2
Axially 122
CHAPTER Loaded Members
Stresses on Inclined Sections
Problem 2.6-1 A steel bar of rectangular cross section (1.5 in. 2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 15,000 psi and 7,000 psi, respectively. Determine the maximum permissible load Pmax. Solution 2.6-1 Rectangular bar in tension
2.0 in. P P 1.5 in. 2.0 in. P P
1.5 in.
Maximum shear stress: tmax
allow
sx 2
P 2A
15,000 psi
allow
7,000 psi
allow,
Because allow is less than one-half of shear stress governs. Pmax 2
allow
the
A
1.5 in. 3.0 in.2
2.0 in.
A
2(7,000 psi) (3.0 in.2)
42,000 lb
Maximum Normal Stress: sx P A
Problem 2.6-2 A circular steel rod of diameter d is subjected to a tensile force P 3.0 kN (see figure). The allowable stresses in tension and shear are 120 MPa and 50 MPa, respectively. What is the minimum permissible diameter dmin of the rod? Solution 2.6-2 Steel rod in tension
P d
P
d
P = 3.0 kN
P
P
3.0 kN
A
d2 4 P A sx 2
allow
Because allow is less than one-half of shear stress governs. tmax P 2A 50 MPa P 2A or 50 MPa 3.0 kN d2 (2) 4
allow,
the
Maximum normal stress: sx Maximum shear stress: tmax
allow
Solve for d: dmin
6.18 mm
120 MPa
SECTION 2.6
Stresses on Inclined Sections
123
Problem 2.6-3 A standard brick (dimensions 8 in. 4 in. 2.5 in.) is compressed lengthwise by a force P, as shown in the figure. If the ultimate shear stress for brick is 1200 psi and the ultimate compressive stress is 3600 psi, what force Pmax is required to break the brick?
P
8 in.
4 in.
2.5 in.
Solution 2.6-3
Standard brick in compression
P
Maximum shear stress: tmax sx 2 P 2A
ult
8 in.
4 in.
2.5 in.
ult
3600 psi
1200 psi
ult,
Because ult is less than one-half of stress governs. tmax A 2.5 in. 4.0 in. 10.0 in.2 Pmax Maximum normal stress: sx P A P 2A or Pmax 2Atult
the shear
2(10.0 in.2 )(1200 psi) 24,000 lb
Problem 2.6-4 A brass wire of diameter d 2.42 mm is stretched tightly between rigid supports so that the tensile force is T 92 N (see figure). What is the maximum permissible temperature drop T if the allowable shear stress in the wire is 60 MPa? (The coefficient of thermal expansion for the wire is 20 10 6/C and the modulus of elasticity is 100 GPa.)
T
d
T
Probs. 2.6-4 and 2.6-5
Solution 2.6-4
T
Brass wire in tension MAXIMUM SHEAR STRESS
d T
tmax d A 2.42 mm d2 4 20 4.60 mm2 10 6/ C E 100 GPa 92 N T A
x allow
sx 2
Solve for temperature drop T: T 60 MPa 2tmax T A E tmax tallow
1 T B 2 A
E (T )R
Initial tensile force: T
SUBSTITUTE NUMERICAL VALUES: T E ( T) 2(60 MPa) (92 N) (4.60 mm2 ) (100 GPa)(20 10 6 C) 120 MPa 20 MPa 2 MPa C 50 C
Stress due to initial tension: sx Stress due to temperature drop: (see Eq. 2-18 of Section 2.5) T Total stress: sx E (T ) A
124
CHAPTER 2
Axially Loaded Members
Problem 2.6-5 A brass wire of diameter d 1/16 in. is stretched between rigid supports with an initial tension T of 32 lb (see figure). (a) If the temperature is lowered by 50F, what is the maximum shear stress max in the wire? (b) If the allowable shear stress is 10,000 psi, what is the maximum permissible temperature drop? (Assume that the coefficient of thermal expansion is 10.6 10 6/F and the modulus of elasticity is 15 106 psi.)
Solution 2.6-5
Brass wire in tension (a) MAXIMUM SHEAR STRESS WHEN TEMPERATURE DROPS 50 F tmax sx 2 1 T B 2 A E (T )R (Eq. 1)
T
d
T
d A
1 in. 16 d2 4 0.003068 in.2 10.6 10 6/ F 106 psi 32 lb T A
x
Substitute numerical values: tmax 9,190 psi
E
15
(b) MAXIMUM PERMISSIBLE TEMPERATURE DROP IF 10,000 psi allow Solve Eq. (1) for T: 2tmax T A tmax tallow E Substitute numerical values: T T 60.2 F
Initial tensile force: T
Stress due to initial tension: sx Stress due to temperature drop: (see Eq. 2-18 of Section 2.5) T E (T ) Total stress: sx A
E ( T)
Problem 2.6-6 A steel bar with diameter d tensile load P 9.5 kN (see figure).
12 mm is subjected to a
P d = 12 mm P = 9.5 kN
(a) What is the maximum normal stress max in the bar? (b) What is the maximum shear stress max? (c) Draw a stress element oriented at 45 to the axis of the bar and show all stresses acting on the faces of this element.
SECTION 2.6
Stresses on Inclined Sections
125
Solution 2.6-6
P
Steel bar in tension
d = 12 mm P = 9.5 kN
(c) STRESS ELEMENT AT
9,000 9,000 = 45 x 9,000 9,000 9,000
45
P
9.5 kN
y 9,000 0
(a) MAXIMUM NORMAL STRESS sx smax P A 9.5 kN 2 4 (12 mm) 84.0 MPa 84.0 MPa
NOTE: All stresses have units of MPa.
(b) MAXIMUM SHEAR STRESS The maximum shear stress is on a 45 plane and equals x /2. tmax sx 2 42.0 MPa
Problem 2.6-7 During a tension test of a mild-steel specimen (see figure), the extensometer shows an elongation of 0.00120 in. with a gage length of 2 in. Assume that the steel is stressed below the proportional limit and that the modulus of elasticity E 30 106 psi. (a) What is the maximum normal stress max in the specimen? (b) What is the maximum shear stress max? (c) Draw a stress element oriented at an angle of 45 to the axis of the bar and show all stresses acting on the faces of this element.
2 in. T T
126
CHAPTER 2
Axially Loaded Members
Solution 2.6-7
Tension test
2 in. T T
Elongation:
0.00120 in.
(b) MAXIMUM SHEAR STRESS The maximum shear stress is on a 45 plane and equals x /2. 0.00060 tmax 106 psi)(0.00060) sx 2 9,000 psi 45
(2 in. gage length) Strain: e L 0.00120 in. 2 in.
x
Hooke's law :
Ee
(30
18,000 psi (a) MAXIMUM NORMAL STRESS
x
(c) STRESS ELEMENT AT
NOTE: All stresses have units of psi.
9,000 y 9,000 0 x 9,000 = 45 9,000 9,000 9,000
is the maximum normal stress. 18,000 psi
smax
Problem 2.6-8 A copper bar with a rectangular cross section is held without stress between rigid supports (see figure). Subsequently, the temperature of the bar is raised 50C. Determine the stresses on all faces of the elements A and B, and show these stresses on sketches of the elements. (Assume 17.5 10 6/C and E 120 GPa.)
45 A B
Solution 2.6-8
Copper bar with rigid supports
45 A B
STRESSES ON ELEMENTS A AND B
105 A 105 52.5 52.5 y 0 x 52.5 52.5 B 52.5 52.5 = 45
T E
50 C (Increase) 17.5 10 6/ C 120 GPa
STRESS DUE TO TEMPERATURE INCREASE
x
E ( T) (See Eq. 2-18 of Section 2.5) 105 MPa (Compression)
NOTE: All stresses have units of MPa.
MAXIMUM SHEAR STRESS tmax sx 2 52.5 MPa
SECTION 2.6
Stresses on Inclined Sections
127
Problem 2.6-9 A compression member in a bridge truss is fabricated from a wide-flange steel section (see figure). The cross-sectional area A 7.5 in.2 and the axial load P 90 k. P Determine the normal and shear stresses acting on all faces of stress elements located in the web of the beam and oriented at (a) an angle 0, (b) an angle 30, and (c) an angle 45. In each case, show the stresses on a sketch of a properly oriented element. Solution 2.6-9
P
P
Truss member in compression
P
x
sin
cos
( 12.0 ksi)(sin 120 ) (cos 120 )
5.2 ksi P A sx 90 k 7.5 in.2 P A 90 k 7.5 in.2
9.0 3.0 5.2 y 0 5.2 x 3.0 9.0 = 30
12.0 ksi (Compression) (a) 0
y 12.0 ksi 0 x
x
NOTE: All stresses have units of ksi.
12.0 ksi
(c)
45 cos2
x
( 12.0 ksi)(cos 45 )2 cos
6.0 ksi
(b)
30
sin
( 12.0 ksi)(sin 45 ) (cos 45 )
Use Eqs. (2-29a) and (2-29b):
x
6.0 ksi
6.0 6.0 6.0 y 0 6.0 x 6.0 = 45
cos2
( 12.0 ksi)(cos 30 )2
9.0 ksi
x
sin cos
( 12.0 ksi)(sin 30 )(cos 30 )
5.2 ksi 30
x
90 cos2
120 ( 12.0 ksi)(cos 120 )2 3.0 ksi
6.0
NOTE: All stresses have units of ksi.
128
CHAPTER 2
Axially Loaded Members
Problem 2.6-10 A plastic bar of diameter d 30 mm is compressed in a testing device by a force P 170 N applied as shown in the figure. Determine the normal shear and stresses acting on all faces of stress elements oriented at (a) an angle 0, (b) an angle 22.5, and (c) an angle 45. In each case, show the stresses on a sketch of a properly oriented element.
P = 170 N 100 mm 300 mm
Plastic bar d = 30 mm
Solution 2.6-10
100 mm
Plastic bar in compression
300 mm P = 170 N
x
sin
cos
( 962.0 kPa)(sin 22.5 )(cos 22.5 ) 340 kPa
Plastic bar d = 30 mm
22.5
x
90
112.5 ( 962.0 kPa)(cos 112.5 )2
cos2
141 kPa
x
sin
cos
FREE-BODY DIAGRAM
( 962.0 kPa)(sin 112.5 )(cos 112.5 ) 340 kPa
P = 170 N 141 y 0 x 141 821
100 mm F
300 mm
340
821
= 22.5
F F
Compressive force in plastic bar 4P 4(170 N) 680 N
340
PLASTIC BAR (ROTATED TO THE HORIZONTAL)
y F F
NOTE: All stresses have units of kPa. (c) 45
x
0
d = 30 mm
x
cos2
( 962.0 kPa)(cos 45 )2
481 kPa
x
sx
F A
680 N 2 4 (30 mm)
sin
cos 481 kPa
( 962.0 kPa)(sin 45 )(cos 45 )
481 481 y 0 481 481 x 481 = 45
962.0 kPa (Compression) (a) 0
962 kPa y 0 x 962 kPa
(b)
22.5
481
Use Eqs. (2-29a) and (2-29b)
x
NOTE: All stresses have units of kPa.
cos2
( 962.0 kPa)(cos 22.5 )2
821 kPa
SECTION 2.6
Stresses on Inclined Sections
129
Problem 2.6-11 A plastic bar fits snugly between rigid supports at room temperature (68F) but with no initial stress (see figure). When the temperature of the bar is raised to 160F, the compressive stress on an inclined plane pq becomes 1700 psi. (a) What is the shear stress on plane pq? (Assume 60 10 6/F 3 psi.) and E 450 10 (b) Draw a stress element oriented to plane pq and show the stresses acting on all faces of this element.
p
q Probs. 2.6-11 and 2.6-12
Solution 2.6-11
Plastic bar between rigid supports
p
q
60
10 6/ F E
450
103 psi
(b) STRESS ELEMENT ORIENTED TO PLANE pq 34.18 1700 psi 90 124.18 1150 psi
Temperature increase: T 160 F 68 F 92 F
34.18
x
NORMAL STRESS
x x
x IN THE BAR
cos2
( 2484 psi)(cos 124.18 )2
E ( T ) (See Eq. 2-18 in Section 2.5) (450 103 psi)(60 10 6/ F)(92 F)
784 psi
x
sin
cos
2484 psi (Compression) ANGLE
x
( 2484 psi)(sin 124.18 )(cos 124.18 ) 1150 psi
TO PLANE
pq 1700 psi psi)(cos2 )
1700
784 1150 1700 y 0 x = 34.18
cos2
For plane pq: 1700 psi ( 2484 0.6844 34.18
Therefore, cos2u cos
1700 psi 2484 psi 0.8273
1150 784
NOTE: All stresses have units of psi.
(a) SHEAR STRESS ON PLANE pq
x
sin
cos
( 2484 psi)(sin 34.18 )(cos 34.18 ) 1150 psi (Counter clockwise)
130
CHAPTER 2
Axially Loaded Members
Problem 2.6-12 A copper bar is held snugly (but without any initial stress) between rigid supports (see figure). The allowable stresses on the inclined plane pq, for which 55, are specified as 60 MPa in compression and 30 MPa in shear. (a) What is the maximum permissible temperature rise T if the allowable stresses on plane pq are not to be exceeded? (Assume 17 10 6/C and E 120 GPa.) (b) If the temperature increases by the maximum permissible amount, what are the stresses on plane pq?
Solution 2.6-12
Copper bar between rigid supports
p
Shear stress governs.
x
63.85 MPa
Due to temperature increase T:
q
x
E ( T) (See Eq. 2-18 in Section 2.5) (120 GPa)(17 31.3 C 10 6/ C)( T)
17 E
10 6/ C
63.85 MPa T
120 GPa 55
Plane pq:
(b) STRESSES ON PLANE pq
x
Allowable stresses on plane pq:
allow allow
63.85 MPa
x
60 MPa (Compression) 30 MPa (Shear) cos2 ( 63.85 MPa)(cos 55 )2
(a) MAXIMUM PERMISSIBLE TEMPERATURE RISE T
x x
21.0 MPa (Compression)
x
cos2
60 MPa
x
(cos 55 )2
sin
cos
182.4 MPa
x
( 63.85 MPa)(sin 55 )(cos 55 ) 30.0 MPa (Counter clockwise)
sin
x
cos (sin 55 )(cos 55 )
30 MPa
x
63.85 MPa
Problem 2.6-13 A circular brass bar of diameter d is composed of two segments brazed together on a plane pq making an angle 36 with the axis of the bar (see figure). The allowable stresses in the brass are 13,500 psi in tension and 6500 psi in shear. On the brazed joint, the allowable stresses are 6000 psi in tension and 3000 psi in shear. If the bar must resist a tensile force P 6000 lb, what is the minimum required diameter dmin of the bar?
P
p q
d
P
SECTION 2.6
Stresses on Inclined Sections
131
Solution 2.6-13
Brass bar in tension
n
Tensile stress:
= 54 P
P
p d q
x
cos2
sx
sallow cos2u
6000 psi (cos 54 ) 2 (3)
x
36 90 P A 6000 lb d2 4
x BASED UPON ALLOWABLE STRESSES
17,370 psi 54 Shear stress: sx ` tallow ` sin ucos u sin cos
3,000 psi (sin 54 )(cos 54 ) (4)
6,310 psi ALLOWABLE STRESS Compare (1), (2), (3), and (4). 0 ):
allow
STRESS
IN THE BRASS
Tensile stress (
x
13,500 psi (1)
Shear stress on the brazed joint governs.
x
13,500 psi 45 ):
allow
6310 psi
Shear stress ( tmax
x
6500 psi DIAMETER OF BAR A P sx d2 4 dmin 6000 lb 6310 psi d2 4A 0.951 in.2 dmin 4A B
sx 2 2
allow
13,000 psi STRESS
(2)
A
x BASED UPON ALLOWABLE STRESSES ON THE
1.10 in.
BRAZED JOINT
allow allow
(
54 )
6000 psi (tension) 3000 psi (shear)
Problem 2.6-14 Two boards are joined by gluing along a scarf joint, as shown in the figure. For purposes of cutting and gluing, the angle between the plane of the joint and the faces of the boards must be between 10 and 40. Under a tensile load P, the normal stress in the boards is 4.9 MPa. (a) What are the normal and shear stresses acting on the glued joint if 20? (b) If the allowable shear stress on the joint is 2.25 MPa, what is the largest permissible value of the angle ? (c) For what angle will the shear stress on the glued joint be numerically equal to twice the normal stress on the joint?
P
P
132
CHAPTER 2
Axially Loaded Members
Solution 2.6-14
y P
Two boards joined by a scarf joint
x
P
10
40
x
33.34 4.9 MPa 20 90 Since
or <
56.66 56.66 or 33.34
Due to load P:
(a) STRESSES ON JOINT WHEN
n = 90 a
must be between 10 and 40 , we select
33.3 Note: If | | If 90
x
is between 10 and 33.3 ,
2.25 MPa. is between 33.3 and 40 , 2.25 MPa.
WHAT IS
70 cos2 (4.9 MPa)(cos 70 )2
| | (c)
if
2 ?
0.57 MPa
x
Numerical values only: cos ` | | tu ` su
x x
sin
sin
cos
| |
x
cos2
( 4.9 MPa)(sin 70 )(cos 70 ) 1.58 MPa (b) LARGEST ANGLE
allow x sin
2 cos 2 cos 63.43 2 xcos2 or tan 90 2
sin
IF allow
2.25 MPa
sin
cos a
The shear stress on the joint has a negative sign. Its numerical value cannot exceed allow 2.25 MPa. Therefore, 2.25 MPa sin cos (4.9 MPa)(sin )(cos ) or 0.4592 1 sin 2u 2 0.9184
26.6
NOTE: For 26.6 and 63.4 , we find 0.98 MPa and 1.96 MPa. Thus, ` tu ` su
From trigonometry: sin u cos u Therefore: sin 2 Solving : 2 2(0.4592) or
2 as required.
66.69
113.31
SECTION 2.6
Stresses on Inclined Sections
133
Problem 2.6-15 Acting on the sides of a stress element cut from a bar in uniaxial stress are tensile stresses of 10,000 psi and 5,000 psi, as shown in the figure. (a) Determine the angle and the shear stress and show all stresses on a sketch of the element. (b) Determine the maximum normal stress max and the maximum shear stress max in the material.
5,000 psi
= 10,000 psi
10,000 psi
5,000 psi
Solution 2.6-15
Bar in uniaxial stress
5,000 psi
From Eq. (1) or (2): 15,000 psi tu sx sin u cos u ( 15,000 psi)(sin 35.26 )(cos 35.26 ) 7,070 psi Minus sign means that for which 35.26 . acts clockwise on the plane sx
10,000 psi
10,000 psi 5,000 5,000 psi y
AND SHEAR STRESS
x
10,000
(a) ANGLE
= 35.26
7,070 0 10,000 5,000 x 7,070
cos2
10,000 psi sx su cos2u 10,000 psi cos2u 90 90 ) ] 2 sx [ sin u] 2 (1)
PLANE AT ANGLE su
90
NOTE: All stresses have units of psi. (b) MAXIMUM NORMAL AND SHEAR STRESSES smax sx sx 2 15,000 psi 7,500 psi
sx [cos(u sx sin2u
su
90
5,000 psi su 90 sin2u 5,000 psi sin2u (2)
sx
tmax
Equate (1) and (2): 10,000 psi cos2u tan2u 1 2 5,000 psi sin2u tan u 1 2 u 35.26