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.