: ©Fx = 0;
01 Solutions
+ 46060
5/6/10
+ c ©Fy = 0;
2:43 PM
30.0 - A x = 0
0.75 m
Page 32
A x = 30.0 kN
Ay - 8 = 0
0.75 m
0.75 m
P
A y = 8.00 kN
© 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.
Equations
of Equilibrium:
For
pointSaddle
C River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
© 2010 Pearson
Education, Inc.,
Upper
exist.
No
portion
of
this
material
may
be reproduced, in any form or by any means, without permission in writing from the publisher.
+
-NC - 30.0 = 0
: ©Fx = 0;
•1–5.
Determine the resultant internal loadings in the
= -30.0
kN E. Point E is
beam
at
cross sections throughNpoints
D and
1–47. Determine the average Cshear stress developed in pins
just to
the
right
of
the
3-kip
load.
A and B of the smooth two-tine grapple that supports the log
+ c ©Fy = 0;
VC + 8.00 = 0
having a mass of 3 Mg. Each pin has a diameter of 25 mm and
is subjected to double shear.
VC = -8.00 kN
a++c©M
0;
©FyC == 0;
3 kip
Ans.
1.5 kip/ ft
20!
Ans.
A
MC =kN6.00 kN # m
F = 29.43
Ans.
C
E
B
D
8.00(0.75)
C = 0= 0
2(F
sin 30°)--M
29.43
A
6 ft
D
0.2
4 ftm
6 ft
C
E
B
4 ft
1.2 m
Negative
that NC and
VC act
the opposite
direction
to thatcos
shown
= 0; indicate
P cos 20°(0.2)
(29.43
cosin30°)(1.2)
+ (29.43
sin 30°)(0.4
30°)
a + ©MEsigns
on FBD.
= 0
Support Reactions: For member AB
P = 135.61 kN
9.00(4) - A y(12) = 0
A y = 3.00 kip
3
1–7. The cable will135.61(10
fail when
subjected to a tension
of 2 kN.
)
V
2
+ Determine
the
largest
vertical
load
P
the
frame
will
support
=
t
=
=
=
138
MPa
t
0;
: ©F
Ax = B
p Bx = 0
(0.025)2 normal force, shear force, and
and calculateA the 4internal
at the cross
loading.
+ c moment
©F = 0;
B + section
3.00 -through
9.00 =point
0 CBfor =this
6.00
kip
30!
a + ©MB = 0;
y
y
0.1 m
+
: ©Fx = 0;
0.75 m
P(2.25) - 2(0.6) = 0
ND = 0
P
3.00 - 2.25 - VD = 0
2 - Ax = 0
A x = 2.00 kN
VD = 0.750 kip
Equations of Equilibrium: For point E
+ c ©Fy = 0;
A
0.75 m
0.75 m
Ans.
Ans.
P = 0.5333 kN = 0.533 kN
*1–48. The beam is supported by a pin at A and a short
+ c ©Fy = 0;
A y - 0.5333 = 0
A y = 0.5333 kN
average
a + link
= If
0; P = 15
MDkN,
+ determine
2.25(2) - the
3.00(6)
= 0shear stress
©MDBC.
developed in the pins at A, B, and C. All pins are in double
Equations of Equilibrium: For point C
shear as shown, and
has akip
diameter
# ft of 18 mm.
MDeach
= 13.5
+
01 Solutions :
46060
5/6/10
2:43
PM
Page
-NC - 2.00 = 043
©Fx = 0;
+
: ©Fx = 0;
0.5 m
C
Support of
Reactions:
Equations
Equilibrium: For point D
+ c ©Fy = 0;
B
Ans.
y
+ a + ©M = 0;
: ©Fx = A0;
30!
0.4 m
Ans.
P
4P
0.5m Ans.
1m
C
30!
NC = -2.00 kN
B
NE = 0
VC + 0.5333 = 0
4P
1.5 m
2P
0.5 m
1.5 m
Ans.
Ans.
A
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
= 0;No portion
-6.00
- 3material
- VE may
= 0 be reproduced, in any form or by any means, without permission in writing from the publisher.
+ c ©Fyexist.
of V
this
Ans.
C = -0.533 kN
For pins B and C:
Ans.
kip
V0.5333(0.75)
E = -9.00
a + ©MC = V
0; 82.5
- MC = 0
(103)
Ans.
t
=
t
=
=
=
324
MPa
B
C
p 18maximum
(1000
)2 = 0.400
a +•1–61.
0; A M4Ethe
+M
6.00(4)
=magnitude
0kN # m P of the load
©ME =Determine
Ans.
Caverage shear
the beam will support if the
stress in each pin
For
pin
A:
is not to allowed toM
exceed
60 MPa.
# ftpins are subjected to
Ans.
-24.0
kipAll
E =that
Negative
signs
NC and
act in theofopposite
double
shear
asindicate
has VaCdiameter
18 mm. direction to that shown
2shown, and2 each
= 2 (82.5) + (142.9) = 165 kN
FAFBD.
on
C
Negative signs indicate that ME and VE act in the opposite direction to that shown
Referring to the FBD
3 of member AB, Fig. a,
on FBD. V 82.5 (10 )
tA = 46060
= p 5/6/10
= 2:43
324 MPa
01 Solutions
PM Page 65
( 18 )2
a + ©MA
A = 0;4 1000FBC sin 30°(6) - P(2) - P(4) = 0
+
: ©Fx = 0;
A x - 2P cos 30° = 0
+ c ©Fy = 0;
FBC = 2P
30!
Ans.
A
B
2m
A x = 1.732P
2m
P
2m
P
A y - P - P + 2P sin 30° = 0
Ay = P
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights 4
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.
Thus, the force acting on pin A is
FA = 2 A x 2 + A y 2 = 2 (1.732P)2 + P2 = 2P
*1–96. If the allowable bearing stress for the material under
the are
supports
at Ato
and
B isforce
1sb2and
= 1.5 MPa,
allow double
All pins
subjected
same
shear.determine
Referring to the FBD of the
the
maximum
load
P
that
can
be
applied
to
the beam. The
pin, Fig. b,
bearing plates A¿ and B¿ have square cross sections of
F
2P
V = 150=mm *=150
P mm and 250 mm * 250 mm, respectively.
2
2
The cross-sectional area of the pin is A =
p
(0.0182) = 81.0(10 - 6)p m2. Thus,
4
V
P
6
tallowReferring
=
; to60(10
) = of the beam,
the FBD
Fig. a,
A
81.0(10 - 6)p
32
a + ©MA = 0;
NPB(3)
+ 268
40(1.5)(0.75)
- P(4.5) = 0
= 15
N = 15.3 kN
NB = 1.5P - Ans.
15
a + ©MB = 0;
40(1.5)(3.75) - P(1.5) - NA(3) = 0
N
3A = 75 - 0.5P
For plate A¿ ,
NA
(sb)allow =
;
A A¿
1.5(106) =
(75 - 0.5P)(103)
0.15(0.15)
40 kN/m
A
1.5 m
P
A¿
B¿
3m
B
1.5 m