© 2008 by R.C. Hibbeler. Published by Pearson Prentice ...users.rowan.edu/~sukumaran/solidmechanics/solutions/hw1-soln-2010.… · Published by Pearson Prentice Hall, Pearson Education,

© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under allcopyright 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.

*t-28. Determine the resultant internal loadings actingon the cross section of the frame at points F and G. Thecontact at E is smooth.

Member lil-.l':0; Ne (5)-80(9) = 0

Nc = 144 Ib

Member BCE :

ZM,=0; F,,c<-X3)-144sin30°(6) = 0

FAC = 180lb

=0; B, + 180(-)- 144cos30° = 0

B, = 1670816

F, =0, -fl, + !80<-)-t44sin30° = 0

B, = 72.0 Ib

For point F :, = 0; Nf = 0

, =0;

=0;

Ans

80 Ib Ans

Hf = 160 Ib tl Ans

For point G :

- » I f i = 0 : 16.708-WC=0: Afc = 16.7lb Ans

+ T £ F, = 0; Vc - 72.0 = 0; \fc = 72.0 Ib Ans

( + I A f c = 0 ; 72(1 5)-A/6-=0; Afc = 108 tt) ft Ans

^.o I',

1-29. The bolt shank is subjected to a tension of 80 Ib.Determine the resultant internal loadings acting on thecross section at point C.

Segment AC:

Z F, - 0; Mr + 80 = 0; Nc = -80 Ib

M , - » X ( K 6 )

Ans

Ans

-4X0 Ib in Ans

16

© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under allcopyright 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.

Xl-44. The 50-lb lamp is supported by three steel rodsconnected by a ring at A. Determine the angle of orientation0 of AC such that the average normal stress in rod AC istwice the average normal stress in rod AD. What is themagnitude of stress in each rod? The diameter of each rod isgiven in the figure.

r~

^=(0.098175X7^ f

- » Z f ; = 0 ; - TAD cos 45° + T4C cos6 = 0 (1)

+ tiF,=Q; T*c sin 6 + 7*050145°-50 = 0 (2)

Thus-(0.070686)<T/u,(cos 45°) + (0.098175)ffAC(cos 9) = 0

0 = 59.39° = 59 4° AnsFrom Eq. (2):

(0.098175)CT,u, sin 59.39° + (0.070686)(T/U) sin 45° - 50 = 0

OAD = 371.8 psi = 372 psi AnsHence,

0AC = 2(371.8) = 744 psi Ans

And,

, T*' =——— = 520psi Ans!(0.35)2 J(0.35)2

1-45. The shaft is subjected to the axial force of 30 kN. Ifthe shaft passes through the 53-mm diameter hole in thefixed support A, determine the bearing stress acting on thecollar C. Also, what is the average shear stress acting alongthe inside surface of the collar where it is fixed connected tothe 52-mm diameter shaft?

52 mm

30 kN

Bearing Slrest : 40 mm

30O03)A J(0.06Z-0.053*)

Average Shear Stress :

30(10')

= 48.3 MPa

X *(0.052X0.01)= 18.4 MPa

Ans

Ans

23

© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under allcopyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

A-65. Member A of the timber step joint for a truss is

subjected to a compressive force of 5 kN. Determine the

average normal stress acting in the hanger rod C which has

a diameter of 10 mm and in member B which has a

thickness of 30 mm.

5 k N

Equations of Equilibrium :

4 !/=; = 0; 5cos 60° - F, = 0 F, - 2.50 kN

+ T£/y=0; /v-5sin60° = 0 F<= 4.330 kN

Average Normal Stress :

33

© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under allcopyright 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.

/1-87. The frame is subjected to the load of 1.5 kip.Determine the required diameter of the pins at A and B ifthe allowable shear stress for the material is rallow = 6 ksi.Pin A is subjected to double shear, whereas pin B issubjected to single shear.

Support Reaction; : From FED (a).

f+IMD = 0; /ic(sin45°)(5)- 1.5(7) =0F,c - 2.970 lap

From FED (b).r+LMi=0; D, (10)- 1.5(7) »0 Dt = 1.05kip

Az / i=0 ; X, -1.5*0 A, -1.50fcp

+ T l^=0; 1.05-A, =0 4, = 1.05kip

Allowable Shear Stress : Design of pin sues

For pin APin A is subjected to double shear and

Ft = /l.SP+l.OS1 ' 1.831 kip.

Therefore. V. « y " °-"55 kip

VA 0.9155

-J, = 0 4 4 1 in. Ans

For fin BPin B is subjected to single shear. Therefore,V, = F, = f,c = 2.970 kip

*̂ 2.970

**i W</,= 0.794 in. Ans

15 kip

*•/>

t* 2-974 Kif

*l-88. The two steel wires AB and AC are used to supportthe load. If both wires have an allowable tensile stress of""allow = 200 MPa, determine the required diameter of eachwire if the applied load is P = 5 kN.

+ If, = 0; -Ftc - Fa sin 60° = 0

, = 0; -FAC+ £u cos 60° - 5 = 0

Solving Eqs. (1) and (2) yields :Fa = 4.34% kN; F4C = 4.7086 kN

Applying <T.HO, = -

For wire AB.

(1)

(2)

200(10', = 4-3496(1°3)

2

dtt = 0.00526 m - 5. 2ft mm Ans

For wir

Am4,c = 0.00548 m = 5.48 mm

42

© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under allcopyright 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.

1-101. The hanger assembly is used to support a distributedloading of w = 0.8 kip/ft. Determine the average shearstress in the 0.40-in.-diameter bolt at A and the averagetensile stress in rod AB, which has a diameter of 0.5 in. If theyield shear stress for the bolt is ry = 25 ksi, and the yieldtensile stress for the rod is ay = 38 ksi, determine the factorof safety with respect to yielding in each case.

For boltA :V 3

r =- = -A J(0.4J)

23.9 ksi

F.S.23.9

= 1.05 Am

For rod AB :

Pa = -

A f (0.52)

JL30.6

30 6 ksi Ans

• 1.24 Ans

/1-102. Determine the intensity w of the maximumdistributed load that can be supported by the hangerassembly so that an allowable shear stress of rallow =13.5 ksiis not exceeded in the 0.40-in.-diameter bolts at A and B, andan allowable tensile stress of <ranow = 22 ksi is not exceededin the 0.5-in.-diameter rod AB.

Assume failure of pin A or B :3.75w

T.MO. = 13.5 =

w = 0.452 kip/ft (controls) Ans

Assuming failure of rod AS :7.5w

375W

22 =|(0.52)

w = 0.576 kip/ft

49