01 Solutions 46060 5/6/10 2:43 PM Page 1
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–1. Determine the resultant internal normal force acting 5 kip 8 kN
on the cross section through point A in each column. In
(a), segment BC weighs 180 lb>ft and segment CD weighs
250 lb>ft. In (b), the column has a mass of 200 kg>m.
B
200 mm 200 mm
6 kN 6 kN
(a) + c ©Fy = 0; FA - 1.0 - 3 - 3 - 1.8 - 5 = 0
10 ft 3m
FA = 13.8 kip Ans. 8 in. 8 in.
200 mm 200 mm
(b) + c ©Fy = 0; FA - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0 3 kip 3 kip
4.5 kN 4.5 kN
C
FA = 34.9 kN Ans.
4 ft
A A
4 ft 1m
D
(a) (b)
1–2. Determine the resultant internal torque acting on the A
cross sections through points C and D. The support bearings 250 Nm
at A and B allow free turning of the shaft.
150 Nm
©Mx = 0; TC - 250 = 0 300 mm C
400 Nm
200 mm
TC = 250 N # m Ans.
150 mm
©Mx = 0; TD = 0 Ans. 200 mm D B
250 mm
150 mm
1–3. Determine the resultant internal torque acting on the
cross sections through points B and C.
A 600 lbft
©Mx = 0; TB + 350 - 500 = 0 B
350 lbft
TB = 150 lb # ft Ans. 3 ft
C
©Mx = 0; TC - 500 = 0
1 ft 500 lb
TC = 500 lb # ft Ans. 2 ft
2 ft
1
,01 Solutions 46060 5/6/10 2:43 PM Page 2
© 2010 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.
*1–4. A force of 80 N is supported by the bracket as
shown. Determine the resultant internal loadings acting on
the section through point A. 0.3 m
A
30
0.1 m
80 N 45
Equations of Equilibrium:
+
Q©Fx¿ = 0; NA - 80 cos 15° = 0
NA = 77.3 N Ans.
a+ ©Fy¿ = 0; VA - 80 sin 15° = 0
VA = 20.7 N Ans.
a+ ©MA = 0; MA + 80 cos 45°(0.3 cos 30°)
- 80 sin 45°(0.1 + 0.3 sin 30°) = 0
MA = - 0.555 N # m Ans.
or
a+ ©MA = 0; MA + 80 sin 15°(0.3 + 0.1 sin 30°)
-80 cos 15°(0.1 cos 30°) = 0
MA = - 0.555 N # m Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
2
,01 Solutions 46060 5/6/10 2:43 PM Page 3
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•1–5. Determine the resultant internal loadings in the 3 kip
beam at cross sections through points D and E. Point E is
1.5 kip/ ft
just to the right of the 3-kip load.
A
D B E C
6 ft 6 ft 4 ft 4 ft
Support Reactions: For member AB
a + ©MB = 0; 9.00(4) - A y(12) = 0 A y = 3.00 kip
+
: ©Fx = 0; Bx = 0
+ c ©Fy = 0; By + 3.00 - 9.00 = 0 By = 6.00 kip
Equations of Equilibrium: For point D
+
: ©Fx = 0; ND = 0 Ans.
+ c ©Fy = 0; 3.00 - 2.25 - VD = 0
VD = 0.750 kip Ans.
a + ©MD = 0; MD + 2.25(2) - 3.00(6) = 0
MD = 13.5 kip # ft Ans.
Equations of Equilibrium: For point E
+
: ©Fx = 0; NE = 0 Ans.
+ c ©Fy = 0; - 6.00 - 3 - VE = 0
VE = - 9.00 kip Ans.
a + ©ME = 0; ME + 6.00(4) = 0
ME = - 24.0 kip # ft Ans.
Negative signs indicate that ME and VE act in the opposite direction to that shown
on FBD.
3
, 01 Solutions 46060 5/6/10 2:43 PM Page 4
© 2010 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.
1–6. Determine the normal force, shear force, and moment B
at a section through point C. Take P = 8 kN.
0.1 m
Support Reactions: 0.5 m
C A
a + ©MA = 0; 8(2.25) - T(0.6) = 0 T = 30.0 kN
+ 0.75 m 0.75 m 0.75 m
: ©Fx = 0; 30.0 - A x = 0 A x = 30.0 kN
P
+ c ©Fy = 0; Ay - 8 = 0 A y = 8.00 kN
Equations of Equilibrium: For point C
+
: ©Fx = 0; - NC - 30.0 = 0
NC = - 30.0 kN Ans.
+ c ©Fy = 0; VC + 8.00 = 0
VC = - 8.00 kN Ans.
a + ©MC = 0; 8.00(0.75) - MC = 0
MC = 6.00 kN # m Ans.
Negative signs indicate that NC and VC act in the opposite direction to that shown
on FBD.
1–7. The cable will fail when subjected to a tension of 2 kN. B
Determine the largest vertical load P the frame will support
and calculate the internal normal force, shear force, and
moment at the cross section through point C for this loading. 0.1 m
0.5 m
C A
Support Reactions:
0.75 m 0.75 m 0.75 m
a + ©MA = 0; P(2.25) - 2(0.6) = 0
P
P = 0.5333 kN = 0.533 kN Ans.
+
: ©Fx = 0; 2 - Ax = 0 A x = 2.00 kN
+ c ©Fy = 0; A y - 0.5333 = 0 A y = 0.5333 kN
Equations of Equilibrium: For point C
+
: ©Fx = 0; - NC - 2.00 = 0
NC = - 2.00 kN Ans.
+ c ©Fy = 0; VC + 0.5333 = 0
VC = - 0.533 kN Ans.
a + ©MC = 0; 0.5333(0.75) - MC = 0
MC = 0.400 kN # m Ans.
Negative signs indicate that NC and VC act in the opposite direction to that shown
on FBD.
4
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–1. Determine the resultant internal normal force acting 5 kip 8 kN
on the cross section through point A in each column. In
(a), segment BC weighs 180 lb>ft and segment CD weighs
250 lb>ft. In (b), the column has a mass of 200 kg>m.
B
200 mm 200 mm
6 kN 6 kN
(a) + c ©Fy = 0; FA - 1.0 - 3 - 3 - 1.8 - 5 = 0
10 ft 3m
FA = 13.8 kip Ans. 8 in. 8 in.
200 mm 200 mm
(b) + c ©Fy = 0; FA - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0 3 kip 3 kip
4.5 kN 4.5 kN
C
FA = 34.9 kN Ans.
4 ft
A A
4 ft 1m
D
(a) (b)
1–2. Determine the resultant internal torque acting on the A
cross sections through points C and D. The support bearings 250 Nm
at A and B allow free turning of the shaft.
150 Nm
©Mx = 0; TC - 250 = 0 300 mm C
400 Nm
200 mm
TC = 250 N # m Ans.
150 mm
©Mx = 0; TD = 0 Ans. 200 mm D B
250 mm
150 mm
1–3. Determine the resultant internal torque acting on the
cross sections through points B and C.
A 600 lbft
©Mx = 0; TB + 350 - 500 = 0 B
350 lbft
TB = 150 lb # ft Ans. 3 ft
C
©Mx = 0; TC - 500 = 0
1 ft 500 lb
TC = 500 lb # ft Ans. 2 ft
2 ft
1
,01 Solutions 46060 5/6/10 2:43 PM Page 2
© 2010 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.
*1–4. A force of 80 N is supported by the bracket as
shown. Determine the resultant internal loadings acting on
the section through point A. 0.3 m
A
30
0.1 m
80 N 45
Equations of Equilibrium:
+
Q©Fx¿ = 0; NA - 80 cos 15° = 0
NA = 77.3 N Ans.
a+ ©Fy¿ = 0; VA - 80 sin 15° = 0
VA = 20.7 N Ans.
a+ ©MA = 0; MA + 80 cos 45°(0.3 cos 30°)
- 80 sin 45°(0.1 + 0.3 sin 30°) = 0
MA = - 0.555 N # m Ans.
or
a+ ©MA = 0; MA + 80 sin 15°(0.3 + 0.1 sin 30°)
-80 cos 15°(0.1 cos 30°) = 0
MA = - 0.555 N # m Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
2
,01 Solutions 46060 5/6/10 2:43 PM Page 3
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•1–5. Determine the resultant internal loadings in the 3 kip
beam at cross sections through points D and E. Point E is
1.5 kip/ ft
just to the right of the 3-kip load.
A
D B E C
6 ft 6 ft 4 ft 4 ft
Support Reactions: For member AB
a + ©MB = 0; 9.00(4) - A y(12) = 0 A y = 3.00 kip
+
: ©Fx = 0; Bx = 0
+ c ©Fy = 0; By + 3.00 - 9.00 = 0 By = 6.00 kip
Equations of Equilibrium: For point D
+
: ©Fx = 0; ND = 0 Ans.
+ c ©Fy = 0; 3.00 - 2.25 - VD = 0
VD = 0.750 kip Ans.
a + ©MD = 0; MD + 2.25(2) - 3.00(6) = 0
MD = 13.5 kip # ft Ans.
Equations of Equilibrium: For point E
+
: ©Fx = 0; NE = 0 Ans.
+ c ©Fy = 0; - 6.00 - 3 - VE = 0
VE = - 9.00 kip Ans.
a + ©ME = 0; ME + 6.00(4) = 0
ME = - 24.0 kip # ft Ans.
Negative signs indicate that ME and VE act in the opposite direction to that shown
on FBD.
3
, 01 Solutions 46060 5/6/10 2:43 PM Page 4
© 2010 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.
1–6. Determine the normal force, shear force, and moment B
at a section through point C. Take P = 8 kN.
0.1 m
Support Reactions: 0.5 m
C A
a + ©MA = 0; 8(2.25) - T(0.6) = 0 T = 30.0 kN
+ 0.75 m 0.75 m 0.75 m
: ©Fx = 0; 30.0 - A x = 0 A x = 30.0 kN
P
+ c ©Fy = 0; Ay - 8 = 0 A y = 8.00 kN
Equations of Equilibrium: For point C
+
: ©Fx = 0; - NC - 30.0 = 0
NC = - 30.0 kN Ans.
+ c ©Fy = 0; VC + 8.00 = 0
VC = - 8.00 kN Ans.
a + ©MC = 0; 8.00(0.75) - MC = 0
MC = 6.00 kN # m Ans.
Negative signs indicate that NC and VC act in the opposite direction to that shown
on FBD.
1–7. The cable will fail when subjected to a tension of 2 kN. B
Determine the largest vertical load P the frame will support
and calculate the internal normal force, shear force, and
moment at the cross section through point C for this loading. 0.1 m
0.5 m
C A
Support Reactions:
0.75 m 0.75 m 0.75 m
a + ©MA = 0; P(2.25) - 2(0.6) = 0
P
P = 0.5333 kN = 0.533 kN Ans.
+
: ©Fx = 0; 2 - Ax = 0 A x = 2.00 kN
+ c ©Fy = 0; A y - 0.5333 = 0 A y = 0.5333 kN
Equations of Equilibrium: For point C
+
: ©Fx = 0; - NC - 2.00 = 0
NC = - 2.00 kN Ans.
+ c ©Fy = 0; VC + 0.5333 = 0
VC = - 0.533 kN Ans.
a + ©MC = 0; 0.5333(0.75) - MC = 0
MC = 0.400 kN # m Ans.
Negative signs indicate that NC and VC act in the opposite direction to that shown
on FBD.
4
