100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Previously searched by you
Previously searched by you
logo
Hibbeler strength learning effects H1 t / m H7 $3.70
Add to cart
Quickly navigate to
Preview
  2.  
  3. Hogeschool Van Amsterdam (HvA)
  4.  
  5. Mechanische systemen

Answers

Hibbeler strength learning effects H1 t / m H7

25 reviews
 5586 views  117 purchases
  • Course
  • Mechanische systemen
  • Institution
  • Hogeschool Van Amsterdam (HvA)

These are the effects of Hibbeler Sterkteleer, widely used in engineering courses. In English, but should be no problem in finding the correct answer and how to get there from.

Last document update: 8 year ago

Preview 4 out of 531  pages

Document information

  • September 28, 2016
  • September 28, 2016
  • 531
  • 2015/2016
  • Answers
  • Unknown

Subjects

  • hibbeler
  • sterkteleer
  • antwoorden
  • uitwerkingen
  • statica
  • dynamica
  • natuurkunde
  • werktuigbouwkunde
  • krachten
  • krachtenleer
  • hoofstuk 1
  • hoofdstuk 7
  • hoofdstuk 2
  • hoofdstuk 3
  • hoofdstuk 4
  • hoofdstuk 5
  • hoofdstuk 6

Written for

  • Hogeschool van Amsterdam (HvA)
  • Technische Bedrijfskunde
  • Mechanische systemen
All documents for this subject (1)

25  reviews

review-writer-avatar

By: aarninkhof • 3 days ago

review-writer-avatar

By: jensderyck01 • 4 months ago

Translated by Google

exercises are not developed, only provide a solution

review-writer-avatar

By: ikbenreinierkuiper • 3 months ago

review-writer-avatar

By: ebox0252 • 11 months ago

review-writer-avatar

By: erwindupon2 • 11 months ago

review-writer-avatar

By: chielkootstra • 6 months ago

review-writer-avatar

By: 695086X • 1 year ago

Show more reviews  

Seller

avatar-seller
Gl13

Reviews received

Content preview

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

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller Gl13. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $3.70. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

50990 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 15 years now

Start selling
$3.70  117x  sold
  • (25)
Add to cart
Added