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= ⋅