Hibbeler Chapter 6 Part 1 (463-486) - Auburn University
? 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6¨C2. Draw the shear and moment diagrams for the shaft.
The bearings at A and D exert only vertical reaction on the
shaft. The loading is applied to the pulleys at B and C and E.
14 in.
20 in.
15 in.
12 in.
A
E
B
C
D
35 lb
80 lb
110 lb
Ans:
V (lb)
82.2
35
2.24
x
?50
?108
M (lb?in)
1151
1196
x
?420
464
? 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6¨C5. Draw the shear and moment diagrams for the beam.
10 kN
8 kN
15 kN?m
2m
3m
Ans:
V (kN)
18
8
x
M (kN?m)
x
?15
?39
?75
467
? 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6¨C18. Draw the shear and moment diagrams for the beam,
and determine the shear and moment throughout the beam
as functions of x.
10 kip
2 kip/ft
8 kip
40 kip?ft
x
6 ft
Support Reactions: As shown on FBD.
4 ft
Shear and Moment Function:
For 0 ¡ x 6 6 ft:
+ c ?Fy = 0;
30.0 - 2x - V = 0
V = {30.0 - 2x} kip
Ans.
x
a + ?MNA = 0; M + 216 + 2x a b - 30.0x = 0
2
M = {-x2 + 30.0x - 216} kip # ft
Ans.
For 6 ft 6 x ¡ 10 ft:
+ c ?Fy = 0;
a + ?MNA = 0;
V - 8 = 0
V = 8.00 kip
Ans.
-M - 8(10 - x) - 40 = 0
M = {8.00x - 120} kip # ft
Ans.
Ans:
For 0 ¡ x 6 6 ft: V = {30.0 - 2x} kip,
M = {-x2 + 30.0x - 216} kip # ft,
For 6 ft 6 x ¡ 10 ft: V = 8.00 kip,
M = {8.00x - 120} kip # ft
V (kip)
30.0
M (kip?ft)
18.0
8.00
x (ft)
0
6
0
10
?72.0
?216
479
6
10
x (ft)
?40.0
? 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6¨C25. Draw the shear and moment diagrams for the beam
and determine the shear and moment in the beam as
functions of x, where 4 ft < x < 10 ft.
150 lb/ft
200 lb?ft
200 lb?ft
A
B
x
4 ft
+ c ?Fy = 0;
4 ft
-150(x - 4) - V + 450 = 0
V = 1050 - 150x
a + ?M = 0;
6 ft
- 200 - 150(x - 4)
Ans.
(x - 4)
- M + 450(x - 4) = 0
2
M = - 75x2 + 1050x - 3200
Ans.
Ans:
V = 1050 - 150x
M = - 75x2 + 1050x - 3200
V (lb)
450
x
?450
M (lb?ft)
475
x
?200
486
?200
? 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6¨C33. The shaft is supported by a smooth thrust bearing at
A and smooth journal bearing at B. Draw the shear and
moment diagrams for the shaft.
400 N?m
B
A
1m
1m
1m
900 N
Equations of Equilibrium: Referring to the free-body diagram of the shaft shown
in Fig. a,
a + ?MA = 0;
By(2) + 400 - 900(1) = 0
By = 250 N
+ c ?Fy = 0;
Ay + 250 - 900 = 0
Ay = 650 N
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V (N)
650
1
0
2
3
x (m)
?250
M (N?m)
650
400
0
493
x (m)
1
2
3
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