 RB L + 6 wL + C1


1
1
C1 = RB L2 − wL3
2
6
EI y =
1
1
RB ( L − x)3 −
w( L − x) 4 + C1x + C2
6
24
1
RB L3 −
6
1
C2 = − RB L3 +
6
[ x = 0, y = 0]: 0 =
1
wL3 + C2
24
1
wL4
24
[ x = L, y = 0]: 0 = 0 − 0 + C1L + C2
1
1
1
1
RB L3 − wL4 − RB L3 +
wL4 = 0
2
6
6
24
RB =
3
wL ↑ 
8
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.21
For the beam and loading shown, determinne the reaction at the roller
support.
[ x = 0, y = 0]
[ x = L, y = 0]
dy


 x = L, dx = 0


SOLUTION
Reactions are statically indeterm
minate.
Boundary conditions are shown above.
w0
( L − x)
L
w
= − w = − 0 ( L − x)
L
1 
w 
= V = − 0  Lx − x 2  + RA
2 
L 
w=
dV
dx
dM
dx
M =−
EI
w0  1 2 1 3 
 Lx − x  + RA x
6 
L 2
d2y
w 1
1 
= − 0  Lx 2 − x3  + RA x
2
2
6 
L
dx

EI
1 4 1
dy
w 1
x  + RA x 2 + C1
= − 0  Lx3 −
24  2
dx
L 6
EIy = −
1 5 1
w0  1
Lx 4 −
x + RA x3 + C1x + C2

120  6
L  24
[ x = 0, y = 0]
dy


 x = L, dx = 0 


0 = 0 + 0 + 0 + C2
−
w0  1 4
1 4 1
L  + RA L2 + C1 = 0
 L −
L 6
24  2
C1 =
[ x = L, y = 0]
−
C2 = 0
1
1
w0 L3 − RA L2
8
2
w0  1 5
1 5 1
1
1

L −
L  + RA L3 +  w0 L3 − RA L2  L = 0

L  24
120  6
2
8

1
1 
1 1
1
+
 −  RA =  −
 w0 L
2
6
8
24
120




1
11
RA =
w0 L
3
120
RA =
11
w0 L ↑ 
40
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.22
For the beam and loading shown, determine thhe reaction at
the roller support.
[ x = 0, y = 0]
[ x = L, y = 0]
0
dy


 x = 0, dx = 0 


SOLUTION
Reactions are statically indeterminate.
Boundary conditions are shown above.
Using free body JB,
ΣM J = 0: −M + RB ( L − x) +
+
1
2
w0 ( L − x) ( L − x)
2
3
1 w0 x
1
( L − x) ( L − x) = 0
3
2 L
w0
[2 L( L − x) 2 + x( L − x) 2 ]
6L
w
= RB ( L − x) − 0 [2 L3 − 4 L2 x + 2 Lx 2 + xL2 − 2 Lx 2 + x3 ]
6L
w
= RB ( L − x) − 0 ( x3 − 3L2 x + 2 L3 )
6L
M = RB ( L − x) −
d2y
w
= RB ( L − x) − 0 ( x3 − 3L2 x + 2L3 )
2
6L
dx
1
3
dy

 w 1

EI
= RB  Lx − x 2  − 0  x 4 − L2 x 2 + 2L3 x  + C1
2  6L  4
2
dx


EI
1  w  1
1

1
EIy = RB  Lx 2 − x3  − 0  x5 − L2 x3 + L3 x 2  + C1x + C2
6  6L  20
2

2
[x = 0, y = 0]
→ C2 = 0
dy


 x = 0, dx = 0  → C1 = 0


4
1
1 1 w L  1

[ x = L, y = 0] 0 = RB L3  −  − 0 
− + 1
2
6
6
20
2




1
 1  11 
RB =    w0 L
3
 6  20 
RB =
11
w0 L ↑ 
40
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.23
For the beam shown, determine the reaction at the roller support when
w0 = 15 kN/m.
SOLUTION
Reactions are statically indeterminate.
Boundary conditions are shown at left.
Using free body JB,
MJ = 0 :
[ x = 0, y = 0]
[ x = L, y = 0]
L
− M + x
dy
[ x = 0,
= 0]
dx
w0 2
ξ (ξ − x)dξ + RB ( L − x) = 0
L2
w0 L 2
 ξ (ξ − x)dξ − RB ( L − x)
L2 x
M =
L
EI
w0  1 4 1 3 
 ξ − xξ  − RB ( L − x)
3
L2  4
x
=
w0  1 4 1 3
1 4
x  − RB ( L − x)
 L − Lx+
3
12 
L2  4
d2y
w 1
1
1 4
x  − RB ( L − x)
= 20  L4 − L3 x +
2
4
3
12
dx
L 

EI
dy
w 1
1
1 5
1 

= 20  L4 x − L3 x 2 +
x  − RB  Lx − x 2  + C1
6
60 
2 
dx
L 4

EIy =
dy


 x = 0, dx = 0 


[ x = 0, y = 0]
[ x = L, y = 0]
Data:
=
1 3 3
1 6
1 
w0  1 4 2
1
Lx −
Lx +
x  − RB  Lx 2 − x3  + C1x + C2
2 
18
360 
6 
L 8
2
0 = 0 + 0 + C1
0 = 0 + 0 + 0 + C2
C1 = 0
C2 = 0
1 
1 1
1 1
4
3
+
 −
 w0 L −  −  RB L = 0
 8 18 360 
2 6
13
1
13
w0 L4 − RB L3 = 0
RB =
w0 L
180
3
60
L = 3m
w0 = 15 kN/m
13
RB =
(15)(3) = 9.75 kN
60
RB = 9.75 kN ↑ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.24
Foor the beam shown, determine the reaction at the roller support
whhen w0 = 65 kN/m.
[ x = 0, y = 0]
[ x = L, y = 0]
dy


 x = L, dx = 0 


SOLUTION
Reactions are statically indeterminate.
Boundary conditions are shown at left.
w = w0
x2
L2
dV
w
= −w = − 20 x 2
dx
L
dM
w x3
= V = − 20
+ RA
dx
L 3
M =−
EI
d2y
w0 x 4
=
−
+ RA x
dx 2
L2 12
EI
dy
w x5 1
= − 20
+ RA x 2 + C1
dx
L 60 2
EIy = −
dy


 x = L, dx = 0  :


[ x = L, y = 0]:
Data:
w0 x 4
+ RA x
L2 12
w0 x 6
1
+ RA x3 + C1x + C2
2
L 360 6
[ x = 0, y = 0]:
0 = 0 + 0 + 0 + C2
C2 = 0
1
1
1
1
−
w0 L3 + RA L2 + C1 = 0
C1 =
w0 L3 − RA L2
60
2
60
2
−
1
1
1
 1

w0 L4 + RA L3 +  w0 L3 − RA L3  L = 0
360
6
2
 60

1 
1 1
 1
−
 −  RA = 
 w0 L
2 6
 60 360 
1
1
1
w0 L
RA = w0 L
RA =
3
72
18
w0 = 65 kN/m, L = 4 m
1
RA = (65)(4) = 14.44 kN
18

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.25
Determine the reaction at the roller support and draw
d
the bending moment
diagram for the beam and loading shown.
[x =
[ x = 0, y = 0]
dy


 x = 0, dx = 0 


SOLUTION
Reactions are statically indeterm
minate.
ΣFy = 0:
RA + RB = 0
ΣM A = 0:
RA = − RB
− M A − M 0 + RB L = 0
0 <x <
M A = RB L − M 0
L
2
M = RB x + M A = −M 0 + RB L − RB x
d2y
= −M 0 + RB ( L − x)
dx 2
dy
1 

= −M 0 x + RB  Lx − x 2  + C1
EI
dx
2 

EI
1 
1
1
EIy = − M 0 x 2 + RB  Lx 2 − x3  + C1x + C2
2
6 
2

L
< x<L
2
M = RB ( L − x)
d2y
= RB ( L − x)
dx 2
dy
1 

EI
= RB  Lx − x 2  + C3
dx
2 

EI
1 
1
EIy = RB  Lxx 2 − x3  + C3 x + C4
6 
2
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.25 (Continued)
dy


 x = 0, dx = 0  0 + 0 + C1 = 0


C1 = 0
[ x = 0, y = 0] 0 + 0 + 0 + C2 = 0
C2 = 0
L dy
dy 

 x = 2 , dx = dx 


−M 0
L
1 
1 
1
1
+ RB  L2 − L2  = RB  L2 − L2  + C3
2
6 
6 
2
2
C3 = −
M 0L
2
L


x = 2 , y = y


2
L
1
1 3
1 3
1
L
1
L  + C3 + C4
− M 0   + RB  L3 −
L  = RB  L3 −
2
2
48 
48 
8
2
8
1
1
C4 = − M 0 L2 − C3L
8
2
1
 1 1
=  − +  M 0 L2 = M 0 L2
8
 8 4
[ x = L, y = 0]
1  M L
1
1
RB  L3 − L3  + 0 L + M 0 L2 = 0
6 
2
8
2
1 1
1 1
3
2
 2 − 6  RB L =  2 − 8  M 0 L




1
3 M0
RB =
3
8 L
RB =
MA =
9
1
M0 − M0 = M0
8
8
M C − = −M 0 +
9 M0
7
= − M0
8 L
16
L  9 M0  L 
9

=
M C + = RB  L −  =
M0


2  8 L  2  16

9 M0
↑ 
8 L
MA =
1
M0 
8
M C− = −
7
M0 
16
M C+ =
9
M0 
16
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.26
Determine the reaction at the rollerr support and draw the
bending moment diagram for the beam
m and loading shown.
[ x = 0, y = 0]
[ x = L, y = 0]
dy


 x = 0, dx = 0 


SOLUTION
Reactions are statically indeterm
minate.
ΣFy = 0 : RA + RB − P = 0 RA = P − RB
ΣM A = 0: − M A +
M A = RB L −
0< x<
1
PL − RB L = 0
2
1
PL
2
1
L:
2
M = M A + RA x
d2y
= M A + RA x
dx 2
dy
1
E
EI
= M A x + RA x 2 + C1
dx
2
1
1
EIy = M A x 2 + RA x3 + C1x + C2
2
6
EII
1
L< x< L:
2
1 

M = M A + RA x − P  x − L 
2 

EII
d2y
1 

= M = M A + RA x − P  x − L 
2 
dx 2

2
E
EI
1
1 
1 
dy
= M A x + RA x 2 − P  x − L  + C3
2
2 
2 
dx
3
EIy =
1
1
1 
L
M A x 2 + RA x3 − P  x −  + C3 x + C4
2
6
6 
2
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROB
BLEM 9.26 (Continued)
dy


 x = 0, dx = 0


[ x = 0,
0 + 0 + C1 = 0
C1 = 0
y = 0 ] 0 + 0 + 0 + C2 = 0
C2 = 0
L dy
dy 

 x = 2 , dx = dx 


1
1
1
1
M A L + RA L2 + 0 = M A L + RA L2 − 0 + C3
2
8
2
8
C3 = 0
L


x = 2 , y = y


1
1
1
1
M A L2 +
RA L3 + 0 + 0 = M A L2 +
RA L3 − 0 + 0 + C4 C4 = 0
8
48
8
48
0
[ x = L, y = 0]
1
1
1
M A L2 + RA L3 −
PL3 + 0 + 0 = 0
2
6
48
1
1  3 1
1
3
PL3 = 0
 RB L − P  L + ( P − RB ) L −
2
2 
6
48
RB =
5
P↑ 
16
5
P
16
RA =
7
P↑
16
5
1
PL − PL
16
2
MA = −
3
PL 
16
MC =
5
PL 
32
RA = P −
MA =
 L   5  L 
M C = RB   =  P  
 2   16  2 
MB = 0 
Bending moment diagram
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.27
Determine the reaction at the roller support and draw the bending moment
diagram for the beam and loading shown.
SOLUTION
Reactions are statically indeterminate.
0< x<
EI
L
2
d2y
= M = RA x
dx 2
EI
(1)
dy 1
= RA x 2 + C1
dx 2
EIy =
(2)
1
RA x3 + C1x + C2
6
(3)
L
< x< L
2
EI
d2y
1 
L
= M = RA x − w  x − 
2
2 
2
dx
2
(4)
3
EI
dy 1
1 
L
= RA x 2 − w  x −  + C3
dx 2
6 
2
(5)
4
EIy =
[ x = 0,
y = 0]
1
1 
L
RA x3 −
w  x −  + C3 x + C4
6
24 
2
0 = 0 + 0 + C2
C2 = 0
2
2
L dy dy 

 x = 2 , dx = dx 


1
1
L
L
RA   + C1 = RA   + 0 + C3
2
2
2
2
L


x = 2 , y = y


L
1 L
1 L
L
RA   + C1 + C2 = RA   − 0 + C3 + C4
2
6 2
6 2
2
dy


 x = L, dx = 0 


1
1 L
Rx L2 − w   + C3 = 0
2
6 2
[ x = L,
1
1 L  1
1

RA L2 −
w   +  wL3 − RA L2  L + 0 = 0
6
24  2   48
2

y = 0]
(6)
3
C1 = C3
3
3
C3 =
C2 = C4 = 0
1
1
wL3 − RA L2
48
2
4
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.27 (Continued)
1  4
1 1
3  1
 2 − 6  RAL =  48 − 384  wL




From (1), with x =
L
,
2
1
7
RA =
wL
3
384
RA =
7
L
M C = RA   =
wL2
2
256
 
7
wL ↑ 
128
M C = 0.02734 wL2 
2
From (4), with x = L,
M B = RA L −
1 L
1
9
 7
w  = 
−  wL −
wL2
2 2
128
 128 8 
M B = − 0.07031wL 
Location of maximum positive M:
L
< x<L
2
From (4), with x = xm ,
L

Vm = RA − w  xm −  = 0
2

L
7
71
xm = +
L=
L
2 128
128
M m = RA xm −
1 
L
w  xm − 
2 
2
xm −
L
R
7
= A =
L
2
w 128
2
 7
 71  1  7 
=
wL 
L  − w
L
 128
 128  2  128 

2
M m = 0.02884 wL2 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.28
Determine the reaction at the roller support and draw the bending moment
diagram for the beam and loading shown.
SOLUTION
Reactions are statically indeterminate.
ΣFy = 0 : RA + RB −
w0 L
=0
4
RB =
w0 L
− RA
4
 w L  2L 
ΣM B = 0 : − RA L +  0 
 + MB = 0
 4  3 
0≤ x ≤
w=
M B = RA L −
L
:
2
L
≤ x≤L:
2
2w0
x
L
M = RA x −
w0 2
x
L
w
M = RA x − 0 x3
3L
V = RA −
EI
d2y
w
= RA x − 0 x 3
2
3L
dx
dy
1
1 w0 4
= RA x 2 −
EI
x + C1
dx
2
12 L
1
1 w0 5
EIy = RA x3 −
x + C1x + C2
6
60 L
w0 L 
L
x − 
4 
3
1 
d2y
1
= RA x − w0 L  x −
L
2
12 
dx
4
EI
EI
[ x = 0, y = 0] : 0 = 0 − 0 + 0 + C2
w0 L2
6
1
1
dy
1

= RA x 2 − w0 L  x 2 −
Lx  + C3
2
12 
dx
8
EIy =
1
1
 1

RA x3 − w0 L  x3 −
Lx 2  + C3 x + C4
6
24
 24

∴ C2 = 0
L dy
dy  1
1
1
1
1

2
3
2
3
 x = 2 , dx = dx  : 8 RA L − 192 w0 L + C1 = 8 RA L + 96 w0 L + C3 ∴ C3 = C1 − 64 w0 L


L
1
1

 1
3
4
 x = 2 , y = y  : 48 RA L − 1920 w0 L + 2 C1L + 0


=
1
1 3
 1 3
RA L3 − w0 L 
L −
L 
48
192
96


1

 L 
+  C1 −
w0 L3   + C4
64

 2 
∴ C4 =
1
w0 L4
480
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROB
BLEM 9.28 (Continued)
dy
1
1

 1
1 2 1 2
2
2
3
 x = L, dx = 0 : 2 RA L − w0 L  8 L − 12 L  + C3 = 0 ∴ C3 = − 2 RAL + 24 w0 L




[ x = L, y = 0] :
Over
1
1 3  1
1
1
 1

RA L3 − w0 L  L3 −
L  +  − RAL2 +
w0 L3  (L) +
w0 L4 = 0
6
24   2
24
480
 24

RB =
w0 L
21
−
w0 L
160
4
MB =
21
w L2
w0 L2 − 0
160
6
0< x<
21
w0 L ↑ 
1160
RB =
119
w0 L ↑
160
17
w0 L2 = −0.0354w0 L2 
480
w
21
w
L
, V = RA − 0 x 2 =
w0 L − 0 x 2
2
L
160
L
V = 0 at
M =
MB = −
RA =
x = xm = 0.36228L
21
w
w0 Lx − 0 x3
160
3L
M A = M ( x = 0) = 0
L

M C = M  x =  = 0.0240w0 L2
2

M m = M ( xm = 0.36228L) = 0.00317w0 L2 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.29
Determine the reaction at the roller support and the
deflection at point C.
[ x = 0, y = 0]
[ x = L, y = 0]
dy


 x = 0, dx = 0 


L

x = 2 , y =


y

L dy
dy 

 x = 2 , dx = dx 


SOLUTION
Reactions are statically indeterm
minate.
ΣFy = 0: RA +
1
1
wL − wL + RB = 0 RA = − RB
2
2
1
L
ΣM A = 0: − M A −  wL  + RB L = 0
2

2
M A = RB L −
0< x≤
From A to C:
1 2
wL
4
L
2
d2y
1
= M = M A + RA x + wx 2
2
2
dx
dy
1
1
EI
= M A x + RA x 2 + wx3 + C1
dx
2
6
1
1
1
EIy = M A x 2 + RA x3 +
wx 4 + C1x + C2
2
6
24
EI
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.29 (Continued)
L
≤ x< L
2
From C to B:
d2y
1
L 1 
L

EI 2 = M = M A + RA x + wL  x −  − w  x − 
2
4 2 
2
dx

2
EI
2
3
dy
1
1
L
1 
L

= M A x + RA x 2 + wL  x −  − w  x −  + C3
2
4
4
6
2
dx



3
EIy =
4
1
1
1
L
1 
L

M A x 2 + RA x 3 +
wL  x −  −
w  x −  + C3 x + C4
2
6
12
4
24 
2

dy


 x = 0, dx = 0 


0 + 0 + 0 + C1 = 0
C1 = 0
[ x = 0, y = 0] 0 + 0 + 0 + 0 + C2 = 0
C2 = 0
L dy
dy 

 x = 2 , dx = dx 


2
MA
3
L
1 L
1 L
L
1 L
− RA   + w   = M A + RA  
2 2
6 2
2
2 2
2
2
2
1
L
+ wL   − 0 + C3
4
4
1  3
1
 1
C3 = 
−
wL =
wL3

192
 48 64 
L


x = 2 , y = y


2
3
1
1 L
1 L
L
M A   + RA   +
w 
2
6 2
24  2 
2
2
=
3
4
1
1 L
1
L
L
M A   + RA   +
wL  
2
2
6
2
12
 
 
4
−0+
3
1
L
wL3   + C4
192
2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.29 (Continued)
1
1  4
1
 1
C4 = 
−
−
wL = −
wL4

768
 384 768 384 
[ x = L, y = 0]
3
1
1
1
1 L
 3L 
M A L2 + RA L3 +
wL   −
w 
2
6
12
4
24
 
2
1
1
+
wL3 (L) −
wL4 = 0
192
768
4
1
1
1
1
1
1  4

 27
RB L − wL2  L2 + (− RB )L3 + 
−
+
−

 wL = 0
2
4
6

 768 384 192 768 
7  4
1 1
1
3
 2 − 6  RB L =  8 − 192  wL




RA = − RB = −
M A = RB L −
Deflection at C:
1
17
RB =
wL
3
192
RB =
17
wL ↑ 
64
17
wL
64
1 2  17 1  2
1
wL = 
wL2
−  wL =
4
64
 64 4 
L

 y at x = 2 


2
EIyC =
3
1
1 L
1 L
L
M A   + RA   +
w
2
6 2
24  2 
2
2
=
4
3
1 1
1  17
1 L
 L 
2  L 
w 
 wL   +  − wL   +
2  64
6  64
24  2 
 2 
 2 
17
1  4
1
 1
=
−
+
wL4
 wL = −
1024
 512 3072 384 
4
yC =
1 wL4
↓ 
1024 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.30
Determine the reaction at the roller support and the deflection at
point C.
SOLUTION
Reactions are statically indeterminate.
0 < x<
L
2
d2y
1
= M = RA x − wx 2
2
2
dx
dy
1
1
EI
= RA x 2 − wx3 + C1
2
6
dx
1
1
EIy = RA x3 −
wx 4 + C1x + C2
6
24
EI
L
< x< L
2
(See free body diagram.)
1
L

 M K = 0: − RA x + wL  x −  + M = 0
2
EI

4
d2y
1
1 

= M = RA x − wL  x − L 
2
2
4 
dx

2
dy
L
1
1

= RA x 2 − wL  x −  + C3
EI
dx
2
4
4

3
EIy =
L
1
1

R A x3 −
wL  x −  + C3 x + C4
6
12
4

[ x = 0, y = 0] :
0 − 0 + 0 + C2 = 0
L dy
dy 

 x = 2 , dx = dx  :


1 L
1 L
1 L
1
L
RA   − w   + C1 = RA   − wL   + C3
2 2
6 2
2 2
4
4
2
3
C1 = C3 +
3
C2 = 0
4
2
2
1
1
1
wL3 −
wL3 = C3 +
wL3
48
64
192
3
3
L
1 L
1
1 L
1
L

 1 L

L
3 L
 x = 2 , y = y  : 6 RA  2  − 24 w  2  +  C3 + 192 wL  2 = 6 RA  2  − 12 wL  4  + C3 2 + C4


 
 


 
 
C4 = −
1
1
1
1
wL4 +
wL4 +
wL4 =
wL4
384
384
768
768
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.30 (Continued)
dy


 x = L, dx = 0  :


2
1
1
 3L 
RA L2 − wL   + C3 = 0
2
4
 4 
C3 =
9
1
wL3 − RA L2
64
2
3
[ x = L, y = 0] :
1
1
1
1
 3L 
 9

RA L3 −
wL   +  wL3 − RA L2  L +
wL4 = 0
6
12
2
768
 4 
 64

27
1  4
1 1
 9
3
 2 − 6  RA L =  64 − 768 + 768  wL




C3 =
1
41
RA =
wL
3
384
41
wL ↑ 
128
9
1 41 3
5
wL3 −
wL = −
wL3
64
2 128
256
C1 = −
5
1
11
wL3 +
wL3 = −
wL3
256
192
768
L

 y at x = 2 


Deflection at C:
yC =
3
4

wL4  1 41  1 
1 1
11 1
⋅  −
⋅  −
⋅ + 0
 ⋅
EI  6 128  2 
24  2 
768 2

1
11  wL4
19 wL4
 41
=
−
−
=
−

6144 EI
 6144 384 1536  EI
or
RA =
yC =
wL4
EI
 1 41  1 3
1
 ⋅
  −
12
 6 128  2 
yC =
19 wL4
↓ 
6144 EI
3
5 1
1 
1
⋅  +
⋅ +

256 2 768 
4
 41
1
5
1  wL4
19 wL4
=
−
−
+
=−

768  EI
6144 EI
 6144 768 512
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.31
Determine the reaction at the roller support and the deflection at
point D if a is equal to L/3.
SOLUTION
ΣFy = 0: RA + RB = 0 RA = − RB
ΣM A = 0: M 0 − M A + RB L = 0 M A = RB L + M 0
0 ≤ x ≤ a:
M = M A + RA x
d2y
= M = M A + RA x
dx 2
1
dy
EI
= M A x + RA x 2 + C1
2
dx
1
1
EIy = M A x 2 + RA x3 + C1x + C2
2
6
EI
a ≤ x ≤ L:
M = M A + RA x − M 0
d2y
= M = M A + RA x − M 0
dx 2
1
dy
EI
= M A x + RA x 2 − M 0 x + C3
2
dx
1
1
1
EIy = M A x 2 + RA x3 − M 0 x 2 + C3 x + C4
2
6
2
EI
dy


 x = 0, dx = 0  : 0 + 0 + C1 = 0


∴ C1 = 0
[ x = 0, y = 0] : 0 + 0 + 0 + C2 = 0
∴ C2 = 0
dy dy 

 x = a, dx = dx  :


M Aa +
1
1
RAa 2 = M Aa + RAa 2 − M 0a + C3
2
2
∴ C3 = M 0a
[ x = a, y = y ] :
1
1
1
1
1
1
M Aa 2 + RAa3 = M Aa 2 + RAa3 − M 0a 2 + (M 0a)(a) + C4 ∴ C4 = − M 0a 2
2
6
2
6
2
2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.31 (Continued)
[ x = L, y = 0] :
1
1
1
1
M A L2 + RA L3 − M 0 L2 + (M 0a)( L) − M 0a 2 = 0
2
6
2
2
1
1
1
1
( RB L + M 0 ) L2 + (− RB ) L3 − M 0 L2 + M 0aL − M 0a 2 = 0
2
6
2
2
RB =
Deflection at D:
3M 0a
3M 0  L  L
5M 0

(a − 2L) =
− 2L  = −
3
3  
6L
2L
2L  3  3

RB =
5M 0
↓ 
6L
L

 y at x = a = 3 


yD =
=
1 1
1
2
3
 M A x + RA x 
EI  2
6

1
EI
=−
2
3
 1  5M
1  5M 0  L  
 L 
0
L
M
−
+
+
+
 
0  

  
6  6L  3  
 3 
 2  6L
7M 0 L2
486EI
yD =
7M 0 L2
↑ 
486EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.32
Determine the reaction at the roller support and the deflection at
point D if a is equal to L/3.
SOLUTION
0 ≤ x ≤ a:
M = RA x
d2y
= M = RA x
dx 2
1
dy
EI
= RA x 2 + C1
2
dx
1
EIy = RA x3 + C1x + C2
6
EI
a ≤ x ≤ L:
M = RA x − P( x − a)
d2y
= M = RA x − P( x − a )
dx 2
dy
1
1
EI
= RA x 2 − P( x − a) 2 + C3
dx
2
2
1
1
3
EIy = RA x − P( x − a)3 + C3 x + C4
6
6
EI
[ x = 0, y = 0] : 0 + 0 + C2 = 0
∴ C2 = 0
dy dy 

 x = a, dx = dx  :


1
1
RAa 2 + C1 = RAa 2 − 0 + C3
2
2
∴ C1 = C3
[ x = a, y = y ] :
1
1
RAa3 + C1a + 0 = RAa3 − 0 + C1a + C4
6
6
∴ C4 = 0
dy


 x = L, dx = 0  :


1
1
RAL2 − P( L − a) 2 + C3 = 0
2
2
∴ C3 =
1
1
P( L − a) 2 − RA L2
2
2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.32 (Continued)
[ x = L, y = 0] :
1
1
1
1

RA L3 − P( L − a)3 +  P( L − a)2 − RA L2  (L) + 0 = 0
6
6
2
2


RA =
Deflection at D:
P
P  3
L3 
3
2
3
3
L
aL
a
L
L
(2
−
3
+
)
=
2
−
+


9 
2L3
2L3 
RA =
14
P ↑ 
27
L

 y at x = a = 3 


yD =
1
EI
 1  L 3
 L  
 RA   + C1   
 3  
 6  3 
=
1
EI
2
 1  14  L 3  1 
L
1  14  2   L  
  P   +  P  L −  −  P  L    
3
2  27    3  
 6  27  3 
 2 

=−
20 PL3
2187 EI
yD =
20 PL3
↓ 
2187 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.33
Determine the reaction at A and draw the bending moment
diagram for the beam and loading shown.
SOLUTION
Reactions are statically indeterminate.
Because of symmetry,
[ x = 0, y = 0]
[ x = L, y = 0]
L
dy
= 0 and V = 0 at x = .
2
dx
L

Use portion AC of beam.  0 < x ≤ 2 


dy
dy

 

 x = 0, dx = 0   x = L, dx = 0 

 

L


 x = 2 , V = 0


L dy


 x = 2 , dx = 0


dV
w
= − w = −2 0 x
dx
L
EI
dM
w
= V = − 0 x 2 + RA
dx
L
(1)
d2y
1 w0 3
= M =−
x + RA x + M A
2
3 L
dx
(2)
EI
dy
1 w0 4 1
=−
x + RA x 2 + M A x + C1
dx
12 L
2
EIy = −
(3)
1 w0 2 1
1
x + RA x3 + M A x 2 + C1x + C2
60 L
6
2
dy


 x = 0, dx = 0  :


0 = 0 + 0 + 0 + C1
C1 = 0
[ x = 0, y = 0] :
0 = 0 + 0 + 0 + 0 + C2
C2 = 0
L


 x = 2 , V = 0 :


−
w0  L 
  + RA = 0
L 2
L dy


 x = 2 , dx = 0  :


−
1 w0  L 
11
L
L
  +  w0 L    + M A + 0 = 0
12 L  2 
2 4
2
 2
2
4
RA =
(4)
wL

4
2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.33 (Continued)
1 
5
 1
M A = −2 
−
w0 L2 = − w0 L2

96
 32 192 
From (2), with x =
M A = −0.05208w0 L2 
L
,
2
3
MC = −
1 w0  L 
1
 L  5
w0 L12
  +  w0 L   −
3 L 2
4
 2  96
1
5 
1
 1
= −
+ −
w0 L2 =
w0 L2

32
 24 8 96 
M C = 0.03125w0 L2 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.34
Determine the reaction at A and draw the bendding moment
diagram for the beam and loading shown.
[ x = 0, y = 0]
[ x = L,
dy


 x = 0, dx = 0


y = 0]
dy


 x = L, dx = 0 


SOLUTION
Reactions are statically indeterminate.
By symmetry,
RB = RA ; M B = M A
dy
L
= 0 at x =
2
dx
RB = RA =
ΣFy = 0 : RA + RB − wL = 0
Over entire beam,
M = M A + RA x −
1
wL 
2
1 2
wx
2
d2y
1
1
= M A + wLx − wx 2
2
2
2
dx
dy
1
1
EI
= M A x + wLx 2 − wx3 + C1
dx
4
6
EI
dy


 x = 0, dx = 0  : 0 + 0 − 0 + C1 = 0 ∴ C1 = 0


L dy
1
1

 1
3
3
 x = 2 , dx = 0  : 2 M A L + 16 wL − 48 wL + 0 = 0


MA = −
M =−
1
1
1
wL2 + wLx − wx 2
12
2
2
1
wL2 
12
M = w[6 x ( L − x ) − L2 ] / 12 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.35
For the beam and loading shown, determine (a) the
equation of the elastic curve, (b) the slope at end A, (c) the
deflection of point C.
[ x = 0, y = 0]
[ x = L, y = 0]
SOLUTION
Reactions:
RA =
M0
↑,
L
RB =
M0
↓
L
0< x<a
M = RA x
a< x< L
M = RA x − M 0
Using singularity functions,
d2y
= M = R A x − M 0  x − a 0
dx 2
dy
1
= RA x 2 − M 0  x − a1 + C1
EI
dx
2
1
1
EIy = RA x3 − M 0  x − a 2 + C1x + C2
6
2
EI
[ x = 0, y = 0]
0 = 0 − 0 + 0 + C2
C2 = 0
[ x = L, y = 0]
1
1
RA L3 − M 0 ( L − a)2 + C1L + 0 = 0
6
2
1 M0 3 1
L + M 0b 2
6 L
2
M
C1 = 0 (3b 2 − L2 )
6L
C1L = −
(a)
Elastic curve:
y=
1 1 M0 3 1
M

x − M 0  x − a 2 + 0 (3b 2 − L2 ) x 

EI  6 L
2
6L

{
{
}
M0
x3 − 3L x − a 2 + (3b2 − L2 ) x 
6EIL
y =
}
dy
M0
=
3x 2 − 6L x − a1 + (3b 2 − L2 )
dx
6EIL
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.35 (Continued)
(b)
Slope at A:
 dy

 dx at x = 0 


θA =
(c)
Deflection at C:
M0
{0 − 0 + 3Lb2 − L3}
6EIL
θA =
M0
(3b2 − L2 )
6EIL

( y at x = a)
M0 3
{a − 0 + (3b2 − L2 )a}
6 EIL
M 0a 2
{a + 3b 2 − (a + b)2}
=
6 EIL
M 0a 2
=
{ a + 3b 2 − a 2 − 2ab − b2}
6 EIL
M 0a
=
{2b 2 − 2ab}
6 EIL
yC =
yC =
M 0ab
(b − a) ↑ 
3EIL
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.36
For the beam and loading shown, determine (a) the
equation of the elastic curve, (b) the slope at end A, (c) the
deflection of point C.
[ x = 0, M = 0]
[ x = L, M = 0]
[ x = 0,
[ x = L, y = 0]
y = 0]
SOLUTION
ΣM B = 0: − RA L + Pb = 0
RA =
Pb
L
dM
Pb
= V = R A − P  x − a 0 =
− P  x − a 0
dx
L
Pb
M =
x − P x − a1 + M A
L
d2y
Pb
=
x − P x − a
2
L
dx
dy
Pb 2 1
EI
=
x − P x − a 2 + C1
dx
2L
2
Pb 3 1
EIy =
x − P x − a 3 + C1x + C2
6L
6
EI
[ x = 0, y = 0]
C2 = 0
[ x = L, y = 0]
(a)
y=
Elastic curve:
Pb 3 1
L − P( L − a)3 + C1L = 0
6L
6
1R
1 Pb 2
C1 = −
(bL2 − b3 ) = −
( L − b2 )
6L
6 L
P
EI
1b 2
b 3 1

3
( L − b2 ) x 
 x −  x − a −
6
L
6
6
L


y =
(b)
P
{bx3 − L x − a 3 − b( L2 − b 2 ) x} 
6EIL
Slope at end A:
EI
dy
dx
x =0
= C1 = −
θA = −
Pb 2
(L − b2 )
6L
Pb 2
( L − b2 )
6 EIL
θA =
Pb 2
(L − b2 )
6EIL

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.36 (Continued)
(c)
Deflection at C:
Pb 3
Pba3 Pb 2
−
a + C1a =
( L − b 2 )a
6L
6L
6L
Pba 2
=
(a − L2 − b 2 )
6L
Pab 2
yC = −
( L − a 2 − b2 )
6EIL
Pab
=−
a 2 + 2ab + b 2 − a 2 − b 2
6EIL
EIyC =
{
=−
Pa 2b 2
3EIL
}
yC =
Pa 2b 2
↓
3EIL
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.37
For the beam and loading shown, determine the deflection at
(a) point B, (b) point C, (c) point D.
SOLUTION
ΣFy = 0: RA − P − P − P = 0
RA = 3P
ΣM A = 0: − M A − Pa − P(2a) − P(3a) = 0
M A = −6 Pa
dM
= V = 3P − P  x − a 0 − P  x − 2a 0
dx
d2y
EI 2 = M = 3Px − P  x − a1 − P x − 2a1 − 6 Pa
dx
dy 3 2 1
1
= Px − P  x − a 2 − P  x − 2a 2 − 6 Pax + C1
EI
dx 2
2
2
dy


∴ C1 = 0
 x = 0, dx = 0  : 0 − 0 − 0 − 0 + C1 = 0


1
1
1
EIy = Px3 − P x − a 3 − P x − 2a 3 − 3Pax 2 + C2
2
6
6
[ x = 0,
0 y = 0]: 0 − 0 − 0 − 0 + C2 = 0
Elastic curve:
y=
∴ C2 = 0
P  3
3x −  x − a3 −  x − 2a 3 − 18ax 2  
6 EI 
Pa3
[3 − 0 − 0 − 18]
6 EI
yB =
5Pa3

2 EI
(a)
x = a : yB =
(b)
x = 2a : yC =
Pa3
[24 − 1 − 0 − 72]
6 EI
yC =
49 Pa3

6 EI
(c)
x = 3a : yD =
Pa 3
[81 − 8 − 1 − 162]
6 EI
yD =
15Pa3

EI
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.38
For the beam and loading shown, determine the deflection at
(a) point B, (b) point C, (c) point D.
SOLUTION
dM
3P
=V =
− P  x − a 0 − P x − 2a 0 − P  x − 3a 0
dx
2
d2 y
3P
EI 2 = M =
x − P  x − a1 − P  x − 2a1 − P x − 3a1
2
dx
dy 3P 2 1
1
EI
=
x − P  x − a 2 − P  x − 2a 2
dx
4
2
2
1
− P  x − 3a 2 + C1
2
P 3 1
1
EIy = x − P x − a 3 − P  x − 2a 3
4
6
6
1
− P x − 3a 3 + C1 x + C2
6
[ x = 0,
y = 0 ] : 0 − 0 − 0 − 0 + 0 + C2 = 0
[ x = 4a, y = 0] : 16 Pa 3 −
Elastic curve:
y=
∴ C2 = 0
9
4
1
Pa 3 − Pa 3 − Pa 3 + 4aC1 = 0
2
3
6
∴ C1 = −
5
Pa 2
2
P  3
3x − 2 x − a3 − 2 x − 2a 3 − 2 x − 3a 3 − 30a 2 x  
12 EI 
Pa3
[3 − 0 − 0 − 0 − 30]
12 EI
(a)
x = a : yB =
(b)
x = 2a : yC =
Pa3
[24 − 2 − 0 − 0 − 60]
12 EI
(c)
x = 3a : yD =
Pa3
[81 − 16 − 2 − 0 − 90]
12 EI
yB =
yC =
9 Pa3

4 EI
19 Pa3

6 EI
yD =
9 Pa3

4 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.39
For the beam and loading shown, determine (a) the deflection
end A, (b) the deflection at point C, (c) the slope at end D.
SOLUTION
Since loads self-equilibrate,
RB = 0, RD = 0
(0 < x < 2a): M = −M 0
[ x = a, y = 0] [ x = 3a, y = 0]
(2a < x < 3a): M = −M 0 + M 0 = 0
Using singularity functions,
d2y
0
= M = − M 0 + M 0 x − 2a
2
dx
dy
1
EI
= −M 0 x + M 0 x − 2a + C1
dx
1
1
2
EIy = − M 0 x 2 + M 0 x − 2a + C1x + C2
2
2
EI
[ x = 3a, y = 0]
1
1
− M 0 (3a) 2 + M 0a 2 + C1(3a) + C2 = 0
2
2
3aC1 + C2 = 4M 0a 2
[ x = a, y = 0]
1
− M 0a 2 + 0 + C1a + C2 = 0
2
aC1 + C2 =
2aC1 =
Subtracting,
7
M 0a 2
2
C1 =
1
M 0a 2
2
7
M 0a
4
1
5
M 0a 2 − aC1 = − M 0a 2
2
4
M  1
1
7
5 
2
y = 0 − x 2 +
x − 2a + ax − a 2 
EI  2
2
4
4 
C2 =
dy
M 
= 0 − x + x − a
dx
EI 
(a)
Deflection at A:
yA =
1
+
7 
a
4 
( y at x = 0)
M 0a 2 
5
5 M 0a 2
,
 −0 + 0 + 0 −  = −
EI 
4
4 EI
yA =
5 M 0a 2
↓ 
4 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.39 (Continued)
(b)
Deflection at C:
yC =
(c)
M 0a 2  1 2
7
5  1 M 0a 2
 − (2) + 0 + (2) −  =
EI  2
4
4  4 EI
Slope at D:
θD =
( y at x = 2a)
yC =
1 M 0a 2
↑ 
4 EI
 dy

 dx at x = 3a 


M 0a 
7
1 M 0a
,
−3 + 1 +  = −
EI 
4
4 EI
θD =
1 M 0a
4 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.40
For the beam and loading shown, determine (a) the deflection at
end A, (b) the deflection at point C, (c) the slope at end D.
[ x = a, y = 0]
[ x = 3a, y = 0]
SOLUTION
RD = 0
RB = 2 P ↑,
Reactions:
(0 < x < a):
V = −P
(a < x < 2a):
V = − P + 2P
(2a < x < 3a):
V = − P + 2P − P
Using singularity functions,
dM
= V = − P + 2P x − a 0 − P x − 2a 0
dx
M = − Px + 2P x − a1 − P x − 2a1 + M A
MA = 0
M = 0 at x = 0
But
EI
d2y
= M = − Px + 2P x − a
dx 2
EI
dy
1
= − Px 2 + P x − a
dx
2
2
1
1
EIy = − Px3 + P x − a
6
3
1
−
3
− P x − 2a
1
P x − 2a
2
−
2
1
P x − 2a
6
1
(1)
+ C1
3
(2)
+ C1x + C2
(3)
[ x = a, y = 0]
1
− Pa3 + 0 − 0 + C1a + C2 = 0
6
aC1 + C2 =
[ x = 3a, y = 0]
1
1
1
− P(3a)3 + P(2a)3 − Pa3 + C1(3a) + C2 = 0
6
3
6
3aC1 + C2 = 2Pa 2 (5)
Eq (5) – Eq (4)
2C1a +
C2 =
y =
11 2
Pa
6
C1 +
1 3
Pa (4)
6
11 2
Pa
12
1 2
3
Pa − aC1 = − Pa3
6
4
P  1 3 1
− x + x − a
EI  6
3
dy
P  1 2
=
− x + x − a
dx
EI  2
2
3
−
−
1
x − 2a
6
1
x − 2a
2
2
3
+
+
11 2
3 
a x − a3 
12
4 
11 2 
a 
12 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.40 (Continued)
(a)
Deflection at A:
( y at x = 0)
yA =
(b)
Deflection at C:
Slope at D:
yA =
3 Pa3
↓ 
4 EI
( y at x = 2a)
yC =
(c)
Pa 3 
3
3 Pa 3
 −0 + 0 − 0 + 0 −  = −
EI 
4
4 EI
Pa 3  1 3 1 3
11
3
(2) − 
 − (2) + (1) − 0 +
EI  6
3
12
4
yC =
1 Pa3
↑ 
12 EI
 dy

 dx at x = 3a 


θD =
Pa 2  1 2
1 2 11 
1 Pa 2
2
 − (3) + (2) − (1) +  = −
2
12 
12 EI
EI  2
θD =
1 Pa 2
12 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.41
For the beam and loading shown, determine (a) the equation of the
elastic curve, (b) the deflection at the midpoint C.
SOLUTION
By symmetry,
RA = RB = wa
w( x) = w − w x − a 0 + w x − 3a 0
dV
= −w( x) = −w + w x − a 0 − w x − 3a 0
dx
dM
= V = RA − wx + w x − a1 − w x − 3a1
dx
1
1
1
M = M A + RA x − wx 2 + w x − a 2 − w x − 3a 2
2
2
2
(with M A = 0)
1
1
1
d2y
= M = wax − wx 2 + w x − a 2 − w x − 3a 2
2
2
2
dx 2
1
dy
1
1
1
EI
= wax 2 − wx3 + w x − a 3 − w x − 3a 3 + C1
6
dx
2
6
6
1
1
1
1
EIy = wax3 −
wx 4 +
w x − a 4 −
w x − 3a 4 + C1x + C2
6
24
24
24
EI
[ x = 0, y = 0] :
0 − 0 + 0 − 0 + 0 + C2 = 0 ∴ C2 = 0
[ x = 4a, y = 0] :
(a)
1
1
1
1
wa(4a)3 −
w(4a) 4 +
w(3a) 4 −
w(a) 4 + C1(4a) = 0
6
24
24
24
5
∴ C1 = − wa3
6
Equation of elastic curve:
y=
(b)
w 1 3
1 4
1
1
5

ax −
x +
 x − a 4 −
 x − 3a 4 − a3 x  

EI  6
24
24
24
6

( y at x = 2a)
Deflection at C:
yC =
wa 4
EI
=−
1
1 4
5 
1 3
4
 6 (2) − 24 (2) + 24 (1) − 0 − 6 (2) 


23 wa 4
24 EI
yC =
23 wa 4
↓ 
24 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
P
PROBLEM
9.42
Foor the beam and loading shown, determine (a) the eqquation of the
ellastic curve, (b) the deflection at point B, (c) the deflection
d
at
pooint D.
SOLUTION
Use free body ABCD with the distributed loaads replaced by equivalent concentrated loads.
 wL  3L   wL  L 
ΣM C = 0: − RA L + 
  − 
  = 0
 2  4   2  4 
1
RA = wL
4
 wL  L   wL  5L 
ΣM A = 0: RC L − 
  − 
  = 0
 2  4   2  4 
3
RC = wL
4
0
dV
L
= −w = −w + w x −
dx
2
−w x−L
0
Integrating and adding terms to account for thhe reactions,
dM
L
= V = −wx + w x −
2
dx
EI
1
− w x − L1 + RA + RC  x − L 0
2
1
1
L
d2y
= M = − wx 2 + w x −
2
2
2
2
dx
EI
1
1
dy
L
= − wx3 + w x −
6
6
2
dx
EIy = −
3
L
1
1
wx 4 +
w x−
24
24
2
−
1
w x − L 2 + RA x + RC  x − L1
2
1
1
1
w x − L 3 + RA x 2 + RC  x − L 2 + C1
6
2
2
−
4
−
1
1
1
w x − L 4 + RA x3 + RC  x − L 3 + C1x + C2
24
6
6
[ x = 0, y = 0] − 0 + 0 − 0 + 0 + 0 + 0 + C2 = 0
[ x = L, y = 0] −
1
1 L
w 
wL4 +
24
24  2 
−0+
+
4
1  wL  3

 L + 0 + C1L + 0 = 0
6 4 
L
1
1
EIy = − wx 4 +
w x−
24
24
2
4
+
C2 = 0
C1 = −
1
wL3
384
1
w x − L 4
244
1  wL  3 1  3wL 
1
3
wL3 x

x + 
  x − L −
6 4 
6 4 
384
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.42 (Continued)
(a)
Elastic curve:
y =
(b)
(c)
w
24 EI
w
24EI
=−
4
−  x − L 4 + Lx3 + 3L x − L 3 −
3
  L 4
L
 1  L  
−
+
−
+
+ 0 −  L3   
0
0
(
L
)
  


2
 16  2  
  2 
1 3 
L x  
16

yB =
wL4
↑ 
768EI
yD =
5wL4
↓ 
256EI
3L 

 y at x = 2 


Deflection at D:
yD =
 4
L
− x + x −
2

L

 y at x = 2 


Deflection at B:
yB =
w
24 EI
4
3
3
  3L 4
L
 3L 
L
 1  3L  
4
−
+
−
+
+
L
(
L
)
(3
L
)
  
 
 
  −  L   
2
 2 
2
 16  2  
  2 
5wL4
256EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.43
For the beam and loading shown, determine (a) the equation of the elastic
curve, (b) the deflection at the midpoint C.
SOLUTION
k =
Distributed loads:
(1)
w1( x) = w0 − kx
2w0
L
(2) w2 ( x) = k x −
 w L  5 
ΣM B = 0:  0  L  − RA L = 0
 4  6 
L
2
w( x) = w0 − kx + k x −
1
= w0 −
RA =
L
2
5
w0 L ↑
24
2w0
2w0
L
x+
x−
L
L
2
dV
2w0
2w0
L
= −w = −w0 +
x−
x−
dx
L
L
2
5
1
1 w0 3 1 w0
d2y
L
=M =
w0 Lx − w0 x 2 +
x −
x−
2
24
2
3 L
3 L
2
dx
dy
5
1
1 w0 4
1 w0
L
EI
=
w0 Lx 2 − w0 x3 +
x −
x−
dx
48
6
12 L
12 L
2
EIy =
2
3
4
+ C1
L
5
1
1 w0 5
1 w0
w0 Lx3 −
w0 x 4 +
x −
x−
144
24
60 L
60 L
2
5
+ C1x + C2
C2 = 0
[ x = 0, y = 0] :
5
5
1
1
1 w0  L 
w0 L4 −
w0 L4 +
w0 L4 −
  + C1L = 0
144
24
60
60 L  2 
Equation of elastic curve:
[ x = L, y = 0] :
(a)
(b)
Deflection at C:
yC =
w0 L4
5760EI
1
1
dM
5
w
w
L
=V =
w0 L − w0 x + 0 x 2 − 0 x −
dx
24
L
L
2
EI
1

L
y = w0 96 x5 − 96 x −
2

L

 y at x = 2 


5
C1 = −
53
w0 L3
5760

− 240 Lx 4 + 200 L2 x3 − 53L4 x  / 5760 EIL 

240 200 53 
3w0 L4
 96
−
0
−
+
−
=
−


16
8
2 
1280 EI
 32
yC =
3w0 L4
↓
1280EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.44
For the beam and loading shown, determine (a) the equation of the elastic
curve, (b) the deflection at the midpoint C.
SOLUTION
k1 =
Distributed loads:
2w0
L
(1)
w1( x) = w0 − k1x
(2)
w2 ( x) = k2 x −
L
2
k2 =
4w0
L
1
 w L  5   w L  L 
ΣM B = 0:  0  L  +  0   + RA L = 0
 4  6   4  6 
wL
RA = 0 ↑
4
w( x) = w0 − k1x + k2 x −
= w0 −
L
2
1
2w0
4w0
L
x+
x−
2
L
L
1
2w0
4w0
dV
L
= −w = −w0 +
x−
x−
dx
L
L
2
1
2w0
dM
wL
w
L
= V = 0 − w0 x + 0 x 2 −
x−
4
2
dx
L
L
d2y
1
1
1 w0 3 2 w0
L
EI 2 = M = w0 Lx − w0 x 2 +
x −
x−
4
2
3 L
3 L
2
dx
EI
dy 1
1
1 w0 4 1 w0
L
x −
x−
= w0 Lx 2 − w0 x3 +
6
12 L
6 L
2
dx 8
EIy =
2
3
4
+ C1
1
1
1 w0 5
1 w0
L
w0 Lx3 −
w0 x 4 +
x −
x−
24
24
60 L
30 L
2
5
+ C1x + C2
[ x = 0, y = 0] :
C2 = 0
[ x = L, y = 0] :
1
1
1
1 w0  L 
w0 L4 −
w0 L4 +
w0 L4 −
  + C1L = 0
24
24
60
30 L  2 
5
C1 = −
1
w0 L3
64
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.44 (Continued)
(a)
Equation of elastic curve:

L
y = w0 16 x5 − 32 x −
2

(b)
Deflection at C:
yC =

 y at

x=
5

− 40Lx 4 + 40L2 x3 − 15L4 x  960 EIL 

L
2 
w0 L4  1
5
15 
3w0 L4
 −0− +5−  = −
960EI  2
2
2
640EI
yC =
3w0 L4
↓ 
640EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.45
For the beam and loading shown, determine (a) the slope at
end A, (b) the deflection at point C. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN, lengths in m
M D = 0:
− 1.6 RA + (9.6)(0.8) + (20)(0.4) = 0
RA = 9.8 kN
w( x) = 12 x − 0.4
0
− 12 x − 1.2
dV
= −w( x) = −12 x − 0.4
dx
dM
1
1
= V = 9.8 − 12 x − 0.4 + 12 x − 1.2 − 20 x − 1.2
dx
0
0
kN/m
+ 12 x − 1.2
0
0
kN/m
kN
d2y
2
2
1
= M = 9.8 x − 6 x − 0.4 + 6 x − 1.2 − 20 x − 1.2
kN ⋅ m
2
dx
dy
EI
= 4.9 x 2 − 2 x − 0.4 3 + 2 x − 1.2 3 − 10 x − 1.2 2 + C1
kN ⋅ m 2
dx
1
1
10
EIy = 1.63333x3 −  x − 0.4 4 +  x − 1.2 4 −
 x − 1.2 3 + C1x + C2 kN ⋅ m3
2
2
3
EI
[ x = 0, y = 0] : 0 − 0 + 0 − 0 + 0 + C2 = 0
[ x = 1.6, y = 0] : (1.63333)(1.6)3 −
C2 = 0
1
1
10
(1.2) 4 + (0.4) 4 − (0.4)3 + C1(1.6) + 0 = 0
2
2
3
C1 = −3.4080 kN ⋅ m 2
Data:
E = 200 × 109 Pa, I = 6.83 × 106 mm 4 = 6.83 × 10−6 mm 4
EI = (200 × 104 )(6.83 × 10−6 ) = 1.366 × 106 N ⋅ m 2 = 1366 kN ⋅ m 2
(a)
Slope at A:
EI
 dy
 dx

at

x = 0

dy
= 0 − 0 + 0 − 0 − 3.4080 kN ⋅ m 2
dx
θA = −
3.4080
= − 2.49 × 10−3 rad
1366
θ A = 2.49 × 10−3 rad

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.45 (Continued)
(b)
x = 1.2 m)
Deflection at C:
( y at
EIyC = (1.63333)(1.2)3 −
1
(0.8) 4 + 0 − 0 − (3.4080)(1.2) + 0
2
= −1.4720 kN ⋅ m3
yC = −
1.4720
= −1.078 × 10−3 m
1366
yC = 1.078 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.46
For the beam and loading shown, determine (a) the slope at
end A, (b) the deflection at the midpoint C. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN, lengths in meters.
Σ M B = 0: −5.4 RA − (1.8) (6.2 + 10.8) = 0
RA = 5.6667 kN
w( x) = 3 x − 1.8 0
dV
= − w( x) = −3 x − 1.8 0
dx
dM
= V = 5.6667 − 3 x − 1.81 − 6.2 x − 3.6 0
dx
d2y
3
= M = 5.6667 x −  x − 1.8 2 − 6.2 x − 3.61
2
dx 2
dy
1
EI
= 2.8333 x 2 −  x − 1.8 3 − 3.1 x − 3.6 2 + C1
2
dx
1
3
EIy = 0.9444 x −  x − 1.8 4 − 1.0333 x − 3.6 3 + C1 x + C2
8
[ x = 0, y = 0]:
0 − 0 − 0 + 0 + C2 = 0
C2 = 0
1
3
4
[ x = 5.4, y = 0]: (0.9444)(5.4) − (3.6) − 1.0333(1.8)3 + C1 (5.4) + 0 = 0
8
kN ⋅ m
EI
kN ⋅ m 2
kN ⋅ m 3
C1 = −22.535 kN ⋅ m 2
E = 200 × 109 Pa, I = 129 × 106 mm 4 = 129 × 10−6 m 4
Data:
EI = (200 × 109 )(129 × 10−6 ) = 25.8 × 106 N ⋅ m 2 = 25.8 × 103 kN ⋅ m 2
(a)
 dy

 dx at x = 0 


Slope at A:
EI
dy
= 0 − 0 − 0 − 22.535 kN ⋅ m 2
dx
22.535
θA = −
= −873 × 10−6
25.8 × 103
θ A = 0.873 × 10−3 rad

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.46 (Continued)
(b)
( y at x = 2.7 m)
Deflection at C:
1
EIyC = (0.9444)(2.7)3 − (0.9) 4 − 0 − (22.585)(2.7) + 0
8
= −42.337 kN ⋅ m3
yC = −
42.337
= −1.641 × 10−3 m
25.8 × 103
yC = 1.641 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.47
For the beam and loading shown, determine (a) the slope at
end A, (b) the deflection at the midpoint C. Use E = 200 GPa.
SOLUTION
Distributed loads:
(1) w1( x) = w0 − kx
(2) w2 = k x − 1
1
w0 = 48 kN/m, k = 48 kN/m 2
 2
ΣM B = 0: − 2RA + (24) 1  + (8)(1) = 0
 3
RA = 24 kN ↑
1
w( x) = w0 − kx + k x − 1 = 48 − 48x + 48 x − 1
1
dV
1
= −w = −48 + 48x − 48 x − 1
kN/m
dx
dM
2
0
= V = 24 − 48x + 24 x 2 − 24 x − 1 − 8 x − 1
kN
dx
d2y
3
1
kN ⋅ m
= M = 24 x − 24 x 2 + 8 x3 − 8 x − 1 − 8 x − 1
2
dx
dy
= 12 x 2 − 8 x3 + 2 x 4 − 2 x − 1 4 − 4 x − 1 2 + C1 kN ⋅ m 2
EI
dx
2
2
4
EIy = 4 x3 − 2 x 4 + x5 −  x − 1 5 −  x − 1 3 + C1x + C2 kN ⋅ m3
5
5
3
EI
[ x = 0, y = 0]: 0 − 0 + 0 − 0 − 0 + 0 + C2 = 0
[ x = 2, y = 0]: 4(2)3 − 2(2) 4 +
Data:
∴ C2 = 0
2 5 2 5 4 3
83
(2) − (1) − (1) + C1(2) = 0 ∴ C1 = − kN ⋅ m 2
5
5
3
15
I = 5.12(106 ) mm 4 = 5.12(10−6 ) m 4
E = 200(106 ) kN/m 2
EI = (200 × 106 )(5.12 × 10−6 ) = 1024 kN ⋅ m 2
(a)
Slope at A:
 dy
 dx

at

x = 0

EIθ A = 0 − 0 + 0 − 0 − 0 −
θA = −
83
kN ⋅ m 2
15
83/15
= −5.4036 × 10−3 rad
1024
θ A = −5.40 × 10−3 rad

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.47 (Continued)
(b)
Deflection at C:
( y at
EIyC = 4(1)3 − 2(1) 4 +
yC = −
x = 1 m)
2 4
83
(1) − 0 − 0 −
(1) = −3.1333 kN ⋅ m3
5
15
3.1333
= −3.0599 × 10−3 m
1024
yC = 3.06 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.48
For the timber beam and loading shown, determine (a) the slope at
end A, (b) the deflection at the midpoint C. Use F = 12 GPa.
SOLUTION
Units:
Forces in kN, lengths in meters.
I=
1
(50)(150)3 = 14.0625 × 106 mm 4
12
= 14.0625 × 10 −6 m 4
EI = (12 × 109 )(14.0625 × 10 −6 )
= 168.75 × 103 N ⋅ m 2 = 168.75 kN ⋅ m 2
M D = 0: − 2 RA + (1.5)(4) + (0.5)(5) = 0
RA = 4.25 kN
w( x ) = 5 x − 1 0
kN ⋅ m
dV
= − w = −5 x − 1 0 kN/m
dx
dM
= V = −5 x − 11 + 4.25 − 4 x − 0.5 0
dx
kN
d2y
5
= M = −  x − 1 2 + 4.25 x − 4 x − 0.51
2
2
dx
dy
5
EI
= −  x − 1 3 + 2.125 x 2 − 2 x − 0.5 2 + C1
dx
6
5
2.125 3 2
EIy = −  x − 1 4 +
x −  x − 0.5 3 + C1 x + C2
24
3
3
[ x = 0, y = 0]
− 0 + 0 − 0 + 0 + C2 = 0
EI
[ x = 2m, y = 0]
kN ⋅ m
kN ⋅ m 2
kN ⋅ m3
C2 = 0
 5 
 2.125  3  2 
3
−   (1) 4 + 
 (2) −   (1.5) + 2C1 = 0
 24 
 3 
3
C1 = −1.60417 kN ⋅ m 2
(a)
Slope at end A:
 dy

 dx at x = 0 


 dy 
EI   = −0 + 0 − 0 + C1
 dx  A
C1 −1.60417
 dy 
=
= −9.51 × 10−3
  =
168.75
 dx  A EI
θ A = 9.51 × 10−3 rad

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.48 (Continued)
(b)
Deflection at midpoint C:
( y at x = 1 m)
 2.125  3  2 
3
EIyC = −0 + 
 (1) −   (0.5) + (−1.60417)(1)
3
3


 
= −979.17 × 10−3 kN ⋅ m3
yC =
−979.17 × 10−3
= −5.80 × 10−3 m
168.75
yC = 5.80 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.49
For the beam and loading shown, determine (a) the reaction at
the roller support, (b) the deflection at point C.
[ x = 0, y = 0]
dy


 x = L, dx = 0 


[ x = L, y = 0]
SOLUTION
1
,
2
For
0≤x≤
For
L
≤ x ≤ L,
2
M = RA x
M = RA x − M 0
Then
EI
d2y
L
= M = RA x − M 0 x −
2
2
dx
EI
1
dy
L
= RA x 2 − M 0 x −
2
2
dx
EIy =
1
1
1
L
R A x3 − M 0 x −
6
2
2
+ C1
2
+ C1x + C2
[ x = 0, y = 0] 0 − 0 + 0 + C2 = 0
C2 = 0
dy


 x = L, dx = 0 


1
L
RA L2 − M 0   + C1 = 0
2
2
C1 =
[ x = L, y = 0]
1
1
1
L
RA L3 − M 0   + (M 0 L − RA L2 ) L + 0 = 0
6
2
2
2
 
1
( M 0 L − RA L2 )
2
2
−
(a)
0
1
3
RA L3 + M 0 L2 = 0
3
8
M A = 0,
Reaction at A:
C1 =
EIy =
RA =
9M 0
↑ 
8L
1
1
 9M 0  2 
M 0L − 
 (L ) = − M 0L
2
8
L
16



1  9M 0  3 1
L

 x − M0 x −
6  8L 
2
2
2
−
1
M 0 Lx + 0
16
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.49 (Continued)
y =
Elastic curve:
(b)
M 0  9 3 1
L
 x − L x−
EIL  8
2
2
2
−
1 2 
L x
16

L

 y at x = 2 


Deflection at point C:
3
M 0  1  9  L 
 1 2  L  
yC =
    − 0 −  L   
EIL  6  8  2 
 16  2  
=−
M 0 L2
128EI
yC =
M 0 L2
↓ 
128 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.50
For the beam and loading shown, determiine (a) the reaction at the
roller support, (b) the deflection at point C.
[ x = 0, y = 0]
[ x = L, y = 0]
dy


 x = 0, dx = 0 


SOLUTION
RA = P − RB
ΣFy = 0: RA + RB − P = 0
ΣM A = 0: − M A − P
L
+ RB L = 0
2
M A = RB L −
1
PL
2
Reactions are statically indeterm
minate.
dM
L
= V = RA − P x −
dx
2
EI
0
1
d2y
L
= M = M A + RA x − P x −
2
dx 2
EI
2
dy
L
1
1
= M A x + RA x 2 − P x −
dx
2
2
2
EIy =
+ C1
L
1
1
1
M A x 2 + RA x3 − P x −
2
6
6
2
dyy


0 + 0 + 0 + C1 = 0
 x = 0, dxx = 0


[ x = 0, y = 0] 0 + 0 + 0 + 0 + C2 = 0
3
+ C1x + C2
C1 = 0
C2 = 0
3
1
1
1 L
[ x = L, y = 0]
M A L2 + RA L3 − P   + 0 + 0 = 0
2
6
6 2
1
1
1
 2 1
3
PL3 = 0
 RB L − PL  L + ( P − RB )L −
2
2
6
48

1  3
1 1
1 1
3
 2 − 6  RB L =  4 − 6 + 48  PL




1
5
RB =
P
3
48
(a) RB =
5
P↑ 
16
5
11
P=
P
16
16
5
1
3
=
PL − PL = − PL
16
2
16
RA = P −
MA
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.50 (Continued)
(b)
L

 y at x = 2 


Deflection at C:
1
yC =
EI
=
2
3
 1
1 L
L
 M A   + RA   + 0 + 0 +
6 2
2
 2

0

PL3  1  3  1   1  11  1  
7 PL3
  −   +      = −
EI  2  16  4   6  16  8  
168 EI
yC =
7 PL3
↓
168 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.51
For the beam and loading shown, determine (a) the reaction at the roller
support, (b) the deflection at point B.
SOLUTION
ΣFy = 0: RA + RD = 0
RA = − RD
ΣM A = 0: − M A + M 0 − M 0 + RL = 0
M A = RD L
M ( x) = M A + R A x − M 0 x −
L
4
L
d2y
EI 2 = RD L − RD x − M 0 x −
4
dx
0
+ M0 x −
0
+ M0
1
L
dy
= RD Lxx − RD x 2 − M 0 x −
2
4
dx
EI
dyy


 x = 0, dxx = 0 


3L
4
3L
x−
4
1
+ M0 x −
0
0
3L
4
1
+ C1
0 − 0 − 0 + 0 + C1 = 0
C1 = 0
EIy =
1
1
1
L
RD Lx 2 − RD x3 − M 0 x −
2
6
2
4
1
+
1
3L
M0 x −
2
4
+ C2
C2 = 0
[ x = 0, y = 0]
0 − 0 − 0 + 0 + C2 = 0
[ x = L, y = 0]
1
1
1
1
 3L 
L
RD L3 − RD L3 − M 0   + M 0   = 0
2
6
2
2
 4 
4
2
(a)
1
2
RD =
Reaction at D:
EIy =
1  3M 0  2 1  3M 0  3 1
L

 Lx − 
 x − M0 x −
2  4L 
6  4L 
2
4
2
+
1
3L
M0 x −
2
4
3M0
↑ 
4L
2
Elastic curve:
y =
(b)
M 0  3 2 1 3 1
L
 Lx − x − L x −
EIL  8
8
2
4
2
+
1
3L
L x−
2
8
2



L

 y at x = 4 


Deflection at point B:
yB =
2
3

M 0  3   L 
 1  L 
 L    −     − 0 + 0 
EIL  8   4 
 8  4 

yB =
11 M 0 L2
↑
512EI
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBL
LEM 9.52
For the beeam and loading shown, determine (a) the reaction at
a the roller
support, (b) the deflection at point B.
SOLUTION
ΣFy = 0:
ΣM A = 0:
RA − P − P + RD = 0
−M A −
RA = 2P − RD
PL
L 2PL
−
+ RD L = 0
3
3
M A = RD L − PL
L
dM
= V = RA − P x −
dx
3
EI
0
2L
−P x−
3
L
d2y
= M = M A + RA x − P x −
2
3
dx
EI
L
dy
1
1
= M A x + RA x 2 − P x −
dx
2
3
2
0
1
−P x−
2
−
L
1
1
1
M A x 2 + R A x3 − P x −
2
6
6
3
3
−
[ x = 0, y = 0] 0 + 0 − 0 − 0 + C2 = 0
EIy =
1
1
1
L
M A x 2 + RA x3 − P x −
2
6
6
3
1
2
1
2L
P x−
2
3
dy


 x = 0, dx = 0 0 + 0 − 0 − 0 + C1 = 0


EIy =
2L
3
+ C1
C1 = 0
1
2L
P x−
6
3
3
+ C2
C2 = 0
3
−
1
2L
P x−
6
3
3
3
[ x = L, y = 0]
3
1
1
1  2L 
1 L
( RD L − PL) L2 + (2P − RD ) L3 − P 
 − P  = 0
2
6
6  3 
6 3
1
2 3
RD L3 − PL
L =0
3
9
(a)
RD =
Reaction at D:
MA =
2
1
PL − PL = − PL
3
3
EIy =
1 1
L
 2 14  2 1
 − PL  x +  P  x − P x −
2 3 
6 3 
6
3
RA = 2P −
2
P↑ 
3
2
4
P= P
3
3
3
−
1
2L
x−
6
3
3
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.52 (Continued)
(b)
P
EI
 1 2 2 3 1
L
x−
 − Lx + x −
6
9
6
3

Elastic curve:
y =
Deflection at B:
L

 y at x = 3 


yB =
P
EI
=−
3
−
1
2L
x−
6
3
3



  1  L 2 2  L 3

−  L   +   − 0 − 0 
9 3 
  6  3 

5 PL3
486 EI
yB =
5 PL3
↓ 
486 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.53
For the beam and loading shown, determine (a) thhe reaction at
point C, (b) the deflection at point B. Use E = 20
00 GPa.
[ x = 0, y = 0]
[ x = 8, y = 0]
dy


 x = 0, dx = 0


SOLUTION
Units:
Forces in kN; lengths in m.
ΣFy = 0: RA − 70 + RC = 0
RA = 70 − RC
kN
M A = 0: − M A − (700)(2.5) + 8RC = 0
M A = 8RC − 175
kN
N⋅m
Reactions are statically indeterminate.
w( x) = 14 − 14  x − 5  0
kN/m
dV
4  x − 5 0
= −w = −14 + 14
dx
dM
= V = RA − 14 x + 14 x − 51
dx
EI


d2y
= M = M A + RA x − 7 x 2 + 7 x − 5 2
dx 2
EI
dy
1
7
7
= M A x + RA x 2 − x3 +  x − 5 3 + C1
dx
3
3
2
EIy =
1
1
7 4
7
M A x 2 + RA x 3 −
x +  x − 5 4 + C1x + C2
2
6
12
12

dy


 x = 0, dx = 0



[ x = 0, y = 0]

[ x = 8, y = 0]
0 + 0 + 0 + 0 + C1 = 0
0 + 0 + 0 + 0 + 0 + C2 = 0
kN/m
kN
kN ⋅ m
kN ⋅ m 2 
kN ⋅ m3 
C1 = 0 
C2 = 0 
1
1
7
7
M A (8) 2 + RA (8)3 − (8) 4 + (3) 4 + 0 + 0 = 0 
2
6
12
12
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.53 (Continued)
32(8 RC − 175) +


170.667 RC = 5600 −
(a)
Reaction at C:
512
28105
(70 − RC ) −
= 0
6
12
35840 28105
+
= 1968.75
6
12
RC = 11.536 kN ↑ 
RC = 11.546 kN ↑ 
M A = (8) (11.536) − 175 = −82.715 kN ⋅ m

RA = 70 − 11.536 = 58.464 kN
E = 200 × 109 Pa
Data:

I = 216 × 106 mm 4 = 216 × 10−6 m 4
EI = (200 × 109 )(216 × 10−6 ) = 43.2 × 106 N ⋅ m 2 = 43200 kN ⋅ m 2
(b)
( y at x = 5 m)
Deflection at B:
EIyB =
1
1
7
(−82.715)(5)2 + (58.464)(5)3 − (5)4 = −180.52 kN ⋅ m3
2
6
12
yB = −
180.52
= −4.18 × 10−3 m
43200
yB = 4.18 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.54
For the beam and loading shown, determine (a) thhe reaction at
point A, (b) the deflection at point C. Use E = 200 GPa.
SOLUTION
36 kN/m
= 20 kN/m 2
1.8 m
w( x) = 20 x − 20 x − 1.81 kN/m
m
k=
dV
dx
dM
dx
d2y
EI 2
dx
dy
EI
dx
= − w( x) = −20 x + 20 x − 1.81 kN/m
= V = RA − 10 x 2 + 10 x − 1.8 2 kN
10 3 10
x +  x − 1.8 3 kN ⋅ m
3
3
1
5
5
= RA x 2 − x 4 +  x − 1.8 4 + C1 kN ⋅ m3
2
6
6
1
1
1
EIy = RA x3 − x5 +  x − 1.8 5 + C1 x + C2 kN ⋅ m3
6
6
6
= M = RA x −
[ x = 0, y = 0] : 0 − 0 + 0 + 0 + C2 = 0
∴ C2 = 0
dy
5
5

 1
2
4
4
 x = 3.6 m, dx = 0  : 2 RA (3.6) − 6 (3.6) + 6 (1.8) + C1 = 0


∴ C1 = 131.22 − 6.48RA kN ⋅ m 2
1
1
1
RA (3.6)3 − (3.6)5 + (1.8)5 + (131.22 − 6.48 RA ) (3.6) = 0
[ x = 3.6 m, y = 0]:
6
6
6
N
RA = 24.097 kN
(7.776 − 23.328) RA = −374.76
(a)
RA = 24.1 kN 
Reaction at A:
C1 = 131.22 − 6.48(24.097) = −24.929 kN ⋅ m
Data:
E = 200 × 1006 kPa
2
I = 48.9 × 10−6 m 4
EI = (200 × 106 )(48.9 × 10 −6 ) = 9780 kN ⋅ m 2
(b)
( y at x = 1.8 m)
Deflection at C:
1
1
(24.0997)(1.8)3 − (1.8)5 + 0 − 24.929(1.8)
6
6
= −24.5999 kN ⋅ m3
24.59
99
yC = −
= −2.515 × 10−3 m
97800
EIyC =
yC = 2.5 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.55
For the beam and loading shoown, determine (a) the
reaction at point A, (b) the defflection at point C. Use
E = 200 GPa.
[ x = 0, y = 0]
[ x = 3, y = 0]
[ x = 0, M = 0]
dy


 x = 3, dx = 0 


SOLUTION
Units:
Forces in kN, lengths in m
0
w( x) = 70 x − 0.75 − 70 x − 2.25
0
dV
0
= − w( x) = −70 x − 0.75 + 70 x − 2.25
dx
0
kN 2 /m
dM
1
1
= V = RA − 70 x − 0.75 + 70 x − 2.25 kN
dx
d2y
2
2
EI 2 = M = RA x − 35 x − 0.75 + 35 x − 2.25 kN ⋅ m
dx
dy 1
35
35
EI
= RA x 2 −  x − 0.75 3 +  x − 2.25 3 + C1 kN ⋅ m 2
dx 2
3
3
1
35
35
EIy = RA x3 −  x − 0.75 4 +  x − 2.25 4 + C1 x + C2 kN ⋅ m3
6
12
12
[ x = 0, y = 0] : 0 + 0 + 0 + 0 + C2 = 0 ∴ C2 = 0
dy
35
35
3

 1
2
3
 x = 3, dx = 0  : 2 R A (3) − 3 (2.25) + 3 0.75 + C1 = 0


C1 = 128 − 4.5 R A kN ⋅ m 2
[ x = 3, y = 0]:
1
35
35
R A (3)3 − (2.25) 4 + (0.75)9 + (128 − 4.5 RA )(3) + 0 = 0
6
12
12
9 RA = 310.2
C1 = 128 − 4.5(344.5) = −27.25 kN ⋅ m
Data:
R A = 34.5 kN 
2
8 × 10 −6 m 4
E = 200 GPa, I = 82.7
EI = (200 × 106 )(82.7 × 10−6 ) = 16540 kN ⋅ m 2
(b)
Deflection at C:
( y at x = 1.5 m)
1
355
(34.5)(1.5)3 − (0.75) 4 + 0 − (27.25)(1.5) = −22.39
6
122
22.39
yC = −
= −1.35 × 10−3 m
16540
EIyC =
yC = 1.35 mm 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.56
For the beam shown and knowing that P = 40 kN,
determine (a) the reaction at point E, (b) the deflection at
point C. Use E = 200 GPa.
SOLUTION
Units: Forces in kN; lengths in m.
 Fy = 0: RA − 40 − 40 − 40 + RE = 0
RA = 120 − RE
kN
M A = 0: − M A − 20 − 40 − 60 + 2RB = 0
kN
N⋅m
M A = 2RE − 120
Reactions are statically indeterminate.
dM
= V = RA − 40 x − 0.5
dx
0
− 40 x − 1
0
− 40 x − 1.5
0
d2y
1
1
1
= M = M A + RA x − 40 x − 0.5 − 40 x − 1 − 40 x − 1.5
dx 2
1
dy
2
2
2
= M A x + RA x 2 − 20 x − 0.5 − 20 x − 1 − 20 x − 1.5 + C1
EI
2
dx
1
1
20
20
20
3
3
3
2
x − 0.5 −
x −1 −
x − 1.5 + C1x + C2
EIy = M A x + RA x3 −
2
6
3
3
3
EI
dy
= 0]
dx
[ x = 0, y = 0]
[ x = 0,
[ x = 2, y = 0]
(a)
0 + 0 + 0 + 0 + 0 + C1 = 0
C1 = 0
0 + 0 + 0 + 0 + 0 + 0 + C2 = 0
C2 = 0
1
1
20
20 3 20
M A (2)2 + RA (2)3 −
(1.5)3 −
(1) −
(0.5)3 + 0 + 0 = 0
2
6
3
3
3
Reaction at E:
1
1
(2 RE − 120)(2) 2 + (1200 − RE )(2)3 = 30
2
6
2.66667 RE = 30 + 240 − 160 = 110
RE = 41
1.25 kN ↑ 
M A = (2)(41.25) − 120 = −37.5 kN ⋅ m
RA = 120 − 41.25 = 78..25 kN
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.56 (Continued)
I = 45.8 × 106 mm 4 = 45.8 × 10−6 m 4
E = 200 × 109 Pa,
Data:
EI = (200 × 109 )(45.8 × 10−6 ) = 9.16 × 106 N ⋅ m 2 = 9160 kN ⋅ m 2
(b)
( y at x = 1 m)
Deflection at C:
EIyC =
1
1
20
(−37.5)(1)2 + (78.75)(1)3 −
(0.5)3 − 0 − 0 + 0 + 0
2
6
3
= −6.4583 kN ⋅ m3
yC = −
6.4583
= −0.705 × 10−3 m
9160
yC = 0.705 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBL
LEM 9.57
For the beam
b
and loading shown, determine (a) the reactionn at point A,
(b) the deeflection at midpoint C.
SOLUTION
dM
L
= V = RA − P x −
dx
3
EI
0
d2y
L
= M = M A + RA x − P x −
3
dx 2
EI
dy
1
L
1
= M A x + RA x 2 − P x −
dx
2
2
3
EIy =
1
2
+ C1
3
L
1
1
1
M A x 2 + RA x 3 − P x −
2
6
6
3
dy


 x = 0, dx = 0


[ x = 0, y = 0]
dy


 x = L, dx = 0 


+ C1x + C2
∴ C1 = 0
0 + 0 − 0 + C1 = 0
0 + 0 − 0 + C2 = 0
∴ C2 = 0
2
M AL +
1
1  2L 
RA L2 − P 
 =0
2
2  3 
(1)
3
1
1
1  2L 
M A L2 + RA L3 − P 
 =0
2
6
6  3 
Solving Eqs. (1) and (2) simultaneouslly,
[ x = L, y = 0]
(a)
RA =
20
P
27
MA = −
4
PL
27
(2)
MA
Elastic curve:
(b)
Deflection at midpoint C:
yC =
y =
 2
10 3 1
L
Lx 2 +
x −
x−
−
2
81
6
3
 27
L

 y at x = 2 


P
EI
2
3
3
P  2 L
10  L 
1 L 
5PL3
L  +   −    = −
−
EI  27  2 
81  2 
6  6  
1296 EI
20
P ↑ 
27
4
=
PL 
27
RA =
3


yC =
5PL3
↓ 
1296EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.58
For the beam and loading shown, determine (aa) the reaction at point A,
(b) the deflection at midpoint C.
SOLUTION
L
w( x) = w x −
2
0
dV
L
= −w( x) = −w x −
dx
2
dM
L
= V = RA − w x −
dx
2
EI
0
1
d2y
1
L
= M = M A + RA x − w x −
2
2
2
dx
EI
dy
1
1
L
= M A x + RA x 2 − w x −
2
6
2
dx
EIy =
2
3
+ C1
1
1
1
L
M A x 2 + RA x 3 −
w x−
2
6
24
2
dyy


 x = 0, dx = 0 


4
+ C1x + C2
0 + 0 − 0 + C1 = 0
C1 = 0
[ x = 0, y = 0]
0 + 0 − 0 + 0 + C2 = 0
C2 = 0
d
dy


= 0
 x = L, d
dx


M AL +
3
1
1 L
RA L2 − w   = 0
2
6 2
(1)
4
1
1
1 L
M A L2 + RA L2 −
w
=0
2
6
24  2 
[ x = L, y = 0]
(2)
Solving Eqs. (1) and (2) simultaaneously,
(a)
RA =
3wL
32
MA = −
5wL2
192
RA =
MA =
EIyy = −
5
3
1
L
wL2 x 2 +
wLx3 −
w x−
384
192
24
2
3wL
↑ 
32
5wL2
192

4
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.58 (Continued)
Elastic curve:
y =
(b)
w  5 2 2
3
1
L
Lx +
Lx3 −
x−
−
EI  384
192
24
2



L

 y at x = 2 


Deflection at midpoint C:
yC =
4
w
EI
=−
2
3
 5 2  L 

 3  L 
L   + 
L   − 0 
 −
 192  2 
 384  2 

wL4
768EI
yB =
wL4
↓ 
768EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.59
For the beam and loading of Prob. 9.45, determine the
magnitude and location of the largest downward deflection.
SOLUTION
See solution to Prob. 9.45 for the derivation of the equations used in the following:
EI = EI = 1366 kN ⋅ m 2
dy
3
3
2
= 4.9 x 2 − 2 x − 0.4 + 2 x − 1.2 − 10 x − 1.2 − 3.4080 kN ⋅ m 2
dx
1
1
10
4
4
3
EIy = 1.63333x3 −
x − 0.4 +
x − 1.2 −
x − 1.2 − 3.4080 x kN ⋅ m3
2
2
3
EI
To find the location of maximum y , set
dy
= 0. Assume 0.4 < x < 1.2.
dx
4.9 x 2 − 2( x − 0.4)3 − 3.4080 = f ( x) = 0
x = 0.8
Solve by iteration:
0.858 0.857 0.8570
xm = 0.8570 m 
df /dx = 6.88 7.123 7.145
EIym = (1.63333)(0.8570)3 −
1
(0.8570 − 0.4)4 − (3.4080)(0.8570)
2
= −1.9144 kN ⋅ m
ym = −
1.9144
= −1.401 × 10−3 m
1366
ym = 1.401 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.60
For the beam and loading indicated, determine the magnitude
and location of the largest downward deflection.
SOLUTION
See solution to Prob. 9.47 for the derivation of the equations used in the following:
EI = 25.8 × 103 kN ⋅ m 2
dy
1
2
2
EI
x − 1.8 − 3.1 x − 3.6 − 22.535
= 2.8333 x 2 −
dx
2
1
3
3
3
EIy = 0.9444 x − x − 1.8 − 1.03333 x − 3.6 − 22.535 x
8
To find the location of maximum y , set
EI
Solving by iteration:
dy
= 0. Assume 1.8 ≤ xm < 3.6
dx
dy
1
= 2.8333xm2 − ( xm − 1.8) 2 − 22.535 = 0
dx
2
xm = 3, 2.86, 2.855
xm = 2.855 m 
df /dx = 15.8, 15.15
1
EIym = 0.9444 xm3 − ( xm − 1.8)3 − 22.535 xm
8
1
= (0.9444)(2.855)3 − (2.855 − 1.8)3 − (22.535)(2.855) = −42.507 kN ⋅ m3
8
42.507
ym = −
= −1.648 × 10−3 m
ym = 1.648 mm 
25.8 × 103
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.61
For the beam and loading of Prob.9.47, determine the magnitude
and location of the largest downward deflection.
SOLUTION
See solution to Prob. 9.47 for the derivation of the equations used in the following:
EI = 1024 kN ⋅ m 2
83
dy
kN ⋅ m 2
= 12 x 2 − 8x3 + 2 x 4 − 2 x − 1 4 − 4 x − 1 2 −
15
dx
2 5 2
4
83
3
4
5
3
EIy = 4 x − 2 x + x −  x − 1 −  x − 1 −
x kN ⋅ m3
5
5
3
15
EI
To find location of maximum y , set
EI
dy
= 0. Assume 0 < x < 1 m.
dx
dy
83
= 12 x 2 − 8x3 + 2 x 4 −
=0
dx
15
xm = 0.942 m 
x = 0.94166 m
Solving:
EIym = 4(0.94166)3 − 2(0.94166) 4 +
2
83
(0.94166)5 −
(0.94166)
5
15
= −3.1469 kN ⋅ m3

ym = −
3.1469
= −3.0731 × 10−3 m
1024
ym = 3.07 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.62
For the beam and loading of Prob. 9.48, determine the
magnitude and location of the largest downward deflection.
SOLUTION
See solution to Prob. 9.48 for the derivation of equations used in the following:
EI = 168.75 kN ⋅ m 2
C1 = −1.60417 kN ⋅ m 2 , C2 = 0
5
dy
3
2
= − x − 1 + 2.125 x 2 − 2 x − 0.5 + C1 kN ⋅ m 2
6
dx
5
2.125 3 2
4
2
EIy = −
x −1 +
x − x − 0.5 + C1 x + C2 kN ⋅ m3
24
3
3
EI
Compute slope at C.
 dy

at x = 1 m 

 dx

 dy 
EI   = 0 + (2.125)(1)2 − 2(0.5)2 − 1.60417 = 20.83 × 10−3 kN ⋅ m 2
 dx C
Since the slope at C is positive, the largest deflection occurs in portion BC, where
dy
= 2.125 x 2 − 2( x − 0.5)2 − 1.60417
dx
2.125 3 2
EIy =
x − ( x − 0.5)3 − 1.60417 x
3
3
dy
To find the location of the largest downward deflection, set
= 0.
dx
EI
2.125 xm2 − 2( xm2 − xm + 0.25) − 1.60417
= 0.125 xm2 + 2 xm − 2.10417 = 0
x = 1.0521 − 0.0625 x 2
Solve by iteration.
xm = 1 0.989 0.991
xm = 0.991 m 
2
 2.125 
3
3
EIym = 
 (0.991) − 3 (0.991 − 0.5) − (1.60417)(0.991)
3


= −0.97927 kN ⋅ m3
−0.97927
= −5.80 × 10 −3 m
ym =
168.75
ym = 5.80 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.63
The rigid bars BF and DH are
a welded to the rolledsteel beam AE as shown. Determine for the loading
p
B, (b) the deflection
shown (a) the deflection at point
at midpoint C of the beam. Usse E = 200 GPa.
SOLUTION
Use joint G as a free body,
By symmetry, FGH = FFG
 Fy = 0: 2 FGHy − 100 = 0
FGHy = 50 kN
FGHx = 2 FGH
Hy = 100 kN.
Forces in kN; lengths in m.
V = 50 − 50 x − 0.5 0 − 50
0 x − 1.1 0
kN
M = 50 x − 50 x − 0.51 − 50 x − 1.1 0
+ 40 x − 0.5 0 − 40 x − 1.1 0
kN ⋅ m
dy
= 25 x 2 − 25 x − 0.5
0  2 − 25 x − 1.1 2 − 40 x − 0.51 + 40 x − 1.11 + C1
kN ⋅ m 2
dx
25 3 25
25
 x − 0.5 3 −
 x − 1.1 3 − 20 x − 0.5 2 + 20 x − 1.1 2 + C1x + C2 kN ⋅ m3
EIy =
x −
3
3
3
C2 = 0
[ x = 0, y = 0]
EI
[ x = 1.6, y = 0]
 25 
 25 
 25 
3
3
3
2
2
  (1.6) −   (1.1) −   (0.5) − (20)(1.1) + (20)(0.5) + C1(1.6) + 0 = 0
3
3
3
 
 
 
C1 = −1.75 kN ⋅ m3
For EIyB,
x = 0.5m
0
 25 
EIyB =   (0.5)3 − 0 − 0 + 0 − 0 − (1.75)(0.5) = 0.1667 kN
N ⋅ m3
 3 
For EIyC,
x = 0.8
0 m
 25 
 25 
EIyC =   (0.8)3 −   (0.3)3 − 0 − (20)(0.3) 2 − 0 − (1.775)(0.8) + 0
 3 
 3 
= −0.88417 kN ⋅ m3
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.63 (Continued)
For W 100 × 19.3 rolled-steel shape, I = 4.70 × 106 mm 4 = 4.70 × 10−6 m 4
EI = (200 × 109 )(4.70 × 10 −6 ) = 940 × 103 N ⋅ m 2 = 940 kN ⋅ m 2
(a)
yB =
0.1667
= 0.177 × 10−3 m
940
y B = 0.177 mm ↑ 
(b)
yC =
0.8417
= 0.895 × 10−3 m
940
yC = 0.895 mm ↑ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.64
The rigid bar DEF is welded at point D to the rolled-steel beam AB.
For the loading shown, determine (a) the slope at point A, (b) the
deflection at midpoint C of the beam. Use E = 200 GPa.
SOLUTION
Units: Forces in kN; lengths inn meters.
M B = 0: − 4.8RA + (30)(2.4)(3.6)
50)(2.4) = 0
+ (5
RA = 79 kN ↑
I = 212 × 106 mm 4 = 212 × 10−6 m 4
EI = (200 × 109 )(212 × 10−6 ) = 42.4 × 106 N ⋅ m 2
= 42400 kN ⋅ m 2
w( x) = 30 − 30 x − 2.4 0
dV
= −w = −30 + 30 x − 2.4 0
dx
kN/m
dM
= V = 79 − 30 x + 30 x − 2.41 − 50 x − 3.6 0
dx
EI
kN
d2y
= M = 79 x − 15x 2 + 15 x − 2.4 2 − 50 x − 3.61 − 60 x − 3.6 0
dx
kN ⋅ m
dy
79 2
x − 5x3 + 5 x − 2.4 3 − 25 x − 3.6 2 − 60 x − 3.61 + C1
=
2
dx
kN ⋅ m 2
EI
EIy =
79 3 5 4 5
25
x − x +  x − 2.4 4 −
 x − 3.6 3 − 30 x − 3.6 2 + C1x + C2
6
4
4
3
[ x = 0, y = 0]
0 − 0 + 0 − 0 − 0 + 0 + C2 = 0
[ x = 4.8, y = 0]
7 
4
 79
5
5
3
4
  (4.8) −   ( 4.8 ) +   (2.4)
6
4
4
 
 
 
kN ⋅ m3
C2 = 0
 25 
−   (1.2)3 − (30)(1.2) 2 + 4.8C1 = 0
 3 
C1 = −161.7
76 kN ⋅ m 2
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.64 (Continued)
(a)
 dy

 dx at x = 0 


Slope at point A:
 dy 
EI   = 0 − 0 + 0 − 0 − 0 − 161.76
 dx  A
= −161.76 kN ⋅ m 2
−161.76
 dy 
−3
 dx  = 42400 = −3.82 × 10
 A
(b)
Deflection at midpoint C:
θ A = 3.82 × 10−3 rad.

( y at x = 2.4)
 79 
5
EIyC =   (2.4)3 −   (2.4) 4 + 0 − 0 − 0 − (161.76)(2.4) + 0
 6 
4
= −247.68 kN ⋅ m3
yC =
−247.68
= −5.84 × 10−3 m
42400
yC = 5.84 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.65
Foor the cantilever beam and loading shown, determinee the slope and deflection
at the free end.
SOLUTION
Loading I:
Counterclockwise co
ouple PL at B.
Case 3 of Appendix D applied to po
ortion BC.
θ B′ = −
y′B =
( PL)( L / 2) 1 PL2
=
EI
2 EI
( PL)( L / 2)2
1 PL3
=
2 EI
8 EI
AB remains straight.
θ A′ = θ B′ =
1 PL2
2 EI
1 PL3 1 PL3
3 PL3
L
−
=−
y′A = y′B −  θ B′ = −
8 EI
4 EI
8 EI
2
Loading II:
Case 1 of Appenddix D.
θ A′′ =
PL2
,
2EL
y′′A −
PL3
3EI
By superposition,
θ A = θ ′A + θ ′′A =
1 PL2 1 PL2
PL2
+
=
2 EI
2 EI
EI
y A = y′A + y′′A = −
3 PL3 1 PL3
17 PL3
−
=−
8 EI
3 EI
24 EI
θA =
yA =
PL2
EI

17 PL3
↓ 
24 EI
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.66
For thhe cantilever beam and loading shown, determine thhe slope and
deflecttion at the free end.
SOLUTION
Loading I:
P downward at B.
Case 1 of Appendix D applied to portion AB.
θ B′ = −
P( L/2) 2
1 PL2
=−
2 EI
8 EI
y′B = −
P( L/2)3
1 PL3
=−
3 EI
24 EI
BC remains straight.
θC′ = θ B′ = −
1 PL2
8 EI
1 PL3
1 PL3
L
yC′ = y′B −  θ B′ = −
−
24 EI
16 EI
2
=−
Loading II:
5 PL3
48 EI
P downward at C.
Case 1 of Appendix D.
θC′′ = −
PL2
2EI
yC′′ = −
PL3
3EI
By superposition,
θC = θC′ + θC′′ = −
1 PL2 1 PL2
5 PL2
−
=−
8 EI
2 EI
8 EI
yC = yC′ + yC′′ = −
5 PL3 1 PL3
21 PL3
−
=−
48 EI
3 EI
48 EI
5PL2
8EI

7 PL3
↓ 
16EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.67
For the cantilever beam and loading shown, determine the slope and
deflection at point B.
SOLUTION
Consider portion AB as a cantilever beam subjected to the three loadings shown.
P = wa
By statics:
a 3

M = ( wa)  a +  = wa 2
2 2

Slope at B.
Case 2 of App. D.
(θ B ) w = −
wa3
6 EI
Case 1 of App. D.
(θ B ) p = −
(wa)a 2
wa3
=−
2 EI
2 EI
Case 3 of App. D.
(θ B ) M = −
θB = −
Deflection at B:
θ B = (θ B ) w + (θ B ) P + (θ B ) M
13wa
6 EI
( 32 wa 2 ) a
EI
=−
3wa3
2 EI
3
θB =
13wa3
6 EI

yB = ( yB ) w + ( yB ) P + ( yB ) M
Case 2 of App. D.
( yB ) w = −
wa 4
8EI
Case 1 of App. D.
( yB ) P = −
( wa)a3
wa 4
=−
3EI
3EI
Case 3 of App. D.
( yB )M = −
yB = −
4
29 wa
24 EI
( 32 wa 2 )a
2 EI
=−
3wa 4
4 EI
yB =
29wa 4

24EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.68
For the cantilever beam and loading shown, determine the slope and
deflection at point B.
SOLUTION
Consider portion AB as a cantilever beam subjected to the three loadings shown.
By statics:
Slope at B.
wL wL
=
2
2
2
 wL  1 L  wL
M0 = 
−
=


8
 2  2 2 
Q=P−
θ B = (θ B ) w + (θ B )Q + (θ B ) M
3
w L
wL3
=
−
 
6 EI  2 
48EI
Case 2 of App. D.
(θ B ) w = −
Case 1 of App. D.
( wL /2)  L 
wL3
(θ B )Q = +
  =
2 EI  2  16 EI
Case 3 of App. D.
(θ B ) M = −
2
θB = −
Deflection at B:
( wL 2/8)  L 
wL3
=
−
 
EI  2 
16 EI
wL3
48EI
θB =
wL3
48EI

y B = ( y B ) w + ( y B )Q + ( y B ) M
4
Case 2 of App. D.
( yB ) w = −
W L
wL4
  =−
8EI  2 
128EI
Case 1 of App. D.
( y B )Q = +
( wL /2)  L 
wL4
  =
3EI  2 
48 EI
Case 3 of App. D.
( yB ) M = −
( wL 2/8)  L 
wL4
=
−
 
2 EI  2 
64 EI
3
2
yB = −
wL4
384 EI
yB =
wL4

384EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.69
For the beam and loading shown, determine (a) the deflection at C, (b) the
slope at end A.
SOLUTION
Loading I:
Downward load P at B.
Use Case 5 of Appendix D with
P = P, a =
y =
For x < a, given elastic curve is
L
2L
2L
, b=
, L = L, x =
3
3
3
Pb 3
[ x − ( L2 − b2 ) x]
EIL
To obtain elastic curve for x > a, replace x by L − x and interchange a and b to get
y =
yC =
2L
Pa
at point C.
[( L − x)3 − ( L2 − a 2 )( L − x)] with x =
3
6EIL
P( L/3)
6 EIL
5 PL2
Pb( L2 − b 2 )
P(2 L/3)[ L2 − (2 L/3) 2 ]
=−
=−
6 EIL
6 EIL
81 EI
θA = −
Loading II:
2
  L 3 
L  L 
7 PL3
  −  L2 −       = −

486 EI
 3 
 3    3  

Upward load at C.
P = − P, a =
Use Case 5 of Appendix D with
2L
L
2L
, b = , L = L, x = a =
3
3
3
yC = −
(− P)(2 L / 3) 2 ( L / 3) 2
4 PL3
=
3EIL
243 EI
θA = −
(− P)( L / 3)( L2 − ( L / 3) 2 )
4 PL2
=
6 EIL
81 EI
(a)
Deflection at C:
yC = −
7 PL3
4 PL3
+
486 EI
243 EI
(b)
Slope at A:
θA = −
5 PL2
4 PL2
+
81 EI
81 EI
yC =
θA =
1 PL3
↑ 
486 EI
1 PL2
81 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.70
For the beam and loading shown, determine (a) the deflection at C, (b) the
slope at end A.
SOLUTION
Loading I:
Load at B. Case 5 of Appendix D.
a=
L
2L
2L
, b=
, x=
3
3
3
For x > a, replace x by L − x and interchange a and b.
y =
yC =
Pa
[( L − x)3 − ( L2 − a 2 )( L − x)]
6EIL
3
 2 L2  
P( L /3) 
2L 
2L  
7 PL3
 L −
  L −
 −  L −
 = −
6EIL 
3 
9 
3  
486 EI

θA = −
Loading II:
Pb( L2 − b2 )
P(2L /3)( L2 − 4L2 /9)
5 PL2
=−
=−
6EIL
6 EIL
81 EI
Load at C. Case 5 of Appendix D.
a=
yC =
2L
L
2L
, b= , x=
3
3
3
Pb 3
P( L /3)  8L3  2 L2   2L  
[ x − ( L2 − b2 ) x] =
− L −



6EIL
6EIL  27 
9   3  
=−
θA = −
8 PL3
486 EI
Pb( L2 − b2 )
P( L /3)( L2 − L2 /9)
4 PL2
=−
=−
6EIL
6EIL
81 EI
(a)
Deflection at C: yC = −
(b)
Slope at A:
7 PL3
8 PL3
15 PL3
−
=−
486 EI
486 EI
486 EI
θA = −
5 PL2
4 PL2
1 PL2
−
=−
81 EI
81 EI
9 EI
yC =
θA =
5 PL3
↓ 
162 EI
1 PL2
9 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.71
For the beam and loading shown, determine (a) the deflection at C,
(b) the slope at end A.
SOLUTION
Loading I:
Case 5:
a=
L
2L
, b=
, P = P, x = a
3
3
2
yC = −
2
Pa 2b 2
P  L   2L 
4 PL3
=−
=
−
  

6EIL
6EIL  3   3 
243 EI
2
Pb( L2 − b2 )
P  2L   2  2L  
5 PL2
=−
θA = −

 L − 
  =−
6EIL
6EIL  3  
81 EI
 3  
Loading II:
(a)
Case 7:
M =−
PL
,
3
x=
L
3
yC = −
3

M
PL / 3  L 
4 PL3
2  L 
( x3 − L2 x) = +
  − L    = −
6EIL
6EIL  3 
243 EI
 3  
θA = +
ML
1 PL2
−( PL / 3)L
=
=−
6EI
6EI
18 EI
Deflection at C:
yC = −
4 PL3
4 PL3
8 PL3
−
=−
243 EI
243 EI
243 EI
yC =
(b)
Slope at A:
θA = −
8 PL3
↓
243 EI
5 PL2
1 PL2
19 PL2
−
=−
81 EI
18 EI
162 EI
θA =
19 PL2
162 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.72
For thhe beam and loading shown, determine (a) the defflection at C,
(b) thee slope at end A.
SOLUTION
Loading I:
Case 6 in Appendix D.
yC = −
Loading II:
MA =
Loading III::
wL3
24EI
3
M A  L 
1 M A L2
 L 
  − L2    =
6 EIL  2 
 2   16 EI
wL2
12
yC =
1 wL4
192 EI
θA =
1 M B L3
16 EI
M L
θA = B
6 EI

MB =
with
wL2
12
M AL
3EI
1 wL3
36 EI
(using Loading II result)
yC =
1 wL4
192 EI
θA =
1 wL3
72 EI
Defleection at C:
yC = −
5 wL4
1 wL4
1 wL4
1 wL4
+
+
=−
3384 EI
384 EI
192 EI
192 EI
yC =
(b)
θA =
Case 7 in Appendix D.
yC =
(a)
θA = −
Case 7 in Appendix D.
yC = −
with
5wL4
384 EI
1 wL4
↓ 
384 EI
Slopee at A:
θA = −
1 wL3
1 wL3
1 wL3
+
+
=0
24 EI
36 EI
72 EI
θ A = 0 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.73
For the cantilever beam and loading shown, determine the
slope and deflection at end C. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in m.
Loading I:
Concentrated load at B
Case 1 of Appendix D applied to portion AB.
θ B′ = −
PL2
(3)(0.75) 2
0.84375
=−
=−
2 EI
2 EI
EI
y′B = −
PL3
(3)(0.75)3
0.421875
=−
=−
3EI
3EI
EI
Portion BC remains straight.
θC′ = θ B′ = −
0.84375
EI
yC′ = y′B − (0.5)θ B′ = −
Loading II:
Concentrated load at C:
0.84375
EI
Case 1 of Appendix D.
θ A′′ = −
PL2
(3)(1.25) 2
2.34375
=−
=−
2 EI
2 EI
EI
y′′A = −
PL3
(3)(1.25)3
1.953125
=−
=−
3EI
3EI
EI
3.1875
EI
2.796875
y A = y′A + y′′A = −
EI
By superposition,
θ A = θ A′ + θ A′′ = −
Data:
E = 200 × 109 Pa, I = 2.52 × 106 mm 4 = 2.52 × 10−6 m4
EI = (200 × 104 )(2.52 × 10−6 ) = 504 × 103 N ⋅ m 2 = 504 kN ⋅ m 2
Slope at C:
θC = −
3.1875
= −6.32 × 10−3 rad
504
Deflection at C:
yC = −
2.796875
= −5.55 × 10−3 m
504
θC = 6.32 × 10−3 rad

yC = 5.55 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.74
For the cantilever beam and loading shown, determine the
slope and deflection at point B. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in m.
The slope and deflection at B depend only on the
deformation of portion AB.
Reduce the force at C to an equivalent force-couple system
at B and add the force already at B to obtain the loadings I
and II shown.
Loading I:
Loading II:
Case 1 of Appendix D.
θ B′ = −
PL2
(6)(0.75) 2
1.6875
=−
=−
2 EI
2 EI
EI
y′B = −
PL3
(6)(0.75)3
0.84375
=−
=−
3EI
3EI
EI
Case 3 of Appendix D.
θ B′′ = −
ML
(1.5)(0.75)
1.125
=−
=−
EI
EI
EI
y′′B = −
ML2
(1.5)(0.75) 2
0.421875
=−
=−
2 EI
EI
EI
By superposition,
2.8125
EI
1.265625
yB = y′B + y′′B = −
EI
θ B = θ B′ + θ B′′ = −
Data:
E = 200 × 109 Pa, I = 2.52 × 106 mm 4 = 2.52 × 10−6 m 4
EI = (200 × 109 )(2.52 × 10 −6 ) = 504 × 103 N ⋅ m 2 = 504 kN ⋅ m 2
Slope at B:
θB = −
2.8125
= −5.58 × 10−3 rad
504
Deflection at B:
yB = −
1.265625
= −2.51 × 10−3 m
504
θ B = 5.58 × 10−3 rad

yB = 2.51 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.75
For the cantilever beam and loading shown, determine the slope
and deflection at end C. Use E = 200 GPa.
SOLUTION
C =
I =
1
1
d = (44) = 22 mm
2
2
π
4
C4 =
π
4
(22) 4 = 183984 mm 4
EI = (200 × 106 )(183984 × 10−12 ) = 36.8 kNm 2
Loading I:
Case 1 of Appendix D.
P = 0.5 kN, L = 1 m
Loading II.
(yC )1 = −
PL3
(0.5)(1)3
=−
= −4.529 × 10−3 m
3EI
(3)(36.8)
(θ C )1 = −
PL2
(0.5)(1) 2
=−
= −6.793 × 10 −3
2 EI
(2)(36.8)
Treat portion AB as a cantilever beam. (Case 2)
w = 2.6 kN/m, L = 0.75 m
(yB ) 2 = −
wL4
(2.6)(0.75) 4
=−
= −2.794 × 10 −3 m
8 EI
(8)(36.8)
(θ B ) 2 = −
wL3
(2.6)(0.75)3
=−
= −4.968 × 10−3
6 EI
(6)(36.8)
Portion BC remains straight for loading II.
LBC = 0.25 m
(yC )2 = ( yB )2 + LBC (θ B )2 = −4.036 × 10−3 m
(θC )2 = (θ B )2 = −4.968 × 10−3
Slope at end C:
θC = −11.761 × 10
Deflection at end C:
θC = (θC )1 + (θC ) 2
By superposition,
−3
By superposition,
yC = −8.565 × 10 −3 m
θC = 11.76 × 10−3 rad

yC = ( yC )1 + ( yC ) 2
yC = 8.57 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.76
For the cantilever beam and loading shown, determine the slope
and deflection at point B. Use E = 200 GPa.
SOLUTION
C=
I=
1
1
d = (44) = 22 mm
2
2
π
C4 =
π
(22)4 = 183984 mm 4
4
4
EI = (200 × 106 )(183984 × 10−6 ) = 36.8 kNm2
Loading I:
Case 1 of Appendix D.
P = 0.5 kN L = 1 m x = 0.75 m
P
y1 =
( x3 − 3Lx 2 )
6 EI
dy
P
θ1 =
=
( x 2 − 2 Lx)
dx 2 EI
0.5
( yB )1 =
[(0.75)3 − (3)(1)(0.75)2 ]
(6)(36.8)
= −2.866 × 10−3 mm
(θ B )1 =
0.5
[(0.75)2 − (2)(1)(0.75)]
(2)(36.8)
= −6.369 × 10 −3
Loading II:
Case 2 of Appendix D.
w = 2.6 kN/m, L = 0.75 m
Slope at point B:
(yB ) 2 = −
wL4
(2.6)(0.75) 4
=−
= −2.794 × 10 −3 m
8 EI
(8)(36.8)
(θ B ) 2 = −
wL3
(2.6)(0.75)3
=−
= −4.968 × 10−3
6 EI
(6)(36.8)
θ B = (θ B )1 + (θ B ) 2
θ B = −11.337 × 10−3
Deflection at point B:
θ B = 11.34 × 10−3 rad

yB = ( yB )1 + ( yB ) 2
yB = −5.66 × 10−3 m
yB = 5.66 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.77
For the beam and loading shown, determine (a) the slope at
end A, (b) the deflection at point C. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in m.
Loading I:
Moment at B.
M = 80 kN ⋅ m, L = 5.0 m, x = 2.5 m
Case 7 of Appendix D:
ML (80)(5.0) 66.667
=
=
6 EI
6 EI
EI
M
80
125
yC = −
( x3 − L2 x) = −
[2.53 − (5.0) 2 (2.5)] =
6 EIL
6 EI (5.0)
EI
θA =
Loading II:
Moment at A:
(Case 7 of Appendix D.)
M = 80 kN ⋅ m, L = 5.0 m, x = 2.5 m
Loading III:
θA =
ML (80)(5.0) 133.333
=
=
3EI
3EI
EI
yC =
125
EI
140 kN concentrated load at C:
(Same as loading I.)
P = 140 kN
PL2
(140)(5.0) 2
218.75
=−
=−
16 EI
16 EI
EI
PL3
(140)(5.0)3
364.583
yC = −
=−
=−
EI
48 EI
48 EI
θA = −
E = 200 × 109 Pa, I = 156 × 106 mm4 = 156 × 10−6 m4
Data:
EI = (200 × 109 )(156 × 10−6 ) = 31.2 × 106 N ⋅ m2 = 31200 kN ⋅ m2
(a)
Slope at A:
θA =
67.667 + 133.333 − 218.75
= −0.601 × 10−3 rad
31200
θ A = 0.601 × 10−3 rad
(b)
Deflection at C:
yC =
125 + 125 − 364.583
= −3.67 × 10−3 m
31200

yC = 3.67 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.78
For the beam and loading shown, determine (a) the slope at
end A, (b) the deflection at point C. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in m.
Loading I:
8 kN/m uniformly distributed.
Case 6:
w = 8 kN/m, L = 3.9 m, x = 1.3 m
WL3
(8)(3.9)3
19.773
=−
=−
24EI
24EI
EI
w
8
yC = −
[ x 4 − 2Lx3 + L3 x] = −
[(1.3) 4 − (2)(3.9)(1.3)3 + (3.9)3 (1.3)]
24EI
24 EI
20.945
=−
EI
θA = −
Loading II:
35 kN concentrated load at C.
Case 5 of Appendix D.
P = 35 kN, L = 3.9 m, a = 1.3 m, b = 2.6 m, x = a = 1.3 m
θA = −
Pb( L2 − b 2 )
(35)(2.6)(3.92 − 2.6) 2
32.861
=−
=−
6EIL
6EI (3.9)
EI
yC = −
Pa 2b 2
(35)(1.3) 2 (2.6)2
34.176
=−
=−
3EIL
3EI (3.9)
EI
E = 200 × 109 , I = 102 × 106 mm 4 = 102 × 10 −6 m 4
Data:
EI = (200 × 109 )(102 × 10 −6 ) = 20.4 × 106 N ⋅ m 2 = 20, 400 kN ⋅ m 2
(a)
Slope at A:
θA = −
θ A = 2.58 × 10−3 rad

(b)
19.773 + 32.861
= −2.58 × 10−3 rad
20, 400
Deflection at C:
yC = −

20.934 + 34.176
= −2.70 × 10−3 m
20, 400
yC = 2.70 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.79
For the uniform beam shown, determine the reaaction at each of the three
supports.
SOLUTION

Consider RC as redundant and replace
r
loading system by I and II.
Loading I:
( yC )1 =
2L
L
L
, b = , x = at C.
3
3
3
3
2
Pb 3
P ( L /3)  L   2  L   L 
  −  L −    
[ x − ( L2 − b2 ) x] =
6 EIL  3  
6 EIL
 3   3 

=−
Loading II:
a=
(Case 5 of Append
dix D).
7 PL3
486 EI
a=
(Case 5 of Appenddix D).
( yC )2 =
2L
L
, b=
3
3
RC 22 b 2 RC ( L /3)2 (2 L /3)2 4 RC L3
=
=
3EIL
3EIL
243EI
Superposition and constraint:
yC = ( yC )1 + ( yC ) 2 = 0
−
4 RC
7 PL3
+
=0
486 EI 243EI
RC =
7
P 
8
ΣM D = 0:
 2 L   7  L 
− RA L + P 
 −  P   = 0
 3   8  3 
3
RA = P 
8
7
P=0
8
ΣFy = 0:
RA + RD − P +
RD = P −
7
3
1
P− P=− P
8
8
4
RD =
4
P 
4
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROB
BLEM 9.80
For the uniform
u
beam shown, determine the reaction at each
h of the three
supportss.
SOLUTION

Consider RC as redundant and replace loading system by I and III.
Loading I. (C
Case 4 of Appendix D): YC′ = −
Loading II. (Case 7 of Appendix D): YC′′ = −
Superposition and constraint.
3
3
2
M 0  L 
 L  M L
  − L2    = 0
6 EIL  2 
 2   16 EI
yC = yC′ + yC′′ = 0
2
RC L M 0 L
+
=0
16 EI
48EI
E
 3M 0  L 
ΣM B = 0: − RA L + 
  − M0 = 0
 L  2 
M 0 3M 0
−
+ RB = 0
ΣFy = 0:
2L
L
−
RC L3
48 EI
RC =
3M 0

L
RA =
RB =
M0

2L
5M 0

2L
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.81
For the uniform beam shown, determine (a) the reaction at A, (b) the
reaction at B.
SOLUTION
Beam is indeterminate to first degree. Consider RA as
redundant and replace the given loading by loadings I, II,
and III.
Loading I:
(Case 1 of Appendix D.)
( yA)I =
Loading II:
RA L3
3EI
(Case 2 of Appendix D.)
( y A ) II = −
Loading III:
wL4
8EI
(Case 2 of Appendix D (portion CB).)
(θC ) III = −
( yC ) III =
w( L/2)3
1 wL3
=−
6 EI
48 EI
1 wL4
w( L/2) 4
=
8EI
128 EI
Portion AC remains straight.
( y A ) III = ( yC ) III +
Superposition and constraint:
(a)
L
7 wL4
(θC ) III =
2
384 EI
y A = ( y A ) I + ( y A ) II + ( y A ) III = 0
1 RA L3 1 wL4
7 wL4
1 RA L3
41 wL4
−
+
=
−
=0
3 3EI
8 EI
384 EI
3 EI
384 EI
RA =
41
wL ↑ 
128
RB =
23
wL ↑ 
128
Statics:
(b)
ΣFy = 0 :
41
1
wL − wL + RB = 0
128
2
 41

1
 3L 
ΣM B = 0 : − 
wL  L −  wL   − M B = 0
128
2



 4 
MB =
7
wL2
128

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBL
LEM 9.82
For the uniform
u
beam shown, determine (a) the reaction at
a A, (b) the
reaction at
a B.
SOLUTION
Consider RA as redundant and replace loadin
ng system by I and II.
y ′A =
Loading I:
(Case 1 of Appendix D.)
Loading II:
Portion BC. (Case 3 of Appenndix D.)
yC′′ = −
M 0 ( L − a)2
2 EI
θC′′ =
M 0 ( L − a)
EI
y ′′A = yC′′ − ( a)θ C
Portion AC is straight.
=−
(a)
RA L3
3EI
Superposition and constraint:
M 0 ( L − a) 2 aM 0 ( L − a )
−
2 EI
EI
y A = y′A + y′′A = 0
RA L3 M 0 ( L − a) 2 aM 0 ( L − a)
−
−
=0
3EI
2 EI
EI
2
RA L3 − M 0 ( L − a)( L − a + 2a) = 0
3
2
RA L3 = M 0 ( L2 − a 2 )
3
(b)
ΣFy = 0: RA + RB = 0
RB = − RA
ΣM B = 0: M B + M 0 − RA L = 0
MB =

3M 0  2
2 
L − a 2 − L2 
2 
3 
2L 
M B + M0 −
RA =
3M 0
RB =
3M 0
MB =
M0
2 L3
3
2L
( L2 − a 2 ) ↑ 
( L2 − a 2 ) ↓ 
3 M0 2
(L − a2 ) = 0
2 L2
2
2L
( L2 − 3a 2 )


PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.83
For the beam shown, determine the reaction at B.
SOLUTION
Beam is second degree indeterminate. Choose RB and M B as
redundant reactions.
Loading I:
Case 1 of Appendix D.
( yB ) I =
Loading II:
RB L3
3EI
(θ B ) I =
RB L2
2EI
Case 3 of Appendix D.
( yB ) II = −
M B L2
2EI
(θ B ) II = −
M BL
EI
Loading III: Case 2 of Appendix D.
( yB ) III = −
wL4
8EI
(θ B ) III = −
wL2
6EI
Superposition and constraint:
yB = ( yB )I + ( yB ) II + ( yB )III = 0
L3
L2
wL4
RB −
MB −
=0
3EI
2EI
8EI
(1)
θ B = (θ B )I + (θ B )II + (θ B )III = 0
L2
L
wL3
RB −
MB −
=0
EI
2EI
6EI
(2)
Solving Eqs. (1) and (2) simultaneously,
RB =
MB =
1
wL ↑ 
2
1
wL2
12

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.84
For the beam
b
shown, determine the reaction at B.
SOLUTION
Beam is second degree indeterminate. Choosse RB and M B as redundant reactions.
Loading I:
Case 1 of Appendix D.
( yB )I = −
Loading II:
RB L3
R L2
, (θ B )I = − B
3EI
2 EI
Case 3 of Appendix D.
( yB )II =
M B L2
M L
, (θ B )II = B
2 EI
EI
Loading III: Case 3 applied to portion AC.
M 0 ( L/2) 2 M 0 L2
=
8 EI
2 EI
M 0 ( L/2) M 0 L
=
=
2 EI
EI
( yC ) III =
(θC ) III
Portion CB remains straight.
L
3 M 0 L2
(θC ) III =
2
8 EI
2
1 M0L
=
2 EI
( yB ) III = ( yC ) III +
(θ B ) III = (θC ) III
Superposition and constraint:
yB = ( yB ) I + ( yB ) II + ( yB ) III = 0
−
L2
L3
3 M 0 L2
MB +
=0
RB +
3EI
2 EI
8 EI
(1)
θ B = (θ B ) I + (θ B )III + (θ B ) III = 0
−
L
L2
1 M0L
MB +
=0
RB +
EI
2 EI
2 EI
(2)
Solving Eqs. (1) and (2) simultaneously,
RB =
3 M0
↓ 
2 L
MB =
1
M0
4

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.85
A central beam BD is joined at hinges to two
cantilever beams AB and DE. All beams have the
cross section shown. For the loading shown,
determine the largest w so that the deflection at C
does not exceed 3 mm. Use E = 200 GPa.
SOLUTION
Let
a = 0.4 m.
Cantilever beams AB and CD.
Cases 1 and 2 of Appendix D.
yB = yD = −
( wa)a 3 wa 4
11 wa 4
−
=−
3EI
8EI
24 EI
Beam BCD, with L = 0.8 m, assuming that points B and D do not move.
Case 6 of Appendix D.
yC′ = −
5wL4
384 EI
Additional deflection due to movement of points B and D.
yC′′ = yB = yD = −
Total deflection at C:
yC = yC′ + yC′′
yC = −
Data:
11 wa 4
24 EI
w
EI
 5 L4 11a 4 
+


24 
 384
E = 200 × 109 Pa,
1
(24)(12)3 = 3.456 × 10−3 mm 4 = 3.456 × 10−9 m 4
12
EI = (200 × 109 )(3.456 × 10−9 ) = 691.2 N ⋅ m 2
I=
yC = −3 × 10−3 m
−3 × 10−3 = −
w  (5)(0.8)4 (11)(0.4) 4 
−6
+

 = −24.69 w
691.2  384
24

w = 121.5 N/m 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.86
The two beams shown have the same cross section and
are joined by a hinge at C. For the loading shown,
determine (a) the slope at point A, (b) the deflection at
point B. Use E = 200 GPa.
SOLUTION
Using free body ABC,
ΣM A = 0: 0.45RC − (0.3)(3.5) = 0
RC = 2.33 kN
E = 200 GPa
1
1
I = bh3 = (30)(30)3 = 67500 mm 4
12
12
EI = (200 × 106 )(67500 × 10−12 ) = 13.5 kNm 2
Using cantilever beam CD with load RC ,
Case 1 of Appendix D:
yC = −
RC L3CD
(2.33)(0.3)3
=−
= 0.001555 mm
3EI
(3)(13.5)
Calculation of θ A′ and yB′ assuming that point C does not move.
Case 5 of Appendix D:
P = 3.5 kN, L = 0.45 m, a = 0.3 m, b = 0.15 m
θ A′ = −
Pb( L2 − b 2 )
(3.5)(0.15)(0.452 − 0.152 )
=−
= −0.00259 rad
6 EIL
(6)(13.5)(0.45)
yB′ = −
Pb 2 a 2
(3.5)(0.15) 2 (0.3) 2
=−
= −0.000389 m
3EIL
(3)(13.5)(0.45)
Additional slope and deflection due to movement of point C.
θ A′′ =
yC
0.001555
=−
= −0.003456 rad
0.45
LAC
(0.3)(0.001555)
a
= −0.0010367 mm
yC = −
0.45
L
θ A = θ A′ + θ A′′ = −0.00259 − 0.003456
yB′′ =
(a)
Slope at A:
= −0.006046 rad = 0.00605 rad
(b)

Deflection at B: yB = y B′ + y B′′ = −0.000389 − 0.0010367
= −1.4257 × 10−3 m = 1.43 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.87
Beam DE rests on the cantilever beam
m AC as shown. Knowing
that a square rod of side 10 mm is used for each beam, determine
the deflection at end C if the 25-N ⋅ m coouple is applied (a) to end
E of beam DE, (b) to end C of beam AC.. Use E = 200 GPa.
SOLUTION
E = 200 × 109 Pa
1
I = (10)(10)3 = 833.33 mm 4 = 833.33
8
× 10−12 m 4
12
EI = 166.667 N ⋅ m 2
(a)
Couple applied to beam DE.
Free body DE.
ΣM = 0: 0.180P − 25 = 0 P = 138.889 N
Loads on cantilever beam ABC are P at pointt B and P at point C
as shown.
Due to P at point B.
C
1 of Appendix D,
Using portion AB and applying Case
( yB )1 =
PL3 (1338.889)(0.120)3
=
= 0.480 × 10−3 m
3EI
(3)(166.667)
(θ B )1 =
PL2 (1338.889)(0.120) 2
=
= 6.00 × 10−3
(2)(166.667)
2 EI
−3
( yC )1 = ( yB )1 + LBC
+ (0.180)(6.00 × 10−3 )
B (θ B )1 = 0.480 × 10
= 1.56 × 10−3 m
Due to load P at point C:
(Caase 1 of App. D applied to ABC.)
( yC ) 2 = −
PL3
(138.889)(0.120 + 0.180)3
1 −3 m
=−
= −7.50 × 10
3EI
(3)(166.667)
yC = ( yC )1 + ( yC ) 2 = −5.94 × 10 −3 m
Total deflection at point C:
yC = 5.94 mm 
(b)
Couple applied to beam AC:
Case 3 of Appendix D.
yC = −
ML2
(25)(0.300) 2
=−
= −6.75 × 10−3 m
2 EI
(2)(166.667)
yC = 6.75 mm 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stuudent using this manual is using it
without permission.
PROBLEM 9.88
Beam AC rests on the cantilever beam DE as shown. Knowing that a
W410 × 38.8 rolled-steel shape is used for each beam, determine for the
loading shown (a) the deflection at point B, (b) the deflection at point D.
Use E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in m.
Using free body ABC,
 M A = 0: 4.4 RC − (4.4)(30)(2.2) = 0
RC = 66.0 kN
E = 200 × 109 Pa
I = 125 × 106 mm4 = 125 × 10−6 m4
EI = (200 × 109 )(125 × 10−6 ) = 25.0 × 10−6 N ⋅ m 2
= 25, 000 kN ⋅ m 2
For slope and defection at C, use Case 1 of Appendix D
applied to portion CE of beam DCE.
θC =
RC L2
(66.0)(2.2) 2
=
= 6.3888 × 10−3 rad
2EI
(2)(25, 000)
yC = −
RC L3
(66.0)(2.2)3
=
= −9.3702 × 10−3 m
3EI
(3)(25, 000)
Defection at B, assuming that point C does not move.
Use Case 6 of Appendix D.
( yB )1 = −
5WL4
(5)(30)(4.4) 4
=−
= −5.8564 × 10−3
384 EI
(384)(25, 000)
Additional defection at B due to movement of point C:
(a)
Total deflection at B:
( yB )2 =
1
yC = −4.6851 × 10−3 m
2
yB = ( yB )1 + ( yB ) 2 = −10.54 × 10−3 m
yB = 10.54 mm ↓ 
Portion DC of beam DCB remains straight.
(b)
Deflection at D: yD = yC − aθC = −9.3702 × 10−3 − (2.2)(6.3888 × 10−3 ) = −23.4 × 10−3 m
yD = 23.4 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.89
Before the 30 kN/m load is applied, a gap, δ 0 = 20 mm exists
between the W 410 × 60 beam and thhe support at C. Knowing
that E = 200 GPa, determine the reacction at each support after
the uniformly distributed load is appliied.
SOLUTION
Daata:
δ 0 = 0.02 m
E = 200 × 106 kPa
I = 216 × 10−6 m 4
EI = 43.2 × 103 kN ⋅ m 2
Looading I:
Case 6 of Appendix D.
yC′ = −
5wL4
5(30)(7.2) 4
=−
384 EI
384(43.2 × 103 )
= −24.3 × 10−3 m
Looading II:
yC′′ =
Case 4 of Appendix D.
RC L3
RC (7.2)3
=
48EI 48(43.2 × 103 )
= 0.18 × 10−3 RC
Deflection at C:
yC = yC′ + yC′′ = −δ 0
−24.3 × 10−3 + 0.18 × 10−3 R C = −20 × 10−3
R C = 23.889 kN
RC = 23.9 kN 
ΣM B = 0: (30)(7.2)(3.6) − R A (7.2) − (23.889))(3.6) = 0
RA = 96.055
R A = 96.1 kN 
ΣFy = 0: 96.055 − 30(7.2) + 23.889 + RB = 0
RB = 96.1 kN 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stuudent using this manual is using it
without permission.
PROBLEM 9.90
Th
he cantilever beam BC is attached to the steel cable AB
A as shown.
Knnowing that the cable is initially taut, determine the tension
t
in the
caable caused by the distributed load shown. Use E = 20
00 GPa.
SOLUTION
Let P be the tension developed in member AB
B and δ B be the elongation of that member.
A = 255 mm 2 = 255 × 10−6 m2
Cable AB:
δB =
PL
( P )(3)
=
EA (2000 × 109 )(255 × 10 −6 )
= 58.82 × 100 −9 P
I = 156 × 106 mm
m 4 = 156 × 10−6 m 4
Beam BC:
EI = (200 × 109 )(156 × 10 −6 )
= 31.2 × 106 N ⋅ m 2
Loading I:
20 kN/m downward.
Refer to Case 2 of Appendix D.
( yB )1 = −
wL4
(20 × 103 )(6) 4
=−
8 EI
(8)(31.2 × 106 )
= −103.846 × 10 −3 m
Loading II:
Upward force P at Point B.
Refer to Case 1 of Appendix D.
( yB ) 2 =
PL3
P(6)3
=
= 2.3077 × 10−6 P
3EI (33)(31.2 × 106 )
yB = ( yB )1 + ( yB )2
By superposition,
Also, matching the deflection at B,
yB = −δ B
−103.846 × 10−3 + 2.3077 × 10−6 P = −58.82
5
× 10−9 P
2.3666 × 10−6 P = 1003.846 × 10−3
P = 43.9 × 103 N
P = 43.9 kN 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.91
Before the load P was applied, a gap, δ 0 = 0.5mm, existed
between the cantilever beam AC and the support at B.
Knowing that E = 200 GPa, determine the magnitude of P
for which the deflection at C is 1 mm.
SOLUTION
Let length AB = L = 0.5 m
length BC = a = 0.2 m
Consider portion AB of beam ABC.
The loading becomes forces P and RB at B plus the couple Pa.
The deflection at B is δ 0. Using Cases 1 and 3 of Appendix D,
δ0 =
( P − RB ) L3 PaL2
+
3EI
2EI
 L3 L2a 
L3
RB = EI δ 0
+

 P −
2 
3
 3
(1)
The deflection at C depends on the deformation of beam ABC
subjected to loads P and RB . For loading I, using Case 1 of
Appendix D,
(δ C )1 =
P ( L + a )3
3EI
For loading II, using Case 1 of Appendix D,
yB =
RB L3
3EI
θB =
RB L2
2EI
Portion BC remains straight.
 L3 L2a  RB
yC = yB + aθ B = 
+

2  EI
 3
By superposition, the downward deflection at C is
δC =
P( L + a)3  L3 L2a  RB
− 
+

3EI
2  EI
 3
 L3 L2a 
( L + a )3
P − 
+
 RB = EI δ C
3
2 
 3
(2)
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.91 (Continued)
Data: E = 200 × 109 Pa
I =
1
(60)(60)3 = 1.08 × 106 mm 4 = 1.08 × 10−6 m 4
12
EI = 216 × 103 N ⋅ m 2
δ 0 = 0.5 × 10−3 m
δ C = 1.0 × 10−3 m
Using the data, eqs (1) and (2) become
0.06667 P − 0.04167 RB = 108
(1)′
0.11433 P − 0.06667 RB = 216
(2)′
Solving simultaneously,
P = 5.63 × 103 N
P = 5.63 kN ↓ 
RB = 6.42 × 103 N
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.92
For the loading shown, and knowing thaat beams AB and DE have
the same flexural rigidity, determine the reaction
r
(a) at B, (b) at E.
SOLUTION
Units: Forces in kN; lengths in m.
For beam ACB, using Case 4 of Appendix D.
( yC )1 = −
RC (2a)3
48EI
For beam DCE, using Case 4 of Appendix D.
D
( yC ) 2 =
( RC − P)(2b)3
48EI
Matching deflections at C,
−
Using free body ACB,
RC (2a)3 ( RC − P )(2b)3
=
48EI
48EI
Pb3
(25)(1.5)3
= 3
= 16.53 kN
N
RC = 3
3
a +b
1.2 + 1.53
P − RC = 25 − 16.53 = 8.47 kN
M A = 0: 2aRB − aRC = 0
( a ) RB =
Using free body DCE,
1
RC = 8.265 kN 
2
M D = 0: 2bRE − b( P − RC ) = 0
(b) RE =
1
( P − RC ) = 4.235 kN 
2
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stuudent using this manual is using it
without permission.
PROBLEM 9.93
A 22-mm-diameter rod BC is attached to the lever AB and
to the fixed support at C. Lever AB has a uniform cross
section 10 mm thick and 25 mm deep. For the loading
shown, determine the deflection of point A. Use
E = 200 GPa and G = 77 GPa
SOLUTION
Deformation of rod BC: (Torsion)
C=
J=
1
1
d = (22) = 11 mm
2
2
π
C 4 = 22998 mm 4
2
T = Pa = (350)(0.25) = 87.5 N ⋅ m
L = 0.5 m
TL
(87.5)(0.5)
=
GJ (77 × 109 )(22998 × 10−12 )
= 0.0247 rad
ϕB =
Deflection of point A assuming lever AB to be rigid:
( yB )1 = aϕB = (0.25)(0.0247)
= 0.006175 m
Additional deflection due to bending of lever AB.
Refer to Case 1 of Appendix D.
1
(10)(25)3 = 13021 mm 4
12
(350)(0.25)3
PL3
( yA ) =
=
3EI (3)(200 × 109 )(13021 × 10−12 )
I=
= 2.1 × 10−3 m
Total deflection at point A:
yA = ( y A )1 + ( y A )2 = 8.28 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.94
A 16-mm-diameter rod has been bent into the shape shown.
Determine the deflection of end C after the 200-N force is applied.
Use E = 200 GPa and G = 80 GPa.
SOLUTION
Let 200 N = P.
Consider torsion of rod AB.
TL ( PL) L PL2
=
=
JG
JG
JG
3
PL
yC′ = − LφB = −
JG
φB =
Consider bending of AB. (Case 1 of Appendix D.)
yC′′ = yB = −
PL3
3EI
Consider bending of BC. (Case 1 of Appendix D.)
yC′′′ = −
PL3
3EI
Superposition:
yC = yC′ + yC′′ + yC′′′
=−
PL3 PL3 PL3
PL3  EI 2 
−
−
=−
+
JG 3EI 3EI
EI  JG 3 
Data:
G = 80(109 ) Pa
E = 200(109 ) Pa
EI = 643.40 N ⋅ m 2
yC = −

1
J = π (0.008) 4 = 6.4340(10−9 ) m 4
2
1
I = J = 3.2170(10−9 ) m 4
2
JG = 514.72 N ⋅ m 2
(200)(0.25)3  643.40 2 
+
= −9.3093(10−3 ) m
643.40  514.72 3 
yC = 9.31 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.95
For the uniform cantilever beam and loading shown, determine (a) the slope at
the free end, (b) the deflection at the free end.
SOLUTION
Place reference tangent at B.
Draw M /EI diagram.
A=
1  PL 
PL2
−
=
−
L


2  EI 
2 EI
x=
2
L
3
PL2
2 EI
 PL3
= Ax =  −
 2 EI
θ B /A = A = −
t A/B
(a)
 2 
PL3
  L  = −
3EI
 3 
Slope at end A:
θB = θ A + A
0 = θA −
(b)
PL2
2 EI
θA =
PL2
2 EI

Deflection at A:
y A = t A /B = −
PL3
3EI
yA =
PL3
↓ 
3EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.96
For the uniform cantilever beam and loading shown, determine (a) the slope at
the free end, (b) the deflection at the free end.
SOLUTION
Place reference tangent at B.
Draw M /EI diagram.
M L
M 
A =  0 L = 0
EI
 EI 
1
x= L
2
M0L
EI
 M L  1 
= Ax =  0   L 
 EI   2 
θ B/A = A =
t B/A
=
(a)
M 0 L2
2 EI
Slope at end A:
θB = θ A + A
θA = −
(b)
M0L
EI
0 = θA +
M0L
EI
θA =
M0L
EI

Deflection at A:
y A = t A /B =
M 0 L2
2 EI
yA =
M 0 L2
↑ 
2 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.97
For the uniform cantilever beam and loading shown, determine (a) the slope at
the free end, (b) the deflection at the free end.
SOLUTION
Place reference tangent at B.
θB = 0
1
L
ΣM B = 0:  w0 L  + M B = 0
2
3
1
M B = − w0 L2
6
Draw M /EI curve as cubic parabola.
1  1 w0 L2 
1 w0 L3
A=− 
L=−

4  6 EI 
24 EI
1
4
x = L− L= L
5
5
By first moment-area theorem,
θ B/A = A = −
1 w0 L3
24 EI
θ B = θ A + θ B/A
θ A = θ B − θ B /A = 0 +
1 w0 L3
1 w0 L3
=
24 EI
24 EI
By second moment-area theorem,
3
1 w0 L4
 4   1 w0 L 
t A/B = xA =  L   −
=−

30 EI
 5   24 EI 
y A = t A/B = −
1 w0 L4
30 EI
(a)
(b)
θA =
w0 L3
24 EI
yA =

w0 L4
↓ 
30 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.98
For the uniform cantilever beam and loading shown, determine (a) the slope at
the free end, (b) the deflection at the free end.
SOLUTION
Place reference tangent at B.
θB = 0
Draw M /EI curve as parabola.
1  wL2 
1 wL3
A = − 
 L = −
3  2 EI 
6 EI
1
3
x =L− L= L
4
4
By first moment-area theorem,
θ B/A = A = −
1 wL3
6 EI
θ B = θ A + θ B /A
θ A = θ B − θ B/A = 0 +
1 wL3 1 wL3
=
6 EI
6 EI
By second moment-area theorem,
3
1 wL4
 3   1 wL 
t A/B = xA =  L   −
 = −
8 EI
 4   6 EI 
y A = t A/B = −
1 wL4
8 EI
(a)
(b)
θA =
wL3
6 EI
yA =

wL4
↓ 
8EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.99
For the uniform cantilever beam and loading shown, determine (a) the slope
and deflection at (a) point B, (b) point C.
SOLUTION
(a) At point B:
θ B = θ B /A = A1 + A2 = −
Pa 2 Pa 2
3Pa 2
−
=−
2 EI
EI
2 EI
θB =
3Pa 2
2EI

 2a 
 3a 
yB = t B /A = A1   + A2  
3
 
 2 
 Pa 2   2a   Pa 2   3a 
11Pa3
=  −
   +  −
   = −
6 EI
 2 EI   3   EI   2 
yB =
11Pa3

6EI
θC =
Pa 2
EI
(b) At point C:
θC = θC /A = A2 = −
A1 =
Pa 2
EI
1
Pa 2
 Pa 
(a )  −
=−
2
2 EI
 EI 
Pa 2
 Pa 
A2 = ( a )  −
=
−

EI
 EI 

a
yC = tC /A = A2  
2
 Pa 2   a 
Pa3
=  −
   = −
2EI
 EI   2 
yC =
Pa3

2EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.100
For the uniform cantilever beam and loadinng shown, determine the
slope and deflection at (a) point B, (b) point C.
C
SOLUTION
Place reference tangent at A. θ A = 0
Draw
M
diagram.
EI
 M  L  1 M 0 L
A1 =  0   =
 EI  2  2 EI
1 M 0L
 M  L 
A2 =  − 0   = −
2 EI
 EI  2 
(a)
Slope at B:
θ B /A = A1 + A2 =
1 M 0L 1 M 0L
−
=0
2 EI
2 EI
θB = 0 
θB = θ A + θB / A = 0
Deflection at B:
L 1 L
1 L
yB = tB /A = A1  + ⋅  + A2  ⋅ 
2 2 2
2 2
=
(b)
3 M 0 L2 1 M 0 L2
1 M 0 L2
−
=
8 EI
8 EI
4 EI
yB =
1 M 0 L2
↑ 
4 EI
θC =
1 M 0L
2 EI
yC =
1 M 0 L2
↑ 
8 EI
Slope at C:
θC / A = A1 =
1 M0L
θC = θ A + θC / A
2 EI

Deflection at C:
2
 1 L  1 M 0L
yC = tC / A = A1  ⋅  =
 2 2  8 EI
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stuudent using this manual is using it
without permission.
PROBLEM 9.101
Two C150 × 12.2 channels are welded back to back and loaded
as shown. Knowing that E = 200 GPa., determine (a) the slope
at point D, (b) the deflection at point D.
SOLUTION
Units: Forces in kN, lengths in m.
E = 200 × 109 Pa
I = (2)(5.35 × 106 ) = 10.7 × 106 mm 4
EI = (200 × 109 )(10.7 × 10 −6 ) = 2140 kN ⋅ m 2
Draw
M
diagram by parts.
EI
M1 (5)(1.8)
9.0 −1
=
=−
m
EI
EI
EI
1  9.0 
81
(1.8) = −
A1 = 
2  EI 
EI
1
x1 = (1.8) = 0.6 m
3
M2
(5)(1.2)
6
=−
=−
m −1
EI
EI
EI
1 6 
3.6
A2 =  −
 (1.2) = −
2  EI 
EI
1
x 2 = (1.2) = 0.4 m
3
M3
(5)(0.6)
3.0 −1
=−
=−
m
EI
EI
EI
1  3.0 
A3 =  −
 (0.6) = −0.9
2  EI 
1
x 3 = (0.6) = 0.2 m
3
Place reference tangent at A.
θA = 0
θ D / A = A1 + A2 + A3 = −
12.6
12.6
=−
= −5.84 × 10 −3 rad
EI
2140
θ D = θ A + θ D / A = −5.88 × 10−3 rad 

19.56
19.56
 81 
 3.6  1.4   3.0 
−3
tD /A =  −
 (1.2) +  − EI  3  +  − EI  (1.6) = − EI = − 2140 = 9.14 × 10 m 
EI




 



yD = t D / A = 9.14 × 10−3 m = 9.14 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.102
For the cantilever beam and loading shown, determine (a) the
slope at point A, (b) the deflecction at point A. Use
E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in m.
E = 200 × 109 Pa
I = 28.7 × 106 mm 4 = 28.7 × 100−6 m 4
EI = (200 × 109 )(28.7 × 106 )
= 5.74 × 106 N ⋅ m 2
= 5740 kN ⋅ m 2
Draw M/EI diagram by parts:
(5)(3.5)
M1
=−
= −3.0488 × 10−3 m −1
5740
EI
1
A1 = (−3.0488 × 10−3 )(3.5) = −5.3354 × 10−3
2
1
x1 = (3.5) = 1.16667 m
3
(4)(2.5) 2
M2
1 −3 m −1
=−
= −2.1777 × 10
EI
(2)(5740)
1
(−2.1777 × 10−3 )(2.5) = −1.81475 × 10−3
3
1
x 2 = (2.5) = 0.625m
4
A2 =
Place reference tangent at C.
θC = 0
θC / A = A1 + A2 = −7.1502 × 10−3
(a)
Slope at A:
θ A = θC − θC / A = 7.1502 × 10−3
θ A = 7.15 × 10−3 rad

t A / C = (2.3333)(−5.3354 × 10−3 ) + (2.875)(−1.81475
1
× 10−3 )
= −17.6665 × 10−3 m
(b)
Deflection at A:
y A = t AC = −17.67 × 10−3 m
y A = 17.67 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stuudent using this manual is using it
without permission.
PROBLEM 9.103
For the cantilever beam and loading shown, determine (a) the
slope at point C, (b) the deflection at point C. Use E = 200 GPa.
SOLUTION
Units: Forces in kN, lengths in m.
E = 200 × 109 Pa
I=
π d 
4
4
π  75 
=   = 1.55 × 106 mm 4
4  2 
4 2 
EI = (200 × 109 )(1.55 × 10 −6 ) = 310 kN ⋅ m 2
Draw
M
diagram by parts.
EI
M1
(7)(0.6)
=−
= −13.5 × 10−3 m −1
310
EI
1
A1 = ( −13.5 × 10−3 )(0.6) = −4.05 × 10−3
2
1
x1 = (0.6) = 0.2 m
3
1
1
M 2 2 (60)(0.4) ( 3 ⋅ 0.4 )
=
= −5.16 × 10−3 m −1
EI
310
1
A2 = ( −5.16 × 10−3 )(0.4) = −0.516 × 10−3
4
1
x 2 = ⋅ (0.4) = 0.1 m
4
Place reference tangent at A.
θA = 0
θC /A = A1 + A2 = −4.566 × 10−3 rad
θC = θ A + θC /A = −4.566 × 10−3 rad

tC /A = (0.4)(−4.05 × 10−3 ) + (0.5)(−0.516 × 10−3 )

= −1.878 × 10−3 m
yC = y A + (2)(θ A ) + tC /A

= 0 + 0 − 1.878 × 10 −3 m 

= 1.878 mm 


PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.104
For the cantilever beam and loading shown, determine (a) the slope
at point A, (b) the deflection at point A. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in meters.
I = 178 × 106 mm 4 = 178 × 10−6
EI = (200 × 109 )(178 × 10−6 ) = 35600 kN ⋅ m 2
Draw
M
EI
diagram by parts.
M 1 (20)(2.1)
=
= 1.17978 × 10−3 m −1
EI
35600
1
A1 =   (2.1)(1.17978 × 10−3 ) = 1.23876 × 10−3
2
M2 = −
( 12 ) (120) (3)(1) = −5.0562 × 10−3 m−1
35600
1
A2 =   (3)(−5.0562 × 10−3 ) = −3.7921 × 10−3
4
Place reference tangent at C.
θC = 0
(a)
Slope at A:
θ A = −θC/A = − A1 − A2
θ A = 2.55 × 10−3 rad
(b)

Deflection at A:
y A = t A/C
3
yC = A1 (3 − 0.7) + A2 (3 − ) = −6.25 × 10−3 m
5
yC = 6.25 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.105
For the cantilever beam and loading shown, determine (a) the slope at
point C, (b) the deflection at point C.
SOLUTION
(a)
A1 =
1  PL  L 
PL2
−
=−



2  EI  2 
4 EI
A2 =
1  PL  L 
PL2
−
=
−

2  3EI 
12 EI
 2 
A3 =
1  PL  L 
PL2
−
=
−

2  2 EI 
8 EI
 2 
Slope at C:
θC = A1 + A2 + A3
=−
PL2  1 1 1 
11PL2
+ + =−

EI  4 12 8 
24 EI
θC =
(b)
Deflection at C:
5 
2 
1 
yC = tC/A = A1  L  + A2  L  + A3  L 
6 
3 
3 
2
2
 PL   5   PL   2   PL2
=  −
  L  +  −
  L  +  −
 4 EI   6   12 EI   3   8 EI
11PL2
24 EI

 1 
22 PL3
  L  = −
72 EI
 3 
yC =
11PL3
↓ 
36 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.106
For the cantilever beam and loading shown, dettermine the deflection and
slope at end A caused by the moment M0.
SOLUTION
Draw
M
diagram.
EI
M 0a
EI
M 0a
A2 = +
2 EI
M 0a
A3 = +
3EI
θ 0 = 0, yD = 0
A1 = +
Place reference tangent at D.
Deflection at A
y A = t A/ D
1 
3 
 5  25M 0a 2
y A = A1  a  + A2  a  + A3  a  =
12EI
2 
2 
2 

Slope at A. θ A = −θC /A
θ A = − A1 − A2 − A3 = −
11M 0 a
6 EI
θA =
11M 0a
6 EI

PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stuudent using this manual is using it
without permission.
PROBLEM 9.107
Two cover plates are welded to the rolled-steel beam as shown.
Using E = 200 GPa, determine the slope and deflection at end C.
SOLUTION
Use units of kN and m
I = 129 × 10−6 m 4
Over portion BC
EI = (200 × 106 )(129 × 10−6 )
= 25800 kN ⋅ m 2
Portion AB:
Top plate
A(mm 2 )
d (mm)
2700
145.5
W310 × 60
Bot. plate
0
2700
145.5
Σ
Ad 2 (mm 4 )
57.16×106
I (mm 4 )
32400
129 × 106
0
57.16 × 106
114.32 × 10
6
32400
129.065 × 106
I = (106 )(114.32 + 129.065) = 243.385 × 106 mm 4
EI = 48677 kN ⋅ m 2
Draw
M
diagram.
EI
M1
(120)(0.9)
=−
= −2.219 × 10−3 m −1
48677
EI
M2
(80)(1.5)
=−
= −2.465 × 10−3 m −1
48677
EI
M4
(80)(0.6)
=−
= −1.86 × 10−3 m −1
25800
EI
1
A1 =   (0.9)(−0.002219) = −0.9986 × 10−3
2
1
A2 =   (0.9)(0.002465) = −1.10925 × 10−3
 2
 0.6 
−3
A3 = A2 
 = −0.4437 × 10
 1.5 
1
A4 =   (0.6)( −0.00186) = −0.558 × 10−3
 2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.107 (Continued)
Place reference tangent at A.
Slope at C:
θ C = θ C /A = A1 + A2 + A3 + A4 = −3.11 × 10 −3
Deflection at C:
yC = tC /A

yC = (1.2)( A1 ) + (1.2)( A2 ) + (0.9)( A3 ) + (0.4)( A4 )
= −3.152 × 10−3 m = 3.15 m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.108
Two cover plates are welded to the rolled-steel beam as
shown. Using E = 200 GPa, determine (a) the slope at
end A, (b) the deflection at end A.
SOLUTION
Portion AB: I = 216 × 106 mm4
EI = (200 × 106 kPa)(216 × 10−6 m 4 ) = 43, 200 kN ⋅ m 2
Portion BC:
A(mm 2 )
Top plate
2400
d (mm)
209
Ad 2 (mm 4 )
104.834 × 106
W410 × 60
Bot. plate
I (mm 4 )
28,800
216 × 106
2400
209
Σ
104.834 × 106
28,800
209.67 × 106
216.06 × 106
I = 209.67 × 106 + 216.06 × 106 = 425.73 × 106 mm 4
EI = (200 × 106 kPa)(425.73 × 10−6 m 4 ) = 85,146 kN ⋅ m 2
Draw
M
diagram:
EI
(40)(0.6)
M1
=−
= −0.55556 × 10−3 m −1
43200
EI
(40)(2.7)
M2
=−
= −1.26841 × 10−3 m −1
EI
85146
M4
(90)(2.1)(1.05)
=−
= −2.3307 × 10−3 m −1
EI
85146
1
(0.6)(−0.55556 × 10−3 ) = −0.166668 × 10−3
2
1
A2 = (2.1)(−1.26841 × 10−3 ) = −1.33183 × 10−3
2
0.6
A3 =
A2 = −0.29596 × 10−3
2.7
1
A4 = (2.1)(−2.3307 × 10−3 ) = −1.63149 × 10−3
3
A1 =
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.108 (Continued)
Place reference tangent at C.
θC = 0
(a)
Slope at A:
θ A = θC − θ A/C = 0 − ( A1 + A2 + A3 + A4 )
(b)
Deflection at A:
yA = t A/ C
θ A = 3.43 × 10−3 rad

y A = (0.4)( A1) + (2)( A2 ) + (1.3)( A3 ) + (2.175)( A4 ) = −6.66
6 × 10−3 m
y A = 6.66 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stuudent using this manual is using it
without permission.
PROBLE
EM 9.109
For the prrismatic beam and loading shown, determine (a) the slope at
end A, (b) the deflection at the center C of the beam.
SOLUTION
Symmetric beam and loading.
Place reference tangent at C.
θC = 0, yC = −t A/C
Reactions:
Bending moment at C:
RA = RB =
MC =
A=
(a)
Slope at A:
1
PL
4
1  1 PL
P  L  1 PL2
=
2  4 EI
E 
 2  16 EI
θ A = θC − θC/A
θA = 0 −
(b)
Deflection at C:
1
P
2
1 PL2
16 EI
θA = −
1 PL2
1 EI
16

 1 PL 2   L 
L
yC = −t A/C = − A   = − 
 16 EI   3 
3


yC =
1 PL3
↓ 
48 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.110
For the prismatic beam and loading shown, determine (a) the slope at
end A, (b) the deflection at the center C of the beam.
SOLUTION




(a)
A1 =
1  3PL  L  3PL2

  =
2  4 EI  2  16 EI
A2 =
1  PL  L 
PL2
−
=
−

 
2  4 EI  4 
32 EI
Slope at A:
θC = θ A + θC/A ; θ A = 0 − θC/A
θ A = −θC/A = −( A1 + A2 )

 3PL2
PL2 
= −
−

 16 EI 32 EI 

=−
5PL2
32 EI
θA =
(b)
5PL2
32 EI

Deflection at C:
L
 5L 
t A/C = A1   + A2 

3
 12 
 3PL2
= 
 16 EI
=
  L   PL2
   +  −
  3   32 EI
  5L 
  
  12 
19 PL3
384 EI
yC = −t A / C = −
19 PL3
384 EI
yC =
19 PL3
↓ 
384 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.111
For the prismatic beam and loading shown, determine (a) the slope at
end A, (b) the deflection at the center C of the beam.
SOLUTION
Symmetric beam and loading.
Place reference tangent at C.
θC = 0
Reactions:
RA = RE = wa
Bending moment:
Over AB:
M = wax −
Over BD:
M=
1 2
wa
2
1 2
wa
2
Draw M /EI diagram by parts:
M1 wa 2
=
EI
EI
M2
1 wa 2
=−
EI
2 EI
M 3 1 wa 2
=
EI 2 EI
1 M1
1 wa3
a=
2 EI
2 EI
1 M2
1 wa3
A2 = −
a=−
3 EI
6 EI
2
M L
 1 wa
A3 = 3  − a  =
( L − 2a )
EI  2
 4 EI
A1 =
(a)
Slope at A:
θ A = θC − θC / A = 0 − ( A1 + A2 + A3 )
1 wa3 1 wa 3 1 wa 2
( L − 2a )
+
−
2 EI
6 EI
4 EI
wa 2  1
1 
=−
L − a

6 
EI  4
=−
=−
1 wa 2
(3L − 2a )
12 EI
θA =
wa 2
(3L − 2a )
12 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.111 (Continued)
(b)
Deflection at C:
yC = −tC/A
2
a,
3
3
x2 = a,
4
1L
 1
x3 = a +  − a  = ( L + 2a)
2 2
 4
x1 =
yC = −tC/A = − A1 x1 − A2 x2 − A3 x3
 1 wa3   2   1 wa3   3  1  wa 2
= − 
  a  + 
  a  − 
 2 EI   3   6 EI   4  4  EI

1
 ( L − 2a) ( L + 2a)
4

1 wa3 1 wa3 1 wa 2 2
+
−
( L − 4a 2 )
3 EI 8 EI 16 EI
1 wa 2
wa 2  1 2 1 2 
=−
−
=
−
(3L2 − 2a 2 )
L
a


24 
48 EI
EI  16
=−
yA =
wa 2
(3L2 − 2a 2 ) ↓ 
48EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.112
For the prismatic beam and loading shown, determine (a) the slope at
end A, (b) the deflection at the center C of the beam.
SOLUTION
Symmetric beam and loading.
Place reference tangent at C.
θC = 0
Reactions:
RA = RB =
Draw
w0 L
4
M
diagram by parts.
EI
M 1 RA L w0 L2
=
=
2
8EI
EI
1  L   M  w L3
A1 =   1  = 0
2  2  EI  32 EI
w0 L2
M2
1 1  w0 L  1 L 
=
⋅ −
⋅
=
−
EI
EI 2  2   3 2 
24 EI
A2 =
(a)
Slope at A:
1  L   w0 L2
−
4  2   24 EI

w0 L3
 = −
192 EI

θ A = −θC/A

1
1  w L3
0
θ A = − A1 − A2 =  − +

EI
32
192


(b)
Deflection at C:
θA =
5w0 L3
192 EI

yC = t A/C
 2  L  
 4  L    1  1   2  1   w0 L4
t A/C = A1     + A2     =    −  

 3  2  
 5  2    3  32   5  192   EI
=
w0 L4
120 EI
yc =
w0 L4
↓ 
120 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.113
For the prismatic beam and loading shown, determine (a) the slope at
end A, (b) the deflection at the center C of the beam.
SOLUTION
Symmetric beam and loading.
θC = 0.
Place reference tangent at C.
Draw
(a)
M
diagram.
EI
θA = 0
Slope at A:
L
 1 M0
 2 − a  = 2 EI ( L − 2a )


1M
θ A = θ C − θ C/A = 0 − A = −
( L − 2a )
2 EI
Α=
M0
EI
θA =
(b)
1 M0
( L − 2a )
2 EI

Deflection at C:
1 L
 1
x = a +  − a  = ( L + 2a )
2 2
 4
yC = −tC/A = Ax
1 M0
1
( L − 2a ) ( L + 2a )
2 EI
4
1 M0 2
=−
( L − 4a 2 )
8 EI
=−
yC =
1 M0 2
( L − 4a 2 ) ↓ 
8 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.114
For the prismatic beam and loading shown, determine (a) the slope at end A, (b)
the deflection at the center C of the beam.
SOLUTION

Symmetric beam and loading.

Place reference tangent at C.

θC = 0

RA = RE =
Reactions:

1
P
2
Draw V (shear) and M/EI diagrams.
A1 = A2 =
(a)
1  1 PL  L 1 PL2

 =
2  8 EI  4 64 EI
Slope at A:
θ A = θC − θ A /C = 0 − A1 − A2
=−
(b)
1 PL2
32 EI
θA =
PL2
32 EI

Deflection at C:
L
 L
yC = −t A/C = −  A1 + A2 
6
3

 1 PL3 L 1 PL3 L 
= − 
⋅ +
⋅ 
 64 EI 6 64 EI 3 


=−
1 PL3
128 EI
yC =
PL3
↓ 
128 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.115
For the beam and loading shown, determine (a) the slope at end A,
(b) the deflection at the center C of the beam.
SOLUTION
Symmetric beam and loading.
1
P
2
1 
=  P  (2a) = Pa
2 
RA = RE =
M max
Draw M and M /EI diagrams.
A1 =
1  Pa 
1 Pa 2

a =
2  2 EI 
4 EI
A2 =
1  Pa

2  4 EI
1 Pa 2

a =
8 EI

A3 =
1  Pa
2  2 EI
1 Pa 2

 a = 4 EI

Place reference tangent at C.
θC = 0
(a)
Slope at A:
θ A = θC − θC /A = 0 − ( A1 + A2 + A3 )
5 Pa 2
=−
8 EI
(b)
θA =
5Pa 2
8EI

Deflection at C:
2 
4 
5 
| yC | = t A/ C = A1  a  + A2  a  + A3  a 
3 
3 
3 
=
1 Pa3 1 Pa3 5 Pa3 3 Pa3
+
+
=
6 EI
6 EI 12 EI
4 EI
yC =
3Pa 3
↓ 
4 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.116
For the beam and loading shown, determine (a) the slope at end A,
(b) the deflection at the center C of the beam.
SOLUTION
Symmetric beam and loading.
RA = RE = 2P.
Draw V, M, and M/EI diagrams.
A1 =
1  2 Pa 
Pa 2
a=


2  EI 
EI
A2 =
1  2 Pa 
1 Pa 2
=
a
2  3 EI 
3 EI
A3 =
1  Pa 
1 Pa 2
=
a
2  EI 
2 EI
Place reference tangent at C.
θC = 0
(a)
Slope at A:
θ A = θC − θC /A = 0 − ( A1 + A2 + A3 )
=−
(b)
11 Pa 2
6 EI
θA =
11 Pa 2
6 EI

Deflection at C:
| yC | = t A/ C
2 
4 
5 
= A1  a  + A2  a  + A3  a 
3 
3 
3 
=
35 Pa3
18 EI
yC =
35Pa3
↓ 
18EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.117
For the beam and loading shown and knowing that
w = 8 kN/m, determine (a) the slope at end A, (b) the
deflection at midpoint C. Use E = 200 GPa.
SOLUTION
E = 200 × 109 Pa
I = 128 × 106 mm 4 = 128 × 10−6 m 4
EI = (200 × 109 )(128 × 10−6 ) = 25.6 × 106 N ⋅ m 2
= 25, 600 kN ⋅ m 2
Symmetrical beam and loading.
RA = RB =
1
(8)(10) = 40 kN
2
Bending moment:
1
M = 40 x − 40 − (8) x 2
2
At x = 5,
M = 200 − 40 − 100
Draw
M
diagram by parts.
EI
M1
200
=
= 7.8125 × 10−3 m −1
EI 25, 600
M2
−40
=
= −1.5625 × 10−3 m −1
EI 25, 600
M3
−100
=
= −3.9063 × 10−3 m −1
EI 25, 600
A1 =
1
(7.8125 × 10−3 )(5) = 19.5313 × 10−3
2
2
x1 =   (5) = 3.3333 m
3
A2 = −(1.5625)(5) = −7.8125 × 10−3
1
x2 =   (5) = 2.5 m
2
1
A3 = − (3.9063)(5) = −6.5105 × 10−3
3
 3
x3 =   (5) = 3.75 m
 4
Place reference tangent at C.
θC = 0
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.117 (Continued)
(a)
Slope at A:
θ A = θ C − θC /A = 0 − ( A1 + A2 + A3 )
θ A = −(19.5313 × 10−3 − 7.8125 × 10−3 − 6.5105 × 10−3 ) = −5.21 × 10−3
θ A = 5.21 × 10−3 rad
(b)

Deflection at C:
| yC | = t A/ C
= (19.5313 × 10 −3 )(3.3333) − (7.8125 × 10−3 )(2.5) − (6.5105 × 10 −3 )(3.75)
= 21.2 × 10−3 m
yC = 21.2 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.118
For the beam and loading shown, determine (a) the
slope at end A, (b) the deflection at the midpoint of
the beam. Use E = 200 GPa.
SOLUTION
Use units of kN and m.
For S250 × 37.8
I = 51.2 × 106 mm4 = 51.2 × 10−6 m4
EI = (200 × 109 )(51.2 × 10−6 )
= 10.24 × 106 N ⋅ m 2 = 10, 240 kN ⋅ m 2
Place reference tangent at midpoint C.
Reactions: RA = RE =
1
(40)(3.6 − 1.2) = 48 kN ↑
2
Draw bending moment diagram of left half of beam by parts.
M1 = (48)(1.8) = 86.4 kN ⋅ m
A1 =
1
(1.8)(86.4) = 77.76 kN ⋅ m 2
2
A2 = (1.8)(−10) = −18 kN ⋅ m 2
1
(40)(1.8 − 0.6) 2 = −28.8 kN ⋅ m
2
1
A3 = (1.2)(−28.8) = −11.52 kN ⋅ m 2
3
1
x = (1.2) = 0.30 m
4
M3 =
(a) Slope at end A. θ A = −θ A / C
θA =
1
−77.76 + 18 + 11.52
{− A1 − A2 − A3} =
10, 240
EI
= −4.71 × 10−3 rad
θ A = 4.71 × 10−3 rad

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.118 (Continued)
(b)
Deflection at midpoint C:
yC = −t A / C
1
{1.2 A1 + 0.9 A2 + (1.8 − 0.3) A3}
EI
(1.2)(77.76) − (0.9)(18) − (1.5)(11.52)
=
= 5.84 × 10 −3 m
10, 240
tA/ C =
yC = −5.84 × 10−3 m
yC = 5.84 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.119
For the beam and loading shown, determine (a) the
slope at end A, (b) the deflection at the midpoint of
the beam. Use E = 200 GPa.
SOLUTION
Use units of kN and m.
For W460 × 74,
I = 333 × 106 mm 4 = 333 × 10−6 m 4
EI = (200 × 109 )(333 × 10−6 )
= 66.6 × 106 N ⋅ m 2 = 66600 kN ⋅ m 2
Symmetric beam and loading. Place reference tangent at midpoint C where θC = 0.
RA = RE = 150 kN ↑
Reactions:
Draw bending moment diagram of left half of beam by parts.
M 1 = (2)(150) = 300 kN ⋅ m
1
A1 =   (2)(300) = 300 kN ⋅ m 2
2
A2 = (0.5)(300) = 150 kN ⋅ m 2
M 3 = −60 kN ⋅ m
A3 = (2.5)( −60) = −150 kN ⋅ m 2
(a)
Slope at end A:
θ A = −θC/A
1
{− A1 − A2 − A3 }
EI
−300 − 150 + 150
=
66600
θA =
= −4.50 × 10−3 rad
(b)
Deflection at midpoint C:
θ A = 4.50 × 10−3 rad

yC = −t A/C
1  2 
0.5 

 2.5  
 ⋅ 2  A1 +  2 +
 A2 + 
 A3 
EI  3 
2


 2  
400 + 337.5 − 187.5
=
= 8.26 × 10−3 m
66600
yC = −8.26 × 10−3 m
tA/ C =
yC = 8.26 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.120
Knowing that P = 8 kN, determine (a) the slope at end A,
(b) the deflection at midpoint C. Use E = 200 GPa .
SOLUTION
E = 200 × 109 Pa
I = 16.6 × 106 mm 4 = 16.6 × 10−6 m 4
EI = (200 × 109 )(16.6 × 10−6 ) = 3.32 × 106 N ⋅ m 2
= 3320 kN ⋅ m 2
Symmetric beam and loading:
RA = RB = P + 5 = 8 + 5 = 13 kN
Bending moment:
Over AB:
M = − Px = −8 x
Over BC:
M = −8 x + 13( x − 1) = 5( x − 1) − 8
Draw
M
diagram by parts.
EI
A1 =
1  8.5 
7.225
(1.7) =


2  EI 
EI
1 8 
4
A2 = − 
 (1) = −
2  EI 
EI
13.600
 8 
A3 = − 
(1.7) = −

EI
 EI 
Place reference tangent at C.
(a)
θC = 0
Slope at A: θ A = θC − θC /A = 0 − ( A1 + A2 + A3)
 7.225
4
θB = − 
−
−
EI
 EI
(b)
Deflection at C:
13.600  10.375 10.375
=
=
= 3.125 × 10−3 rad
EI 
EI
3320

yC = −t B / C
= −( A1x1 + A3 x 3 )
 7.225   2
  13.600   17   3.3717 3.3717
= − 
=
  (1.7)  − 
   =
EI
3320
  EI   2  
 EI   3
= 1.016 × 10−3 m = 1.016 mm

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.121
For the beam and loading of Prob.
P
9.117, determine the
value of w for which the deflection
d
is zero at the
midpoint C of the beam. Use E = 200 GPa.
SOLUTION
Symmetric beam and loading:
RA = RB = 5w
( w in kN/m)
Bending moment in kN⋅m:
M = 5wx − 40 −
1 2
wx
2
At x = 5 m,
M = 25w − 40 − 12.5w
Draw M /EI diagram by parts.
1  25w 
62.5w
(5) =


2  EI 
EI
(40)(5)
200
A2 = −
=−
EI
EI
1  12.5w 
20.833w
A3 = − 
(5) = −
EI
3  EI 
2
x1 = (5) = 3.3333 m
3
1
x2 = (5) = 2.5 m
2
3
x3 = (5) = 3.75 m
4
A1 =
Place reference tangent at C.
Deflection at C is zero.
t A/C = y A − yC = 0
A1x1 + A2 x2 + A3 x3 = 0
 200 
 20.833 w 
 62.5 w 
 (3.75) = 0
 EI  (3.3333) −  EI  (2.5) − 
EI






130.21 w 500
−
=0
EI
EI
500
w=
= 3.84 kN/m
130.21
w = 3.84 kN/m 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.122
For the beam and loading of Prob. 9.120, determine the
magnitude of the forces P for which the deflection is zero at
end A. Use E = 200 GPa.
SOLUTION
Symmetric beam and loading:
RA = RB = P + 5
(P in kN)
Bending moment:
Over AB:
M = − Px kN ⋅ m
Over BC:
M = − Px + ( P + 5)( x − 1)
= 5( x − 1) − P (1)
x = 2.7 m
M = 8.5 − P(1)
At
Draw
M
diagram by parts.
EI
1  8.5 
7.225
A1 = 
(1.7) =
EI
2  EI 
1 P 
0.5P
A2 = − 
 (1) = −
EI
2  EI 
1.7 P
 P 
A3 = − 
(1.7) = −

EI
 EI 
Place reference tangent at C.
y A = yB = 0
y A − yB = 0
t A/ C − t B / C = 0
2


 1

2
2

1

A1 1 + ⋅ 1.7  + A3 1 + ⋅ 1.7  + A3   − A1  ⋅ 1.7  − A3  ⋅ 1.7  = 0
3
2
3
3
2




 




2
A1 (1) + A3 (1) + A2   = 0
3
7.225 1.7 P 0.33333P
7.225
−
−
=0 P=
= 3.55 kN
2.0333
EI
EI
EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.123*
A uniform rod AE is to be supported at two pointts B and D. Determine the
distance a for which the slope at ends A and E is zero.
z
SOLUTION
Let w = weight per unit length of
o rod.
Symmetric beam and loading:
RB = RD =
1
wL
2
Bending moment:
Over AB:
1
M = − wx 2
2
Over BCD:
1
1
M = − wx 2 + wL( x − a )
2
2
Draw M /EI diagram by parts.
M1 1 wL( L2 − a ) 1 wL( L − 2a)
=
=
EI 2
EI
4
EI
2
M 2 1 w( L2 )
1 wL2
=
=−
EI 2 EI
8 EI
2
1 M1  L
 1 wL( L − 2a )
A1 =
a
−
=
 16
EI
2 EI  2

1 M  L
1 wL3
A2 =  2  = −
3  EI  2
48 EI
Place reference tangent at C.
θC = 0
θ A = θC − θC/A = 0 − ( A1 + A2 ) = 0
−
Let u =
1 wL( L − 2a)2 1 wL3
+
=0
16
EI
48 EI
wL3
a
and divide by
.
48EI
L
1 − 3(1 − 2u ) 2 = 0
3
3
1
3
u = 1 −
 = 0.21132
2 
3 
a
= 0.211
L
1 − 2u =
a = 0.211 L 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBL
LEM 9.124*
A uniform
m rod AE is to be supported at two points B and D. Determine
D
the
distance a from the ends of the rod to the points of suppport, if the
downwardd deflections of points A, C, and E are to be equal.
SOLUTION
Let w = weight per unit length of rod.
Symmetric beam and loading:
RB = RD =
1
wL
2
Bending moment:
Over AB:
1
M = − wx 2
2
Over BCD :
1
1
M = − wx 2 + wL( x − a )
2
2
Draw M /EI diagram by parts.
wL( L2 − a) 1 wL( L − 2a )
M1 1 w
=
=
EI 2
EI
4
EI
2
2
L
M2
1 w( 2 )
1 wL
=−
=−
EI
2 EI
8 EI
2
M
1 1L
 1 wL ( L − 2a)
A1 =
a
−
=


2 EI  2
EI
 16
1 wL3
1  M  L 
A2 =  2    = −
48 EI
3  EI   2 
2 L
 1
x1 = a +  − a  = ( L + a)
3 2
 3
x2 =
L 1 L 3
−  = L
2 4 2  8
Place reference tangent at C.
y A − yc = t A/C = 0
A1 x1 + A2 x2 = 0
1 wL3 3
1 wL( L − 2a) 1
( L + a) −
L=0
3
48 EI 8
16
EI
2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.124* (Continued)
Let u =
wL4
a
.
. Divide by
L
48EI
3
=0
8
5
4u 3 − 3u + = 0
8
(1 − 2u ) 2 (1 + u ) −
Solving for u,
u = 0.22315
a
= 0.223
L
a = 0.223 L 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.125
For the prismatic beam and loading shown, determine (a) the deflection
at point D, (b) the slope at end A.
SOLUTION
From Statics,
Draw
RA = RB = 0
M
diagram.
EI
 M  L  1 M 0 L
A =  0   =
 EI  4  4 EI
Place reference tangent at A.
5 M 0 L2
L L
t B /A = A  +  =
 2 8  32 EI
1 M 0 L2
L
t D /A = A   =
 8  32 EI
(a)
Deflection at D:
xD
1
t B /A = t D /A − t B / A
L
2
1 M 0 L2 5 M 0 L2
3 M 0 L2
=
−
=−
32 EI
64 EI
64 EI
y D = t D /A −
yD =
(b)
3M 0 L2

64 EI
Slope at A:
θA = −
t B /A
5 M0L
=−
32 EI
L
θA =
5M 0 L
32 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.126
For the prismatic beam and loading shown, determine (a) the deflection
at point D, (b) the slope at end A.
SOLUTION
From Statics,
RA =
1
2
P , RB = P
3
3
2  2
M D = RA  L  = PL
3  9
Draw
M
diagram.
EI
A1 =
1  2L  2PL  2PL2


=
2  3  9EI  27 EI
A2 =
1  L  2PL 
PL2
 
=
2  3  9EI  27 EI
Place reference tangent at A.
 L 1 2L 
 2 L  4PL3
tB /A = A1  + ⋅
 + A2  ⋅  =
3 3 3 
 3 3  81EI
4PL3
 L 2L 
tD /A = A1  ⋅
=
 3 3  243EI
(a)
Deflection at D:
xD
t B /A
L
4 PL3  2  4 PL3
yD =
−
243EI  3  81EI
y D = t D /A −
= −
(b)
Slope at end A:
θA = −
4 PL3
243EI
t B /A
4 PL2
=−
81 EI
L
yD =
θD =
4 PL3

243EI
4 PL2
81EI

PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.127
For the prismaatic beam and loading shown, determine (a) the deflection at
point D, (b) thee slope at end A.
SOLUTION
1  1 w0 L2

2  6 EI
t B /A =
2

 L  1  1 w0 L
 ( L)   +  −
 3  4  6 EI

7 w0 L4
360 EI
1  1 w0 L2
= 
2  12 EI
 L
 ( L)  
 5
=
tD/A
=
(a)
  L  L  1  1 w0 L2
    +  −
  2  6  4  48 EI
  L  L 
   
  2  10 
37 w0 L4
11520 EI
Deflection at D:
1
t B/A − t D/A
2
1  7 w L4  37 w0 L4
=  0 −
2  360 EI  11520 EI
yD =
=
75w0 L4
11520 EI
yD =
(b)
5w0 L4
↓ 
768EI
Slope at A:
θA = −
t B/A
7 w L3
=− 0
L
360 EI
θA =
7 w0 L3
36
60 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.128
For the prismatic beam and loading shown, deterrmine (a) the deflection at
point D, (b) the slope at end A.
SOLUTION
ΣM A = 0: RB L −
wL  L 
=0
2  4 
1
RB = wL ↑
8
Draw M /EI diagram by parts.
M 1 RB L wL2
=
=
EII
EI
8 EI
3
M2
wL
=−
8 EI
EII
2
wL3
 1  L  wL
A1 =   
=
 2  2  16 EI 64 EI
2
wL3
 1  L  wL
A2 =   
=−
48EI
 3  2  8EI
2
wL3
 1  L  wL
A3 =   
=
 2  2  16 EI 64 EI
2
wL3
 1  L  wL
A4 =   
=
 2  2  8 EI 32 EI
(a)
Deflection at D:
Place reference tangent att B.
L
tA/ B
2
wL4
1 L
=  ⋅  A1 =
384 EI
3 2
L
7 wL4
1 L
= ( A1 + A3 + A4 ) +  ⋅  A4 =
3
384 EI
4 2
yD = tD / B −
t D/ A
t B/A
yD =
wL4
1 7 wL4
5wL4
− ⋅
=−
384 EI
E
2 384 EI
768EI
yD =
5wL4
↓ 
768EI
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.128 (Continued)
(b)
Slope at A:
Place reference tangent at A.
1
L
1 L 
 1   2 L 

= −   
 ( A1 + A3 + A4 ) +  L − ⋅  A2 
4 2 
 L   3 

θ A = − t B /A
=−
3wL3
128EI
θA =
3wL3
128EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.129
For the beam and loading shown, determine (a) the slope at
end A, (b) the deflection at point D. Use E = 200 GPa.
SOLUTION
E = 200 × 109 Pa
I = 70.8 × 106 mm 4 = 70.8 × 10 −6 m 4
EI = (200 × 109 )(70.8 × 10−6 ) = 19.16 × 106 N ⋅ m 2
= 14,160 kN ⋅ m 2
ΣM B = 0: − 6RA + (4.5)(40) + (3)(20) = 0
RA = 40 kN
M
diagrams.
Draw shear and
EI
1  60 
45
A1 = 
 (1.5) =
EI
2  EI 
90
 60 
A2 = 
 (1.5) =
EI
 EI 
1  60 
90

 (3) =
2  EI 
EI
Place reference tangent at A.
tB /A = A1(4.5 + 0.5) + A2 (3 + 0.75) + A3 (2.0)
A3 =
742.5
m
EI
= A1(1.5 + 0.5) + A2 (0.75)
=
t D /A
157.5
m
EI
t
742.5
123.75
123.75
θ A = − B/A = −
=−
=−
L
6EI
EI
14,160
=
(a)
Slope at A:
= −8.74 × 10−3
(b)
θ A = 8.74 × 10−3 rad

Deflection at D:
yD = tD/A −
xD
157.5  3  742.5 
213.75
−  
tB/A =
=−
L
EI
EI
EI
6
 

213.75
=−
= −15.10 × 10−3 m
14,160
yD = 15.10 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PR
ROBLEM 9.130
Foor the beam and loading shown, determine (a) the sloope at end A,
(b) the deflection at point D. Use E = 200 GPa.
SOLUTION
Units: Forces in kN; lengths in meters.
I = 13.4 × 106 mm4
For W150 × 24,
= 13.4 × 10−6 m 4
EI = (200 × 109 )(13.4 × 10−6 ) = 2.68 × 106 N ⋅ m 2
= 2680 kN ⋅ m 2
ΣM B = 0: − 2.4 RA + (0.8)(30) + (1.2)(2.4)(20) = 0
RA = 34 kN ↑
Draw bending moment diagram by parts.
M1 = (1.6)(34) = 54.4 kN
k ⋅m
M 2 = (2.4)(34) = 81.6 kN
k ⋅m
1
M 3 = − (20)(1.6) 2 = −25.6 kN ⋅ m
2
1
M 4 = − (20)(2.4) 2 = −57.6 kN ⋅ m
2
M 5 = −(0.8)(30) = −24 kN ⋅ m
1
(1.6)(54.4) = 433.52 kN ⋅ m 2
2
1
= (2.4)(81.6) = 977.92 kN ⋅ m 2
2
1
= (1.6)(−25.6) = −13.6533 kN ⋅ m 2
3
1
= (2.4)(−57.6) = −46.08 kN ⋅ m 2
3
1
= (0.8)(−24) = −9.6
9 kN ⋅ m 2
2
A1 =
A1 + A2
A3
A3 + A4
A5
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.130 (Continued)
(a)
Slope at A: Place reference tangent at A.
1
L

1 
1
1
1
=
( A1 + A2 )   (2.4) + ( A3 + A4 )   (2.4) + A5   (0.8) 
EI 
3
4
3

θ A = − t B /A
t B /A
48.128
= 17.9582 × 10−3 m
2680
17.9582 × 10−3
= −7.48258 × 10−3
θA = −
2.4
=
θ A = 7.48 × 10−3 rad.
(b)

Deflection at point D:
y D = t D /A + θ A xD

1  1
1
 A1   (1.6) + A2   (1.6) 
EI   3 
4

17.7493
=
= 6.62289 × 10−3 m
2680
yD = 6.62289 × 10 −3 + ( −7.48258 × 10−3 )(1.6)
t D /A =
= −5.3492 × 10−3 m
yD = 5.35 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
P
PROBLEM
9.131
Foor the beam and loading shown, determine (a) the
t slope at
pooint A, (b) the deflection at point E. Use E = 200 GPaa.
SOLUTION
Units:
Forces in kN; lengths in m.
For
W310 × 28.7,
I = 85.1 × 10−6 m4
EI = (200)(885.1) = 17020 kN ⋅ m2
ΣM B = 0: − 3RA + (75)(1.2)(1.8) + (120)(1.2)(0.6) = 0
RA = 82.8 kN
Consider loading as 75 kN/m from D to B plus 45 kN/m from E to B. Draw bending momentt diagram by
parts.
M 1 = 3RA = 248.2 kN ⋅ m
M 2 = 1.8 RA = 149 kN ⋅ m
1
M 3 = − (75)(2.4) 2 = −216 kN ⋅ m
2
1
M 4 = − (75)(1.2)2 = −54 kN ⋅ m
2
1
M 5 = − (45)(1.2) 2 = −32.4 kN ⋅ m
2
1
A1 + A2 = (3)(248.4) = 392.3 kN ⋅ m 2
2
1
A1 = (1.8)(149) = 134 kN ⋅ m 2
2
1
A3 + A4 = (2.4)(−216) = −172.8 kN ⋅ m 2
3
1
A3 = (1.2)( −54) = −21.6 kN ⋅ m 2
3
1
A5 = (1.2)( −32.4) = −13
3
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.131 (Continued)
(a)
yB = y A + θ A L + t B/A
Slope at A:
y A = yB = 0
θ A = −t B /A / L

1 
1
1
1
( A1 + A2 )   (3) + ( A3 + A4 )   (2.4) + ( A5 )   (1.2) 
EI 
3
4
4

264.72
=
= 0.01555 m
17020
0.01555
θA = −
= −5.183 × 10−3
3
t B /A =
θ A = 5.18 × 10−3 rad
(b)
Deflection at E:

yE = xEθ A + tE /A

 73.92
1
1
= 4.343 × 10−3 m
( A1 )   (1.8) + ( A3 )   (1.2)  =
3
4
17020
 
 


yE = (1.8)( −0.005183) + 0.004343 = −0.00499 m
t E /A =
1
EI
= 5 mm

[I
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.132
For the timber and loading shown, determine (a) the slope at
end A, (b) the deflection at the midpoint C. Use E = 12 GPa.
SOLUTION
Units:
Forces in kN, lengths in meters.
I=
1
(50)(150)3 = 14.0625 × 106 mm 4
12
= 14.0625 × 10−6 m 4
EI = (12 × 109 )(14.0625 × 10 −6 ) = 168.75 × 103 N ⋅ m 2 = 168.75 kN ⋅ m 2
ΣM D = 0: − 2RA + (1.5)(4) + (0.5)(5) = 0
RA = 4.25 kN
w( x) = 5 x − 1 0 kN ⋅ m
dV
= − w = −5 x − 1 0 kN/m
dx
dM
= V = −5 x − 11 + 4.25 − 4 x − 0.5 0 kN
dx
d2y
5
= M = −  x − 1 2 + 4.25 x − 4 x − 0.51 kN ⋅ m
2
2
dx
dy
5
EI
= −  x − 1 3 + 2.125 x 2 − 2 x − 0.5 2 + C1 kN ⋅ m 2
dx
6
5
2.125 3 2
EIy = −  x − 1 4 +
x −  x − 0.5 3 + C1 x + C2 kN ⋅ m3
24
3
3
EI
[ x = 0, y = 0]:
−0 + 0 − 0 + 0 + C2 = 0
∴C2 = 0
 5 
 2.125  3  2 
3
[ x = 2 m, y = 0] : −   (1) 4 + 
 (2) −   (1.5) + 2C1 = 0
 24 
 3 
3
C1 = −1.60417 kN ⋅ m2
(a)
Slope at end A:
 dy

 dx at x = 0 


 dy 
EI   = −0 + 0 − 0 + C1
 dx  A
C1 −1.60417
 dy 
=
= −9.51 × 10−3
  =
168.75
 dx  A EI
θ A = 9.51× 10−3 rad

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.132 (Continued)
(b)
Deflection at midpoint C:
( y at x = 1 m)
 2.125  3  2 
3
−3
3
EIyC = −0 + 
 (1) −   (0.5) + (−1.60417)(1) = −979.17 × 10 kN ⋅ m
 3 
3
yC =
−979.17 × 10 −3
= −5.80 × 10−3 m
168.75
yC = 5.80 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.133
For the beam and loading shown, determine (a) the slope at point A,
(b) the deflection at point D.
SOLUTION
 MC = 0 :
Draw
− RA L + P
L
L
−P =0
2
2
RA = 0
M
diagram.
EI
1  PL  L 
1 PL2
=
−
A1 = − 
 
2  2EI  2 
8 EI
1  PL  L 
1 PL2
A2 = − 
=
−
 
2  2EI  2 
8 EI
Place reference tangent at A.
1 PL3
1 L
tC / A = A1 ⋅  ⋅  = −
48 EI
3 2
(a)
Slope at A:
θA = −
tC/A
L
θA =
1 PL2

48 EI


(b)
Deflection at D:
1 PL3
L L
2 L
t D / A = A1  +  + A2  ⋅  = −
8 EI
2 6
3 2
yD = t D / A −
xD
1 PL3  3   1 PL3 
tC / A = −
−    −

L
8 EI
 2   48 EI 
yD = −
3 PL3

32 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.134
For the beam and loading shown, determine (a) the slope at point A,
(b) the deflection at point A.
SOLUTION
M 0a
EI
M0L
A2 = −
2 EI
 2L 
tC/B = A2 

 3 
 M L  2 L 
=  − 0 

 2 EI   3 
A1 = −
=−
(a)
M 0 L2
3EI
Slope at A:
t C/B M 0 L
=
3EI
L
θ B = θ A + θ B/A = θ A + A1
θB =
M a
M0L
= θA − 0
3EI
E
EI
(b)
θA =
M0
( L + 3a )
3EI

Deflection at A:
M a2
a
t A/B = A1   = − 0
2 EI
2
a
y A = tC/B + t A/B
L
a  M L2  M a 2
= − 0  − 0
L  3EI  2 EI
yA =
M 0a
(2 L + 3a ) ↓ 
6 EI
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.135
For the beam and loading shown, determine (a) the slope at
G
point C, (b) the deflection at point D. Use E = 200 GPa.
SOLUTION

Freee Body AD:
ΣM C = 0: (64)(1.8) − (144)(0.6) − 3.6
3 RA = 0

RA = 8 kN
ΣFy = 0: 8 − 64 + RB − 144 = 0
RB = 200 kN
For W310 × 44.5:
I = 99.2 × 10−6 m4
EI = (200 × 106 )(99.2 × 10−6 ) = 19840 kN ⋅ m2
(a) Slope at C:
1
(28.8)(3.6) = 51.84 kN ⋅ m 2
2
1
A2 = (−115.2)(1.8) = −103.68 kN
N ⋅ m2
2
= A1 (2.4 m) + A2 (3 m)
A1 =
EIt A/C



= (51.84)(2.4) + ( − 103.68)(3) = −186.624 kN ⋅ m3
186.624
t A /C = −
= −9.406 × 10 −3 m
19840
t
9.406 × 10−3 m
θ C = A /C = −
L
3.6 m
θC = 2.61× 10−3 rad

(b) Deflection at D:
1
EItD /C = A1 (0.9 m) = ( −86.4)(1.2)(0.9) = −31.1 kN
N⋅m
3
31.1
t D /C = −
= −1.5675 × 10−3 m
19840
1.2
1
y D = t D /C +
t A/C = −1.5675 × 10−3 + (−9.4066 × 10−3 )
36
3
= −4.703 × 10−3 m
yC = 4.7 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.136
For the beam and loading shown, determinne (a) the slope at point B,
(b) the deflection at point D. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in
i meters.
I = 462 × 106 mm 4 = 462 × 10−6 m 4
EI = (200 × 109 )(462 × 10−6 )
= 92.4 × 106 N ⋅ m 2 = 92400 kN ⋅ m 2
ΣM B = 0: − 4.8 RA + (40)(4.8)(2.4) − (160)(1.8) = 0
RA = 36 kN
Draw bending moment diagram by parts.
1
(4.8)(172.8) = 414.72 kN ⋅ m 2
2
1
A2 = (4.8)(−460.8) = −737.28 kN ⋅ m 2
3
1
A3 = (1.8)(−288) = −259.2 kN ⋅ m 2
2
A1 =
Place reference tangent at B.
(a)
Slope at B.
y A = yB − Lθ B + t A/B
θB =
=
(b)

t B/A
1  2
3
=
 A1   (4.8) + A2   (4.8) 
L
EIL   3 
4

−1327.104
= −2.99922 × 10 −3
(92400)(4.8)
θ B = 2.99 × 10−3 rad

Deflection at D.
yD = yB + aθ B + tD/B
= 0 + (1.8)(−2.9922 × 10
1 −3 ) −
= −5.3860 × 10−3 −
= −8.75 × 10−3 m
1
EI
 2

A3   (1.8) 
 3

311.04
922400
yD = 8.75 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.137
Knowing that the beam AB is made of a solid steel rod of
diameter d = 18 mm, determine for the loading shhown (a) the
slope at point D, (b) the deflection at point A. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in m.
C=
I=
1 1
= (18) = 9 mm
m
2 2
π
π
5
mm 4
c 4 = (9)4 = 5153
4
4
EI = (200 × 106 )(5153 × 10−12 ) = 1.0306 kN ⋅ m 2
Draw
M
EI
diagram by parts by considering thee bending moment diagram due to each of the appliedd loads.
M1 (0.6)(0.1)
=
= 0.0582 m −1
EI
1.0306
M2
(1.2)(0.15)
=−
= −0.11747 m −1
EI
1.0306
1
A1 = (0.6)(0.0582) = 0..01746
2
1
A2 = (0.6)(−0.1747) = −0.05241
2
1
A3 = (0.1)(0.0582) = 0.00291
2
Place reference tangent at D.
(a)
Slope at point D:
−t E /D
L
−3
= 0.4 A1 + 0.2 A2 = −3.498 × 10
yE = yD + Lθ D + tE /D
t E /A
θD =
(b)
θD =
−0.003498
= −5.83 × 10−3
0.6
θ D = 5.83 × 100−3 rad

Deflection at A:
y A = y D − aθ D + t A /D = t A /D − aθ D
2
y A = A3   (0.1) − (0.1)(0.00583) = −0.389 × 10−3 m
3
y A = 0.39 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.138
Knowing that the beam AD is maade of a solid steel bar,
determine the (a) slope at point B, (b) the deflection at
point A. Use E = 200 GPa.
SOLUTION
E = 200 × 109 Pa
I=
1
(30)(30)3 = 67.5 × 103 mm 4 = 67.5 × 10−9 m 4
12
EI = (200 × 109 )(67.5 × 100−9 ) = 13500 N ⋅ m 2 = 13.5 kN ⋅ m 2
ΣM B = 0: −(0.2)(1.2)) − (3)(0.25)(0.125) + 5RD = 0
Draw
RD = 0.6675 kN
M
diagram by parts.
EI
M 1 = (0.6675)(0.5) = 0.333375 kN ⋅ m
k ⋅m
M 2 = (1.2)(0.2) = 0.240 kN
1
M 3 = − (3)(0.25)2 = −0.09375 kN ⋅ m
2
1
(0.33375)(0.5)/EII = 0.0834375/EI
2
1
A2 = (0.240)(0.2)/EI = 0.024/EI
2
1
A3 = (−0.09375)(0.25)//EI = −0.0078125/EI
3
A1 =
Place reference tangent at B.

2

3
tD/B = A1  ⋅ 0.5  + A3  ⋅ (0.25) + 0.25  = 0.024395/EI
3

4

(a)
Slope at B:
θB = −
t D/B
0.024395
0.048789
=−
=−
0.5EI
L
EI
= −3.6140
3
× 10−3
θ B = 3.61 × 10−3 rad

2

t A/B = A2  (0.20)  = 0.0032/EI = 0.23704 × 10−3 m
3

(b)
Deflection at A:
y A = t A/B − LABθ B
= 0.23704 × 10 −3 − (0.2)(−3.6140 × 10−3 ) = 0.960 × 10−3 m
y A = 0.960 mm ↑ 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLE
EM 9.139
For the beeam and loading shown, determine the deflection (aa) at point D,
(b) at pointt E.
SOLUTION
A1 =
1  PL  L  PL2
=
2  6 EI 
 3  36 EI
2
 PL  L  PL
A2 = 
  =
 6 EI  3  18EI
A3 =
1  PL  L  PL2
=
2  3EI 
 3  18 EI
2
 L   PL
tD/A = A1   = 
 9   36 EI
 L 
PL3
   =
  9  324 EI
L L
L
tE/A = A1  +  + A2  
9 3
6
 PL3

 36 EI
=
  4 L   PL2
 
 + 
  9   18 EI
 L 
  
 6 
7 PL3
324 EI
L
 2L 
 7L 
+ A2   + A3 
tB/A = A1 


 9 
2
 9 
 PL3
= 
 36 EI
=
  7 L   PL2
 
 + 
  9   18EI
  L   PL2
   + 
  2   18EI
5PL3
81EI
(a)
Deflection at D:

1
1  5PL3 
PL3
17 PL3
yD = t B/A − t D/A = 
=

 −
3
3  81EI  324 EI 972 EI
(b)
Deflection at E:
yE =
  2L 
 

 9 
2
2  5PL3  7 PL3
19 PL3
t B/A − t E/A = 
=
 −
3
3  81EI  324 EI 972 EI
yD =
17 PL3
↓ 
972 EI
yE =
19 PL3
↓ 
972 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.140
For the beam and loading shown, determine (a) thhe slope at end A, (b) the
slope at end B, (c) the deflection at the midpoint C.
SOLUTION
RA = RB =
Reactions:
1
wL
2
EI diagrams by parts as shown.
Draw bending moment and M/E
1 L wL2
wL3
⋅ ⋅
=
2 2 4 EI 16 EI
1 L wL2
wL3
A2 = − ⋅ ⋅
=−
3 2 8EI
48EI
2
1 L wL
wL3
A3 = ⋅ ⋅
=−
2 2 8EI
32 EI
2
1 L wL
wL3
A4 = − ⋅ ⋅
=−
3 2 16 EI
96 EI
A1 =
Place reference tangent at A.
(a)
Slope at end A:
yB = y A + Lθ A + t B/A
θ A = −t B/A /L
L
3L
L L
L L
tB/A =  +  A1 +  +  A2 + A3 +
A4
3
8
2 6
2 8
=
wL4  1
5
1
1  9wL4
−
+
−
=

EI  24 384 96
9
256  256 EI
θA = −
(b)
9wL4 1
9wL
w 3
⋅ =−
256 EI L
256 EI
θA =
9wL3
256 EI

θB =
7 wL3
256 EI

Slope at end B:
θ B = θ A + θ B/A = −
θB =
7 wL3
256 EI
9wL3
+ A1 + A2 + A3 + A4
256 EII
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.140 (Continued)
(c)
Deflection at midpoint C:
L
θ A + tC/A
2
wL4
L
L
=   A1+   A2 =
128 EI
6
8
y A = yC +
t C /A
3
 L   9wL
yC = 0 +    −
 2   256 EI
 wL4
5wL4
=−
 +
512 EI
 128EI
yC =
5wL4
↓ 
512 EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.141
For the beam and loading shown, determine thhe magnitude and location
of the largest downward deflection.
SOLUTION
From the solution to Problem 9.126,
1
4 PL2
P , θA = −

3
81 EI
M R0 x Px
=
=
EI
3EI
EI
RA =
Over portion AD,
Ak =
1  Pxk  Fxk2
=
xk
2  3EI  6 EI
θ K = θ A + AK = 0
4 PL
P 2 PxK2
+
=0
E
6 EI
81EI
8 2
xK2 =
L
27
xK = 0.54433L
−
AK = −θ A =
4 PL2
81EI
PL
2 
t A/ K =  xk  Ak = 0.01792
EI
3 
y A = yk + t A / K = 0
3
yK = −t A/ K = −0.01792


PL3
EI
yK = 0.01792
PL3

EI
xK = 0.544 L 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.142
For the beam and
a loading of Prob. 9.127, determine the magnitude and location
of the largest doownward deflection.
SOLUTION
θA = −
From Prob. 9.127:
7 w0 L3
360 EII
A1 =
w0 Lxm2
1  w0 L 
(
)
=
x
x
m
m

12 EI
2  6 EI

A2 =
w0 xm4
1  w0 xm3 
 −
 ( xm ) = −
E 
4  6 EIL
24 EIL
Maximum deflection occurs at K, where θ K = 0.
θ K = θ A + θ K/A = θ A + A1 + A2
0=−
0=
Rearranging:
7 w0 L3 w0 Lxm2
w x4
+
− 0 m
24 EIL
360 EI 12 EI
w0 L2
360 EI
2
4

x 
x  
 −7 + 30  m  − 15  m  
 L 
 L  

2
Solving biquadratic:
 xm 
 L  = 0.26970


xm = 0.51933L

ym is 0.519L from A. 
t A/K = A1



 w Lx 2
2 xm
4x
+ A2 m =  0 m
 12 EI
3
5

w L4
= 0
90 EI
 2 xm  w0 xm4  4 xm
+−

 24 EIL  5
 3


5
4
  x 3
x   wL
5  m  − 3  m   = 0 5(0.51933)3 − 3(0.51933)5 
 L   90 EI
  L 
= 0.0065222
w0 L4
EI

ym = |t A/K | 
ym = 6.52 × 10
1 −3
w0 L4

EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.143
For the beam and loading of Prob. 9.129, determine the
magnitude and location of the largest downward deflection.
SOLUTION
Referring to the solution to Prob. 9.129,
EI = 14,160 kN ⋅ m 2
RA = 40 kN,
A1 =
45
EI
742.5
m
EI
123.75
θA = −
EI
t B/A =
Let K be the location of the maximum deflection. Assume
that K lies between C and D.
θ K = θ A + θ K/ A
123.75
+ A1 + A4
EI
123.75 45 60u
=−
+
+
=0
EI
EI
EI
=−
123.75 − 45
= 1.3125 m
60
xK = 1.5 + u = 2.8125 m
u =
1 
t K/A = A1(u + 0.5) + A4  u 
2 
=
(60)(1.3125) ( 12 ) (1.3125) 133.242
45
=
(1.8125) +
EI
EI
EI
xK
t B/A
L
133.242 2.8125  742.5 
214.80
214.80
=
−
=−

=−
EI
EI
6  EI 
14,160
yK = t K/A −
= −15.17 × 10−3 m
yK = 15.17 mm ↓ 

xK = 2.81 m 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.144
For the beam and loading shown, determine the magnitude
m
and
location of the largest downward deflection.
PROBLEM 9.144 Beam and loading of Prob. 9.131.
SOLUTION
From the solution to Problem 9.131
EI = 170220 kNm 2
RA = 82.88 kN
A1 = 134 kNm 2
A3 = −21.6 kNm 2
θ A = −0.0005183
Slope at E:
θ E = θ A + θ E/A
1
278.767
= 6.604 × 10−3
{ A1 + A3 } =
EI
41083
θ E = 1.4211 × 10−3
θ E /A =
Since θ E > 0, the point K of zero slope lies too the left of point E. Let xK be the coordinate of poinnt K.
1
RA xK2 = 41.4 xK2
2
1
A7 = − (75)( xK − 0.6)3
6
A6 =
θ K = θ A + θ K/ A = θ A +
1
{ A6 + A7 } = 0
EI
A6 + A7 + EIθ A = 0
f ( xK ) = 41.44 xK2 −
75
( xK − 0.6)3 − 88.2 = 0
6
df
= 82.88 xK − 37.5( xK − 0.6) 2
dxK
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.144 (Continued)
Solve for xK by iteration.
xK = ( x K ) 0 −
xK
f
df/dxK
1.5
–4.1625
93.825
f
df/dxK
1.5442
1.54575
–0.00157
xK = 1.545 m
0.145
94.428
A6 = 98.823 kN ⋅ m2 , A7 = −10.55 kN ⋅ m2
Maximum deflection:
y A = yK + t A/K = 0
yK = −t A/K
3x + 0.6
2
3
xK x7 = 0.6 + ( xK − 0.6) = K
3
4
4
1
87.98
y7 = − { A6 x6 + A7 x7 } = −
= −0.005169 m
17020
EI
x6 =
xK = 1.545 m 
yK = 5.2 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
P
PROBLEM
9.145
For the beam and loading of Prob. 9.136, determinne the largest
F
u
upward
deflection in span AB.
SOLUTION
Units:
Forces in kN; lengths in meters.
I = 4662 × 106 mm 4 = 462 × 10−6 m 4
EI = (2200 × 109 )(462 × 10−6 )
2.4 × 106 N ⋅ m 2 = 92400 kN ⋅ m
= 92
M B = 0:: − 4.8RA + (40)(4.8)(2.4) − (160)(1.8) = 0
RA = 366 kN
1
x(36 x) = 18 x 2
2
20
1
A2 = x( −20 x 2 ) = − x3 
3
3
A1 =

Place reference tangent at A.
yB = y A + Lθ A + t B/A = 0
t B/A
L
( A1 ) B = (18)(4.8) 2 = 414.72 kN ⋅ m 2
θA = −
2 
 20
( A2 ) B =   (4.8)3 = −737.28 kN ⋅ m 2
 3 
θA = −


1 
1
1
( A1 ) B   (4.8) + ( A2 ) B   (4.8) 
EIL 
3
4

−221.184
= 0.49870 × 10−3
=−
(92400)(4.8)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.145 (Continued)
Locate Point K of maximum deflection.
θ K = θ A + θ K /A = 0
EIθ A + A1 + A2 = 0
20 3
f = 46.08 + 18 xK2 −
xK = 0
3
df
= 36 xK − 20 xK2
dx
Solve by iteration.
xK = ( x K ) 0 −
xK 
df /dx
f
3
f
df /dx
3.39
−72
−107.8
28.08
−6.78
3.327
−101.6
−0.188
3.32514 ←
3.3251
−101.42
0.005
Place reference tangent at K.
y A = y K + t A /K
y A − y K = −t A /K

1
2 
 3 
12 xK3 + 5 xK4
( A1 )  xK  + A2  xK   = −
EI
3
4





170.064
=−
= −1.841 × 10−3 m
92400
=−

1
EI
{
}
yK = 1.841 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
P
PROBLEM
9.146
For the beam and loading of Prob. 9.137, determinne the largest
F
u
upward
deflection in span DE.
SOLUTION
Units:
Forces in kN; lengths in m.
From the solution to Problem 9.137.
EI = 1.0306 kN ⋅ m 2
M1
= 0.0582 m −1
EI
M2
= −0.1747 m −1
EI
θ D = 5.83 × 10−3
Location of maximum deflection:
M 3 M1 
U 
=
1 −

EI  0.6 
EI
M4 M2 U
=
EI
EI 0.6
1 M1
⋅ U = 0.0291U
A5 =
2 EI
1 M3  U 
U 

A6 =
= 0.02911 −


U
2 EI  0.6 
 0.6 
1 M4 U
U 
A7 =
= −0.0873 
U
2 EI 0.6
 0.6 
θ K = θ D + A5 + A6 + A7 = 0
U 
U

0.005833 + 0.0291U + 0.02911 −
U =0
U − (0.0873)
0.6
0.6


Multiply by103 :
0.00583 + 0.0582U − 0.097 U 2 = 0
U = 0.3807 m = 381 mm
A5 = 0.01
111, A6 = 0.00405, A7 = −0.0211
Maximum deflection in portion DE:
yD = yK + tD/K = 0
 U 
 2U 
 2U  
yK = −tD /K = −  A5   + A6 
+ A7 


 3 
 3 
 3
= − {−0.000292} = 2.9 mm

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.147
For the beam and loading shown, determine the
t reaction at the roller
support.
SOLUTION
Remove support B and treat RB as redundant.
Replace loading by equivalent shhown at left.
Draw M/EI diagram for load w0 and RB .
Use parts as shown.
1  RB L 
1 RB L2
( L) =


2  EI 
2 EI
1
= − w0 L2
2
1  1 w0 L2 
1 w0 L3
L=−
= −

3  2 EI 
6 EI
1 w0 3 1
=
L = w0 L2
6 L
6
2
1  1 w0 L 
1 w0 L3
= 
L=

4  6 EI 
24 EI
A1 =
M2
A2
M3
A3
Place reference tangent at A.
2 
3 
4 
t B/A = A1  L  + A 2  L  + A 3 L 
3 
4 
5 
=
RB =
1 RB L3 1 w0 L4 1 w0 L4
−
+
=0
3 EI
8 EI
30 EI
11
w0 L ↑
40
RB = 0.275 w0 L ↑ 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLE
EM 9.148
For the beam
b
and loading shown, determine the reaction at the roller
support.
SOLUTION
Remove support A and treat RA as redundantt.
Draw the M /EI diagram by parts.
1 RA L RA L2
=
L
2 EI
2 EI
PL2
1 L PL
=−
A2 = −
8EI
22 2
A1 =
Place reference tangent at B.
y A = yB − θ B L + t A/B = 0
t A/B = 0
 2L 
L L
A1
 + A2  2 + 3  = 0
3




RA L3 5PL3
−
=0
48EI
3EI
E
RA =
5
P ↑ 
16
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.149
For the beam and loading shown, determine the reaction at the roller
support.
SOLUTION
Remove support A and treat RA as redundant.
Draw M/EI diagram for loads RA and w.
2
1 L
1
M 2 = − w   = − wL2
2 2
8
A1 =
1  RA L 
1 RA L2
L=


2  EI 
2 EI
1  1 wL2   L 
1 wL3
A2 =  −
   = −
3  8 EI   2 
48 EI
Place reference tangent at B.
2 
L 3 L
t A/B = A1 L  + A2  +

3 
2 4 2
=
RA =
1 RA L3
7 wL4
−
=0
3 EI
384 EI
7
wL
128
RA =
7
wL ↑ 
128
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLE
EM 9.150
For the beam
b
and loading shown, determine the reaction at the roller
support.
SOLUTION
Remove support B and treat RB as redundantt.
Draw M/EI diagram.
1 RA L RA L2
L
=
2 EI
E
2 EI
M
L
M
L2
L
A2 = ⋅ 0 = 0
2 EI
E
2 EI
A1 =
Place reference tangent at A.
yB = y A + Lθ A + t B/A = 0
t B/A = 0
 2L 
L L
A1 
 + A2  2 + 4  = 0
3




RA L3 3M 0 L2
−
=0
3EI
8EI
RA =
9 M0
↑ 
8 L
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.151
For the beam and loading shown, determine the reaction at each support.
SOLUTION
Remove support B and consider RB as redundant.
Consider loads RB and w separately.
Place reference tangent at A.
A1 =
1  1 RB L  L 
1 RB L2
⋅−
=
−


2  3 EI   2 
12 EI
A2 =
1  1 RB L 
1 RB L2
−
L=−


2  3 EI 
6 EI
1  1 wL2  3L 3 wL2
A3 = 
=

2  2 EI  2 8 EI
1  1 wL2 
1 wL3
A4 =  −
 L = −
3  2 EI 
6 EI
1 L

2 
tC /A = A1  L +
 + A2  L 
3 2

3 
 1 3L 
1 
+ A3  ⋅  + A4  L 
3
2


4 
7 RB L3 1 RB L3
−
72 EI
9 EI
4
3 wL
1 wL4
+
−
16 EI 24 EI
5 RB L3 7 wL3
=−
+
24 EI
48 EI
=−
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBL
LEM 9.151 (Continued)
A5 =
1  1 wL2  L 1 wL3

 =
2  6 EI  2 24 EI
1 RB L3
1 wL4
1 L
1 L
+
=
−
+
tB /A = A1 
A
5


72 EI
144 EI
3 2
3 2
yB = t B / A −
L /2
5  R L3  1
7 wL4 
 1
tC /A =  − +  B + 
−

3L /2
 72 72  EI
 144 144 EI 
1 RB L3 1 wL
w 3
3
−
= 0 RB = wL
4
18 EI
24 EI
1
2
1
1
1
1
RA = wL − RB = wL − wL = − wL = wL
3
3
3
2
6
6
2
1
2
1
5
RC = wL − RB = wL − wL = wL
3
3
3
4
12
=



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.152
For the beam and loading shown, determiine the reaction at each
support.
SOLUTION
Choose RB ↓ as the redundant reaction.
r
Draw M /EI diagram for the loadds RB and M 0 .
A1 =
A2 =
2
1
R L R L
( L)  B  = B
2
 3EI  6 EI
1  L   RB L  RB L2
=
2  2 
 3EI  12 EI
M0L

 = − 2 EI

M L
1
 1  M 
A4 = ( L )   − 0  = − 0
2
6 EI
 3  EI 
A3 =
1
 M
( L)  − 0
2
 EI
1  3L   M
A 3 + A 4 + A 5 =   − 0
2  2  EI
3M 0 L

 = − 4 EI

yB = y A + Lθ A + tB/A
yC = y A +
3L
θ A + tC/A = 0
2
θ A = −t B/A /L
3
− t B /A + t C/ A = 0
2
3
2
L
 2L 
 L  RB L 7 M 0 L
+
=
−
A
t B/A = ( A1)   + A 3
4

 
18 EI
3
 3 
 3  18EI
R L3 3M 0 L2
L L
L
tC/A = ( A1)  +  + A 2   + ( A3 + A4 + A5 )( L ) = B −
4 EI
6 EII
2 3
3
R L3 M L2
3
− t B /A + t C / A = B − 0 = 0
2
12 EI
6 EI
ΣM C = 0: M 0 +
RA =
RB =
2M 0
↓ 
L
RA =
4M 0
↑ 
3L
RC =
2M 0
↑ 
3L
L
3L
RB −
RA = 0
2
2
2
[M 0 + M 0 ]
3L
ΣFy = 0: R A + RB + RC = 0
4 M 0 2M 0
−
+ RC = 0
3 L
L
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.153
Deetermine the reaction at the roller support and draw
w the bending
mooment diagram for the beam and loading shown.
SOLUTION
Units: Forces in kN; lengths in meters.
Let RA be the redundant reaction.
Remove support at A and add reaction RA ↑ .
Draw bending moment diagram by parts.
M 1 = 3.6 RA kN ⋅ m
M 2 = −(75)((0.3 + 2.4) = −202.5 kN ⋅ m
1
M 3 = − (40)(2.4) 2 = −115.2 kN ⋅ m
2
1
(3.66)(3.6 RA ) = 6.48 kN ⋅ m 2
2
1
A2 = (2.77)(−202.5) = −273.375 kN ⋅ m 2
2
1
A3 = (2.4)( −115.2) = −92.16 kN ⋅ m 2
3
A1 =
Place reference tangent at B, where
θ B = 0 annd yB = 0.
Then
y A = t A/B = 0
t A/B =
=
1
EI
 2
2
3




 
 ⋅ 3.6  A1 +  0.9 + ⋅ 2.7  A2 +  0.9 + 0.3 + ⋅ 2.4  A3 
3
4




 
 3
1
{155.552 RA − 1014.5925} = 0
EI
RA = 65.24 kN ↑ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.153 (Continued)
Draw shear diagram.
A to D : V = RA = 65.24 kN
D to E : V = 65.24 − 75 = −9.76 kN
E to B : V = −9.76 − 40( x − 1.2) kN
At B, VB = −105.76 kN
Bending moment diagram:
MA = 0
M D = M A + 58.72 = 58.72 kN ⋅ m
M E = 58.72 − 2.93 = 55.79 kN ⋅ m
M B = 55.79 − 138.62 = −82.83 kN ⋅ m

PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
P
PROBLEM
9.154
Determine the reaction at the roller support and draw
D
w the bending
m
moment
diagram for the beam and loading shown.
SOLUTION
Unitts: Forces in kN, lengths in m.
Let RB be the redundant reaction.
Rem
move support B and add load RB.
Draw
w bending moment diagram by parts.
M1 = 3.6 RB kN ⋅ m
M 2 = −(1.35 + 0.9)(40) = −90 kN ⋅ m
M 3 = −(1.35)(120) = −162 kN ⋅ m
1
(3.6)(3.6RB ) = RB 6.48 kN ⋅ m2
2
1
A2 = (2.25)(−90) = −101.25 kN ⋅ m2
2
1
A3 = (1.35)(−162) = −109.35 kN ⋅ m2
2
yB = y A + 3.6θ A + t B /A = 0
A1 =
t B /A = 0
t B /A =
1
{(6.48RB )(2.4) + (−101.25)(2.85)
EI
+ (−109.35)(3.15)} = 0
15.552RB − 633.015 = 0
RB = 40.7 kN 
w shear diagram working from right to left.
Draw
Areas of shear diagram:
B to E
V = −RB = −40.7 kN
E to D
V = −40.7 + 40 = −0.7 kN
D to A
V = −0.7 + 120 = 119.3 kN
AAD = (1.35)(119.3) = 161.055 kN ⋅ m
ADE = (0.9)(−0.7) = −0.63 kN ⋅ m
AEB = (1.35)(−40.7) = 54.945 kN ⋅ m
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.154 (Continued)
Bending moments:
M A = M1 + M 2 + M 3 = −105.5 kN ⋅ m

M D = M A + M AD = 55.6 kN ⋅ m

M E = M D + ADE = 55 kN ⋅ m

M B = M E + AEB = 0

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.155
For the beeam and loading shown, determine the spring constannt k for which
the force in the spring is equal to one-third of the total load on the
t beam.
SOLUTION
Symmetric beam and loading:
Spring force:
RC = R A
1
2
F = (2 wL ) = wL
3
3
ΣFy = 0: RA + F − 2wL + RC = 0
RA = RC =
2
wL
3
Draw M/EI diagram by parts.
1  2 wL2 
1 wL3
A1 = 
 L =
2  3 EI 
3 EI
2
1  1 wL 
1 wL3
A2 = − 
=
−
L

3  2 EI 
6 EI
Place reference tangent at B.
θB = 0
y B = −t A /B
2
3 

= −  A1 ⋅ L + A2 ⋅ L 
3
4 

=−
7 wL4
72 EI
F = − kyB
k=−
F
=
yB
2 wL
3
7 wL4
72 EI
k=
48 EI

7 L3
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.156
For the beam and loading shown, determine the sppring constant k for which
the bending moment at B is M B = − wL2 /10.
SOLUTION
Using free body AB,
L 1
M B = 0: −RA L + ( wL)   − wL2 = 0
 2  10
RA =
Symmetric beam and loading:
Using free body ABC,
2
wL ↑
5
RC = RA
ΣFy = 0:
2
2
wL + F + wL − 2 wL = 0
5
5
F=
6
wL
5
A1 =
1  2 wL2 
1 wL3

 L =
2  5 EI 
5 EI
Draw M /EI diagram by parts.
1  1 wL2 
1 wL3
A2 = − 
 L = −
3  2 EI 
6 EI
Place reference tangent at B.
θB = 0
y B = −t A /B
2
3 

= −  A1 ⋅ L + A2 ⋅ L 
3
4 

=−
1 wL4
120 EI
F = − kyB
k =−
F
=
yB
6
wL
5
1 wL4
120 EI
k = 144
EI

L3
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.157
For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the
slope at the free end.
SOLUTION
w( x) =
2 w0
x − w0
L
w
 2w

V ( x) = − w( x) dx = −  0 x − w0  dx = − 0 x 2 + w0 x + C1
L
L




[ x = 0, V = 0] 0 = 0 + 0 + C1 ∴ C1 = 0
w
w
 w

M ( x) = V ( x)dx =  − 0 x 2 + w0 x  dx = − 0 x3 + 0 x 2 + C2
3L
2
 L



[ x = 0, M = 0] 0 = 0 + 0 + C2 ∴ C2 = 0
w
w
d2y
= M = − 0 x3 + 0 x 2
2
3
L
2
dx
w
w
dy
EI
= − 0 x 4 + 0 x3 + C3
12 L
6
dx
EI

 x = L,

w L3 w L3
dy

= 0  0 = − 0 + 0 + C3
dx
12
6

EIy = −
Elastic curve:
(b)
y at x = 0.
(c)
dy
at x = 0.
dx
w0 L4 w0 L4 w0 L4
+
−
+ C4
60
24
12
y=−
yA = +
dy
dx
=−
A
w0 L3
12
w0 5 w0 4 w0 L3
x +
x −
x + C4
60 L
24
12
[ x = L, y = 0] 0 = −
(a)
∴ C3 = −
7 w0 L4
120 EI
w0 L3
12 EI
∴ C4 =
7 w0 L4
120
w0
(2 x5 − 5 Lx 4 + 10 L4 x − 7 L5 ) 
120 EIL
yA =
θA =
7 w0 L4
↑ 
120 EI
w0 L3
12 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.158
(a) Determine the location and magnitude of the maximum absolute
deflection in AB between A and the center of the beam. (b) Assuming
that beam AB is a W460 × 113, M 0 = 224 kN ⋅ m and E = 200 GPa,
determine the maximum allowable length L of the beam if the
maximum deflection if not to exceed 1.2 mm.
SOLUTION
Using AB as a free body
ΣM B = 0
RA =
− 2M 0 − RA L = 0
2M 0
L
Using portion AJ as a free body
ΣM J = 0
−M 0 +
2M 0
x+M =0
L
M0
( L − 2 x)
L
2
d y
M
EI 2 = 0 ( L − 2 x)
L
dx
dy
M0
EI
=
( Lx − x 2 ) + C1
dx
L
M 1
1 
EIy = 0  Lx 2 − x3  + C1x + C2
L 2
3 
M=
[ x = 0, y = 0] 0 = 0 − 0 + 0 + C2
C2 = 0
1
1 3 1 3
C1 = − M 0 L2
 2 L − 3 L  + C1 L + 0
6


M 1
1
1
1 
 dy M 0 
y = 0  Lx 2 − x3 − L2 x 
Lx − x 2 − L2 
=

3
6
6 
EIL  2
 dx EIL 
[ x = L, y = 0] 0 =
To find location maximum deflection set
xm2
1
− Lxm − L2 = 0
6
ym =
M0
L
dy
= 0.
dx
xm =
L−
L2 − (4)
2
( L)
1
6
2
=
1
3
1 −
 L = 0.21132L
2
3 


M 0 L2  1 
M 0 L2
1
1
2
2
  (0.21132) −   (0.21132) −   (0.21132)  = −0.0160375
EI  2 
EI
 3
6

| ym | = 0.0160375
M 0 L2
EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.158 (Continued)
1/2
Solving for L
 EI | ym | 
L=

 0.0160375M 0 
Data:
E = 200 × 109 Pa, I = 556 × 106 mm 4 = 556 × 10−6 m 4
| ym | = 1.2 mm = 1.2 × 10 −3 m. M 0 = 224 × 103 = N ⋅ m
1/ 2
 (200 × 109 )(556 × 10−6 )(1.2 × 10−3 ) 
L=

(0.0160375)(224 × 103 )


= 6.09 m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.159
For the beam and loading shown, determine (a) the equation of the
elastic curve, (b) the slope at end A, (c) the deflection at the midpoint
of the span.
SOLUTION

x2  w
w = w0 1 − 2  = 20 [ L2 − x 2 ]
L 
L

dV
w
= −w = 20 [ x 2 − L2 ]
dx
L

dM
w  x3
= V = 20  − L2 x  + C1
dx
L 3

w0
L2
M =
[ x = 0, M = 0]:
0 = 0 − 0 + 0 + C2
[ x = L, M = 0]: 0 =
EI
d2y
w
= M = 20
2
dx
L
EI
 1 4 1 2 2
12 x − 2 L x  + C1x + C2


w0
L2
∴
 1 4 1 4
12 L − 2 L  + C1L


∴
C1 =
5
w0 L
12
5 3 
1 4 1 2 2
12 x − 2 L x + 12 L x 


dy
w 1
1
5 3 2
L x  + C3
= 20  x5 − L2 x3 +
dx
6
24
L  60

EIy =
w0  1 6
1 2 4
5 3 3
x −
Lx +
L x  + C3 x + C4
2 
24
72
L  360

[ x = 0, y = 0]: 0 = 0 − 0 + 0 + 0 + C4
[ x = L, y = 0]: 0 =
(a)
C2 = 0
∴ C4 = 0
w0  1 6
1 6
5 6
11
L −
L +
L  + C3 L ∴ C3 = −
w0 L3
2 
24
72 
360
L  360
y = w0 ( x6 − 15L2 x 4 + 25L3 x3 − 11L5 x)/360 EIL2 
Elastic curve:
dy
= w0 (6 x5 − 60 L2 x3 + 75L3 x 2 − 11L5 )/360 EIL2
dx
(b)
Slope at end A:
Set x = 0 in
dy dy
.
dx dx
=−
A
11 w0 L3
360 EI
θA =
11 w0 L3
360 EI

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.159 (Continued)
(c)
Deflection at midpoint (say, point C ):
Set x =
15 6
 1
yC = w0  L6 −
L +
64
16

60 6
 1
L +
= w0  L6 −
64
 64

=−
211 w0 L4
23040 EI
L
in. y. 
2
25 6 11 6 
L − L  360 EIL2
8
2 
200 6 352 6 
L −
L  360 EIL2
64
64 
 yC = 0.00916
w0 L4
↓ 
EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.160
Determine the reaction at A and draw the bending
b
moment diagram
for the beam and loading shown.
[ x = 0, y = 0]
dy


 x = 0, dx = 0 


L dyy


 x = 2 , dxx = 0 


SOLUTION
RA = RB
By symmetry,
and
dy
L
= 0 at x = .
2
dx
ΣFy = 0: RA + RB − P = 0
RA = RB =
1
P
2
Moment reaction is statically inddeterminate.
0< x<
L
2
M = M A + RA x = M A +
1
Px
2
d2y
1
= M A + Px
2
2
dx
dy
1
EI
= M A x + Px 2 + C1
dx
4
EI

 x = 0,

dy

= 0  : 0 − 0 + C1 = 0
dx

L dy


 x = 2 , dx = 0  :


C1 = 0
2
MA
L 1 L
+ P  + 0 = 0
2 4 2
1
M A = − PL
8
1
PL
8
1 L
1
1
1
M C = M A + P = − PL + PL = PL
L
2 2
8
4
8
By symmetry, M B = M A =
MA =
1
PL
8



PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PR
ROBLEM 9.161
Foor the beam and loading shown, determine (a) the sloope at end A,
(b)) the deflection at point B. Use E = 200 GPa.
SOLUTION
Units:
Forces in kN lengths in meters.
C=
I=
1
1
d =   (30) = 15 mm
2
2
π
π
C 4 = (15) 4 = 39761 mm 4
4
4
EI = (200 × 106 ) (39761 × 10−12 ) = 7.95 kN ⋅ m 2
Use entire beam ABCD as free body.
ΣM D = 0: − 1.2RA + (0.72)(0.4) + (0.8)(0.2) = 0
RA = 0.373 kN
w( x) = 1.8 x − 0.6 0 − 1.8 x − 1 0 kN/m
dV
= − w = −1.8 x − 0.6 0 + 1.8 x − 1 0 kN/m
dx
dM
= V = −1.8 x − 0.61 + 1.8 x − 11 + 0.373 − 0.8 x − 1 0 kN ⋅ m
dx
d2y
= M = −0.9 x − 0.6 2 + 0.9 x − 1 2 + 0.373 x − 0.8 x − 11 kN ⋅ m
2
dx
dy
EI
= −0.3 x − 0.6 3 + 0.3 x − 1 3 + 0.1865 x 2 − 0.4 x − 1 2 + C1 kN
N ⋅ m2
dx
EI
EIy = −0.075 x − 0.6
0  4 + 0.075 x − 1 4 + 0.0622 x3 − 0.133 x − 1 3 + C1 x + C2 kN ⋅ m3
[ x = 0, y = 0] :
−0 + 0 + 0 + 0 + C2 = 0
∴ C2 = 0
[ x = 1.2, y = 0]: −0.075(0..6) + 0.075(0.2) + 0.0622(1.2)3 − 0.133(0.2)3 + 1.2C
C1 = 0
4
4
C1 = −0.0807 kN ⋅ m2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.161 (Continued)
(a)
 dy

 dx at x = 0 


Slope at end A:
 dy 
EI   = −0 + 0 + 0 + C1
 dx  A
C1 −0.0807
 dy 
 dx  = EI = 7.95 = −0.01015
 A
(b)
Deflection at point B:
θ A = 0.01015 rad

( y at x = 0.6 m)
EIyB = −0 + 0 + 0.0622(0.6)3 − 0 + (−0.0807)(0.6)
= −0.035 kN ⋅ m3
0.035
yB = −
= 4.402 × 10 −3 m
7.95
yB = 4.4 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.162
The rigid bar BDE is welded at point B to thee rolled-steel
beam AC. For the loading shown, determine (a) the slope at
0 GPa.
point A, (b) the deflection at point B. Use E = 200
SOLUTION
M C = 0:
− 4.5RA + (20)(3)(1.5) − (60)(1.5) = 0
Units:
RA = 0
Forces in kN; lengths in m.
1
d2y
2
1
0
= M = 60 x − 1.5 − 90 x − 1.5 − (200) x − 1.5
2
2
dx
dy
1
EI
= 30 x − 1.5 2 − 90 x − 1.51 −   (20) x − 1.5 3 + C1
dx
6
EI
EIy = 10 x − 1.5 3 − 45 x − 1.5 2 −
1
(20) x − 1.5 4
24
+ C1x + C2
Boundary conditions:
Data:
[ x = 0, y = 0]
0 + 0 + 0 + 0 + C2 = 0
[ x = 4.5, y = 0]
(10)(3)3 − (45)(3) 2 −
1
(20)(3) 4 + 4.5C1 + 0 = 0
24
C2 = 0
C1 = 45kN ⋅ m 2
E = 200 × 109 Pa, I = 316 × 106 mm 4 = 316 × 10−6 m 4
EI = (200 × 109 )(316 × 10−6 ) = 63.2 × 106 N ⋅ m 2 = 63, 200 kN ⋅ m 2
(a)
Slope at A:
 dy

 dx at x = 0 


EI
E θ A = C1 = 45 kN ⋅ m2
θA =
(b)
45
= 0.712 × 10 −3 rad
63, 200
θ A = 0.712 × 100−3 rad

( y at x = 1.5)
Deflection at B:
EIyB = (C1)(11.5) = (45)(1.5) = 67.5 kN ⋅ m3
yB =
67.55
= 1.068 × 10−3 m
63, 2000
yB = 1.0068 mm ↑ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.163
Before the uniformly distributed load
d w is applied, a gap,
δ 0 = 1.2 mm, exists between the ends of the cantilever bars AB
and CD. Knowing that E = 105 GPa and w = 30 kN/m, determine
D
(a) the reaction at A, (b) the reaction at D.
SOLUTION
1
(500)(50)3 = 520.833 × 103 mm3 = 520.833 × 10 −9 m
12
EI = (105 × 109 )(520.833 × 10 −6 ) = 54.6875 × 103 N ⋅ m 2
I=
= 54.68875 kN ⋅ m 2
Units: Forces in kN; lengths in meters.
w Case 8 of Appendix D.
Compute deflection at B due to w.
( yB )1 = −
wL4
(30)(0.400) 4
=−
8 EII
(8)(54.6875)
= −1.755543 × 10−3 = −1.7553 mm
The displacement is more than δ 0 , the gap closes.
Let P be the contact force betweeen points B and C.
Compute deflection of B due to P. Use Case 1 of Appendix D.
( yB ) 2 =
P(0.4)3
PL3
=
3EI (3)(54.6875)
= 390.0
095 × 10−6 P
m
Compute deflection of C due to P.
yC = −
PL
L3
P (0.25)3
=−
= −95.238 × 10 −6 P
(3)(54.6875)
3EII
m
Displacement condition:
yB + δ 0 = yC
Using superposition,
( yB )1 + ( yB )2 − δ 0 = yC
−1..75543 × 10−3 + 390.095 × 10−6 P + 1.2 × 10−3 = −95.2388 × 10−6 P
485.333 × 10−6 P = 0.55543 × 10−3
P = 1.14443 kN
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBLEM 9.163 (Continued)
(a)
Reaction at A:
ΣFy = 0: RA − 12 + 1.14443 = 0
RA = 10.86 kN ↑ 
ΣM A = 0: M A − (0.2)(12) + (0.4)(1.14443) = 0
M A = 1.942 kN ⋅ m
(b)
Reaction at D:

ΣFy = 0: RD − 1.14443 = 0
RD = 1.144 kN ↑ 
ΣM D = 0: −M D + (0.25)(1.14443) = 0
M D = 0.286 kN ⋅ m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.164
A central beam BD is joined at hinges to two cantilever
beams AB and DE. All beams have the cross section
shown. For the loading shown, determine the largest w
so that the deflection at C does not exceed 3 mm. Use
E = 200 GPa.
SOLUTION
Let
a = 0.4 m
Cantilever beams AB and CD.
Cases 1 and 2 of Appendix D.
yC = −
( wa ) a 3 wa 4 11 wa 4
−
=
3EI
8 EI 24 EI
Beam BCD, with L = 0.8 m, assuming that points B and D do not move.
Case 6 of Appendix D.
5wL4
384 EI
yC′ = −
Additional deflection due to movement of points B and D.
yC′′ = yB = yD = −
Total deflection at C:
yC = yC′ + yC′′
yC = −
Data:
11 wa 4
24 EI
w  5 L4 11a 4 
+


EI  384
24 
E = 200 × 109 Pa,
1
(24)(12)3 = 3.456 × 10 −3 mm 4 = 3.456 × 10−9 m 4
12
EI = (200 × 109 )(3.456 × 10 −9 ) = 691.2 N ⋅ m 2
I=
yC = −3 × 10−3 m
−3 × 10 −3 = −
w  (5)(0.8) 4 (11)(0.4) 4 
−6
+

 = −24.69 × 10 w
691.2  384
24

w = 121.5 N/m 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.165
For the cantilever beam and loading shown, ddetermine (a)
the slope at point A, (b) the deflection at point
p
A. Use
E = 200 GPa.
SOLUTION
Units:
Forces in kN; lengths in m.
E = 200 × 109 Pa
I = 40.1 × 106 mm 4 = 40.1 × 10−6 m 4
EI = (200 × 109 )(40.1 × 10−6 ) = 8.02 × 106 N ⋅ m 2
= 8020 kN ⋅ m 2
Draw M /EI diagram by parts.
M1
(18)(2.2)
=
= 4.9377 × 10−3 m −1
EI
8020
1
A1 = (4.9377 × 10−3 )(2.2) = 5.4315 × 10−3
2
1
x1 = (2.2) = 0.7333 m
3
M2
(26)(2.7) 2
=−
= −11.8167 × 10−3 m −1
EI
(2)(8020)
1
(−11.8167 × 10−3 )(2.7) = −10.6350 × 10−3
3
1
x2 = (2.7) = 0.675 m
4
A2 =
Draw reference tangent at C.
θC = θ A + θC / A = θ A + A1 + A2 = 0
(a)
Slope at A:
θ A = − A1 − A2 = −5.4315 × 10−3 + 10.6350 × 10−3
= 5.20 × 10−3 rad
θ A = 5.20 × 100−3 rad




PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.165 (Continued)
(b)
Deflection at A:
y A = yC − θC L + t A / C
= 0 − 0 + A1x1 + A2 x2
= 0 − 0 + (5.4315 × 10−3 )(1.9667) − (10.6350 × 10−3 )(2.025)
= −10.85 × 10−3 m

y A = 10.85 mm ↓ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 9.166
Knowing that the magnitude of the load P is 30 kN
N, determine
(a) the slope at end A, (b) the deflection at ennd A, (c) the
deflection at midpoint C of the beam. Use E = 200 GPa.
SOLUTION
Use units of kN and m P = 30 kN
I = 9.11 × 10−6 m
For S150 × 18.6
EI = (200 × 106 )(9.11 × 10−6 ) = 1822 kN ⋅ m
Symmetric beam with symmetric loading. Place referrence tangent
at midpoint C where θC = 0.
RB = RD =
1
(6 + 30 + 6) = 21 kN
2
Draw the bending moment diagram by parts for thee left half of
the beam.
M 1 = (1.4)(21) = 29.4 kN ⋅ m
1
(1.4)(29.4) = 20.58 kN ⋅ m 2
2
M 2 = −(0.6 + 1.4)(6) = −12 kN ⋅ m
A1 =
1
(2)(−12) = −12 kN ⋅ m 2
2
M 3 = −(0.6)(6) = −3.6 kN ⋅ m
A2 =
A3 =
Formulas:
θ A = −θC /A ,
y A − yC = t A/C
yB = y A − 0.6θ A + t B /A = 0,
1
EI
1
=
EI
θ C /A =
t A /C
t B /A =
1
EI
1
(0.6)( −3.6) − 1.08 kN ⋅ m 2
2
yC = y A − t A/C
y A = −2θ A − t B /A
20.58 − 12
= 4.709 × 10−3
1822
2

 15.549
= 8.534 × 10−3 m
(0.6 + 0.433) A1 + (2) A2  =
1822
3


( A1 + A2 ) =
1
 −0.2166
= −0.1186 × 10−3 m
 (0.6) A3  =
3
1822


PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.166 (Continued)
(a)
Slope at end A:
(b)
Deflection at A:
θ A = −4.71 × 10−3 
y A = −(0.6)(−4.709 × 10−3 ) − (−0.1186 × 10−3 )
= 2.944 × 10−3 m
(c)
Deflection at C:
y A = 2.94 mm 
yC = 2.944 × 10−3 − 8.534 × 10−3
yC = −5.59 × 10−3 m
yC = 5.6 mm 
PROPRIETARY MATERIAL. © 2013 The
T McGraw-Hill Companies, Inc. All rights reserved. No part of
o this Manual may be displayed,
reproduced, or distributed in any form or
o by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators perm
mitted by McGraw-Hill for their individual course preparation. A stu
udent using this manual is using it
without permission.
PROBL
LEM 9.167
For the beam
b
and loading shown, determine (a) the slopee at point C,
(b) the deeflection at point C.
SOLUTION
Pa
aL
,
2 EI
E
Paa 2
A2 = −
E
2 EI
PaL2
2 
t A/B = A1  L  = −
3EI
3 
t
PaL
θ B = A /B = −
L
3EI
A1 = −
(a)
Slope at C:
θC = θ B + θC/B
θC = −
=−
(b)
PaaL Pa 2
−
3EI
E
2 EI
Paa (2 L + 3a )
6 EI
θC =
Pa (2 L + 3a )
6 EII

Deflection at point C:
yC = aθ B + tC/B
yC = −
=−
Paa 2 L  Pa 2
+  −
3EI
E
 2 EI
Paa 2 ( L + a )
3EI
 2 
  a 
 3 
yC =
Pa 2 ( L + a)
↓ 
3EI
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may
m be displayed,
reproduced, or distributed in any form or by any means,, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraaw-Hill for their individual course preparation. A student using this manual
m
is using it
without permission.
PROBLEM 9.168
A hydraulic jack can be used to raise point B of the cantilever
beam ABC. The beam was originally straight, horizontal, and
unloaded. A 20-kN load was then applied at point C, causing this
point to move down. Determine (a) how much point B should be
raised to return point C to its original position, (b) the final value
of the reaction at B. Use E = 200 GPa.
SOLUTION
For W130 × 23.8, I x = 8.91 × 106 mm 4
EI = (200 × 106 kPa)(8.91 × 10−6 m 4 ) = 1782 kN ⋅ m 2
Let RB be the jack force in kN.
1
(1.8 RB )(1.8) = 1.62 RB
2
1
A2 = (−60)(3) = −90 kN ⋅ m 2
2
EI tC / A = (2.4) A1 + (2) A2
A1 =
0 = (2.4)(1.62 RB ) + (2)(−90)
RB = 46.296 kN
A1 = 75 kN ⋅ m2
1
(−60)(1.8) = −54 kN ⋅ m 2
2
1
A4 = (−24)(1.8) = −21.6 kN ⋅ m 2
2
EItB/A = (1.2) A1 + (1.2) A3 + (0.6) A4
A3 =
= (1.2)(75) + (1.2)(−54) + (0.6)(−21.6)
= 12.24 kN ⋅ m 2
(a)
Deflection at B:
(b)
Reaction at B:
yB = t B / A =
EI tB / A 12.24
=
= 6.8687 × 10−3 m
EI
1782
yB = 6.87 mm ↑ 
RB = 46.3 kN ↑ 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.