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bot
bot
(500)( 4.778)
155.16
My
I
σ
−
=−=−
bot
15.40 ksi (tension)
σ
=
PROBLEM 11.12
Knowing that for the extruded beam shown the allowable stress is 120
MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
SOLUTION
2
, mmA
0
, mmy
3
0, mmAy
d, mm
2160 27 58320 3
1080 36 38880 3
Σ 3240 97200
97200 30 mm The neutral axis lies 30 mm above the bottom.
3240
Y= =
top bot
3 2 3 2 34
1 11 11
2 2 3 2 34
2 22 22
3 4 94
12
54 30 24 mm 0.024 m 30 mm 0.030 m
11
(40)(54) (40)(54)(3) 544.32 10 mm
12 12
1 11
(40)(54) (40)(54)(6) 213.84 10 mm
36 36 2
758.16 10 mm 758.16 10 m
yy
I bh Ad
I bh Ad
II I
−
=−= = =− =−
= += + = ×
= += + = ×
=+= × = ×
|| | |
σ
σ
= =
My I
M
Iy
Top: (tension side)
69 3
(120 10 )(758.16 10 ) 3.7908 10 N m
0.024
M−
××
= = ×⋅
Bottom: (compression)
69 3
(150 10 )(758.16 10 ) 3.7908 10 N m
0.030
M−
××
= = ×⋅
Choose the smaller as Mall.
3
all
3.7908 10 N mM= ×⋅
all
3.79 kN mM= ⋅
consent of McGraw-Hill Education.
PROBLEM 11.13
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 50 kip ⋅ in., determine the total force acting
(a) on the top flange, (b) on the shaded portion of the web.
SOLUTION
My
consent of McGraw-Hill Education.
PROBLEM 11.14
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is
4 kN m,⋅
determine the total force acting on
the shaded portion of the beam.
SOLUTION
Dimensions in mm:
33
66
6 4 64
11
(12 12)(88) (40)(40)
12 12
1.3629 10 0.213 10
1.5763 10 mm 1.5763 10 m
−
=++
= ×+ ×
=×=×
z
I
For use in Prob. 4.14,
33
66
6 4 64
11
(88)(64) (24 24)(40)
12 12
1.9224 10 0.256 10
1.6664 10 mm 1.6664 10 m
−
= −+
= ×− ×
=×=×
y
I
Bending about horizontal axis.
4 kN m
z
M= ⋅
64
64
(4 kN m)(0.044 m) 111.654 MPa
1.5763 10 m
(4 kN m)(0.020 m) 50.752 MPa
1.5763 10 m
z
Az
z
Bz
Mc
I
Mc
I
σ
σ
−
−
⋅
= = =
×
⋅
= = =
×
SOLUTION Continued
2 62
−
PROBLEM 11.15
Solve Prob. 11.14, assuming that the beam is bent about a vertical axis by a
couple of moment
4 kN m.⋅
PROBLEM 11.14. Knowing that a beam of the cross section shown is bent
about a horizontal axis and that the bending moment is
4 kN m,⋅
determine the
total force acting on the shaded portion of the beam.
SOLUTION
PROBLEM 11.16
Knowing that for the casting shown the allowable stress is 5 ksi in
tension and 18 ksi in compression, determine the largest couple M
that can be applied.
SOLUTION
consent of McGraw-Hill Education.
PROBLEM 11.17
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
SOLUTION
2
, mmA
0
, mmy
3
0, mmAy
Solid rectangle 4608 48 221,184
Square cutout –1296 30 –38,880
Σ 3312 182,304
182,304 55.04 mm
3312
= =Y
Neutral axis lies 55.04 mm above bottom.
top
bot
3 2 3 2 64
1 11 11
3 2 3 2 64
2 22 22
64
12
96 55.04 40.96 mm 0.04096 m
55.04 mm 0.05504 m
11
(48)(96) (48)(96)(7.04) 3.7673 10 mm
12 12
11
(36)(36) (36)(36)(25.04) 0.9526 10 mm
12 12
2.8147 10 mm
=−= =
=−=−
= += + = ×
= += + = ×
=−= × =
y
y
I bh Ad
I bh Ad
II I 64
2.8147 10 m
−
×
||= ∴=+
My I
M
Iy
σ
σ
Top: (tension side)
66
3
(120 10 )(2.8147 10 ) 8.25 10 N m
0.04096
−
××
= =×⋅M
Bottom: (compression)
66
3
(150 10 )(2.8147 10 ) 7.67 10 N m
0.05504
××
= =×⋅M
Mall is the smaller value.
3
7.67 10 N mM=×⋅
7.67 kN m⋅
consent of McGraw-Hill Education.
PROBLEM 11.18
Knowing that for the extruded beam shown the allowable stress is 12 ksi
in tension and 16 ksi in compression, determine the largest couple M that
can be applied.
SOLUTION
A
0
y
0
Ay
2.25 1.25 2.8125
2.25 0.25 0.5625
4.50 3.375
3.375 0.75 in.
4.50
Y= =
The neutral axis lies 0.75 in. above bottom.
top bot
32 3 2 4
1 11 11
22 3 2 4
2 22 22
4
12
2.0 0.75 1.25 in., 0.75 in.
11
(1.5)(1.5) (2.25)(0.5) 0.984375 in
12 12
11
(4.5)(0.5) (2.25)(0.5) 0.609375 in
12 12
1.59375 in
yy
I bh Ad
I bh Ad
II I
My I
M
Iy
σ
σ
=−= =−
= += + =
= += + =
=+=
= =
Top: (compression)
(16)(1.59375) 20.4 kip in.
1.25
M= = ⋅
Bottom: (tension)
(12)(1.59375) 25.5 kip in.
0.75
M= = ⋅
Choose the smaller as Mall.
all
20.4 kip in.M= ⋅
consent of McGraw-Hill Education.
PROBLEM 11.19
The beam shown is made of a nylon for which the allowable stress is
24 MPa in tension and 30 MPa in compression. Determine the largest
couple M that can be applied to the beam.
SOLUTION
2
, mmA
0
, mm
y
3
0, mmAy
600 22.5
3
13.5 10×
300 7.5
3
2.25 10×
Σ 900
3
15.75 10
×
3
0
15.5 10 17.5 mm The neutral axis lies 17.5 mm above the bottom.
900
Y×
= =
top
bot
3 2 3 2 34
1 11 11
3 2 3 2 34
2 22 22
3 4 94
12
30 17.5 12.5 mm 0.0125 m
17.5 mm 0.0175 m
11
(40)(15) (600)(5) 26.25 10 mm
12 12
11
(20)(15) (300)(10) 35.625 10 mm
12 12
61.875 10 mm 61.875 10 m
y
y
I bh Ad
I bh Ad
II I
−
=−= =
=−=−
= += + = ×
= += + = ×
=+= × = ×
|| My I
M
Iy
σ
σ
= =
Top: (tension side)
69
(24 10 )(61.875 10 ) 118.8 N m
0.0125
M−
××
= = ⋅
Bottom: (compression)
69
(30 10 )(61.875 10 ) 106.1 N m
0.0175
M
−
××
= = ⋅
Choose smaller value.
106.1 N mM= ⋅
PROBLEM 11.12
Knowing that for the extruded beam shown the allowable stress is 120
MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
SOLUTION
2
, mmA
0
, mmy
3
0, mmAy
d, mm
2160 27 58320 3
1080 36 38880 3
Σ 3240 97200
97200 30 mm The neutral axis lies 30 mm above the bottom.
3240
Y= =
top bot
3 2 3 2 34
1 11 11
2 2 3 2 34
2 22 22
3 4 94
12
54 30 24 mm 0.024 m 30 mm 0.030 m
11
(40)(54) (40)(54)(3) 544.32 10 mm
12 12
1 11
(40)(54) (40)(54)(6) 213.84 10 mm
36 36 2
758.16 10 mm 758.16 10 m
yy
I bh Ad
I bh Ad
II I
−
=−= = =− =−
= += + = ×
= += + = ×
=+= × = ×
|| | |
σ
σ
= =
My I
M
Iy
Top: (tension side)
69 3
(120 10 )(758.16 10 ) 3.7908 10 N m
0.024
M−
××
= = ×⋅
Bottom: (compression)
69 3
(150 10 )(758.16 10 ) 3.7908 10 N m
0.030
M−
××
= = ×⋅
Choose the smaller as Mall.
3
all
3.7908 10 N mM= ×⋅
all
3.79 kN mM= ⋅
consent of McGraw-Hill Education.
PROBLEM 11.13
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 50 kip ⋅ in., determine the total force acting
(a) on the top flange, (b) on the shaded portion of the web.
SOLUTION
My
consent of McGraw-Hill Education.
PROBLEM 11.14
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is
4 kN m,⋅
determine the total force acting on
the shaded portion of the beam.
SOLUTION
Dimensions in mm:
33
66
6 4 64
11
(12 12)(88) (40)(40)
12 12
1.3629 10 0.213 10
1.5763 10 mm 1.5763 10 m
−
=++
= ×+ ×
=×=×
z
I
For use in Prob. 4.14,
33
66
6 4 64
11
(88)(64) (24 24)(40)
12 12
1.9224 10 0.256 10
1.6664 10 mm 1.6664 10 m
−
= −+
= ×− ×
=×=×
y
I
Bending about horizontal axis.
4 kN m
z
M= ⋅
64
64
(4 kN m)(0.044 m) 111.654 MPa
1.5763 10 m
(4 kN m)(0.020 m) 50.752 MPa
1.5763 10 m
z
Az
z
Bz
Mc
I
Mc
I
σ
σ
−
−
⋅
= = =
×
⋅
= = =
×
SOLUTION Continued
2 62
−
PROBLEM 11.15
Solve Prob. 11.14, assuming that the beam is bent about a vertical axis by a
couple of moment
4 kN m.⋅
PROBLEM 11.14. Knowing that a beam of the cross section shown is bent
about a horizontal axis and that the bending moment is
4 kN m,⋅
determine the
total force acting on the shaded portion of the beam.
SOLUTION
PROBLEM 11.16
Knowing that for the casting shown the allowable stress is 5 ksi in
tension and 18 ksi in compression, determine the largest couple M
that can be applied.
SOLUTION
consent of McGraw-Hill Education.
PROBLEM 11.17
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
SOLUTION
2
, mmA
0
, mmy
3
0, mmAy
Solid rectangle 4608 48 221,184
Square cutout –1296 30 –38,880
Σ 3312 182,304
182,304 55.04 mm
3312
= =Y
Neutral axis lies 55.04 mm above bottom.
top
bot
3 2 3 2 64
1 11 11
3 2 3 2 64
2 22 22
64
12
96 55.04 40.96 mm 0.04096 m
55.04 mm 0.05504 m
11
(48)(96) (48)(96)(7.04) 3.7673 10 mm
12 12
11
(36)(36) (36)(36)(25.04) 0.9526 10 mm
12 12
2.8147 10 mm
=−= =
=−=−
= += + = ×
= += + = ×
=−= × =
y
y
I bh Ad
I bh Ad
II I 64
2.8147 10 m
−
×
||= ∴=+
My I
M
Iy
σ
σ
Top: (tension side)
66
3
(120 10 )(2.8147 10 ) 8.25 10 N m
0.04096
−
××
= =×⋅M
Bottom: (compression)
66
3
(150 10 )(2.8147 10 ) 7.67 10 N m
0.05504
××
= =×⋅M
Mall is the smaller value.
3
7.67 10 N mM=×⋅
7.67 kN m⋅
consent of McGraw-Hill Education.
PROBLEM 11.18
Knowing that for the extruded beam shown the allowable stress is 12 ksi
in tension and 16 ksi in compression, determine the largest couple M that
can be applied.
SOLUTION
A
0
y
0
Ay
2.25 1.25 2.8125
2.25 0.25 0.5625
4.50 3.375
3.375 0.75 in.
4.50
Y= =
The neutral axis lies 0.75 in. above bottom.
top bot
32 3 2 4
1 11 11
22 3 2 4
2 22 22
4
12
2.0 0.75 1.25 in., 0.75 in.
11
(1.5)(1.5) (2.25)(0.5) 0.984375 in
12 12
11
(4.5)(0.5) (2.25)(0.5) 0.609375 in
12 12
1.59375 in
yy
I bh Ad
I bh Ad
II I
My I
M
Iy
σ
σ
=−= =−
= += + =
= += + =
=+=
= =
Top: (compression)
(16)(1.59375) 20.4 kip in.
1.25
M= = ⋅
Bottom: (tension)
(12)(1.59375) 25.5 kip in.
0.75
M= = ⋅
Choose the smaller as Mall.
all
20.4 kip in.M= ⋅
consent of McGraw-Hill Education.
PROBLEM 11.19
The beam shown is made of a nylon for which the allowable stress is
24 MPa in tension and 30 MPa in compression. Determine the largest
couple M that can be applied to the beam.
SOLUTION
2
, mmA
0
, mm
y
3
0, mmAy
600 22.5
3
13.5 10×
300 7.5
3
2.25 10×
Σ 900
3
15.75 10
×
3
0
15.5 10 17.5 mm The neutral axis lies 17.5 mm above the bottom.
900
Y×
= =
top
bot
3 2 3 2 34
1 11 11
3 2 3 2 34
2 22 22
3 4 94
12
30 17.5 12.5 mm 0.0125 m
17.5 mm 0.0175 m
11
(40)(15) (600)(5) 26.25 10 mm
12 12
11
(20)(15) (300)(10) 35.625 10 mm
12 12
61.875 10 mm 61.875 10 m
y
y
I bh Ad
I bh Ad
II I
−
=−= =
=−=−
= += + = ×
= += + = ×
=+= × = ×
|| My I
M
Iy
σ
σ
= =
Top: (tension side)
69
(24 10 )(61.875 10 ) 118.8 N m
0.0125
M−
××
= = ⋅
Bottom: (compression)
69
(30 10 )(61.875 10 ) 106.1 N m
0.0175
M
−
××
= = ⋅
Choose smaller value.
106.1 N mM= ⋅
