consent of McGraw–Hill Education.
PROBLEM 11.67
A vertical force P of magnitude 20 kips is applied
at point C located on the axis of symmetry of the
cross section of a short column. Knowing that
5 in.,y=
determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the
neutral axis.
SOLUTION
Locate centroid.
Part
2
, in
A
, in.y
3
, inAy
76 3.8 in.
20
Ay
yA
Σ
= = =
Σ
12
5
60
8
2
16
Σ
20
76
Eccentricity of load:
5 3.8 1.2 in.e=−=
3 2 4 32 4
12
4
12
11
(6)(2) (12)(1.2) 21.28 in (2)(4) (8)(1.8) 36.587 in
12 12
57.867 in
II
III
= + = = +=
=+=
(a) Stress at A:
3.8 in.
A
c=
20 20(1.2)(3.8)
20 57.867
A
A
P Pec
AI
σ
=−+ =− +
0.576 ksi
A
σ
=
(b) Stress at B:
6 3.8 2.2 in.
B
c=−=
20 20(1.2)(2.2)
20 57.867
B
B
P Pec
AI
σ
=−+ =− −
1.912 ksi
B
σ
= −
(c) Location of neutral axis:
1
00
57.867 2.411 in.
(20)(1.2)
P Pea ea
AI IA
I
aAe
σσ
= =−+ = ∴ =
= = =
Neutral axis lies 2.411 in. below centroid or
3.8 2.411 1.389−=
in. above point A.
Answer:
1.389 in. from point A
consent of McGraw–Hill Education.
PROBLEM 11.68
A vertical force P is applied at point C located
on the axis of symmetry of the cross section of a
short column. Determine the range of values of
y for which tensile stresses do not occur in the
column.
SOLUTION
Locate centroid.
2
, in
A
, in.y
3
, inAy
76 3.8 in.
20
ii
i
Ay
yA
Σ
= = =
Σ
12
5
60
8
2
16
Σ
20
76
Eccentricity of load:
3 2 4 32 4
12
4
12
3.8 in. 3.8 in.
11
(6)(2) (12)(1.2) 21.28 in (2)(4) (8)(1.8) 36.587 in
12 12
57.867 in
ey ye
II
III
=−=+
= + = = +=
=+=
If stress at A equals zero,
3.8 in.
A
c=
1
0
AA
A
P Pec ec
AI IA
σ
=−+ = ∴ =
57.867 0.761 in. 0.761 3.8 4.561 in.
(20)(3.8)
A
I
ey
Ac
= = = = +=
If stress at B equals zero,
6 3.8 2.2 in.
B
c=−=
1
0
57.867 1.315 in.
(20)(2.2)
1.315 3.8 2.485 in.
BB
B
B
P Pec ec
AI I A
I
eAc
y
σ
=−− = ∴ =−
=−=− =−
=− +=
Answer:
2.49 in. 4.56 in.<<y
PROBLEM 11.69
The C–shaped steel bar is used as a dynamometer to determine the magnitude P
of the forces shown. Knowing that the cross section of the bar is a square of side
40 mm and that strain on the inner edge was measured and found to be 450
µ
,
determine the magnitude P of the forces. Use
200 GPa.E=
SOLUTION
PROBLEM 11.70
A short length of a rolled–steel column supports a rigid plate
on which two loads P and Q are applied as shown. The strains
at two points A and B on the centerline of the outer faces of the
flanges have been measured and found to be
66
400 10 in./in. 300 10 in./in.
AB
−−
=−× =−×
εε
Knowing that E = 29 × 106 psi, determine the magnitude of
each load.
SOLUTION
PROBLEM 11.71
An eccentric force P is applied as shown to a steel bar of
25
90-mm×
cross section. The strains at A and B have been
measured and found to be
350 70
AB
ε µε µ
=+=−
Knowing that
200E=
GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
3 2 32
3 3 6 4 64
1
15 45 30 90 mm 25 mm 45 mm 0.045 m
2
(25)(90) 2.25 10 mm 2.25 10 m
11
(25)(90) 1.51875 10 mm 1.51875 10 m
12 12
60 45 15 mm 0.015 m 15 45 30 mm 0.030 m
AB
h b ch
A bh
I bh
yy
−
−
=++= = = = =
=−=×=×
== =×=×
=−= = =−=− =−
Stresses from strain gages at A and B:
9 66
96 6
(200 10 )(350 10 ) 70 10 Pa
(200 10 )( 70 10 ) 14 10 Pa
AA
BB
E
E
σε
σε
−
−
==× ×=×
= = × −× =−×
A
A
My
P
AI
σ
= −
(1)
B
BMy
P
AI
σ
= −
(2)
Subtracting,
()
AB
AB
My y
I
σσ
−
−=−
66
()
(1.51875 10 )(84 10 ) 2835 N m
0.045
AB
AB
I
Myy
σσ
−
−××
=− =− =−⋅
−
Multiplying (2) by
A
y
and (1) by
B
y
and subtracting,
()
AB BA A BP
y y yy
A
σσ
−=−
36 6
3
()
(2.25 10 )[(0.015)( 14 10 ) ( 0.030)(70 10 )] 94.5 10 N
0.045
AB BA
AB
Ay y
Pyy
σσ
−
−× − × −− ×
= = = ×
−
(a)
3
2835 0.030 m
94.5 10
M
M Pd d P
−
=− ∴=−=− =
×
30.0 mmd=
(b)
94.5 kNP=
PROBLEM 11.72
Solve Prob. 11.71, assuming that the measured strains are
600 420
AB
εµεµ
=+=+
PROBLEM 11.71 An eccentric force P is applied as shown to a
steel bar of
25 90-mm×
cross section. The strains at A and B have
been measured and found to be
350 70
AB
ε µε µ
=+=−
Knowing that
200E=
GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
3 2 32
3 3 6 4 64
1
15 45 30 90 mm 25 mm 45 mm 0.045 m
2
(25)(90) 2.25 10 mm 2.25 10 m
11
(25)(90) 1.51875 10 mm 1.51875 10 m
12 12
60 45 15 mm 0.015 m 15 45 30 mm 0.030 m
AB
h b ch
A bh
I bh
yy
−
−
=++= = = = =
===×=×
== =×=×
=−= = =−=− =−
Stresses from strain gages at A and B:
96 6
9 66
(200 10 )(600 10 ) 120 10 Pa
(200 10 )(420 10 ) 84 10 Pa
AA
BB
E
E
σε
σε
−
−
==× ×=×
==× ×=×
A
AMy
P
AI
σ
= −
(1)
B
B
My
P
AI
σ
= −
(2)
Subtracting,
()
AB
AB
My y
I
σσ
−
−=−
66
()
(1.51875 10 )(36 10 ) 1215 N m
0.045
AB
AB
I
Myy
σσ
−
−××
=− =− =−⋅
−
Multiplying (2) by
A
y
and (1) by
B
y
and subtracting,
()
AB BA A BP
y y yy
A
σσ
−=−
36 6
3
()
(2.25 10 )[(0.015)(84 10 ) ( 0.030)(120 10 )] 243 10 N
0.045
AB BA
AB
Ay y
Pyy
−
−× × −− ×
= = = ×
−
σσ
(a)
3
3
1215 5 10 m
243 10
M
M Pd d P
−
−
=− ∴=−=− =×
×
5.00 mmd=
(b)
243 kNP=
PROBLEM 11.73
The couple M is applied to a beam of the cross section shown in a plane
forming an angle
β
with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
3 34
3 34
1(0.4)(1.2) 57.6 10 in
12
1(1.2)(0.4) 6.40 10 in
12 0.6 in.
1
z (0.4) 0.2 in.
2
−
−
= = ×
= = ×
==−=
=−=−= =
z
y
AB D
ABD
I
I
yy y
zz
400cos 60 200 lb in., 400 sin 60 346.41lb in.
yz
MM= °= ⋅ =− °=− ⋅
(a)
33
( 346.41)(0.6) (200)(0.2)
57.6 10 6.40 10
yA
zA
Azy
Mz
My
II −−
−
=−+=− +
××
σ
3
9.86 10 psi 9.86 ksi=×=
(b)
63
( 346.41)(0.6) (200)( 0.2)
57.6 10 6.4 10
yB
zB
Bzy
Mz
My
II −−
−−
=−+=− +
××
σ
3
2.64 10 psi 2.64 ksi
=−× =−
(c)
33
( 346.41)( 0.6) (200)( 0.2)
57.6 10 6.40 10
yD
zD
Dzy
Mz
My
II −−
−− −
=−+=− +
××
σ
3
9.86 10 psi 9.86 ksi=−× =−
PROBLEM 11.74
The couple M is applied to a beam of the cross section shown in a plane
forming an angle
β
with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
3 3 4 94
3 6 4 64
1(80)(32) 218.45 10 mm 218.45 10 m
12
1(32)(80) 1.36533 10 mm 1.36533 10 m
12 16 mm
40 mm
z
y
AB D
ABD
I
I
yy y
zzz
−
−
= =×=×
= =×=×
==−=
=−=−=
300cos30 259.81 N m 300sin30 150 N m
yz
MM= °= ⋅ = °= ⋅
(a)
33
96
(150)(16 10 ) (259.81)(40 10 )
218.45 10 1.36533 10
yA
zA
Azy
Mz
My
II
−−
−−
××
=−+=− +
××
σ
6
3.37 10 Pa=−×
3.37 MPa
A
σ
= −
(b)
33
96
(150)(16 10 ) (259.81)( 40 10 )
218.45 10 1.36533 10
yB
zB
Bzy
Mz
My
II
−−
−−
× −×
=−+ =− +
××
σ
6
18.60 10 Pa=−×
18.60 MPa
B
σ
= −
(c)
33
96
(150)( 16 10 ) (259.81)( 40 10 )
218.45 10 1.36533 10
yD
zD
Dzy
Mz
My
II
−−
−−
−× −×
=−+=− +
××
σ
6
3.37 10 Pa= ×
3.37 MPa
D
σ
=
PROBLEM 11.75
The couple M is applied to a beam of the cross section shown in a
plane forming an angle
β
with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
Flange:
32
4
34
1(8)(0.5) (8)(0.5)(4.75)
12
90.333 in
1(0.5)(8) 21.333 in
12
z
y
I
I
= +
=
= =
Web:
34
34
1(0.3)(9) 18.225 in
12
1(9)(0.3) 0.02025 in
12
z
y
I
I
= =
= =
Total:
4
4
(2)(90.333) 18.225 198.89 in
(2)(21.333) 0.02025 42.687 in
z
y
I
I
= +=
= +=
5 in.
4 in.
250 cos30 216.51 kip in
250 sin30 125 kip in
AB D
ABC
z
y
yy y
zzz
M
M
==−=
=−=−=
= °= ⋅
=− °=− ⋅
(a)
(216.51)(5) ( 125)(4)
198.89 42.687
yA
zA
Azy
Mz
My
II
σ
−
=−+=− +
17.16 ksi
A
σ
= −
(b)
(216.51)(5) ( 125)( 4)
198.89 42.687
yB
zB
Bzy
Mz
My
II
σ
−−
=−+=− +
6.27 ksi
B
σ
=
(c)
(216.51)( 5) ( 125)( 4)
198.89 42.687
yD
zD
Dzy
Mz
My
II
σ
−−−
=−+=− +
17.16 ksi
D
σ
=
consent of McGraw–Hill Education.
PROBLEM 11.76
The couple M is applied to a beam of the cross section shown in a
plane forming an angle
β
with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
3 34
334
11
(4.8)(2.4) (4)(1.6) 4.1643 in
12 12
11
(2.4)(4.8) (1.6)(4) 13.5851 in
12 12
1.2 in.
2.4 in.
75sin15 19.4114 kip in.
75cos15 72.444 kip in.
z
y
AB D
ABD
z
y
I
I
yy y
zzz
M
M
= −=
= −=
==−=
=−=−=
= °= ⋅
= °= ⋅
(a)
(19.4114)(1.2) (72.444)(2.4)
4.1643 13.5851
yA
zA
Azy
Mz
My
II
σ
=−+=− +
7.20 ksi
A
σ
=
(b)
(19.4114)(1.2) (72.444)( 2.4)
4.1643 13.5851
yB
zB
Bzy
Mz
My
II
σ
−
=−+=− +
18.39 ksi
B
σ
= −
(c)
(19.4114)( 1.2) (72.444)( 2.4)
4.1643 13.5851
yD
zD
Dzy
Mz
My
II
σ
−−
=−+=− +
7.20 ksi
D
σ
= −
consent of McGraw–Hill Education.
PROBLEM 11.68
A vertical force P is applied at point C located
on the axis of symmetry of the cross section of a
short column. Determine the range of values of
y for which tensile stresses do not occur in the
column.
SOLUTION
Locate centroid.
2
, in
A
, in.y
3
, inAy
76 3.8 in.
20
ii
i
Ay
yA
Σ
= = =
Σ
12
5
60
8
2
16
Σ
20
76
Eccentricity of load:
3 2 4 32 4
12
4
12
3.8 in. 3.8 in.
11
(6)(2) (12)(1.2) 21.28 in (2)(4) (8)(1.8) 36.587 in
12 12
57.867 in
ey ye
II
III
=−=+
= + = = +=
=+=
If stress at A equals zero,
3.8 in.
A
c=
1
0
AA
A
P Pec ec
AI IA
σ
=−+ = ∴ =
57.867 0.761 in. 0.761 3.8 4.561 in.
(20)(3.8)
A
I
ey
Ac
= = = = +=
If stress at B equals zero,
6 3.8 2.2 in.
B
c=−=
1
0
57.867 1.315 in.
(20)(2.2)
1.315 3.8 2.485 in.
BB
B
B
P Pec ec
AI I A
I
eAc
y
σ
=−− = ∴ =−
=−=− =−
=− +=
Answer:
2.49 in. 4.56 in.<<y
PROBLEM 11.69
The C–shaped steel bar is used as a dynamometer to determine the magnitude P
of the forces shown. Knowing that the cross section of the bar is a square of side
40 mm and that strain on the inner edge was measured and found to be 450
µ
,
determine the magnitude P of the forces. Use
200 GPa.E=
SOLUTION
PROBLEM 11.70
A short length of a rolled–steel column supports a rigid plate
on which two loads P and Q are applied as shown. The strains
at two points A and B on the centerline of the outer faces of the
flanges have been measured and found to be
66
400 10 in./in. 300 10 in./in.
AB
−−
=−× =−×
εε
Knowing that E = 29 × 106 psi, determine the magnitude of
each load.
SOLUTION
PROBLEM 11.71
An eccentric force P is applied as shown to a steel bar of
25
90-mm×
cross section. The strains at A and B have been
measured and found to be
350 70
AB
ε µε µ
=+=−
Knowing that
200E=
GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
3 2 32
3 3 6 4 64
1
15 45 30 90 mm 25 mm 45 mm 0.045 m
2
(25)(90) 2.25 10 mm 2.25 10 m
11
(25)(90) 1.51875 10 mm 1.51875 10 m
12 12
60 45 15 mm 0.015 m 15 45 30 mm 0.030 m
AB
h b ch
A bh
I bh
yy
−
−
=++= = = = =
=−=×=×
== =×=×
=−= = =−=− =−
Stresses from strain gages at A and B:
9 66
96 6
(200 10 )(350 10 ) 70 10 Pa
(200 10 )( 70 10 ) 14 10 Pa
AA
BB
E
E
σε
σε
−
−
==× ×=×
= = × −× =−×
A
A
My
P
AI
σ
= −
(1)
B
BMy
P
AI
σ
= −
(2)
Subtracting,
()
AB
AB
My y
I
σσ
−
−=−
66
()
(1.51875 10 )(84 10 ) 2835 N m
0.045
AB
AB
I
Myy
σσ
−
−××
=− =− =−⋅
−
Multiplying (2) by
A
y
and (1) by
B
y
and subtracting,
()
AB BA A BP
y y yy
A
σσ
−=−
36 6
3
()
(2.25 10 )[(0.015)( 14 10 ) ( 0.030)(70 10 )] 94.5 10 N
0.045
AB BA
AB
Ay y
Pyy
σσ
−
−× − × −− ×
= = = ×
−
(a)
3
2835 0.030 m
94.5 10
M
M Pd d P
−
=− ∴=−=− =
×
30.0 mmd=
(b)
94.5 kNP=
PROBLEM 11.72
Solve Prob. 11.71, assuming that the measured strains are
600 420
AB
εµεµ
=+=+
PROBLEM 11.71 An eccentric force P is applied as shown to a
steel bar of
25 90-mm×
cross section. The strains at A and B have
been measured and found to be
350 70
AB
ε µε µ
=+=−
Knowing that
200E=
GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
3 2 32
3 3 6 4 64
1
15 45 30 90 mm 25 mm 45 mm 0.045 m
2
(25)(90) 2.25 10 mm 2.25 10 m
11
(25)(90) 1.51875 10 mm 1.51875 10 m
12 12
60 45 15 mm 0.015 m 15 45 30 mm 0.030 m
AB
h b ch
A bh
I bh
yy
−
−
=++= = = = =
===×=×
== =×=×
=−= = =−=− =−
Stresses from strain gages at A and B:
96 6
9 66
(200 10 )(600 10 ) 120 10 Pa
(200 10 )(420 10 ) 84 10 Pa
AA
BB
E
E
σε
σε
−
−
==× ×=×
==× ×=×
A
AMy
P
AI
σ
= −
(1)
B
B
My
P
AI
σ
= −
(2)
Subtracting,
()
AB
AB
My y
I
σσ
−
−=−
66
()
(1.51875 10 )(36 10 ) 1215 N m
0.045
AB
AB
I
Myy
σσ
−
−××
=− =− =−⋅
−
Multiplying (2) by
A
y
and (1) by
B
y
and subtracting,
()
AB BA A BP
y y yy
A
σσ
−=−
36 6
3
()
(2.25 10 )[(0.015)(84 10 ) ( 0.030)(120 10 )] 243 10 N
0.045
AB BA
AB
Ay y
Pyy
−
−× × −− ×
= = = ×
−
σσ
(a)
3
3
1215 5 10 m
243 10
M
M Pd d P
−
−
=− ∴=−=− =×
×
5.00 mmd=
(b)
243 kNP=
PROBLEM 11.73
The couple M is applied to a beam of the cross section shown in a plane
forming an angle
β
with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
3 34
3 34
1(0.4)(1.2) 57.6 10 in
12
1(1.2)(0.4) 6.40 10 in
12 0.6 in.
1
z (0.4) 0.2 in.
2
−
−
= = ×
= = ×
==−=
=−=−= =
z
y
AB D
ABD
I
I
yy y
zz
400cos 60 200 lb in., 400 sin 60 346.41lb in.
yz
MM= °= ⋅ =− °=− ⋅
(a)
33
( 346.41)(0.6) (200)(0.2)
57.6 10 6.40 10
yA
zA
Azy
Mz
My
II −−
−
=−+=− +
××
σ
3
9.86 10 psi 9.86 ksi=×=
(b)
63
( 346.41)(0.6) (200)( 0.2)
57.6 10 6.4 10
yB
zB
Bzy
Mz
My
II −−
−−
=−+=− +
××
σ
3
2.64 10 psi 2.64 ksi
=−× =−
(c)
33
( 346.41)( 0.6) (200)( 0.2)
57.6 10 6.40 10
yD
zD
Dzy
Mz
My
II −−
−− −
=−+=− +
××
σ
3
9.86 10 psi 9.86 ksi=−× =−
PROBLEM 11.74
The couple M is applied to a beam of the cross section shown in a plane
forming an angle
β
with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
3 3 4 94
3 6 4 64
1(80)(32) 218.45 10 mm 218.45 10 m
12
1(32)(80) 1.36533 10 mm 1.36533 10 m
12 16 mm
40 mm
z
y
AB D
ABD
I
I
yy y
zzz
−
−
= =×=×
= =×=×
==−=
=−=−=
300cos30 259.81 N m 300sin30 150 N m
yz
MM= °= ⋅ = °= ⋅
(a)
33
96
(150)(16 10 ) (259.81)(40 10 )
218.45 10 1.36533 10
yA
zA
Azy
Mz
My
II
−−
−−
××
=−+=− +
××
σ
6
3.37 10 Pa=−×
3.37 MPa
A
σ
= −
(b)
33
96
(150)(16 10 ) (259.81)( 40 10 )
218.45 10 1.36533 10
yB
zB
Bzy
Mz
My
II
−−
−−
× −×
=−+ =− +
××
σ
6
18.60 10 Pa=−×
18.60 MPa
B
σ
= −
(c)
33
96
(150)( 16 10 ) (259.81)( 40 10 )
218.45 10 1.36533 10
yD
zD
Dzy
Mz
My
II
−−
−−
−× −×
=−+=− +
××
σ
6
3.37 10 Pa= ×
3.37 MPa
D
σ
=
PROBLEM 11.75
The couple M is applied to a beam of the cross section shown in a
plane forming an angle
β
with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
Flange:
32
4
34
1(8)(0.5) (8)(0.5)(4.75)
12
90.333 in
1(0.5)(8) 21.333 in
12
z
y
I
I
= +
=
= =
Web:
34
34
1(0.3)(9) 18.225 in
12
1(9)(0.3) 0.02025 in
12
z
y
I
I
= =
= =
Total:
4
4
(2)(90.333) 18.225 198.89 in
(2)(21.333) 0.02025 42.687 in
z
y
I
I
= +=
= +=
5 in.
4 in.
250 cos30 216.51 kip in
250 sin30 125 kip in
AB D
ABC
z
y
yy y
zzz
M
M
==−=
=−=−=
= °= ⋅
=− °=− ⋅
(a)
(216.51)(5) ( 125)(4)
198.89 42.687
yA
zA
Azy
Mz
My
II
σ
−
=−+=− +
17.16 ksi
A
σ
= −
(b)
(216.51)(5) ( 125)( 4)
198.89 42.687
yB
zB
Bzy
Mz
My
II
σ
−−
=−+=− +
6.27 ksi
B
σ
=
(c)
(216.51)( 5) ( 125)( 4)
198.89 42.687
yD
zD
Dzy
Mz
My
II
σ
−−−
=−+=− +
17.16 ksi
D
σ
=
consent of McGraw–Hill Education.
PROBLEM 11.76
The couple M is applied to a beam of the cross section shown in a
plane forming an angle
β
with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
3 34
334
11
(4.8)(2.4) (4)(1.6) 4.1643 in
12 12
11
(2.4)(4.8) (1.6)(4) 13.5851 in
12 12
1.2 in.
2.4 in.
75sin15 19.4114 kip in.
75cos15 72.444 kip in.
z
y
AB D
ABD
z
y
I
I
yy y
zzz
M
M
= −=
= −=
==−=
=−=−=
= °= ⋅
= °= ⋅
(a)
(19.4114)(1.2) (72.444)(2.4)
4.1643 13.5851
yA
zA
Azy
Mz
My
II
σ
=−+=− +
7.20 ksi
A
σ
=
(b)
(19.4114)(1.2) (72.444)( 2.4)
4.1643 13.5851
yB
zB
Bzy
Mz
My
II
σ
−
=−+=− +
18.39 ksi
B
σ
= −
(c)
(19.4114)( 1.2) (72.444)( 2.4)
4.1643 13.5851
yD
zD
Dzy
Mz
My
II
σ
−−
=−+=− +
7.20 ksi
D
σ
= −