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PROBLEM 13.60
Three
1 18-in.×
steel plates are bolted to four
L6 6 1××
angles to form a beam
with the cross section shown. The bolts have a
7
8
-in.
diameter and are spaced
longitudinally every 5 in. Knowing that the allowable average shearing stress in
the bolts is 12 ksi, determine the largest permissible vertical shear in the beam.
(Given:
4
6123 in .
x
I=
)
SOLUTION
Flange:
3 24
1(18)(1) (18)(1)(9.5) 1626 in
12
f
I=+=
Web:
34
1(1)(18) 486 in
12
w
I= =
Angle:
42
24
35.5 in , 11.0 in
1.86 in. 9 1.86 7.14 in.
596.18 in
a
IA
yd
I I Ad
= =
= =−=
=+=
4
2 4 6123 in ,
fw a
III I= ++ =
which agrees with the given value.
Flange:
3
(18)(1)(9.5) 171 in
f
Q= =
Angle:
3
3
(11.0)(7.14) 78.54 in
2 328.08 in
a
fa
Q Ad
QQ Q
= = =
=+=
22
bolt
bolt bolt bolt
bolt
all
70.60132 in
48
2 (2)(12)(0.60132) 14.4317 kips
14.4317 2.8863 kip/s
5
A
FA
F
qs
p
t
= =
= = =
= = =
all
all
(2.8863)(6123)
328.08
VQ q I
qV
IQ
= = =
all
53.9 kipsV=
consent of McGraw-Hill Education.
PROBLEM 13.61
SOLUTION
16
tan 28.07
30
aa
= = °
Side:
2
3 34
(3 sec )(30) 102 mm
1(3 sec )(30) 7.6498 10 mm
12
A
I
a
a
= =
= = ×
Part
2
(mm )
A
0
(mm)y
33
(10 mm )Ay
(mm)d
23 4
(10 mm )Ad
34
(10 mm )I
Top 180 30 5.4 11.932 25.627 neglect
Side 102 15 1.53 3.077 0.966 7.6498
Side 102 15 1.53 3.077 0.966 7.6498
Bot. 84 0 0 18.077 27.449 neglect
Σ 468 8.46 55.008 15.2996
3
0
2 3 4 94
8.46 10 18.077 mm
468
70.31 10 mm 70.31 10 m
Ay
YA
I Ad I
−
Σ×
= = =
Σ
=Σ +Σ = × = ×
(a)
3 3 63
(180)(11.932) 2.14776 10 mm 2.14776 10 m
A
Q−
= =×=×
33
36
6
96
(2)(3 10 ) 6 10 m
(10 10 )(2.14776 10 ) 50.9 10 Pa
(70.31 10 )(6 10 )
A
t
VQ
It
t
−−
−
−−
=×=×
××
= = = ×
××
50.9 MPa
A
t
=
(b)
3 33
63
3
1
(2)(3 sec )(11.932) 11.932
2
2.14776 10 484.06 2.6318 10 mm
2.6318 10 m
6 10 m
mA
QQ
t
a
−
−
=+×
= ×+ = ×
= ×
= ×
C
M=
62.4 MPa
m
t
=
consent of McGraw-Hill Education.
PROBLEM 13.61 (Continued)
33
3
3
(28)(3)(18.077) 1.51847 10 mm
1.51847 10 (50.9)
2.14776 10
36.0 MPa
B
B
BA
A
Q
Q
Q
tt
= = ×
×
= = ×
=
Multiply shearing stresses by
(3 mm 0.003 m)t=
to get shear flow.
consent of McGraw-Hill Education.
PROBLEM 13.60
Three
1 18-in.×
steel plates are bolted to four
L6 6 1××
angles to form a beam
with the cross section shown. The bolts have a
7
8
-in.
diameter and are spaced
longitudinally every 5 in. Knowing that the allowable average shearing stress in
the bolts is 12 ksi, determine the largest permissible vertical shear in the beam.
(Given:
4
6123 in .
x
I=
)
SOLUTION
Flange:
3 24
1(18)(1) (18)(1)(9.5) 1626 in
12
f
I=+=
Web:
34
1(1)(18) 486 in
12
w
I= =
Angle:
42
24
35.5 in , 11.0 in
1.86 in. 9 1.86 7.14 in.
596.18 in
a
IA
yd
I I Ad
= =
= =−=
=+=
4
2 4 6123 in ,
fw a
III I= ++ =
which agrees with the given value.
Flange:
3
(18)(1)(9.5) 171 in
f
Q= =
Angle:
3
3
(11.0)(7.14) 78.54 in
2 328.08 in
a
fa
Q Ad
QQ Q
= = =
=+=
22
bolt
bolt bolt bolt
bolt
all
70.60132 in
48
2 (2)(12)(0.60132) 14.4317 kips
14.4317 2.8863 kip/s
5
A
FA
F
qs
p
t
= =
= = =
= = =
all
all
(2.8863)(6123)
328.08
VQ q I
qV
IQ
= = =
all
53.9 kipsV=
consent of McGraw-Hill Education.
PROBLEM 13.61
SOLUTION
16
tan 28.07
30
aa
= = °
Side:
2
3 34
(3 sec )(30) 102 mm
1(3 sec )(30) 7.6498 10 mm
12
A
I
a
a
= =
= = ×
Part
2
(mm )
A
0
(mm)y
33
(10 mm )Ay
(mm)d
23 4
(10 mm )Ad
34
(10 mm )I
Top 180 30 5.4 11.932 25.627 neglect
Side 102 15 1.53 3.077 0.966 7.6498
Side 102 15 1.53 3.077 0.966 7.6498
Bot. 84 0 0 18.077 27.449 neglect
Σ 468 8.46 55.008 15.2996
3
0
2 3 4 94
8.46 10 18.077 mm
468
70.31 10 mm 70.31 10 m
Ay
YA
I Ad I
−
Σ×
= = =
Σ
=Σ +Σ = × = ×
(a)
3 3 63
(180)(11.932) 2.14776 10 mm 2.14776 10 m
A
Q−
= =×=×
33
36
6
96
(2)(3 10 ) 6 10 m
(10 10 )(2.14776 10 ) 50.9 10 Pa
(70.31 10 )(6 10 )
A
t
VQ
It
t
−−
−
−−
=×=×
××
= = = ×
××
50.9 MPa
A
t
=
(b)
3 33
63
3
1
(2)(3 sec )(11.932) 11.932
2
2.14776 10 484.06 2.6318 10 mm
2.6318 10 m
6 10 m
mA
QQ
t
a
−
−
=+×
= ×+ = ×
= ×
= ×
C
M=
62.4 MPa
m
t
=
consent of McGraw-Hill Education.
PROBLEM 13.61 (Continued)
33
3
3
(28)(3)(18.077) 1.51847 10 mm
1.51847 10 (50.9)
2.14776 10
36.0 MPa
B
B
BA
A
Q
Q
Q
tt
= = ×
×
= = ×
=
Multiply shearing stresses by
(3 mm 0.003 m)t=
to get shear flow.
consent of McGraw-Hill Education.
