PROBLEM 14.75
Determine the range of values of
x
σ
for which the maximum in–plane shearing
stress is equal to or less than 10 ksi.
SOLUTION
For the Mohr’s circle, point Y lies at
(15 ksi, 8 ksi).
The radius of limiting circles is
10 ksi.
R=
Let C1 be the location of the leftmost limiting circle and C2 be that of the rightmost one.
1
2
10 ksi
10 ksi
CY
CY
=
=
Noting right triangles
1
C DY
and
2
,C DY
22 2 2
22
1 11 1
8 10 6 ksiCD DY CY CD CD+ = += =
Coordinates of point C1 are
(0, 15 6) (0, 9 ksi).−=
Likewise, coordinates of point C2 are
(0, 15 6) (0, 21ksi).
+=
Coordinates of point X1:
(9 6, 8) (3 ksi, 8 ksi)− −= −
Coordinates of point X2:
(21 6, 8) (27 ksi, 8 ksi)
+−= −
The point
(, )
x xy
στ
−
must lie on the line X1 X2.
Thus,
3 ksi 27 ksi
x
σ
≤≤
consent of McGraw–Hill Education.
PROBLEM 14.76
For the state of stress shown, it is known that the normal and shearing stresses are
directed as shown and that
14 ksi, 9 ksi,
xy
ss
= =
and
min
5 ksi.
s
=
Determine
(a) the orientation of the principal planes, (b) the principal stress
s
max, (c) the
maximum in–plane shearing stress.
SOLUTION
ave
min ave ave min
2
2
2
2 22
1
14 ksi, 9 ksi, ( ) 11.5 ksi
2
11.5 5 6.5 ksi
2
6.5 2.5 6 ksi
2
x y xy
xy xy
xy
xy
RR
R
R
s s s ss
s s ss
ss τ
ss
τ
= = = +=
=−∴=−
= −=
−
= +
−
=±− =± − =±
But it is given that
xy
τ
is positive, thus
6 ksi.
xy
τ
= +
(a)
2
tan 2
(2)(6) 2.4
5
2 67.38
xy
pxy
p
τ
θss
θ
=−
= =
= °
33.7
a
θ
= °
123.7
b
θ
= °
(b)
max ave R
ss
= +
max
18.00 ksi
s
=
(c)
max(in-plane) R
τ
=
max(in-plane) 6.50 ksi
τ
=
PROBLEM 14.77
Determine the principal planes and the principal
stresses for the state of plane stress resulting
from the superposition of the two states of stress
consent of McGraw–Hill Education.
PROBLEM 14.77 (Continued)
ave
1(6 2) 2
2
σ
= −=+
1(6 2) 4
22
σσ
−= +=
xy
22
(4) (3) 5= +=R
3
tan 2 4
θ
=
p
2 36.87
θ
= °
p
18.4 ,108.4
θ
=°°
p
ave
max
25
σσ
= +=+R
max
7.00 ksi
σ
=
min ave
25
σσ
= −=−R
min 3.00 ksi
σ
= −
consent of McGraw–Hill Education.
PROBLEM 14.78
A standard–weight steel pipe of 12–in. nominal diameter carries water under a pressure of 400 psi.
(a) Knowing that the outside diameter is 12.75 in. and the wall thickness is 0.375 in., determine the maximum
tensile stress in the pipe. (b) Solve part a, assuming an extra–strong pipe is used of 12.75–in. outside diameter
and 0.5-in. wall thickness.
PROBLEM 14.79
Two wooden members of 80 × 120-mm uniform rectangular cross
section are joined by the simple glued scarf splice shown. Knowing
that
22
β
= °
and that the maximum allowable stresses in the joint
are, respectively, 400 kPa in tension (perpendicular to the splice)
and 600 kPa in shear (parallel to the splice), determine the largest
centric load P that can be applied.
PROBLEM 14.80
Two wooden members of 80 × 120–mm uniform rectangular cross
section are joined by the simple glued scarf splice shown. Knowing
that
25
β
= °
and that centric loads of magnitude
10 kNP=
are
applied to the members as shown, determine (a) the in–plane
shearing stress parallel to the splice, (b) the normal stress
perpendicular to the splice.
PROBLEM 14.81
The axle of an automobile is acted upon by the forces and couple
shown. Knowing that the diameter of the solid axle is 32 mm,
determine (a) the principal planes and principal stresses at point H
located on top of the axle, (b) the maximum shearing stress at the
consent of McGraw–Hill Education.
PROBLEM 14.81 (Continued)
2(2)( 54.399)
tan 2 0.77778 2 37.88
139.882
xy
pp
xy
τ
θθ
σσ
−
= = = = °
−−
18.9 and 108.9°
p
θ
= °
(b)
max
88.6 MPaR
τ
= =
max
88.6 MPa
τ
=
consent of McGraw–Hill Education.
PROBLEM 14.82
Square plates, each of 0.5–in. thickness, can be bent and
welded together in either of the two ways shown to form the
cylindrical portion of a compressed–air tank. Knowing that the
allowable normal stress perpendicular to the weld is 12 ksi,
determine the largest allowable gage pressure in each case.
SOLUTION
12
1
12ft 144 in. 71.5 in.
2
2
d r dt
pr pr
tt
σσ
= = = −=
= =
(a)
1
1
12 ksi
(12)(0.5) 0.0839 ksi
71.5
t
pr
σ
σ
=
= = =
83.9 psip=
(b)
ave 1 2
12
ave
13
()
24
1
24
45
cos
3
4
w
pr
t
pr
Rt
R
pr
t
σ σσ
σσ
β
σσ β
= +=
+
= =
=±°
= +
=
4 4 (12)(0.5) 0.1119 ksi
3 3 71.5
w
t
pr
= = =
σ
111.9 psi=
p
consent of McGraw–Hill Education.
PROBLEM 14.83
A torque of magnitude
12 kN mT= ⋅
is applied to the end of a tank containing
compressed air under a pressure of 8 MPa. Knowing that the tank has a 180–mm inner
diameter and a 12–mm wall thickness, determine the maximum normal stress and the
consent of McGraw–Hill Education.
PROBLEM 14.75
Determine the range of values of
x
σ
for which the maximum in–plane shearing
stress is equal to or less than 10 ksi.
SOLUTION
For the Mohr’s circle, point Y lies at
(15 ksi, 8 ksi).
The radius of limiting circles is
10 ksi.
R=
Let C1 be the location of the leftmost limiting circle and C2 be that of the rightmost one.
1
2
10 ksi
10 ksi
CY
CY
=
=
Noting right triangles
1
C DY
and
2
,C DY
22 2 2
22
1 11 1
8 10 6 ksiCD DY CY CD CD+ = += =
Coordinates of point C1 are
(0, 15 6) (0, 9 ksi).−=
Likewise, coordinates of point C2 are
(0, 15 6) (0, 21ksi).
+=
Coordinates of point X1:
(9 6, 8) (3 ksi, 8 ksi)− −= −
Coordinates of point X2:
(21 6, 8) (27 ksi, 8 ksi)
+−= −
The point
(, )
x xy
στ
−
must lie on the line X1 X2.
Thus,
3 ksi 27 ksi
x
σ
≤≤
consent of McGraw–Hill Education.
PROBLEM 14.76
For the state of stress shown, it is known that the normal and shearing stresses are
directed as shown and that
14 ksi, 9 ksi,
xy
ss
= =
and
min
5 ksi.
s
=
Determine
(a) the orientation of the principal planes, (b) the principal stress
s
max, (c) the
maximum in–plane shearing stress.
SOLUTION
ave
min ave ave min
2
2
2
2 22
1
14 ksi, 9 ksi, ( ) 11.5 ksi
2
11.5 5 6.5 ksi
2
6.5 2.5 6 ksi
2
x y xy
xy xy
xy
xy
RR
R
R
s s s ss
s s ss
ss τ
ss
τ
= = = +=
=−∴=−
= −=
−
= +
−
=±− =± − =±
But it is given that
xy
τ
is positive, thus
6 ksi.
xy
τ
= +
(a)
2
tan 2
(2)(6) 2.4
5
2 67.38
xy
pxy
p
τ
θss
θ
=−
= =
= °
33.7
a
θ
= °
123.7
b
θ
= °
(b)
max ave R
ss
= +
max
18.00 ksi
s
=
(c)
max(in-plane) R
τ
=
max(in-plane) 6.50 ksi
τ
=
PROBLEM 14.77
Determine the principal planes and the principal
stresses for the state of plane stress resulting
from the superposition of the two states of stress
consent of McGraw–Hill Education.
PROBLEM 14.77 (Continued)
ave
1(6 2) 2
2
σ
= −=+
1(6 2) 4
22
σσ
−= +=
xy
22
(4) (3) 5= +=R
3
tan 2 4
θ
=
p
2 36.87
θ
= °
p
18.4 ,108.4
θ
=°°
p
ave
max
25
σσ
= +=+R
max
7.00 ksi
σ
=
min ave
25
σσ
= −=−R
min 3.00 ksi
σ
= −
consent of McGraw–Hill Education.
PROBLEM 14.78
A standard–weight steel pipe of 12–in. nominal diameter carries water under a pressure of 400 psi.
(a) Knowing that the outside diameter is 12.75 in. and the wall thickness is 0.375 in., determine the maximum
tensile stress in the pipe. (b) Solve part a, assuming an extra–strong pipe is used of 12.75–in. outside diameter
and 0.5-in. wall thickness.
PROBLEM 14.79
Two wooden members of 80 × 120-mm uniform rectangular cross
section are joined by the simple glued scarf splice shown. Knowing
that
22
β
= °
and that the maximum allowable stresses in the joint
are, respectively, 400 kPa in tension (perpendicular to the splice)
and 600 kPa in shear (parallel to the splice), determine the largest
centric load P that can be applied.
PROBLEM 14.80
Two wooden members of 80 × 120–mm uniform rectangular cross
section are joined by the simple glued scarf splice shown. Knowing
that
25
β
= °
and that centric loads of magnitude
10 kNP=
are
applied to the members as shown, determine (a) the in–plane
shearing stress parallel to the splice, (b) the normal stress
perpendicular to the splice.
PROBLEM 14.81
The axle of an automobile is acted upon by the forces and couple
shown. Knowing that the diameter of the solid axle is 32 mm,
determine (a) the principal planes and principal stresses at point H
located on top of the axle, (b) the maximum shearing stress at the
consent of McGraw–Hill Education.
PROBLEM 14.81 (Continued)
2(2)( 54.399)
tan 2 0.77778 2 37.88
139.882
xy
pp
xy
τ
θθ
σσ
−
= = = = °
−−
18.9 and 108.9°
p
θ
= °
(b)
max
88.6 MPaR
τ
= =
max
88.6 MPa
τ
=
consent of McGraw–Hill Education.
PROBLEM 14.82
Square plates, each of 0.5–in. thickness, can be bent and
welded together in either of the two ways shown to form the
cylindrical portion of a compressed–air tank. Knowing that the
allowable normal stress perpendicular to the weld is 12 ksi,
determine the largest allowable gage pressure in each case.
SOLUTION
12
1
12ft 144 in. 71.5 in.
2
2
d r dt
pr pr
tt
σσ
= = = −=
= =
(a)
1
1
12 ksi
(12)(0.5) 0.0839 ksi
71.5
t
pr
σ
σ
=
= = =
83.9 psip=
(b)
ave 1 2
12
ave
13
()
24
1
24
45
cos
3
4
w
pr
t
pr
Rt
R
pr
t
σ σσ
σσ
β
σσ β
= +=
+
= =
=±°
= +
=
4 4 (12)(0.5) 0.1119 ksi
3 3 71.5
w
t
pr
= = =
σ
111.9 psi=
p
consent of McGraw–Hill Education.
PROBLEM 14.83
A torque of magnitude
12 kN mT= ⋅
is applied to the end of a tank containing
compressed air under a pressure of 8 MPa. Knowing that the tank has a 180–mm inner
diameter and a 12–mm wall thickness, determine the maximum normal stress and the
consent of McGraw–Hill Education.