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PROBLEM 14.12
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
2 ksi
x
s
=
10 ksi
y
s
=
3 ksi
xy
τ
= −
(a)
2 10
tan 2 1.33333
2 (2)( 3)
xy
sxy
ss
θτ
−−
=− =−=−
−
2 53.13
s
θ
=−°
26.6 , 63.4
s
θ
=−° °
(b)
2
2
max
2
xy xy
ss
ττ
−
= +
22
2 10 ( 3)
2
−
= +−
max
5.00 ksi
τ
=
(c)
ave 2 10
22
xy
ss
ss
++
′= = =
6.00 ksi
s
′=
consent of McGraw-Hill Education.
PROBLEM 14.13
For the given state of stress, determine the normal and shearing stresses after the
element shown has been rotated through (a) 25° clockwise, (b) 10° counterclockwise.
SOLUTION
0 8 ksi 5 ksi
4 ksi 4 ksi
22
cos 2 + sin 2
22
sin 2 + cos 2
2
cos 2 sin 2
22
x y xy
xy xy
xy xy
x xy
xy
x y xy
xy xy
y xy
ss τ
ss ss
ss ss
s θτ θ
ss
τ θτ θ
ss ss
s θτ θ
′
′′
′
= = =
+−
= = −
+−
= +
−
= −
+−
=−−
(a)
25 2 50
θθ
=−° =−°
4 4 cos ( 50°) + 5 sin ( 50°)
x
s
′
=−− −
2.40 ksi
x
s
′
= −
4 sin ( 50 ) 5 cos ( 50 )
xy
τ
′′= −°+ −°
0.1498 ksi
xy
τ
′′=
4 4 cos ( 50 ) 5 sin ( 50)
y
s
′= + − °− −
10.40 ksi
y
s
′=
(b)
10 2 20
θθ
=°=°
4 4 cos (20°) + 5 sin (20°)
x
s
′= −
1.951 ksi
x
s
′
=
4 sin (20°) + 5 cos (20°)
xy
τ
′′=
6.07 ksi
xy
τ
′′
=
4 4 cos (20°) 5 cos (20°)
y
s
′=+−
6.05 ksi
y
s
′
=
PROBLEM 14.14
For the given state of stress, determine the normal and shearing stresses after
the element shown has been rotated through (a) 25° clockwise, (b) 10°
counterclockwise.
SOLUTION
8 ksi 12 ksi 6 ksi
2 ksi 10 ksi
22
cos 2 + sin 2
22
sin 2 + cos 2
2
cos 2 sin 2
22
x y xy
xy xy
xy xy
x xy
xy
x y xy
xy xy
y xy
ss τ
ss ss
ss ss
s θτ θ
ss
τ θτ θ
ss ss
s θτ θ
′
′′
′
==−=−
+−
=−=
+−
= +
−
= −
+−
=−−
(a)
25 2 50
θθ
=−° =−°
2 10 cos ( 50 ) 6 sin ( 50 )
x
s
′=−+ −°− −°
9.02 ksi
x
s
′
=
10 sin ( 50 ) 6 cos ( 50 )
xy
τ
′′=− − °− − °
3.80 ksi
xy
τ
′′
=
2 10 cos ( 50 ) 6 sin ( 50 )
y
s
′
=−− −°+ −°
13.02 ksi
y
s
′= −
(b)
10 2 20
θθ
=°=°
2 10 cos (20°) 6 sin (20°)
x
s
′
=−+ −
5.34 ksi
x
s
′=
10 sin (20°) 6 cos (20°)
xy
τ
′′=−−
9.06 ksi
xy
τ
′′= −
2 10 cos (20°) + 6 sin (20°)
y
s
′
=−−
9.34 ksi
y
s
′
= −
PROBLEM 14.15
For the given state of stress, determine the normal and shearing stresses after the
element shown has been rotated through (a) 25° clockwise, (b) 10° counterclockwise.
SOLUTION
0 80 MPa 50 MPa
40 MPa 40 MPa
22
cos 2 sin 2
22
sin 2 + cos 2
2
cos 2 sin 2
22
x y xy
xy xy
xy xy
x xy
xy
x y xy
xy xy
y xy
ss τ
ss ss
ss ss
s θτ θ
ss
τ θτ θ
ss ss
s θτ θ
′
′′
′
==−=−
+−
=−=
+−
=++
−
= −
+−
=−−
(a)
25
θ
=−°
2 50
θ
=−°
40 40 cos ( 50 ) 50 sin ( 50°)
x
s
′=− + − °− −
24.0 MPa
x
s
′=
40 sin ( 50°) 50 cos ( 50 )
xy
τ
′′
=− − − −°
1.498 MPa
xy
τ
′′
= −
40 40 cos ( 50 ) 50 sin ( 50 )
y
s
′=− − − °+ − °
104.0 MPa
y
s
′= −
(b)
10 2 20
θθ
=°=°
40 40 cos (20°) 50 sin (20°)
x
s
′
=−+ −
19.51 MPa
x
s
′
= −
40 sin (20°) 50 cos (20°)
xy
τ
′′=−−
60.7 MPa
xy
τ
′′= −
40 40 cos (20°) + 50 sin (20°)
y
s
′
=−−
60.5 MPa
y
s
′= −
PROBLEM 14.16
For the given state of stress, determine the normal and shearing stresses after
the element shown has been rotated through (a) 25° clockwise, (b) 10°
counterclockwise.
SOLUTION
60 MPa 90 MPa 30 MPa
15 MPa 75 MPa
22
cos 2 + sin 2
22
sin 2 + cos 2
2
cos 2 sin 2
22
x y xy
xy xy
xy xy
x xy
xy
xy xy
xy xy
y xy
s sτ
ss ss
ss ss
s θτ θ
ss
τ θτ θ
ss ss
s θτ θ
′
′
=−= =
+−
= = −
+−
= +
−
= −
+−
=−−
(a)
25 2 50
θθ
=−° =−°
15 75 cos ( 50 ) 30 sin ( 50 )
x
s
′= − −°+ −°
56.2 MPa
x
s
′= −
75 sin ( 50 ) 30 cos ( 50 )
xy
τ
′′
=+ −°+ −°
38.2 MPa
xy
τ
′′= −
15 75 cos ( 50 ) 30 sin ( 50 )
y
s
′= + −°− −°
86.2 MPa
y
s
′=
(b)
10 2 20
θθ
=°=°
15 75 cos (20°) + 30 sin (20°)
x
s
′
= −
45.2 MPa
x
s
′
= −
75 sin (20°) + 30 cos (20°)
xy
τ
′′= +
53.8 MPa
xy
τ
′′=
15 75 cos (20°) 30 sin (20°)
y
s
′
=+−
75.2 MPa
y
s
′=
PROBLEM 14.17
The grain of a wooden member forms an angle of 15° with the vertical. For the
state of stress shown, determine (a) the in-plane shearing stress parallel to the
grain, (b) the normal stress perpendicular to the grain.
SOLUTION
4 MPa 1.6 MPa 0
15 2 30
x y xy
σσ τ
θθ
=−=− =
=−° =−°
(a)
sin 2 + cos 2
2
xy
x y xy
σσ
τ θτ θ
′′
−
= −
4 ( 1.6) sin ( 30 ) 0
2
− −−
=− − °+
0.600 MPa
xy
τ
′′
= −
(b)
cos 2 + sin 2
22
xy xy
x xy
σσ σσ
σ θτ θ
′
+−
= +
4 ( 1.6) 4 ( 1.6) cos ( 30 ) 0
22
−+− −−−
= + − °+
3.84 MPa
x
σ
′= −
consent of McGraw-Hill Education.
PROBLEM 14.18
The grain of a wooden member forms an angle of 15° with the vertical. For the state
of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the
consent of McGraw-Hill Education.
PROBLEM 14.19
Two steel plates of uniform cross section
10 80mm×
are welded
together as shown. Knowing that centric 100-kN forces are applied to
the welded plates and that
25
β
= °
, determine (a) the in-plane shearing
stress parallel to the weld, (b) the normal stress perpendicular to the
weld.
SOLUTION
PROBLEM 14.20
The centric force P is applied to a short post as shown. Knowing that the stresses on
plane a-a are
15 ksi
s
= −
and
5 ksi,=
τ
determine (a) the angle
β
that plane a-a forms
with the horizontal, (b) the maximum compressive stress in the post.
PROBLEM 14.12
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
2 ksi
x
s
=
10 ksi
y
s
=
3 ksi
xy
τ
= −
(a)
2 10
tan 2 1.33333
2 (2)( 3)
xy
sxy
ss
θτ
−−
=− =−=−
−
2 53.13
s
θ
=−°
26.6 , 63.4
s
θ
=−° °
(b)
2
2
max
2
xy xy
ss
ττ
−
= +
22
2 10 ( 3)
2
−
= +−
max
5.00 ksi
τ
=
(c)
ave 2 10
22
xy
ss
ss
++
′= = =
6.00 ksi
s
′=
consent of McGraw-Hill Education.
PROBLEM 14.13
For the given state of stress, determine the normal and shearing stresses after the
element shown has been rotated through (a) 25° clockwise, (b) 10° counterclockwise.
SOLUTION
0 8 ksi 5 ksi
4 ksi 4 ksi
22
cos 2 + sin 2
22
sin 2 + cos 2
2
cos 2 sin 2
22
x y xy
xy xy
xy xy
x xy
xy
x y xy
xy xy
y xy
ss τ
ss ss
ss ss
s θτ θ
ss
τ θτ θ
ss ss
s θτ θ
′
′′
′
= = =
+−
= = −
+−
= +
−
= −
+−
=−−
(a)
25 2 50
θθ
=−° =−°
4 4 cos ( 50°) + 5 sin ( 50°)
x
s
′
=−− −
2.40 ksi
x
s
′
= −
4 sin ( 50 ) 5 cos ( 50 )
xy
τ
′′= −°+ −°
0.1498 ksi
xy
τ
′′=
4 4 cos ( 50 ) 5 sin ( 50)
y
s
′= + − °− −
10.40 ksi
y
s
′=
(b)
10 2 20
θθ
=°=°
4 4 cos (20°) + 5 sin (20°)
x
s
′= −
1.951 ksi
x
s
′
=
4 sin (20°) + 5 cos (20°)
xy
τ
′′=
6.07 ksi
xy
τ
′′
=
4 4 cos (20°) 5 cos (20°)
y
s
′=+−
6.05 ksi
y
s
′
=
PROBLEM 14.14
For the given state of stress, determine the normal and shearing stresses after
the element shown has been rotated through (a) 25° clockwise, (b) 10°
counterclockwise.
SOLUTION
8 ksi 12 ksi 6 ksi
2 ksi 10 ksi
22
cos 2 + sin 2
22
sin 2 + cos 2
2
cos 2 sin 2
22
x y xy
xy xy
xy xy
x xy
xy
x y xy
xy xy
y xy
ss τ
ss ss
ss ss
s θτ θ
ss
τ θτ θ
ss ss
s θτ θ
′
′′
′
==−=−
+−
=−=
+−
= +
−
= −
+−
=−−
(a)
25 2 50
θθ
=−° =−°
2 10 cos ( 50 ) 6 sin ( 50 )
x
s
′=−+ −°− −°
9.02 ksi
x
s
′
=
10 sin ( 50 ) 6 cos ( 50 )
xy
τ
′′=− − °− − °
3.80 ksi
xy
τ
′′
=
2 10 cos ( 50 ) 6 sin ( 50 )
y
s
′
=−− −°+ −°
13.02 ksi
y
s
′= −
(b)
10 2 20
θθ
=°=°
2 10 cos (20°) 6 sin (20°)
x
s
′
=−+ −
5.34 ksi
x
s
′=
10 sin (20°) 6 cos (20°)
xy
τ
′′=−−
9.06 ksi
xy
τ
′′= −
2 10 cos (20°) + 6 sin (20°)
y
s
′
=−−
9.34 ksi
y
s
′
= −
PROBLEM 14.15
For the given state of stress, determine the normal and shearing stresses after the
element shown has been rotated through (a) 25° clockwise, (b) 10° counterclockwise.
SOLUTION
0 80 MPa 50 MPa
40 MPa 40 MPa
22
cos 2 sin 2
22
sin 2 + cos 2
2
cos 2 sin 2
22
x y xy
xy xy
xy xy
x xy
xy
x y xy
xy xy
y xy
ss τ
ss ss
ss ss
s θτ θ
ss
τ θτ θ
ss ss
s θτ θ
′
′′
′
==−=−
+−
=−=
+−
=++
−
= −
+−
=−−
(a)
25
θ
=−°
2 50
θ
=−°
40 40 cos ( 50 ) 50 sin ( 50°)
x
s
′=− + − °− −
24.0 MPa
x
s
′=
40 sin ( 50°) 50 cos ( 50 )
xy
τ
′′
=− − − −°
1.498 MPa
xy
τ
′′
= −
40 40 cos ( 50 ) 50 sin ( 50 )
y
s
′=− − − °+ − °
104.0 MPa
y
s
′= −
(b)
10 2 20
θθ
=°=°
40 40 cos (20°) 50 sin (20°)
x
s
′
=−+ −
19.51 MPa
x
s
′
= −
40 sin (20°) 50 cos (20°)
xy
τ
′′=−−
60.7 MPa
xy
τ
′′= −
40 40 cos (20°) + 50 sin (20°)
y
s
′
=−−
60.5 MPa
y
s
′= −
PROBLEM 14.16
For the given state of stress, determine the normal and shearing stresses after
the element shown has been rotated through (a) 25° clockwise, (b) 10°
counterclockwise.
SOLUTION
60 MPa 90 MPa 30 MPa
15 MPa 75 MPa
22
cos 2 + sin 2
22
sin 2 + cos 2
2
cos 2 sin 2
22
x y xy
xy xy
xy xy
x xy
xy
xy xy
xy xy
y xy
s sτ
ss ss
ss ss
s θτ θ
ss
τ θτ θ
ss ss
s θτ θ
′
′
=−= =
+−
= = −
+−
= +
−
= −
+−
=−−
(a)
25 2 50
θθ
=−° =−°
15 75 cos ( 50 ) 30 sin ( 50 )
x
s
′= − −°+ −°
56.2 MPa
x
s
′= −
75 sin ( 50 ) 30 cos ( 50 )
xy
τ
′′
=+ −°+ −°
38.2 MPa
xy
τ
′′= −
15 75 cos ( 50 ) 30 sin ( 50 )
y
s
′= + −°− −°
86.2 MPa
y
s
′=
(b)
10 2 20
θθ
=°=°
15 75 cos (20°) + 30 sin (20°)
x
s
′
= −
45.2 MPa
x
s
′
= −
75 sin (20°) + 30 cos (20°)
xy
τ
′′= +
53.8 MPa
xy
τ
′′=
15 75 cos (20°) 30 sin (20°)
y
s
′
=+−
75.2 MPa
y
s
′=
PROBLEM 14.17
The grain of a wooden member forms an angle of 15° with the vertical. For the
state of stress shown, determine (a) the in-plane shearing stress parallel to the
grain, (b) the normal stress perpendicular to the grain.
SOLUTION
4 MPa 1.6 MPa 0
15 2 30
x y xy
σσ τ
θθ
=−=− =
=−° =−°
(a)
sin 2 + cos 2
2
xy
x y xy
σσ
τ θτ θ
′′
−
= −
4 ( 1.6) sin ( 30 ) 0
2
− −−
=− − °+
0.600 MPa
xy
τ
′′
= −
(b)
cos 2 + sin 2
22
xy xy
x xy
σσ σσ
σ θτ θ
′
+−
= +
4 ( 1.6) 4 ( 1.6) cos ( 30 ) 0
22
−+− −−−
= + − °+
3.84 MPa
x
σ
′= −
consent of McGraw-Hill Education.
PROBLEM 14.18
The grain of a wooden member forms an angle of 15° with the vertical. For the state
of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the
consent of McGraw-Hill Education.
PROBLEM 14.19
Two steel plates of uniform cross section
10 80mm×
are welded
together as shown. Knowing that centric 100-kN forces are applied to
the welded plates and that
25
β
= °
, determine (a) the in-plane shearing
stress parallel to the weld, (b) the normal stress perpendicular to the
weld.
SOLUTION
PROBLEM 14.20
The centric force P is applied to a short post as shown. Knowing that the stresses on
plane a-a are
15 ksi
s
= −
and
5 ksi,=
τ
determine (a) the angle
β
that plane a-a forms
with the horizontal, (b) the maximum compressive stress in the post.
