© 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 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.
*t-28. Determine the resultant internal loadings acting
on the cross section of the frame at points F and G. The
contact 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
Ans
80 Ib
, =0;
=0;
Ans
Hf = 160 Ib tl
Ans
Afc = 16.7lb
Ans
For point G :
-»Ifi=0:
16.708-W C =0:
+ T £ F, = 0;
Vc - 72.0 = 0;
\fc = 72.0 Ib
Ans
^.o I',
(+IAfc=0;
72(1 5)-A/ 6 -=0;
Afc = 108 tt) ft Ans
1-29. The bolt shank is subjected to a tension of 80 Ib.
Determine the resultant internal loadings acting on the
cross section at point C.
Segment AC:
Z F, - 0;
Mr + 80 = 0;
Nc = -80 Ib
Ans
Ans
M,-»X(K6)
-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 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.
Xl-44. The 50-lb lamp is supported by three steel rods
connected by a ring at A. Determine the angle of orientation
0 of AC such that the average normal stress in rod AC is
twice the average normal stress in rod AD. What is the
magnitude of stress in each rod? The diameter of each rod is
given in the figure.
r~
^=(0.098175X7^
-»Zf;=0;
- TAD cos 45° + T4C cos6 = 0
+ tiF,=Q;
f
(1)
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°
Ans
From Eq. (2):
(0.098175)CT,u, sin 59.39° + (0.070686)(T/U) sin 45° - 50 = 0
OAD = 371.8 psi = 372 psi
Ans
Hence,
0AC = 2(371.8) = 744 psi
Ans
And,
, T*'
=——— = 520psi
!(0.35)2 J(0.35)2
Ans
1-45. The shaft is subjected to the axial force of 30 kN. If
the shaft passes through the 53-mm diameter hole in the
fixed support A, determine the bearing stress acting on the
collar C. Also, what is the average shear stress acting along
the inside surface of the collar where it is fixed connected to
the 52-mm diameter shaft?
30O03)
= 48.3 MPa
J(0.06 Z -0.053*)
Ans
Average Shear Stress :
X
30(10')
= 18.4 MPa
*(0.052X0.01)
30 kN
40 mm
Bearing Slrest :
A
52 mm
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 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.
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.
5kN
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 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.
/1-87. The frame is subjected to the load of 1.5 kip.
Determine the required diameter of the pins at A and B if
the allowable shear stress for the material is rallow = 6 ksi.
Pin A is subjected to double shear, whereas pin B is
subjected to single shear.
15 kip
Support Reaction; : From FED (a).
f+IMD = 0;
/i c (sin45°)(5)- 1.5(7) =0
F,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
+ Tl^=0;
1.05-A, =0
4, = 1.05kip
Allowable Shear Stress : Design of pin sues
For pin A
Pin A is subjected to double shear and
Ft = /l.SP+l.OS 1 ' 1.831 kip.
Therefore. V. « y " °-"55 kip
VA
0.9155
-
*•/>
J, = 0 4 4 1 in.
Ans
For fin B
Pin B is subjected to single shear. Therefore,
V, = F, = f,c = 2.970 kip
t*
*^
**i
2.970
W
</,= 0.794 in.
Ans
*l-88. The two steel wires AB and AC are used to support
the load. If both wires have an allowable tensile stress of
""allow = 200 MPa, determine the required diameter of each
wire if the applied load is P = 5 kN.
+ If, = 0;
, = 0;
-Ftc - Fa sin 60° = 0
-FAC+ £u cos 60° - 5 = 0
(1)
(2)
Solving Eqs. (1) and (2) yields :
Fa = 4.34% kN;
F4C = 4.7086 kN
Applying <T.HO, = For wire AB.
200(10', = 4-3496(1°3)
2
dtt = 0.00526 m - 5. 2ft mm
Ans
For wir
4,c = 0.00548 m = 5.48 mm
Am
42
2-974 Kif
© 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 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.
1-101. The hanger assembly is used to support a distributed
loading of w = 0.8 kip/ft. Determine the average shear
stress in the 0.40-in.-diameter bolt at A and the average
tensile stress in rod AB, which has a diameter of 0.5 in. If the
yield shear stress for the bolt is ry = 25 ksi, and the yield
tensile stress for the rod is ay = 38 ksi, determine the factor
of safety with respect to yielding in each case.
For boltA :
V
3
r = - = - 23.9 ksi
A J(0.4J)
F.S.
23.9
For rod AB :
P
a =A
f (0.52)
JL
30.6
= 1.05
30 6 ksi
• 1.24
Am
Ans
Ans
/1-102. Determine the intensity w of the maximum
distributed load that can be supported by the hanger
assembly so that an allowable shear stress of rallow =13.5 ksi
is not exceeded in the 0.40-in.-diameter bolts at A and B, and
an allowable tensile stress of <ranow = 22 ksi is not exceeded
in 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)
Assuming failure of rod AS :
7.5w
22 =
|(0.52)
Ans
375W
w = 0.576 kip/ft
49