PM
Page 07Ch07.qxd
9/27/08
1:18 571
7
Analysis of Stress and Strain
Plane Stress
1200 psi
Problem 7.2-1 An element in plane stress is subjected to stresses sx 4750 psi, sy 1200 psi, and txy 950 psi, as shown in the figure. Determine the stresses acting on an element oriented at an angle u 60 from the x axis, where the angle u is positive when counterclockwise. Show these stresses on a sketch of an element oriented at the angle u.
950 psi 4750 psi
Solution 7.2-1 sx 4750 psi
u sx1 sx1 60 sx + sy 2 2910 psi +
sy sx 2 ; sy
1200 psi
txy
950 psi
tx1y1 tx1y1
sx 2
sy
sin(2u) + txy cos(2u) ; sx1 ;
2012 psi sx + sy 3040 psi
cos(2u) + txy sin(2u)
sy1 sy1
Problem 7.2-2 Solve the preceding problem for an element in plane stress subjected to stresses sx 100 MPa, sy 80 MPa, and txy 28 MPa, as shown in the figure. Determine the stresses acting on an element oriented at an angle u 30 from the x axis, where the angle u is positive when counterclockwise. Show these stresses on a sketch of an element oriented at the angle u.
80 MPa
28 MPa 100 MPa
571
07Ch07.qxd
9/27/08
1:18 PM
Page 572
572
CHAPTER 7
Analysis of Stress and Strain
Solution 7.2-2
sx u sx1 sx1 100 MPa 30 sx + sy 2 + 119.2 MPa sx 2 ; sy cos (2u) + txy sin (2u) sy 80 MPa txy 28 MPa tx1y1 tx1y1 sy1 sy1 sx 2 5.30 MPa sx + sy 60.8 MPa sy sin(2u) + txy cos(2u) ; sx1 ;
Problem 7.2-3 Solve Problem 7.2-1 for an element in plane stress subjected to 5700 psi, sy 2300 psi, and txy 2500 psi, as shown in the figure. stresses sx Determine the stresses acting on an element oriented at an angle u 50 from the x axis, where the angle u is positive when counterclockwise. Show these stresses on a sketch of an element oriented at the angle u.
2300 psi 2500 psi 5700 psi
Solution 7.2-3 5700 psi sx
u sx1 sx1 50 sx + sy 2 + 1243 psi sx
sy sy 2 ;
2300 psi
txy
2500 psi
tx1y1 tx1y1
sx 2
sy
sin (2u) + txy cos (2u) ; sx1 ;
1240 psi sx + sy 6757 psi
cos (2u) + txy sin (2u)
sy1 sy1
Problem 7.2-4 The stresses acting on element A in the web of a train rail are found to be 40 MPa tension in the horizontal direction and 160 MPa compression in the vertical direction (see figure). Also, shear stresses of magnitude 54 MPa act in the directions shown. Determine the stresses acting on an element oriented at a counterclockwise angle of 52 from the horizontal. Show these stresses on a sketch of an element oriented at this angle.
160 MPa
A A Side View Cross Section
40 MPa 54 MPa
07Ch07.qxd
9/27/08
1:18 PM
Page 573
SECTION 7.2
Plane Stress
573
Solution 7.2-4
sx u sx1 sx1 40 MPa 52 sx + sy 2 + sx 2 sy cos(2u) + txy sin(2u) sy 160 MPa txy 54 MPa tx1y1 tx1y1 sy1 sy1 sx 2 sx + sy 16.6 MPa sy sin (2u) + txy cos (2u) ; sx1 ;
84.0 MPa
136.6 MPa
;
Problem 7.2-5 Solve the preceding problem if the normal and shear stresses acting on element A are 6500 psi, 18,500 psi, and 3800 psi (in the directions shown in the figure). Determine the stresses acting on an element oriented at a counterclockwise angle of 30 from the horizontal. Show these stresses on a sketch of an element oriented at this angle.
18,500 psi
A A Side View Cross Section
6500 psi 3800 psi
Solution 7.2-5 sx 6500 psi
u sx1 sx1 30 sx + sy 2 + 3041 psi
sy sx 2 ;
18500 psi
txy
3800 psi
tx1y1 tx1y1
sx 2 sx + sy
sy
sin (2u) + txy cos (2u) ;
sy
12725 psi sx1 ; 8959 psi
cos (2u) + txy sin (2u)
sy1 sy1
Problem 7.2-6 An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of magnitude 27 MPa in the horizontal direction and tensile stresses of magnitude 5.5 MPa in the vertical direction (see figure). Also, shear stresses of magnitude 10.5 MPa act in the directions shown. Determine the stresses acting on an element oriented at a clockwise angle of 35 from the horizontal. Show these stresses on a sketch of an element oriented at this angle.
5.5 MPa
27 MPa 10.5 MPa
07Ch07.qxd
9/27/08
1:18 PM
Page 574
574
CHAPTER 7
Analysis of Stress and Strain
Solution 7.2-6
sx u sx1 sx1 27 MPa 35 sx + sy 2 + 6.4 MPa sx 2 ; sy cos (2u) + txy sin (2u) sy 5.5 MPa txy 10.5 MPa tx1y1 tx1y1 sy1 sy1 sx sx 2 sy sy sin (2u) + txy cos (2u) ;
18.9 MPa sx1 15.1 MPa
;
Problem 7.2-7 The stresses acting on element B in the web of a wide-flange beam are found to be 14,000 psi compression in the horizontal direction and 2600 psi compression in the vertical direction (see figure). Also, shear stresses of magnitude 3800 psi act in the directions shown. Determine the stresses acting on an element oriented at a counterclockwise angle of 40 from the horizontal. Show these stresses on a sketch of an element oriented at this angle.
2600 psi
B
B
14,000 psi 3800 psi
Side View
Cross Section
Solution 7.2-7 14000 psi sx 3800 psi txy
u sx1 sx1 40 sx + sy 2 + sx 2
sy
2600 psi
tx1y1 tx1y1
sx 2 4954 psi sx + sy
sy
sin (2u) + txy cos (2u) ; sx1 ;
sy ;
cos (2u) + txy sin (2u)
sy1 sy1
3568 psi
13032 psi
Problem 7.2-8 Solve the preceding problem if the normal and shear stresses acting on element B are 46 MPa, 13 MPa, and 21 MPa (in the directions shown in the figure) and the angle is 42.5 (clockwise).
B
13 MPa
21 MPa 46 MPa
07Ch07.qxd
9/27/08
1:18 PM
Page 575
SECTION 7.2
Plane Stress
575
Solution 7.2-8 sx u
sx1 sx1
46 MPa 42.5
sx + sy 2 + sx
sy
sy 2 ;
13 MPa
txy
21 MPa
tx1y1 tx1y1
sx 2 sx + sy
sy
sin (2u) + txy cos (2u) ;
14.6 MPa sx1 7.1 MPa
cos (2u) + txy sin (2u)
sy1 sy1
51.9 MPa
;
Problem 7.2-9 The polyethylene liner of a settling pond is subjected to stresses sx 350 psi, sy 112 psi, 120 psi, as shown by the plane-stress and txy element in the first part of the figure. Determine the normal and shear stresses acting on a seam oriented at an angle of 30 to the element, as shown in the second part of the figure. Show these stresses on a sketch of an element having its sides parallel and perpendicular to the seam.
y 112 psi 30
350 psi O 120 psi x Seam
Solution 7.2-9
Plane stress (angle )
sx u sx1
350 psi 30 sx + sy 2 187 psi +
sy sx 2 ;
112 psi
txy
120 psi
The normal stress on the seam equals 187 psi ; tension. The shear stress on the seam equals 163 psi, acting clockwise against the seam. ;
sy
cos 2u + txy sin 2u
tx1y1
sx 2
sy
sin 2u + txy cos 2u ; sx1 275 psi ;
163 psi sy1 sx + sy
07Ch07.qxd
9/27/08
1:18 PM
Page 576
576
CHAPTER 7
Analysis of Stress and Strain
Problem 7.2-10 Solve the preceding problem if the normal and shear stresses acting on the element are sx 2100 kPa, sy 300 kPa, and txy 560 kPa, and the seam is oriented at an angle of 22.5 to the element (see figure).
y 300 kPa 22.5
O
2100 kPa x 560 kPa Seam
Solution 7.2-10
Plane stress (angle )
sx1 sx + sy 2 1440 kPa tx1 y1 sx 2 1030 kPa sy1 sx + sy + ; sy sin 2u + txy cos 2u ; sx1 960 kPa ; sx 2 sy cos 2u + txy sin 2u
sx u
2100 kPa 22.5
sy
300 kPa
txy
560 kPa
The normal stress on the seam equals 1440 kPa tension. ; The shear stress on the seam equals 1030 kPa, acting clockwise against the seam. ;
Problem 7.2-11 A rectangular plate of dimensions 3.0 in. * 5.0 in. is formed by welding two triangular plates (see figure). The plate is subjected to a tensile stress of 500 psi in the long direction and a compressive stress of 350 psi in the short direction. Determine the normal stress sw acting perpendicular to the line of the weld and the shear tw acting parallel to the weld. (Assume that the normal stress sw is positive when it acts in tension against the weld and the shear stress tw is positive when it acts counterclockwise against the weld.)
350 psi
We
ld
3 in. 5 in.
500 psi
07Ch07.qxd
9/27/08
1:18 PM
Page 577
SECTION 7.2
Plane Stress
577
Solution 7.2-11
Biaxial stress (welded joint)
tx1y1 sy1 sx 2 sx + sy sy sin 2u + txy cos 2u sx1 125 psi 375 psi
STRESSES ACTING ON THE WELD
sx u
500 psi arctan 3 in. 5 in.
sy
350 psi arctan 0.6
txy 30.96
0 sw tw 125 psi 375 psi ; ;
sx1
sx + sy 2 275 psi +
sx 2
sy
cos 2u + txy sin 2u
12.0 MPa
Problem 7.2-12 Solve the preceding problem for a plate of dimensions 100 mm * 250 mm subjected to a compressive stress of 2.5 MPa in the long direction and a tensile stress of 12.0 MPa in the short direction (see figure).
We
ld
100 mm 250 mm
2.5 MPa
Solution 7.2-12
Biaxial stress (welded joint)
tx1y1 sy1 sx 2 sx + sy sy sin 2u + txy cos 2u 10.0 MPa 5.0 MPa
sx1
STRESSES ACTING ON THE WELD
sx u
2.5 MPa arctan 100 mm 250 mm sx +
sy
12.0 MPa arctan 0.4
txy 21.80
0 sw tw 10.0 MPa 5.0 MPa ; ;
sx1
sx + sy 2 0.5 MPa
sy 2
cos 2u + txy sin 2u
07Ch07.qxd
9/27/08
1:18 PM
Page 578
578
CHAPTER 7
Analysis of Stress and Strain
Problem 7.2-13 At a point on the surface of a machine the material is in biaxial stress with sx 3600 psi, and sy 1600 psi, as shown in the first part of the figure. The second part of the figure shows an inclined plane aa cut through the same point in the material but oriented at an angle u. Determine the value of the angle u between zero and 90 such that no normal stress acts on plane aa. Sketch a stress element having plane aa as one of its sides and show all stresses acting on the element.
y 1600 psi a u x a
3600 psi O
Solution 7.2-13
Biaxial stress
STRESS ELEMENT sx1 sy1 sx sy txy 0 3600 psi 1600 psi tx1y1 0 u 56.31 sx1 sy 2 2400 psi 2000 psi ; sx + sy sx
sin 2u + txy cos 2u
Find angle u for s s sx1
0.
normal stress on plane a-a sx + sy 2 + sx 2 sy cos 2u + txy sin 2u
1000 + 2600 cos 2u(psi) For sx1 2u 0, we obtain cos 2u 112.62 and u 1000 2600 56.31
Problem 7.2-14 Solve the preceding problem for sx 50 MPa (see figure). and sy
32 MPa
y 50 MPa a u x a
32 MPa O
07Ch07.qxd
9/27/08
1:18 PM
Page 579
SECTION 7.2
Plane Stress
579
Solution 7.2-14
Biaxial stress
STRESS ELEMENT sx1 sx sy txy 0 0. 32 MPa 50 MPa sy1 tx1y1 0 u 38.66 sx1 sy 2 40 MPa 18 MPa ; sx + sy sx
sin 2u + txy cos 2u ;
Find angles u for s s sx1
normal stress on plane a-a sx + sy 2 + sx 2 sy cos 2u + txy sin 2u
9 + 41 cos 2u ( MPa) For sx1 2u 0, we obtain cos 2u 77.32 and u 9 41 ;
38.66
Problem 7.2-15 An element in plane stress from the frame of a racing car is oriented at a known angle u (see figure). On this inclined element, the normal and shear stresses have the magnitudes and directions shown in the figure. Determine the normal and shear stresses acting on an element whose sides are parallel to the xy axes, that is, determine sx, sy, and txy. Show the results on a sketch of an element oriented at u 0.
y
2475 psi 3950 psi u = 40 14,900 psi O x
Solution 7.2-15 Transform from u
sx txy u sx1 sx1 14900 psi 2475 psi 40 sx + sy 2 +
40 to u sy
0
3950 psi
tx1y1 tx1y1 sy1
sx 2
sy
sin (2u) + txy cos (2u) ; sx1 ;
4962 psi sx + sy 6037 psi
sx 2
sy ;
cos (2u) + txy sin (2u)
sy1
12813 psi
07Ch07.qxd
9/27/08
1:18 PM
Page 580
580
CHAPTER 7
Analysis of Stress and Strain
Problem 7.2-16 Solve the preceding problem for the element shown in the figure.
62.5 MPa
y
24.3 MPa u = 55
O 24.0 MPa
x
Solution 7.2-16 Transform from u
sx txy u sx1 sx1 24.3 MPa 24 MPa 55 sx + sy 2 + 56.5 MPa
55 to sy
u
0
62.5 MPa
tx1y1 tx1y1 sy1
sx 2
sy
sin (2u) + txy cos (2u) ;
32.6 MPa sx + sy sx1 18.3 MPa
sx 2 ;
sy
cos (2u) + txy sin (2u)
sy1
;
Problem 7.2-17 A plate in plane stress is subjected to normal stresses sx and sy and shear stress txy, as shown in the figure. At counterclockwise angles u 35 and u 75 from the x axis, the normal stress is 4800 psi tension. If the stress sx equals 2200 psi tension, what are the stresses sy and txy?
y sy txy O sx = 2200 psi x
Solution 7.2-17 sx 2200 psi
At u 35 and sx + sy 2 + Find sy and txy sx1
sy unknown txy unknown u sy 2 75, sx1 4800 psi
For sx1
u
35 2200 psi + sy + 2200 psi
4800 psi sy 2 2 * cos (70) + txy sin (70) 3323.8 psi (1)
sx
4800 psi cos (2u) + txy sin (2u)
or 0.32899 sy + 0.93969 txy
07Ch07.qxd
9/27/08
1:18 PM
Page 581
SECTION 7.2
Plane Stress
581
(2)
For sx1
u
75: 2200 psi + sy + 2200 psi
or sy
0.93301sy + 0.50000 txy 3805 psi txy 2205 psi
4652.6 psi ;
4800 psi sy 2 2 * cos (150) + txy sin (150)
Solve Eqs. (1) and (2):
4800 psi
Problem 7.2-18 The surface of an airplane wing is subjected to plane stress with normal stresses sx and sy and shear stress txy, as shown in the figure. At a counterclockwise angle u 32 from the x axis, the normal stress is 37 MPa tension, and at an angle u 48, it is 12 MPa compression. If the stress sx equals 110 MPa tension, what are the stresses sy and txy?
O
y sy txy sx = 110 MPa x
Solution 7.2-18 sx 110 MPa
At u At u 32, sx1 48, sx1
sy unknown 37 MPa 12 MPa
sxy unknown (compression)
or For
0.28081sy + 0.89879txy u sx1 48: 12 MPa 110 MPa + sy +
42.11041 MPa (1)
(tension)
110 MPa
Find sy and txy sx1 For sx + sy 2 u sx1 37 MPa + 32 37 MPa 110 MPa + sy 110 MPa sy + 2 2 * cos (64) + txy sin (64) sx 2 sy cos(2u) + txy sin (2u)
12 MPa
sy
2 2 * cos (96) + txy sin (96) 61.25093 MPa (2)
or 0.55226sy + 0.99452txy Solve Eqs. (1) and (2): sy 60.7 MPa txy
27.9 MPa
;
07Ch07.qxd
9/27/08
1:18 PM
Page 582
582
CHAPTER 7
Analysis of Stress and Strain
Problem 7.2-19 At a point in a structure subjected to plane stress, the stresses are 4100 psi, sy 2200 psi, and txy 2900 psi (the sign convention for these sx stresses is shown in Fig. 7-1). A stress element located at the same point in the structure (but oriented at a counterclockwise angle u1 with respect to the x axis) is subjected to the stresses shown in the figure (sb, tb, and 1800 psi). Assuming that the angle u1 is between zero and 90, calculate the normal stress sb, the shear stress tb, and the angle u1
y sb tb 1800 psi u1 O x
Solution 7.2-19 4100 psi sx txy 2900 psi
For sx1 Find Stress sb u u1: 1800 psi sb sx + sy sx + sy 2 +
sy
2200 psi
1800 psi
950 psi 3150 psi cos12u12 + 2900 psi sin 12u12
sy1
sb
tx1y1
tb
SOLVE NUMERICALLY: 2u1 87.32 u1 43.7 ;
sb, tb, and u1 1800 psi sb 3700 psi ;
Shear Stress tb tb sx 2 sy cos ( 2u) + txy sin ( 2u) tb sx 2 3282 psi sy sin 12u12 + txy cos 12u12 ;
Angle u1 sx1
Principal Stresses and Maximum Shear Stresses
When solving the problems for Section 7.3, consider only the in-plane stresses (the stresses in the xy plane).
Problem 7.3-1 An element in plane stress is subjected to stresses sx 4750 psi, sy 1200 psi, and txy figure for Problem 7.2-1). Determine the principal stresses and show them on a sketch of a properly oriented element.
950 psi (see the
Solution 7.3-1 sx 4750 psi
sy
1200 psi
txy
950 psi
up2 s1
up1 + 90 sx + sy 2 sx + sy 2 4988 psi 962 psi + + sx sx
up2 sy 2 sy 2 ; ;
104.08 cos 12up12 + txy sin 12up12
PRINCIPAL STRESSES atana up1 up1 2 txy sx 2 14.08 sy b
s2 s1 s2
cos 12up22 + txy sin 12up22
07Ch07.qxd
9/27/08
1:18 PM
Page 583
SECTION 7.3
Principal Stresses and Maximum Shear Stresses
583
Problem 7.3-2 An element in plane stress is subjected to stresses sx 100 MPa, sy 80 MPa, and txy figure for Problem 7.2-2). Determine the principal stresses and show them on a sketch of a properly oriented element.
28 MPa (see the
sx
Solution 7.3-2 100 MPa
sy
80 MPa
txy
28 MPa s1 s2 s1 s2
sx + sy 2 sx + sy 2 60 MPa + 120 MPa +
sx 2 sx 2 ; ;
sy sy
PRINCIPAL STRESSES atana up1 up1 up2 35.2 up1 + 90 up2 125.17 2 txy sx 2 sy b
cos12up12 + txy sin12up12 cos12up22 + txy sin12up22
Problem 7.3-3 An element in plane stress is subjected to stresses sx 5700 psi, sy 2300 psi, and txy (see the figure for Problem 7.2-3). Determine the principal stresses and show them on a sketch of a properly oriented element.
2500 psi
sx
Solution 7.3-3 5700 psi
sy
2300 psi
txy
2500 psi s1 s2 s1 s2
sx + sy 2 sx + sy 2 + 977 psi 7023 psi +
sx 2 sx 2 ;
sy sy
PRINCIPAL STRESSES atana up2 up2 up1 2txy sx 2 27.89 up2 + 90 up1 62.1 sy b
cos12up12 + txy sin12up12 cos12up22 + txy sin12up22
;
Problem 7.3-4 The stresses acting on element A in the web of a train rail are found to be 40 MPa tension in the horizontal direction and 160 MPa compression in the vertical direction (see figure). Also, shear stresses of magnitude 54 MPa act in the directions shown (see the figure for Problem 7.2-4). Determine the principal stresses and show them on a sketch of a properly oriented element.
07Ch07.qxd
9/27/08
1:18 PM
Page 584
584
CHAPTER 7
Analysis of Stress and Strain
Solution 7.3-4
sx 40 MPa sy 160 MPa txy 54 MPa s1 s2 s1 s2 up2 75.8 sx + sy 2 sx + sy 2 + 53.6 MPa 173.6 MPa + sx 2 ; ; sx 2 sy sy cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22
PRINCIPAL STRESSES atana up1 up1 up2 2txy sx 2 14.2 up1 + 90 sy b
Problem 7.3-5 The normal and shear stresses acting on element A are 6500 psi, 18,500 psi, and 3800 psi (in the directions shown in the figure) (see the figure for Problem 7.2-5). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-5 sx 6500 psi
PRINCIPAL ANGLES atana up1 up1 up2 s1 s2 2txy sx 2 8.45 up1 + 90 sx + sy 2 sx + sy 2 + +
sy
18500 psi
txy
3800 psi
s1 s2
7065 psi 19065 psi
sy
b
MAXIMUM SHEAR STRESSES tmax us1 sx a sy 2 45 2 b + txy2 us1 saver
2
A
tmax 53.4
13065 psi ; ;
;
up1
up2 sx 2 sx 2 sy sy
81.55 cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22 saver
sx + sy
6000 psi
Problem 7.3-6 An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of magnitude 27 MPa in the horizontal direction and tensile stresses of magnitude 5.5 MPa in the vertical direction. Also, shear stresses of magnitude 10.5 MPa act in the directions shown (see the figure for Problem 7.2-6). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
07Ch07.qxd
9/27/08
1:18 PM
Page 585
SECTION 7.3
Principal Stresses and Maximum Shear Stresses
585
Solution 7.3-6
sx 27 MPa sy 5.5 MPa txy 10.5 MPa s1 s2 8.6 MPa 30.1 MPa
PRINCIPAL ANGLES 2txy atana b sx sy up2 2 up2 up1 s1 s2 16.43 up2 + 90 sx + sy 2 sx + sy 2 + + sx 2 sx 2 sy up1 sy 106.43 cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22
MAXIMUM SHEAR STRESSES tmax tmax us1 saver a sx 2 sy 2 b + txy2 ; us1 saver 61.4 10.8 MPa ;
A
19.3 MPa up1 2 45
sx + sy
Problem 7.3-7 The stresses acting on element B in the web of a wide-flange beam are found to be 14,000 psi compression in the horizontal direction and 2600 psi compression in the vertical direction. Also, shear stresses of magnitude 3800 psi act in the directions shown (see the figure for Problem 7.2-7). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-7 14000 psi sx txy 3800 psi
PRINCIPAL ANGLES atana up2 up2 up1 s1 2txy sx 2 16.85 up2 + 90 sx + sy 2 + sx 2 sy b
sy
2600 psi s2 s1 s2
sx + sy 2 + 1449 psi 15151 psi
sx 2
sy
cos12up22 + txy sin12up22
MAXIMUM SHEAR STRESSES tmax sx a 2 sy 2 b + txy2 us1 saver tmax 61.8 8300 psi 6851 psi ; ; ;
A
up1 sy
106.85 cos12up12 + txy sin12up12
us1 saver
up1 2
45
sx + sy
Problem 7.3-8 The normal and shear stresses acting on element B are sx 46 MPa, sy 13 MPa, and txy 21 MPa (see figure for Problem 7.2-8). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
07Ch07.qxd
9/27/08
1:18 PM
Page 586
586
CHAPTER 7
Analysis of Stress and Strain
Solution 7.3-8
sx 46 MPa sy 13 MPa txy 21 MPa s1 s2 b 2.8 MPa 56.2 MPa
PRINCIPAL ANGLES atana up2 up2 up1 s1 s2 2txy sx 2 25.92 up2 + 90 sx + sy 2 sx + sy 2 + + sx 2 sx 2 sy up1 sy 64.08 cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22 sy
MAXIMUM SHEAR STRESSES tmax us1 saver a sx 2 45 sy 2 sy b + txy2 us1 saver
2
A
tmax 19.08
26.7 MPa ; ;
;
up1 sx
29.5 MPa
Problem 7.3-9 A shear wall in a reinforced concrete building is subjected to a vertical uniform load of intensity q and a horizontal force H, as shown in the first part of the figure. (The force H represents the effects of wind and earthquake loads.) As a consequence of these loads, the stresses at point A on the surface of the wall have the values shown in the second part of the figure (compressive stress equal to 1100 psi and shear stress equal to 480 psi).
q H 480 psi A A
1100 psi
(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
sx
Solution 7.3-9 0 sy
Shear wall 1100 psi
txy
480 psi
Therefore, s1 s2
180 psi and up1 1280 psi and up2
20.56
(a) PRINCIPAL STRESSES tan 2up 2up 2up sx1 For 2up For 2up 2txy sx sy 0.87273 20.56 69.44 sy 2 sx1 sx1 cos 2u + txy sin 2u 180 psi 1280 psi
69.44
f
;
41.11 and up 138.89 and up sx + sy 2 + 41.11: 138.89: sx
07Ch07.qxd
9/27/08
1:18 PM
Page 587
SECTION 7.3
Principal Stresses and Maximum Shear Stresses
587
(b) MAXIMUM SHEAR STRESSES tmax us1 us2 saver a sx 2 sy 2 b + t2 xy 730 psi 730 psi 730 psi
A
up1 45 up1+ 45 sx + sy 2
65.56 and t 24.44 and t ;
f
;
550 psi
Problem 7.3-10 A propeller shaft subjected to combined torsion and axial thrust is designed to resist a shear stress of 56 MPa and a compressive stress of 85 MPa (see figure).
(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
85 MPa
56 MPa
sx
Solution 7.3-10 85 MPa
sy
0 MPa
Txy
56 MPa
s1 s2
27.8 MPa 112.8 MPa
; ;
(a) PRINCIPAL STRESSES atana up2 up2 up1 s1 s2 26.4 up2 + 90 sx + sy 2 sx + sy 2 + + sx 2 sx 2 sy up1 sy 116.4 ; 2txy sx 2 sy b
(b) MAXIMUM SHEAR STRESSES tmax tmax us1 saver a sx sy b + txy2 ; us1 saver 71.4 ; ;
2
A 2 70.3 MPa
45 2 sx + sy
up1
cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22
42.5 MPa
07Ch07.qxd
9/27/08
1:18 PM
Page 588
588
CHAPTER 7
Analysis of Stress and Strain
Problems 7.3-11
sx
2500 psi, sy
1020 psi, txy
900 psi
y sy txy sx O x
(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
Probs. 7.3-11 through 7.3-16 Solution 7.3-11 sx 2500 psi
sy
1020 psi
txy
900 psi
Therefore, For up1 25.3: 64.7: s1 s2 2925 psi 595 psi ; ; For up2
(a) PRINCIPAL STRESSES 2txy sx sy atan a up1 2txy sx 2 sy b
(b) MAXIMUM SHEAR STRESSES tmax a sx 2 sy 2 b + txy2
tan(2up) up1 up2 s1 s2
25.29 90 + up1 sx + sy 2 sx + sy 2 + + sx 2 sx 2 sy up2 sy 64.71 cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22
A
tmax us1 t1 us2 t2 saver
1165 psi up1 45 us1 1165 psi ; up1 45 1165 psi sx + sy 2 us2 ; saver 70.3 and 19.71 and
1760 psi
;
Problems 7.3-12
sx
2150 kPa, sy
375 kPa, txy
460 kPa
(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-12
sx 2150 kPa sy 375 kPa txy 460 kPa up1 up2 atana up1 2txy sx 2 sy b s1 s2 sx + sy 2 sx + sy 2 + + sx 2 13.70 90 + up1 sx 2 sy sy up2 76.30
(a) PRINCIPAL STRESSES 2txy sx sy
cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22
tan(2up)
07Ch07.qxd
9/27/08
1:18 PM
Page 589
SECTION 7.3
Principal Stresses and Maximum Shear Stresses
589
Therefore, For up1 For up2 13.70 76.3 s1 s2 2262 kPa 263 kPa ; ;
tmax
145 psi 58.7 ;
us1 us1 up1 45 and t1 1000 kPa
(b) MAXIMUM SHEAR STRESSES tmax a sx 2 sy b + txy2
2
A
us2 up1 + 45 us2 31.3 ; 1000 kPa and t2 sx + sy saver saver 1263 kPa 2
Problems 7.3-13
sx
14,500 psi, sy
1070 psi, txy
1900 psi
(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-13 sx 14500 psi
sy
1070 psi
txy
1900 psi
Therefore, For up1 7.90 97.9 s1 s2 14764 psi 806 psi ; ; For up2
(a) PRINCIPAL STRESSES 2txy sx sy up2 sx + sx + 2 2 sy sy atana up1 2txy sx 2 sy b
(b) MAXIMUM SHEAR STRESSES tmax a sx 2 sy b + txy2 ; ;
2
tan(2up) up1 up2 s1 s2
7.90 90 + up1 sx + sy 2 sx + sy 2 97.90 cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22
A
tmax
6979 psi us1 37.1 52.9
us1 u p1 45 and t1 6979 psi
us2 us2 up1 + 45 and t2 6979 psi saver sx + sy 2 saver
7785 psi
Problems 7.3-14
sx
16.5 MPa, sv
91 MPa, txy
39 MPa
(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
07Ch07.qxd
9/27/08
1:18 PM
Page 590
590
CHAPTER 7
Analysis of Stress and Strain
Solution 7.3-14
sx 16.5 MPa sy 91 MPa txy 2txy sx 2 sy 39 MPa (a) PRINCIPAL STRESSES 2txy sx 17.98 90 + up1 sx + sx + 2 2 sy 2 sx + sy 2 up2 sy 72.02 sy atana up1 b
(b) MAXIMUM SHEAR STRESSES tmax tmax a sx 2 sy b + txy2 ; ;
2
A
9631.7 psi 63.0 27.0
tan(2up) up1 up2 s1 s2
us1 us1 up1 45 and t1 66.4 MPa us2 us2 up1 + 45 and t2 66.4 MPa saver sx + sy 2 saver
sx + sy
cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22
37.3 MPa
Therefore, For up1 For up2 17.98 72.0 s2 s1 29.2 MPa 103.7 MPa
Problems 7.3-15
sx
3300 psi, sy
11,000 psi, txy
4500 psi
(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-15 s 3300 psi
sy
11000 psi
txy 2txy sx 2
4500 psi
(a) PRINCIPAL STRESSES 2txy sx sy atana up1 sy b
Therefore, For up1 24.7 For up2 114.7
s1 s2
1228 psi 13072 psi
(b) MAXIMUM SHEAR STRESSES tmax tmax us1 and t1 us2 and t2 saver a sx 2 sy b + txy2 ; ;
2
tan(2up) up1 up2 s1 s2 2 sx + sy 2
24.73 90 + up1 sx + sx + 2 2 sy sy up2 114.73
A
5922 psi up1 45 5922 psi up1 + 45 5922 psi sx + sy 2 us1 us2 20.3 69.7
sx + sy
cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22
saver
7150 psi
07Ch07.qxd
9/27/08
1:18 PM
Page 591
SECTION 7.3
Principal Stresses and Maximum Shear Stresses
591
Problems 7.3-16
sx
108 MPa, sy
58 MPa, txy
58 MPa
(a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-16 sx 108 MPa
sy
58 MPa
txy
58 MPa
(b) MAXIMUM SHEAR STRESSES tmax a sx 2 sy b + txy2 ;
2
(a) PRINCIPAL STRESSES 2txy sx sy up1 sx + sx + 2 2 sy sy atana up2 2txy sx 2 107.47 cos 12up12 + txy sin 12up12 cos 12up22 + txy sin 12up22 ; ; sy b
A
tmax
14686.1 psi 62.47
tan(2up) up2 up1 s1 s2
17.47 90 + up2 sx + sy 2 sx + sy 2
us1 us1 up1 45 and t1 101.3 MPa
us2 152.47 ; us2 up1 + 45 101.3 MPa and t2 sx + sy saver saver 25.0 MPa 2
Therefore, For up1 For up2 107.47 17.47 s1 s2 76.3 MPa 126.3 MPa
Problem 7.3-17 At a point on the surface of a machine component, the stresses acting on the x face of a stress element are sx 5900 psi and txy 1950 psi (see figure). What is the allowable range of values for the stress sy if the maximum shear stress is limited to t0 2500 psi?
y sy txy = 1950 psi O sx = 5900 psi x
07Ch07.qxd
9/27/08
1:18 PM
Page 592
592
CHAPTER 7
Analysis of Stress and Strain
Solution 7.3-17 sx 5900 psi
sy
unknown
txy
1950 psi 2500 psi
Therefore, 2771 psi sy 9029 psi From Eq. (1): t max (sy1) (1)
Find the allowable range of values for sy if the maximum allowable shear stresses is tmax sx sy 2 tmax a b + txy2 A 2 Solve for sy sy sx a2 2tmax2 22tmax 2
A
a
sx 2
sy1
b + txy2
2
J
txy2 txy2 b
K
sy
a
9029 b psi 2771
2.771 ksi 9.029 ksi
tmax(y1) 2.5 ksi
y1
Problem 7.3-18 At a point on the surface of a machine component the stresses acting on the x face of a stress element are sx 42 MPa and txy 33 MPa (see figure). What is the allowable range of values for the stress sy if the maximum shear stress is limited to t0 35 MPa?
y sy txy = 33 MPa O sx = 42 MPa x
Solution 7.3-18 sx 42 MPa
sy
unknown
txy
33 MPa tmax
Find the allowable range of values for sy if the maximum allowable shear stresses is tmax 35 MPa
A
a
sx 2
sy
b + txy2
2
(1)
07Ch07.qxd
9/27/08
1:18 PM
Page 593
SECTION 7.3
Principal Stresses and Maximum Shear Stresses
593
Solve for sy sy sx
Therefore, 18.7 MPa sy 65.3 MPa
18.7 MPa 65.3 MPa
J
2 2tmax 2 a 22tmax2
From Eq. (1): txy2 txy2 b K sy 65.3 a b MPa 18.7 tmax (sy1)
A
a
sx 2
sy1
b + txy2
2
35 MPa tmax (y1)
y1
Problem 7.3-19 An element in plane stress is subjected to stresses sx 5700 psi 2300 psi (see figure). It is known that one of the principal stresses equals and txy 6700 psi in tension.
(a) Determine the stress sy. (b) Determine the other principal stress and the orientation of the principal planes, then show the principal stresses on a sketch of a properly oriented element.
O
y sy
5700 psi x 2300 psi
Solution 7.3-19 sx 5700 psi
(a) STRESS sy
sy unknown txy
2300 psi
Solve for sy
sy
1410 psi
;
(b) PRINCIPAL STRESSES 2txy sx 23.50 90 + up1 up2 66.50 sy atan a up1 2txy sx 2 sy b
Because sy is smaller than a given principal stress, we know that the given stress is the larger principal stress. s1 s1 6700 psi sx + sy 2 +
A
tan (2up) up1 up2
a
sx 2
sy 2 b + txy2
07Ch07.qxd
9/27/08
1:18 PM
Page 594
594
CHAPTER 7
Analysis of Stress and Strain
s1
sx + sy +
sx
sy
2 2 + txy sin12up12 sx + sy + sx sy 2 2 + txy sin 12up22
cos 12up12 cos 12up22
Therefore, For up1 For up2 23.5 : s1 66.5 : s2 6700 psi 410 psi ; ;
s2
Problem 7.3-20 An element in plane stress is subjected to stresses sx 50 MPa and txy 42 MPa (see figure). It is known that one of the principal stresses equals 33 MPa in tension.
(a) Determine the stress sy. (b) Determine the other principal stress and the orientation of the principal planes, then show the principal stresses on a sketch of a properly oriented element.
O
y sy 42 MPa 50 MPa x
Solution 7.3-20 sx 50 MPa
(a) STRESS sy
sy unknown
txy
42 MPa
up2 up1 s1
26.85 90 + up2 sx + sy 2 + sx up1 sy 63.15 cos 12up12 cos 12up22
Because sy is smaller than a given principal stress, we know that the given stress is the larger principal stress. s1 s1 33 MPa sx + sy 2 +
A
2 + txy sin 12up12 sx + sy + sx sy 2 2 + txy sin 12up22 : s1
a
sx 2
sy 2
b+
s2 txy2 ; Therefore, For up1 63.2
Solve for sy
sy
11.7 MPa
(b) PRINCIPAL STRESSES atan a up2 2txy sx 2 sy b
33.0 MPa 71.3 MPa
; ;
For up2
26.8 : s 2
tan(2up)
2txy sx sy
07Ch07.qxd
9/27/08
1:19 PM
Page 595
SECTION 7.4
Mohrs Circle
595
Mohrs Circle
The problems for Section 7.4 are to be solved using Mohrs circle. Consider only the in-plane stresses (the stresses in the xy plane). sx
y
Problem 7.4-1 An element in uniaxial stress is subjected to tensile stresses 11,375 psi, as shown in the figure. Using Mohrs circle, determine:
(a) The stresses acting on an element oriented at a counterclockwise angle u from the x axis. (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. 24
11,375 psi O x
Solution 7.4-1 sx 11375 psi sy
(a) ELEMENT AT 2u 48 R sc sx1 tx1y1 tx1y1 PointD : sy1
0 psi 24 sx 2 R ;
txy
0 psi
sy1
1882 psi
;
u
(b) MAXIMUM SHEAR STRESSES Point S1: us1 tmax R tmax 90 2 tmax saver 90 2 us1 45 ; 45 5688 psi 5688 psi ; ; ; ;
5688 psi 5688 psi ;
Point C:
R sc 9493 psi
5688 psi us2
Point D: sx1
R + R cos(2u) Point S2: us2 tmax ; saver R R R sin (2u) 4227 psi R R cos (2u)
Problem 7.4-2 An element in uniaxial stress is subjected to tensile stresses sx 49 MPa, as shown in the figure Using Mohrs circle, determine:
(a) The stresses acting on an element oriented at an angle u 27 from the x axis (minus means clockwise). (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
y
49 MPa O x
sx
Solution 7.4-2 49 MPa sy
u
0 MPa 27 sx 2 sc R
txy
0 MPa
Point D:
(a) ELEMENT AT 2u Point C:
sx1 sx1 tx1y1 tx1y1
54.0 R sc R
24.5 MPa 24.5 MPa
R + R cos (|2u|) 38.9 MPa ; R sin (2u) 19.8 MPa ; R R cos ( |2u| ) 10.1 MPa ;
Point D sy1 sy1
07Ch07.qxd
9/27/08
1:19 PM
Page 596
596
CHAPTER 7
Analysis of Stress and Strain
(b) MAXIMUM SHEAR STRESSES Point S1: us1 us1 tmax R tmax 90 2 45.0 ; 24.5 MPa ;
Point S2: us2 tmax saver R R
90 2 tmax saver
us2
45.0
; ; ;
24.5 MPa 24.5 MPa
Problem 7.4-3 An element in uniaxial stress is subjected to compressive stresses of magnitude 6100 psi, as shown in the figure. Using Mohrs circle, determine:
(a) The stresses acting on an element oriented at a slope of 1 on 2 (see figure). (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
y 1 2 O 6100 psi x
sx
Solution 7.4-3 6100 psi sy
0 psi
txy
0 psi
Point D : sy1 sy1
R
R cos (2u) 1220 psi
(a) ELEMENT AT A SLOPE OF 1 ON 2 u 2u 1 atana b 2 53.130 R sc sx1 sx1 tx1y1 R u 26.565 sx 2 sc R ; Point S1: 3050 psi 3050 psi Point S2: Point D: R + R cos (2u) 4880 psi tx1y1 ; 2440 psi ; tmax saver R R tmax
(b) MAXIMUM SHEAR STRESSES us1 R us2 tmax saver 90 2 tmax 90 2 us1 3050 psi us2 3050 psi 3050 psi 45 ; 45 ; ; ; ;
Point C:
R sin (2u)
Problem 7.4-4 An element in biaxial stress is subjected to stresses sx sy 19 MPa, as shown in the figure. Using Mohrs circle, determine:
48 MPa and
y
(a) The stresses acting on an element oriented at a counterclockwise angle u from the x axis. (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
25
19 MPa
48 MPa O x
07Ch07.qxd
9/27/08
1:19 PM
Page 597
SECTION 7.4
Mohrs Circle
597
Solution 7.4-4
sx 48 MPa u sy 25 19 MPa ; R 33.5 MPa tmax R txy 0 MPa (b) MAXIMUM SHEAR STRESSES Point S1: us1 us1 90 2 45.0 tmax ; 33.5 MPa 90 2 45.0 tmax saver ; 33.5 MPa 14.5 MPa ; ; ;
(a) ELEMENT AT 2u
50.0 deg R
|sx| + |sy| 2 sc
Point C: sx Point D: sx1 sx1 tx1y1 tx1y1 Point D : sy1 sy1
sx + R sc
14.5 MPa
R cos(2u) 36.0 MPa R sin(2u) ;
Point S2: us2 us2 tmax R sc saver
25.7 MPa
;
sc + R cos(2u) 7.0 MPa
Problem 7.4-5 An element in biaxial stress is subjected to stresses sx 6250 psi 1750 psi, as shown in the figure. Using Mohrs circle, determine: and sy
(a) The stresses acting on an element oriented at a counterclockwise angle u 55 from the x axis. (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
O
y 1750 psi
6250 psi x
Solution 7.4-5 sx 6250 psi
(a) ELEMENT AT 2u
sy u 60
1750 psi
txy
0 psi
(b) MAXIMUM SHEAR STRESSES Point S1: us1 90 2 45 tmax 90 2 tmax saver ; 4000 psi us2 4000 psi 2250 psi 45 ; ; ; ;
120 R sx
|sx| + |sy| 2 R
R
4000 psi 2250 psi tmax R
us1
Point C: sc Point D: sx1 sx1 tx1y1 tx1y1 Point D : sy1
sc ;
sc + R cos(2u) 250 psi R sin (2u) 3464 psi sc R cos(2u) ; sy1 4250 psi
Point S2: us2 tmax saver R sc
07Ch07.qxd
9/27/08
1:19 PM
Page 598
598
CHAPTER 7
Analysis of Stress and Strain
Problem 7.4-6 An element in biaxial stress is subjected to stresses sx 29 MPa and sy 57 MPa, as shown in the figure. Using Mohrs circle, determine:
(a) The stresses acting on an element oriented at a slope of 1 on 2.5 (see figure). (b) The maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
O
y 57 MPa 1 2.5 29 MPa x
Solution 7.4-6 sx 29 MPa
sy
57 MPa
txy
0 MPa
(b) MAXIMUM SHEAR STRESSES Point S1: us1 90 2 tmax us1 45.0 ; ;
(a) ELEMENT AT A SLOPE OF 1 ON 2.5 u 2u atana 1 b 2.5 u 21.801 R ; 43.0 MPa
tmax
R
43.0 MPa 90 2
43.603 R
|sx| + |sy| 2 sc R cos (2u) 17.1 MPa
Point S2: us2 us2 tmax R sc saver
Point C: sc Point D: sx1 sx1 tx1y1 Point D sy1
:
sx + R sc
14.0 MPa ;
45.0 tmax saver
; 43.0 MPa ; ;
14.0 MPa
R sin (2u)
tx1y1
29.7 MPa
;
sy1
sc + R cos (2u)
45.1 MPa
Problem 7.4-7 An element in pure shear is subjected to stresses txy in the figure. Using Mohrs circle, determine:
2700 psi, as shown 52 from
O
y 2700 psi x
(a) The stresses acting on an element oriented at a counterclockwise angle u the x axis. (b) The principal stresses. Show all results on sketches of properly oriented elements.
07Ch07.qxd
9/27/08
1:19 PM
Page 599
SECTION 7.4
Mohrs Circle
599
Solution 7.4-7
sx 0 psi sy u 0 psi 52 txy R 2700 psi 90) ; 90) ; 90) ; s2 R s2 2700 psi ; Point P2: up2 up2 txy 2700 psi (b) PRINCIPAL STRESSES Point P1: up1 s1 90 2 R 90 2 45 ; up1 s1 45 2700 psi ; ;
(a) ELEMENT AT 2u
104.0 R sx1 tx1y1 tx1y1
Point D: sx1
R cos (2u 2620 psi R sin (2u 653 psi R cos (2u 2620 psi
Point D : sy1
sy1
Problem 7.4-8 An element in pure shear is subjected to stresses txy shown in the figure. Using Mohrs circle, determine:
14.5 MPa, as 22.5
O
y
(a) The stresses acting on an element oriented at a counterclockwise angle u from the x axis (b) The principal stresses. Show all results on sketches of properly oriented elements.
x 14.5 MPa
Solution 7.4-8 sx 0 MPa
(a) ELEMENT AT 2u R 45.00 |txy| sx1 tx1y1 tx1y1 Point D : sy1 sy1
sy u
0 MPa 22.5
txy
14.5 MPa
(b) PRINCIPAL STRESSES Point P1: up1 s1 270 u 2 p1 R s1 270 2 135.0 R 14.50 MPa ; ; 135.0 14.50 MPa ; ;
R
14.50 MPa R cos (2u 10.25 MPa R sin (2u R cos (2u 10.25 MPa 90) ; 90) ; 10.25 MPa 90) ;
Point D: sx1
Point P2: up2 up2 s2 s2
07Ch07.qxd
9/27/08
1:19 PM
Page 600
600
CHAPTER 7
Analysis of Stress and Strain
Problem 7.4-9 An element in pure shear is subjected to stresses txy shown in the figure. Using Mohrs circle, determine:
3750 psi, as
y 4
3
(a) The stresses acting on an element oriented at a slope of 3 on 4 (see figure). (b) The principal stresses. Show all results on sketches of properly oriented elements.
O x 3750 psi
Solution 7.4-9 sy sx 0 psi
0 psi
txy
3750 psi
(b) PRINCIPAL STRESSES Poin P1: up1 s1 90 2 R up1 s1 90 2 45 s2 ; 3750 psi ; 45 3750 psi ; ;
(a) ELEMENT AT A SLOPE OF 3 ON 4 u 2u 3 atana b 4 73.740 sx1 tx1y1 tx1y1 Point D sy1 sy1 3600 psi
:
u R
36.870 txy R 90) ; 90) ; 90) s2 3750 psi
Point P2: up2 up2 R
Point D: sx1
R cos (2u 3600 psi R sin (2u 1050 psi R cos (2u ;
Problem 7.4-10 sx 27 MPa, sy 14 MPa, txy 6 MPa, u 40 Using Mohrs circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
y sy txy sx O x
Probs. 7.4-10 through 7.4-15
07Ch07.qxd
9/27/08
1:19 PM
Page 601
SECTION 7.4
Mohrs Circle
601
Solution 7.4-10
sx u saver R a 27 MPa 40 2(sx atana sx + sy 2 saver 20.50 MPa R 8.8459 MPa sy 14 MPa txy 6 MPa b 2u a b 37.29 Point D: sx1 sx1 tx1y1 tx1y1 Point D : sy1 a 42.71 sy1 saver + R cos (b) 27.5 MPa R sin (b) 5.36 MPa saver ; R cos (b) ; ;
saver)2 + txy 2 txy saver b
sx
13.46 MPa
Problem 7.4-11 sx 3500 psi, sy 12,200 psi, txy 3300 psi, u 51 Using Mohrs circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
Solution 7.4-11 sx 3500 psi
u saver R a 51 2(sx atana 2 ; sx + sy
sy
12200 psi
txy
3300 psi
b
180 + 2u
a
b
40.82
Point D: sx1 saver
2 2
saver + R cos (b) 11982 psi R sin (b) 3569 psi saver 3718 psi ; ; R cos (b) ;
7850 psi R 5460 psi
sx1 tx1y1 tx1y1 Point D sy1
:
saver) + txy txy saver b
sx
a
37.18
sy1
Problem 7.4-12 sx 47 MPa, sy 186 MPa, txy 29 MPa, u 33 Using Mohrs circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
Solution 7.4-12 47MPa sx
u saver R a b 2(sx 33 sx + sy 2
sy
186MPa
txy
29MPa
Point D: sx1 sx1
saver + R cos (b) 61.7 MPa R sin (b) 51.7 MPa saver ; ;
saver
2 2
116.50 MPa R a 75.3077 MPa 22.65
tx1y1 tx1y1 Point D : sy1 sy1
saver) + txy txy sx a saver b
R cos (b) ;
atana ` 2u
`b
171.3 MPa
43.35
07Ch07.qxd
9/27/08
1:19 PM
Page 602
602
CHAPTER 7
Analysis of Stress and Strain
Problem 7.4-13 sx 1720 psi, sy 680 psi, txy 320 psi, u 14 Using Mohrs circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
Solution 7.4-13 1720 psi sx
u saver R a b 14 2(sx atana sx + sy 2
2
sy
680 psi
txy
380 psi
Point D: sx1 sx1
saver
R cos(b) ; 580 psi ; ;
1481 psi R sin (b) tx1y1
saver
2
1200 psi R a 644.0 psi
tx1y1 Point D : sy1
saver + R cos (b) 919 psi
saver) + txy txy saver| b b
sy1
|sx
36.16
2u + a
64.16
Problem 7.4-14 sx 33 MPa, sy 9 MPa, txy 29 MPa, u 35 Using Mohrs circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
Solution 7.4-14 sx 33 MPa sy
u saver R a b 35 2(sx sx + sy 2
2
9 MPa txy
29 MPa
Point D: sx1 sx1
saver + R cos (b) 46.4 MPa R sin (b) 9.81 MPa saver 22.4 MPa ; R cos (b) ; ;
saver txy2
12.00 MPa R a 35.8050 MPa 54.09
tx1y1 tx1y1 Point D : sy1
saver) + txy sx a saver b
atana ` 2u
`b
15.91
sy1
Problem 7.4-15 sx 5700 psi, sy 950 psi, txy 2100 psi, u 65 Using Mohrs circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
07Ch07.qxd
9/27/08
1:19 PM
Page 603
SECTION 7.4
Mohrs Circle
603
Solution 7.4-15
sx u saver R a b 65 2(sx atana 180 sx + sy 2
2
5700 psi sy
950 psi txy
2100 psi
Point D: sx1 sx1
saver + R cos (b) 1846 psi R sin (b) saver 2904 psi ; tx1y1 R cos (b) ; 3897 psi ;
saver
2
2375 psi R a b 3933 psi 32.28
tx1y1 Point D sy1 sy1
:
saver) + txy |txy| saver| b
|sx
2u + a
82.28
Problems 7.4-16 sx 29.5 MPa, sy 29.5 MPa, txy 27 MPa Using Mohrs circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
y sy txy sx O x
Probs. 7.4-16 through 7.4-23
Solution 7.4-16 sx 29.5 MPa sy
saver R a 2(sx sx + sy 2
29.5 MPa txy 0 MPa R a
27 MPa
Point P1: s1 Point P2: s2
R R
s1 s2
40.0 MPa 40.0 MPa
; ;
saver
saver)2 + txy2 txy sx saver
(b) MAXIMUM SHEAR STRESSES 39.9906 MPa us1 42.47 us2 90 a 2 90 + us1 us1 us2 0 MPa 23.8 113.8 ; ; ; ;
atana `
`b
(a) PRINCIPAL STRESSES up1 up2 180 2 up1 a 90 up1 up2 68.8 21.2 ; ;
Point S1: saver tmax R
tmax
40.0 MPa
07Ch07.qxd
9/27/08
1:19 PM
Page 604
604
CHAPTER 7
Analysis of Stress and Strain
Problems 7.4-17 sx 7300 psi, sy 0 psi, txy 1300 psi Using Mohrs circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
sx
Solution 7.4-17 7300 psi sy
2(sx sx + sy 2
2
0 psi txy saver txy2 a
1300 psi
Point P1: s1 s1 Point P2: s2 s2
R + saver 7525 psi R + saver 225 psi ;
saver R a
3650 psi R 3875 psi 19.60
saver) + txy sx saver
atana `
`b
(b) MAXIMUM SHEAR STRESSES us1 90 + a 2 90 + us1 us1 us2 3650 psi 3875 psi 35.2 54.8 ; ;
(a) PRINCIPAL STRESSES up1 up2 a 2 up1 9.80 up2 ;
us2
Point S1: saver 99.8 tmax R
a + 180 2
tmax
Problems 7.4-18 sx 0 MPa, sy 23.4 MPa, txy 9.6 MPa Using Mohrs circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-18 sx 0 MPa sy
saver R a 2(sx sx + sy 2
23.4 MPa txy saver 11.70 MPa R a
9.6 MPa
Point P1: s1 s1 Point P2: s2 s2
R + saver 3.43 MPa R + saver 26.8 MPa ; ;
saver)2 + txy2 txy sx saver
15.1344 MPa 39.37
atana `
`b
(b) MAXIMUM SHEAR STRESSES us1 90 2 a us1 us2 64.7 25.3 ;
(a) PRINCIPAL STRESSES up1 up2 a 2 us2 up1 19.68 up2 70.32 ; ; tmax
90 + us1 R tmax
Point S1: saver
11.70 MPa 15.13 MPa ;
up1 + 90
07Ch07.qxd
9/27/08
1:19 PM
Page 605
SECTION 7.4
Mohrs Circle
605
Problems 7.4-19 sx 2050 psi, sy 6100 psi, txy 2750 psi Using Mohrs circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-19 sx 2050 psi sy
saver R a 2(sx sx + sy 2
2
6100 psi txy saver txy2 a 4075 psi R
2750 psi
Point P1: s1 s1 Point P2: s2
R + saver 7490 psi R + saver 660 psi ; ;
saver) + txy sx saver
3415 psi 53.63
s2
atana `
`b
(b) MAXIMUM SHEAR STRESSES us1 90 + a 2 90 + us1 us1 us2 4075 psi 3415 psi 18.2 71.8 ; ; ;
(a) PRINCIPAL STRESSES up1 up2 180 2 a 2 a up2 up1 63.2 26.8 ; ;
us2
Point S1: saver tmax R
tmax
Problems 7.4-20 sx 2900 kPa, sy 9100 kPa, txy 3750 kPa Using Mohrs circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-20 sx 2900 kPa sy
saver R a 2(sx sx + sy 2
9100 kPa txy saver 6000 kPa R a
3750 kPa
Point P1: s1 s1 Point P2: s2 s2
R + saver 10865 KPa R + saver 1135 kPa ; ;
saver)2 + txy2 txy sx saver
4865.4393 kPa 50.42
atana `
`b
(b) MAXIMUM SHEAR STRESSES us1 90 + a 2 90 + us1 us1 us2 6000 kPa R tmax 70.2 160.2 ; 4865 kPa ; ; ;
(a) PRINCIPAL STRESSES up1 up2 a + 180 2 a 2 up2 us2 up1 25.2 115.2 ; ;
Point S1: saver tmax
07Ch07.qxd
9/27/08
1:19 PM
Page 606
606
CHAPTER 7
Analysis of Stress and Strain
Problems 7.4-21 sx 11,500 psi, sy 18,250 psi, txy 7200 psi Using Mohrs circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-21 11500 psi sx
txy saver R a 7200 psi 2(sx sx + sy 2
sy
18250 psi
Point P1: s1 s1
R + saver 6923 psi R + saver ; ;
saver
14875 psi R a 7952 psi
Point P2: s2 s2 22827 psi
saver)2 + txy2 txy sx saver
atana `
`b
(b) MAXIMUM SHEAR STRESSES 64.89 us1 us2 32.4 ; ; 270 2 a us1 us2 102.6 192.6 ; 7952 psi ; ;
(a) PRINCIPAL STRESSES up1 up2 a 2 180 2 up1 a up2
90 + us1
Point S1: saver tmax R
14875 psi tmax
57.6
Problems 7.4-22 sx 3.3 MPa, sy 8.9 MPa, txy 14.1 MPa Using Mohrs circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-22 3.3 MPa sx
txy saver R a 14.1 MPa 2(sx sx + sy 2
sy
8.9 MPa
(a) PRINCIPAL STRESSES up1 a + 180 2 a 2 up2 s1 s1 Point P2: s2 up1 33.3 R + saver 18.2 MPa R + saver ; 123.3 ;
saver
2
2.8 MPa up2 R a 15.4 MPa 66.6
saver) + txy sx saver
txy2
Point P1:
atana `
`b
07Ch07.qxd
9/27/08
1:19 PM
Page 607
SECTION 7.4
Mohrs Circle
607
s2
12.6 MPa
;
us2
90 + us1 saver tmax
us2
168.3 ; ;
(b) MAXIMUM SHEAR STRESSES us1 90 + a 2 us1 78.3
Point S1: tmax R
2.8 MPa 15.4 MPa
Problems 7.4-23 sx 800 psi, sy 2200 psi, txy 2900 psi Using Mohrs circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-23 sy 2200 psi txy sx 800 psi sx + sy saver saver 700 psi 2
R a 2(sx saver)2 + txy2 txy sx saver R a 3265 psi 62.65
2900 psi
Point P1: s1 s1 Point P2: s2 s2
R + saver 2565 psi R + saver 3965 psi ; ;
atana `
`b
(b) MAXIMUM SHEAR STRESSES us1 90 + a 2 90 + us1 us1 us2 700 psi R tmax 13.7 76.3 ; 3265 psi ; ;
(a) PRINCIPAL STRESSES a up1 31.3 up1 2 up2 180 + a 2 up2
;
us2 ;
Point S1: saver 121.3 tmax
07Ch07.qxd
9/27/08
1:20 PM
Page 608
608
CHAPTER 7
Analysis of Stress and Strain
Hookes Law for Plane Stress
When solving the problems for Section 7.5, assume that the material is linearly elastic with modulus of elasticity E and Poissons ratio n.
sy
Problem 7.5-1 A rectangular steel plate with thickness t 0.25 in. is subjected to uniform normal stresses x and y, as shown in the figure. Strain gages A and B, oriented in the x and y directions, respectively, are attached to the plate. The gage readings give normal strains ex 0.0010 0.0007 (shortening). (elongation) and ey 0.3, determine the stresses Knowing that E 30 106 and psi x and y and the change t in the thickness of the plate.
y B A O x sx
Probs. 7.5-1 and 7.5-2
t
Solution 7.5-1 Rectangular plate in biaxial stress 0.0007 0.25 in. x 0.0010 y
30 * 106 psi 0.3
Eq. (7-39c): z t E zt (sx + sy) 32.1 * 10 128.5 * 10
6 6
E
SUBSTITUTE NUMERICAL VALUES: Eq. (7-40a): sx E (1 E (1 )2 )2 (x + y) 26,040 psi ;
in.
;
(Decrease in thickness)
Eq. (7-40b): sy (y + x) 13,190 psi ;
Problem 7.5-2 Solve the preceding problem if the thickness of the steel plate is t 10 mm, the gage readings are ex 480 10 6 (elongation) and ey 130 10 6 (elongation), the modulus is E 200 GPa, and Poissons ratio is
0.30.
Solution 7.5-2 Rectangular plate in biaxial stress t 10 mm x 480 * 10 6
y E 130 * 10 200 GPa
6
Eq. (7-40b): sy E (1 )2 (y + x) 60.2 MPa ;
0.3
SUBSTITUTE NUMERICAL VALUES: Eq. (7-40a): sx E (1 )2 (x + y) 114.1 MPa ;
Eq. (7-39c): z t E z t (sx + sy) 2610 * 10 261.4 * 10
6 6
mm
;
(Decrease in thickness)
07Ch07.qxd
9/27/08
1:20 PM
Page 609
SECTION 7.5
Hookes Law for Plane Stress
609
Problem 7.5-3 Assume that the normal strains Px and Py for an element in plane stress (see figure) are measured with strain gages.
(a) Obtain a formula for the normal strain Pz in the z direction in terms of Px, Py, and Poissons ratio . (b) Obtain a formula for the dilatation e in terms of Px, Py, and Poissons ratio .
z
y
sy txy sx
O x
Solution 7.5-3
Given: x, y,
Plane stress
(b) DILATATION Eq. (7-47): e E (1 E (1
2
(a) NORMAL STRAIN z Eq. (7-34c): z Eq. (7-36a): sx Eq. (7-36b): sy (sx + sy) (x + y) (y + x)
1 E
2
(sx + sy)
Substitute sx and sy from above and simplify: e 1 1 2 (x + y) ;
E
2
) )
Substitute sx and sy into the first equation and simplify: z 1 (x + y) ;
sy
Problem 7.5-4 A magnesium plate in biaxial stress is subjected to tensile stresses sx 24 MPa and sy 12 MPa (see figure). The corresponding strains in the plate are x 440 10 6 and 80 10 6. y Determine Poissons ratio and the modulus of elasticity E for the material.
y O x sx
Probs. 7.5-4 through 7.5-7
Solution 7.5-4 Biaxial stress sx 24 MPa sy 12 MPa
x 440 * 10
6
Substitute numerical values:
6
y 1 (s Ex 1 (s Ey
80 * 10
E (440 * 10 E (80 * 10 0.35
6
6
)
24 MPa 12 MPa 45 GPa
(12 MPa) (24 MPa) ;
POISSONS RATION AND MODULUS OF ELASTICITY Eq. (7-39a): x Eq. (7-39b): y sy) sx)
)
Solve simultaneously: E
07Ch07.qxd
9/27/08
1:20 PM
Page 610
610
CHAPTER 7
Analysis of Stress and Strain
Problem 7.5-5 Solve the preceding problem for a steel plate with sx 10,800 psi (tension), sy 300 10 6 (shortening). sion), ex 420 10 6 (elongation), and ey
5400 psi (compres-
Solution 7.5-5 Biaxial stress sx 10,800 psi sy 5400 psi
x 420 * 10
6
Substitute numerical values:
6
y
300 * 10
E (420 * 10 E(
6
)
6
10,800 psi ) 5400 psi
(
5400 psi) (10,800 psi)
POISSONS RATIO AND MODULUS OF ELASTICITY Eq. (7-39a): x 1 (s Ex 1 (s Ey sy)
300 * 10 1/3 E
Solve simultaneously: 30 * 106 psi ;
Eq. (7-39b): y
sx)
Problem 7.5-6 A rectangular plate in biaxial stress (see figure) is subjected to normal stresses x 90 MPa (tension) and 20 MPa (compression). The plate has dimensions 400 * 800 * 20 mm and is made of steel with E 200 GPa and y 0.30.
(a) Determine the maximum in-plane shear strain gmax in the plate. (b) Determine the change t in the thickness of the plate. (c) Determine the change V in the volume of the plate.
Solution 7.5-6 Biaxial stress sx 90 MPa sy 20 MPa E 200 GPa 0.30 Dimensions of Plate: 400 mm * 800 mm * 20 mm Shear Modulus (Eq. 7-38): E G 76.923 GPa 2(1 + )
(a) MAXIMUM IN-PLANE SHEAR STRAIN Principal stresses: s1 Eq. (7-26): tmax Eq. (7-35): gmax s1 2 tmax G 90 MPa s2 s2 20 MPa
(b) CHANGE IN THICKNESS Eq. (7-39c): z t z t E (sx + sy)
6
105 * 10 ;
6
2100 * 10
mm
(Decrease in thickness) (c) CHANGE IN VOLUME From Eq. (7-47): V V0 (400)(800)(20) 1 E 2 V0 a 1 E 2 b(sx + sy)
55.0 MPa
6
6.4 * 106 mm3 140 * 10
6 6
715 * 10
;
Also, a V
b(sx + sy)
3
(6.4 * 106 mm3)(140 * 10 896 mm ;
)
(Increase in volume)
07Ch07.qxd
9/27/08
1:20 PM
Page 611
SECTION 7.5
Hookes Law for Plane Stress
611
Problem 7.5-7 Solve the preceding problem for an aluminum plate with sx 12,000 psi (tension), sy 0.33. (compression), dimensions 20 30 0.5 in., E 10.5 106 psi, and
3,000 psi
Solution 7.5-7 Biaxial stress 3,000 psi sx 12,000 psi sy
E 10.5 * 10 psi
6
(b) CHANGE IN THICKNESS Eq. (7-39c): z E (sx + sy)
6
0.33
Dimensions of Plate: 20 in. * 30 in. * 0.5 in. Shear Modulus (Eq. 7-38): G E 2(1 + ) 3.9474 * 106 psi
282.9 * 10 t z t 141 * 10
6
in.
;
(Decrease in thickness) (c) CHANGE IN VOLUME From Eq. (7-47): V V0 (20)(30)(0.5) 1 E 2 V0 a 1 E 2 b(sx + sy)
(a) MAXIMUM IN-PLANE SHEAR STRAIN Principal stresses: s1 s2 Eq. (7-26): tmax Eq. (7-35): gmax s1 2 tmax G 12,000 psi 3,000 psi s2 7,500 psi
6
300 in.3 291.4 * 10
6 6
1,900 * 10
;
Also, a V
b(sx + sy)
3
(300 in.3)(291.4 * 10 0.0874 in. ;
)
(Increase in volume)
Problem 7.5-8 A brass cube 50 mm on each edge is compressed in two perpendicular directions by forces P 175 kN (see figure). Calculate the change V in the volume of the cube and the strain energy U stored in the cube, assuming E 100 GPa and 0.34.
P = 175 kN
P = 175 kN
Solution 7.5-8
Biaxial stress-cube
sx sy P b
2
(175 kN) (50 mm)2
70.0 MPa
CHANGE IN VOLUME Eq. (7-47): e V0 V Side b E 50 mm P 175 kN 0.34 ( Brass) 100 GPa 1 E b3 eV0 2 (sx + sy) 448 * 10 125 * 103mm3 ;
6
(50 mm)3 56 mm3
(Decrease in volume)
07Ch07.qxd
9/27/08
1:20 PM
Page 612
612
CHAPTER 7
Analysis of Stress and Strain
STRAIN ENERGY Eq. (7-50): u 1 (s2 + s2 y 2E x 0.03234 MPa 2 sxsy)
U
uV0 4.04 J
(0.03234 MPa)(125 * 103 mm3) ;
0.1) Problem 7.5-9 A 4.0-inch cube of concrete (E 3.0 * 106 psi, is compressed in biaxial stress by means of a framework that is loaded as shown in the figure. Assuming that each load F equals 20 k, determine the change V in the volume of the cube and the strain energy U stored in the cube.
F
F
Solution 7.5-9
Biaxial stress concrete cube
CHANGE IN VOLUME 1 E b3 eV0 2
b E F
4 in. 3.0 * 10 psi 0.1 20 kips
6
Eq. (7-47): e V0 V
(sx + sy) 64 in.3
0.0009429
(4 in.)3
0.0603 in.3
;
(Decrease in volume) STRAIN ENERGY
Joint A:
Eq. (7-50): u
P
F12 28.28 kips P b2
U uV0
1 (s2 + s2 y 2E x 0.9377 psi ;
2 sxsy)
60.0 in.-lb
sx
sy
1768 psi
Problem 7.5-10 A square plate of width b and thickness t is loaded by normal forces Px and Py, and by shear forces V, as shown in the figure. These forces produce uniformly distributed stresses acting on the side faces of the place. Calculate the change V in the volume of the plate and the strain energy U stored in the plate if the dimensions are b 600 mm and t 40 mm, the plate is made of magnesium with E 45 GPa and v 0.35, and the forces are Px 480 kN, Py 180 kN, and V 120 kN.
Py t V y Px V V Py bO b V x Px
Probs. 7.5-10 and 7.5-11
07Ch07.qxd
9/27/08
1:20 PM
Page 613
SECTION 7.5
Hookes Law for Plane Stress
613
Solution 7.5-10 b 600 mm
E Px Py V 45 GPa 480 kN 180 kN 120 kN
Square plate in plane stress t 40 mm
v sx sy txy 0.35 (magnesium) Px bt Py bt V bt 20.0 MPa 7.5 MPa 5.0 MPa
V0 V
b 2t eV0
14.4 * 106 mm3 2640 mm3 ;
(Increase in volume) STRAIN ENERGY Eq. (7-50): u G 12 (sx + s2 y 2E E 2(1 + ) 2 sxsy) + 16.667 GPa t2 xy 2G
CHANGE IN VOLUME Eq. (7-47): e 1 E 2 (sx + sy) 183.33 * 10
6
Substitute numerical values: u U 4653 Pa uV0 67.0 N # m 67.0 J ;
Problem 7.5-11 Solve the preceding problem for an aluminum plate with b Px 90 k, Py 20 k, and V 15 k.
12 in., t
1.0 in., E
10,600 ksi,
0.33,
Solution 7.5-11 b 12.0 in.
E Px Py V 10,600 ksi 90 k 20 k 15 k
Square plate in plane stress t 1.0 in.
0.33 (aluminum) sx sy txy Px bt Py bt V bt 7500 psi
STRAIN ENERGY Eq. (7-50): u G 1 (s2 + s2 y 2E x E 2(1 + ) 2 sxsy) + t2 xy 2G
3985 ksi
1667 psi 1250 psi
Substitute numerical values: u U 2.591 psi uV0 373 in.-lb ;
CHANGE IN VOLUME Eq. (7-47): e V0 V 1 E b 2t eV0 2 (sx + sy) 294 * 10
6
144 in.3 0.0423 in.3 ;
(Increase in volume)
07Ch07.qxd
9/27/08
1:20 PM
Page 614
614
CHAPTER 7
Analysis of Stress and Strain
Problem 7.5-12 A circle of diameter d 200 mm is etched on a brass plate (see figure). The plate has dimensions 400 400 20 mm. Forces are applied to the plate, producing uniformly distributed normal stresses x 42 MPa and y 14 MPa. Calculate the following quantities: (a) the change in length ac of diameter ac; (b) the change in length bd of diameter bd; (c) the change t in the thickness of the plate; (d) the change V in the volume of the plate, and (e) the strain energy U stored in the plate. (Assume E 100 GPa and v 0.34.)
z
y
sy
sx
d a b x sy c sx
sx
Solution 7.5-12 Plate in biaxial stress 42 MPa sy 14 MPa
(c) CHANGE IN THICKNESS Eq. (7-39c): z (s + sy) Ex 190.4 * 10 6 0.00381 mm ;
Dimensions: 400 * 400 * 20 (mm) Diameter of circle: d 200 mm E 100 GPa 0.34 (Brass)
t z t (decrease) (d) CHANGE IN VOLUME Eq. (7-47): e V0 V 1 E 2 (sx + sy)
(a) CHANGE IN LENGTH OF DIAMETER IN x DIRECTION Eq. (7-39a): x ac 1 (s Ex sy) 372.4 * 10 ;
6
x d 0.0745 mm (increase)
179.2 * 10 3.2 * 106 mm3
6
(b) CHANGE IN LENGTH OF DIAMETER IN y DIRECTION Eq. (7-39b): y bd 1 (s Ey y d 560 * 10 (decrease)
6
(400)(400)(20) eV0
sx)
2.80 * 10
6
573 mm3 ; (increase)
(e) STRAIN ENERGY mm ; Eq. (7-50): u 12 (s + s2 y 2E x 7.801 * 10 U uV0
3
2 sx sy) MPa 25.0 J ;
25.0 N # m
07Ch07.qxd
9/27/08
1:22 PM
Page 615
SECTION 7.6
Triaxial Stress
615
Triaxial Stress
When solving the problems for Section 7.6, assume that the material is linearly elastic with modulus of elasticity E and Poissons ratio n.
y a c b O x
Problem 7.6-1 An element of aluminum in the form of a rectangular parallelepiped (see figure) of dimensions a 6.0 in., b 4.0 in, and c 3.0 in. is subjected to triaxial stresses sx 12,000 psi, sy 4,000 psi, and sz 1,000 psi acting on the x, y, and z faces, respectively. Determine the following quantities: (a) the maximum shear stress tmax in the material; (b) the changes a, b, and c in the dimensions of the element; (c) the change V in the volume; and (d) the strain energy U 0.33.) stored in the element. (Assume E 10,400 ksi and
z
Probs. 7.6-1 and 7.6-2
Solution 7.6-1 Triaxial stress 4,000 psi sx 12,000 psi sy
sz a E 1,000 psi 6.0 in. b 10,400 ksi 4.0 in. c 0.33 3.0 in. ( aluminum)
a b c
ax by cz
0.0079 in. ( increase) 0.0029 in. ( decrease) 0.0011 in. ( decrease)
(a) MAXIMUM SHEAR STRESS s1 s3 tmax 12,000 psi s2 4,000 psi s1 2 s3 8,000 psi ; 1,000 psi
(c) CHANGE IN VOLUME Eq. (7-56): e V V 1 E abc e (abc) 2 (sx + sy + sz)
M
;
228.8 * 10
6
0.0165 in.3 ( increase)
;
(b) CHANGES IN DIMENSIONS Eq. (7-53 a): x sx E (d) STRAIN ENERGY E (sy + sz) Eq. (7-57a): u
6
1312.5 * 10 Eq. (7- 53 b): y sy E sz E E E
1 (s + sy y + sz z) 2 xx 9.517 psi
(sz + sx)
6
U
u (abc)
685 in.-lb
;
733.7 * 10 Eq. (7-53 c): z
(sx + sy)
6
350.0 * 10
07Ch07.qxd
9/27/08
1:22 PM
Page 616
616
CHAPTER 7
Analysis of Stress and Strain
Problem 7.6-2 Solve the preceding problem if the element is steel (E = 200 GPA, 0.30) with dimensions a = 300 mm, b = 150 mm, and c = 150 mm and the stresses are sx 60 MPa, sy 40 MPa, and sz 40 MPa.
Solution 7.6-2 Triaxial stress 60 MPa sy 40 MPa sx
sz a E 40 MPa 300 mm 200 GPa b 150 mm 0.30 c (steel) 150 mm
a b c
ax by cz
0.0540 mm (decrease) 0.0075 mm (decrease) 0.0075 mm. (decrease)
(a) MAXIMUM SHEAR STRESS s1 s3 tmax 40 MPa 60 MPa s1 2 s3 s2 40 MPa
(c) CHANGE IN VOLUME Eq. (7-56): e V 1 E abc e(abc) 2 (sx + sy + sz)
M
;
280.0 * 10
6
10.0 MPa
; V
1890 mm3 (decrease)
;
(b) CHANGES IN DIMENSIONS Eq. (7-53 a): x Eq. (7-53 b): y Eq. (7-53 c): z sx E sy E sz E E E E (sy + sz) (sz + sx) (sx + sy) 180.0 * 10 50.0 * 10 50.0 * 10
6 6
(d) STRAIN ENERGY Eq. (7-57 a): u 1 (s + sy y + sz z) 2 xx 0.00740 MPa 50.0 N # m 50.0 J ;
U
6
u (abc)
Problem 7.6-3 A cube of cast iron with sides of length a = 4.0 in. (see figure) is tested in a laboratory under triaxial stress. Gages mounted on the testing machine show that the compressive strains in the material are Px 225 * 10 6 and Py Pz 37.5 * 10 6. Determine the following quantities: (a) the normal stresses sx, sy, and sz acting on the x, y, and z faces of the cube; (b) the maximum shear stress tmax in the material; (c) the change V in the volume of the cube; and (d) the strain energy U stored in the cube. (Assume E = 14,000 0.25.) ksi and
z
y a a a O x
Probs. 7.6-3 and 7.6-4
07Ch07.qxd
9/27/08
1:22 PM
Page 617
SECTION 7.6
Triaxial Stress
617
Solution 7.6-3
x z E
Triaxial stress (cube)
6 6
225 * 10 37.5 * 10 14,000 ksi
y a 0.25
37.5 * 10 4.0 in. (cast iron)
6
(c) CHANGE IN VOLUME Eq. (7-55): e V V a
3
x + y + z
0.000300 ;
ea 3
0.0192 in.3 ( decrease)
(a) NORMAL STRESSES Eq. (7-54a): sx E (1 + )(1 4200 psi 2) ; [(1 )x + (y + z)]
(d) STRAIN ENERGY Eq. (7-57a): u 1 (sx x + sy y + sz z) 2 0.55125 psi U ua
3
In a similar manner, Eqs. (7-54 b and c) give 2100 psi sz 2100 psi ; sy (b) MAXIMUM SHEAR STRESS s1 s3 tmax 2100 psi s2 4200 psi s1 2 s3 1050 psi ; 2100 psi
35.3 in.-lb
;
Problem 7.6-4 Solve the preceding problem if the cube is granite (E 60 GPa, 720 * 10 6 and Py Pz 270 * 10 6. compressive strains Px
0.25) with dimensions a = 75 mm and
Solution 7.6-4
x z
Triaxial stress (cube)
6 6
720 * 10 270 * 10 0.25
y a
270 * 10 75 mm E
6
tmax 60 GPa
s1 2
s3
10.8 MPa
;
(Granite)
(c) CHANGE IN VOLUME Eq. (7-55): e V V 2) [(1 )x + (x + z)] (d) STRAIN ENERGY Eq. (7-57 a): u 1 (s + syy + szz) 2 xx 0.03499 MPa U 43.2 MPa ua
3
(a) NORMAL STRESSES Eq.(7-54a): sx E (1 + )(1 64.8 MPa
x + y + z
1260 * 10 ;
6
a3 ea 3 532 mm3 ( decrease)
;
In a similar manner, Eqs. (7-54 b and c) give 43.2 MPa sz 43.2 MPa ; sy (b) MAXIMUM SHEAR STESS s1 s3 43.2 MPa s2 64.8 MPa
34.99 kPa ;
14.8 N # m
14.8 J
07Ch07.qxd
9/27/08
1:22 PM
Page 618
618
CHAPTER 7
Analysis of Stress and Strain
Problem 7.6-5 An element of aluminum in triaxial stress (see figure) is subjected to stresses sx 5200 psi (tension), sy 4750 psi (compression), and sz 3090 psi (compression). It is also known that the normal strains in the x and y directions are Px 7138.8 * 10 6 (elongation) and Py 502.3 * 10 6 (shortsx ening). What is the bulk modulus K for the aluminum?
z
y
sy sz sx
O x sy
sz
Probs. 7.6-5 and 7.6-6
Solution 7.6-5 Triaxial stress (bulk modulus) sx 5200 psi sy 4750 psi
sz y Find K. Eq. (7-53 a): x Eq. (7-53 b): y sx E sy E E E (sy + sz) (sx + sy) 3090 psi x 502.3 * 10
6
Substitute numerical values and rearrange: (713.8 * 10
6
713.8 * 10
6
)E
6
5200 + 7840 4750 2110
(1) (2)
( 502.3 * 10 Units: E = psi
)E
Solve simultaneously Eqs. (1) and (2): E 10.801 * 106 psi E 3(1 2) 0.3202 10.0 * 10
6
Eq. (7-16): K
psi
;
Problem 7.6-6 Solve the preceding problem if the material is nylon subjected to compressive stresses sx 4.5 MPa , sy 3.6 MPa , and sz 2.1 MPa, and the normal strains are Px 740 * 10 6 and 320 * 10 6 (shortenings). Py
Solution 7.6-6 Triaxial stress (bulk modulus) sx 4.5 MPa sy 3.6 MPa
sz y Find K. Eq. (7-53 a): x Eq. (7-53 b): y sx E sy E E E (sy + sz) (sz + sx) 2.1 MPa x 320 * 10
6
Substitute numerical values and rearrange: ( 740 * 10 ( 320 * 10
6 6
740 * 10
6
)E )E
4.5 + 5.7 3.6 + 6.6
(1) (2)
Units: E = MPa Solve simultaneously Eqs. (1) and (2): E 3,000 MPa 3.0 GPa E 3(1 2) 0.40 5.0 GPa ;
Eq. (7-16): K
07Ch07.qxd
9/27/08
1:22 PM
Page 619
SECTION 7.6
Triaxial Stress
619
Problem 7.6-7 A rubber cylinder R of length L and cross-sectional area A is compressed inside a steel cylinder S by a force F that applies a uniformly distributed pressure to the rubber (see figure).
(a) Derive a formula for the lateral pressure p between the rubber and the steel. (Disregard friction between the rubber and the steel, and assume that the steel cylinder is rigid when compared to the rubber.) (b) Derive a formula for the shortening d of the rubber cylinder.
S
F
F
S
R
L
Solution 7.6-7
Rubber cylinder
Solve for p: p (b) SHORTENING Eq. (7-53 b): y sx sz x z p sy p 0 y F A sy E F EA E E (sz + sx) ( 2p) v 1 F ab vA ;
Substitute for p and simplify: F (1 + )( EA 1 1+2)
(a) LATERAL PRESSURE Eq. (7-53 a): x sx E p E (sy + sz)
(Positive y represents an increase in strain, that is, elongation.) d d yL (1 + )(1 2 ) FL a b (1 ) EA ;
or 0
a
F A
pb
(Positive d represents a shortening of the rubber cylinder.)
Problem 7.6-8 A block R of rubber is confined between plane parallel walls of a steel block S (see figure). A uniformly distributed pressure p0 is applied to the top of the rubber block by a force F.
(a) Derive a formula for the lateral pressure p between the rubber and the steel. (Disregard friction between the rubber and the steel, and assume that the steel block is rigid when compared to the rubber.) (b) Derive a formula for the dilatation e of the rubber. (c) Derive a formula for the strain-energy density u of the rubber.
F
F
S S
R
07Ch07.qxd
9/27/08
1:22 PM
Page 620
620
CHAPTER 7
Analysis of Stress and Strain
Solution 7.6-8
Block of rubber
(b) DILATATION Eq. (7-56): e 1 E 1 E Substitute for p: sx sy x 0 p p0 sz 0 (c) STRAIN ENERGY DENSITY Eq. (7-57b): u p0 ; 1 (s2 + s2 + s2) x y z 2E
2
2
(sx + sy + sz) 2 ( p p 0)
e
(1 + )(1 E
2 )p0
;
y Z 0 z Z 0
(a) LATERAL PRESSURE Eq. (7-53 a): x
OR
sx E (
E p 0)
(sy + sz) p
v (sx sy + sx sz + sy sz) E
0
p
Substitute for sx, sy, sz, and p: u (1 2E )p 2 0 ;
Problem 7.6-9 A solid spherical ball of brass (E 15 * 106 psi 0.34) is lowered into the ocean to a depth of 10,000 ft. The diameter of the ball is 11.0 in. Determine the decrease d in diameter, the decrease V in volume, and the strain energy U of the ball.
Solution 7.6-9
E 15 * 10
6
Brass sphere
psi 0.34 10,000 ft DECREASE IN VOLUME Eq. (7-60): e V0 V 4431 psi eV0 30 43 pr 3 283.6 * 10
6
Lowered in the ocean to depth h Diameter d Sea water: g Pressure: s0 11.0 in. 63.8 lb/ft gh
3
4 11.0 in. 3 (p)a b 3 2 ;
696.9 in.3
638,000 lb/ft2
0.198 in.3 (decrease)
DECREASE IN DIAMETER Eq. (7-59): 0 d s0 (1 E 2) 94.53 * 10
3 6
STRAIN ENERGY Use Eq. (7-57 b) with sx u U 3(1 uV0 2 2E )s2 0 sy sz s0:
0d 1.04 * 10 (decrease)
in.
;
0.6283 psi ;
438 in.-lb
07Ch07.qxd
9/27/08
1:22 PM
Page 621
SECTION 7.6
Triaxial Stress
621
Problem 7.6-10 A solid steel sphere (E reduced by 0.4%.
210 GPa,
= 0.3) is subjected to hydrostatic pressure p such that its volume is
(a) Calculate the pressure p. (b) Calculate the volume modulus of elasticity K for the steel. (c) Calculate the strain energy U stored in the sphere if its diameter is d
150 mm
E
Solution 7.6-10 210 GPa
0.004 V0
Steel sphere 0.3
(b) VOLUME MODULUS OF ELASTICITY Eq. (7-63): K s0 E 700 MPa 0.004 175 GPa ;
Hydrostatic Pressure. V0 = Initial volume V
Dilatation: e (a) PRESSURE Eq.(7-60): e or s0
V V0
0.004
(c) STRAIN ENERGY (d = diameter) d = 150 mm r = 75 mm From Eq. (7-57b) with sx sy u V0 U ; 3(1 2 )s2 0 2E 1.40 MPa
6
sz
s0:
3s0(1 2) E Ee 3(1 s0 2) 700 MPa
4pr 3 3 uV0
1767 * 10 2470 N # m
m3 ;
2470 J
Pressure p
700 MPa
Problem 7.6-11 A solid bronze sphere (volume modulus of elasticity K 14.5 106 psi) is suddenly heated around its outer surface. The tendency of the heated part of the sphere to expand produces uniform tension in all directions at the center of the sphere. If the stress at the center is 12,000 psi, what is the strain? Also, calculate the unit volume change e and the strain-energy density u at the center.
Solution 7.6-11
K s0
6
Bronze sphere (heated)
UNIT VOLUME CHANGE AT THE CENTER Eq. (7-62): e s0 K 828 * 10
6
14.5 * 10 psi 12,000 psi (tension at the center)
;
STRAIN AT THE CENTER OF THE SPHERE s0 (1 2) Eq. (7-59): 0 E E Eq. (7-61): K 3(1 2) Combine the two equations: 0 s0 3K 276 * 10
6
STRAIN ENERGY DENSITY AT THE CENTER Eq. (7-57b) with sx u u 2 2E 4.97 psi ; 3(1 )s2 0 sy
2 s0
sz
s0:
2K
;
07Ch07.qxd
9/27/08
1:24 PM
Page 622
622
CHAPTER 7
Analysis of Stress and Strain
Plane Strain
When solving the problems for Section 7.7, consider only the in-plane strains (the strains in the xy plane) unless stated otherwise. Use the transformation equations of plane strain except when Mohrs circle is specified (Problems 7.7-23 through 7.7-28).
y
sy
h b
sx x
Problem 7.7-1 A thin rectangular plate in biaxial stress is subjected to stresses sx and sy, as shown in part (a) of the figure on the next page. The width and height of the plate are b 8.0 in. and h 4.0 in., respectively. Measurements show that the normal strains in the x and y directions are Px 195 * 10 6 and Py 125 * 10 6, respectively. With reference to part (b) of the figure, which shows a two-dimensional view of the plate, determine the following quantities: (a) the increase d in the length of diagonal Od; (b) the change f in the angle f between diagonal Od and the x axis; and (c) the change c in the angle c between diagonal Od and the y axis.
z (a) y d c h f O b (b) x
Probs. 7.7-1 and 7.7-2
Solution 7.7-1
Plate in biaxial stress
For u d f x1L d 26.57, x1 130.98 * 10 ;
6
0.00117 in.
(b) CHANGE IN ANGLE f Eq. (7-68): a For u b y f Ld 8.0 in. h arctan b 1b 2 + h2 125 * 10 h
6
(x 26.57: a
y) sinu cosu 128.0 * 10
gxy sin2u
6
f
rad
4.0 in. gxy 0
x
195 * 10
6
Minus sign means line Od rotates clockwise (angle f decreases). f 128 * 10
6
rad (decrease)
;
26.57 8.944 in.
(c) CHANGE IN ANGLE c Angle c increases the same amount that f decreases. c gxy 2 sin 2u 128 * 10
6
(a) INCREASE IN LENGTH OF DIAGONAL x1 x + y 2 + x 2 y cos 2u +
rad (increase)
;
07Ch07.qxd
9/27/08
1:24 PM
Page 623
SECTION 7.7
Plane Strain
623
6
Problem 7.7-2 Solve the preceding problem if b
160 mm, h
60 mm, Px
410 * 10
6
, and Py
320 * 10
.
Solution 7.7-2
Plate in biaxial stress
For u d f 20.56: x1 319.97 * 10 ;
6
x1 L d
0.0547 mm
(b) CHANGE IN ANGLE f Eq. (7-68): a For u b y f Ld 160 mm 320 * 10 h
6
(x 20.56: a
y) sin u cos u
gxy sin2u
6
f
240.0 * 10
rad
60 mm gxy 0
x
410 * 10
6
Minus sign means line Od rotates clockwise (angle f decreases.) f 240 * 10
6
h 20.56 arctan b 1b 2 + h2 170.88 mm
rad (decrease)
;
(c) CHANGE IN ANGLE c Angle c increases the same amount that f decreases. c gxy 2 sin 2u 240 * 10
6
(a) INCREASE IN LENGTH OF DIAGONAL x1 x + y 2 + x 2 y cos 2u +
rad (increase)
;
Problem 7.7-3 A thin square plate in biaxial stress is subjected to stresses sx and sy, as shown in part (a) of the figure . The width of the plate is b 12.0 in. Measurements show that the normal strains in the x and y directions are Px 427 * 10 6 and Py 113 * 10 6, respectively. With reference to part (b) of the figure, which shows a two-dimensional view of the plate, determine the following quantities: (a) the increase d in the length of diagonal Od; (b) the change f in the angle f between diagonal Od and the x axis; and (c) the shear strain g associated with diagonals Od and cf (that is, find the decrease in angle ced).
y
sy
y c sx d
b b
b f
e f x
x z (a)
O
b (b)
PROBS. 7.7-3 and 7.7-4
07Ch07.qxd
9/27/08
1:24 PM
Page 624
624
CHAPTER 7
Analysis of Stress and Strain
Solution 7.7-3
Square plate in biaxial stress
(b) CHANGE IN ANGLE f Eq. (7-68): a For u f (x 45: a y) sin u cos u 157 * 10
6
gxy sin2u
rad
Minus sign means line Od rotates clockwise (angle f decreases.) f 157 * 10
6
rad (decrease)
;
b y f Ld
12.0 in. 113 * 10 45 b 12
6
x gxy 0
427 * 10
6
(c) SHEAR STRAIN BETWEEN DIAGONALS Eq. (7-71b): For u g gxy 2 sin 2u f gx1y1 2 45: gx1y1
6
x 2
y
sin 2u +
gxy 2
6
cos 2u
314 * 10 ;
rad
16.97 in.
(Negative strain means angle ced increases) 314 * 10 rad
(a) INCREASE IN LENGTH OF DIAGONAL x1 For u d x + y 2 f x1L d + 45: x1 x 2 y cos 2u +
6
270 * 10
0.00458 in.
;
Problem 7.7-4 Solve the preceding problem if b
225 mm, Px
845 * 10
6
, and Py
211 * 10
6
.
Solution 7.7-4
Square plate in biaxial stress
(a) INCREASE IN LENGTH OF DIAGONAL x1 For u d x + y 2 f x1L d + 45: x1 x 2 y cos 2u +
6
gxy 2
sin 2u
528 * 10 ;
0.168 mm
(b) CHANGE IN ANGLE f b 12 Eq. (7-68): a x
6
(x 45: a
y) sin u cos u 317 * 10
6
gxy sin2u rad
b y Ld
225 mm 211 * 10
845 * 10 f 45
6
For u 0
f
gxy
318.2 mm
Minus sign means line Od rotates clockwise (angle f decreases.) f 317 * 10
6
rad (decrease)
;
07Ch07.qxd
9/27/08
1:24 PM
Page 625
SECTION 7.7
Plane Strain
625
(c) SHEAR STRAIN BETWEEN DIAGONALS Eq. (7- 71b): g x1y1 2 x 2 y sin 2u + gxy 2 cos 2u
For u g
f
45: gx1y1
6
634 * 10
6
rad
(Negative strain means angle ced increases) 634 * 10 rad
Problem 7.7-5 An element of material subjected to plane strain (see figure) has strains as follows: Px 220 * 10 6, Py 480 * 10 6 and , gxy 180 * 10 6. Calculate the strains for an element oriented at an angle u 50 and show these strains on a sketch of a properly oriented element.
y ey gxy 1
O
1
ex
x
Probs. 7.7-5 through 7.7-10
Solution 7.7-5
x gxy x1 gx1 y1 2 y1 For u x1 y1 50:
Element in plane strain
6 6
220 * 10 180 * 10 x + y 2 x 2 x + y 461 * 10 239 * 10 + y
y y 2
480 * 10
6
x
cos 2u + g xy 2
gxy 2
sin 2u
sin 2u + x1
6 6
cos 2u
gx1y1
225 * 10
6
Problem 7.7-6 Solve the preceding problem for the following data: Px and u 37.5.
420 * 10
6
, Py
170 * 10
6
, gxy
310 * 10 6,
07Ch07.qxd
9/27/08
1:24 PM
Page 626
626
CHAPTER 7
Analysis of Stress and Strain
Solution 7.7-6 Element in plane strain x 420 * 10 6 y 170 * 10
gxy x1 gx1y1 2 y1 For u x1 y1 310 * 10 x + y 2 x 2 x + y 37.5: 351 * 10
6 6 6
6
x + y 2
y
cos 2u + gxy 2
gxy 2
sin 2u
sin 2u + x1
cos 2u
gx1y1
490 * 10
6
101 * 10
Problem 7.7-7 The strains for an element of material in plane strain (see figure) are as follows: Px 480 * 10 6, Py 140 * 10 6, and gxy 350 * 10 6. Determine the principal strains and maximum shear strains, and show these strains on sketches of properly oriented elements.
Solution 7.7-7
x gxy 480 * 10
Element in plane strain
6 6
y
140 * 10
6
350 * 10
PRINCIPAL STRAINS 1, 2 x + y 2 ; 310 * 10 1 tan 2up 2up up For up x1 554 * 10 gxy x y
A
6 6
a
x
gxy 2 y 2 b +a b 2 2
6 6
; 244 * 10 2 1.0294 134.2 67.1 y 2
66 * 10
MAXIMUM SHEAR STRAINS gmax 2 gmax a x 2 y b +a
2
A
gxy 2
b
2
45.8 and 22.9 and 22.9: x + y 2 22.9 67.1 2 +
6
244 * 10 488 * 10 up 1 45 488 * 10 us1 + 90 us1 sin 2u gmax us2 ; ; gmin aver
6 6
x
cos 2u +
gxy 2
6
67.9 or 112.1
6
; 22.1
6
554 * 10 up1 up2
1
554 * 10
6
488 * 10 x + y 2
;
6
66 * 10
310 * 10
07Ch07.qxd
9/27/08
1:24 PM
Page 627
SECTION 7.7
Plane Strain
627
Problem 7.7-8 Solve the preceding problem for the following strains: Px gxy 360 * 10 6.
120 * 10
6
, Py
450 * 10
6
, and
Solution 7.7-8
x gxy 120 * 10
Element in plane strain
6 6
y
450 * 10
6
MAXIMUM SHEAR STRAINS gmax 2 a x 2
6 6
360 * 10
PRINCIPAL STRAINS 1,2 x + y 2 ;
A
6
A
y
b +a
2
gxy 2
b
2
a
6
x
gxy 2 y 2 b +a b 2 2
6 6
337 * 10 gmax us1 gmax us2 gmin aver 674 * 10 up 1 us1 45 674 * 10 90
165 * 10 1 tan 2up 2up up For up x1 172 * 10 gxy x y
; 377 * 10 2
118.9
6
502 * 10
; 28.9
6
0.6316
674 * 10 x + y 2
; 165 * 10
6
327.7 and 147.7 163.9 and 73.9 163.9: x + y 2 163.9 73.9 +
6
x 2 1 2
y
cos 2u +
gxy 2
6 6
sin 2u
172 * 10 up1 up2
172 * 10
; ;
502 * 10
07Ch07.qxd
9/27/08
1:24 PM
Page 628
628
CHAPTER 7
Analysis of Stress and Strain
Problem 7.7-9 A element of material in plane strain (see figure) is subjected to strains Px 480 * 10 6, Py 70 * 10 6, and gxy 420 * 10 6. Determine the following quantities: (a) the strains for an element oriented at an angle u 75, (b) the principal strains, and (c) the maximum shear strains. Show the results on sketches of properly oriented element.
Solution 7.7-9
x gxy x1 gx1y1 2 y1 For u x1 y1 75:
Element in plane strain
6 6
480 * 10 420 * 10 x + y 2 x 2 x + y 202 * 10 348 * 10 + y
y y 2
70 * 10
6
For up x1
22.85: x + y 2 22.8 112.8
x +
6
y 2
cos 2u +
gxy 2
sin 2u
x
cos 2u + gxy 2
gxy 2
sin 2u up1
568 * 10 up2
1 2
568 * 10 18 * 10
6 6
; ;
sin 2u + x1
6 6
cos 2u
gx1y1
569 * 10
6
MAXIMUM SHEAR STRAINS gmax 2 gmax us1 PRINCIPAL STRAINS 1,2 x + y 2 ; 275 * 10 1 tan 2up 2up up 568 * 10 gxy x y gmax us2 a x gxy 2 y 2 b +a b 2 2
6
A
a
x 2
y
6
b +a
2
gxy 2
b
2
293 * 10
6
587 * 10 up1 45 587 * 10 us1 + 90
22.2 or 157.8
6
A
6 6
; 67.8
6
gmin aver
6
587 * 10 x + y 2
;
6
; 293 * 10 2
275 * 10
18 * 10
1.0244
45.69 and 225.69 22.85 and 112.85
07Ch07.qxd
9/27/08
1:24 PM
Page 629
SECTION 7.7
Plane Strain
629
Problem 7.7-10 Solve the preceding problem for the following data: Px gxy 780 * 10 6, and u 45.
1120 * 10
6
, Py
430 * 10
6
,
Solution 7.7-10
x gxy x1 gx1y1 2 y1 For u x1 y1 45:
Element in plane strain
6
1120 * 10 780 * 10 x + y 2 x 2 x + y + y
6
y y
430 * 10
6
up1 up2
65.7 155.7
1 2
254 * 10 1296 * 10
6 6
; ;
x 2
cos 2u + gxy 2
gxy 2
sin 2u
sin 2u + x1
6 6
cos 2u
385 * 10 1165 * 10
gx1y1
690 * 10
6
MAXIMUM SHEAR STRAINS gmax 2 gmax PRINCIPAL STRAINS x + y x gxy 2 y 2 1, 2 ; a b +a b 2 A 2 2 775 * 10 1 tan 2up 2up up For up x1 x 254 * 10 gxy y
6 6
A
a
x 2
y
6
b +a
2
gxy 2
b
2
521 * 10 1041 * 10 up1 45 1041 * 10 us1 + 90 x + y 2 us1 gmax us2 gmin
6
6
20.7
6
;
6
110.7 ;
6
1041 * 10
; 521 * 10 2
6
1296 * 10
aver
775 * 10
1.1304
131.5 and 311.5 65.7 and 155.7 65.7: x + y 2 +
6
x 2
y
cos 2u +
gxy 2
sin 2u
254 * 10
07Ch07.qxd
9/27/08
1:24 PM
Page 630
630
CHAPTER 7
Analysis of Stress and Strain
Problem 7.7-11 A steel plate with modulus of elasticity E 30 * 106 psi and 0.30 is loaded in biaxial stress by normal stresses sx and sy Poissons ratio (see figure). A strain gage is bonded to the plate at an angle f 30. If the stress sx is 18,000 psi and the strain measured by the gage is P 407 * 10 6, what is the maximum in-plane shear stress (tmax)xy and shear strain (gmax)xy? What is the maximum shear strain (gmax)xz in the xz plane? What is the maximum shear strain (gmax)yz in the yz plane?
y
sy
f
sx
x z
Probs. 7.7-11 and 7.7-12
Solution 7.7-11 Steel plate in biaxial stress gxy 0 sy ? sx 18,000 psi
E 30 * 106 psi 30 0.30 407 * 10
6
MAXIMUM IN-PLANE SHEAR STRESS (tmax)xy sx 2 sy 7800 psi ;
Strain gage: f
UNITS: All stresses in psi. STRAIN IN BIAXIAL STRESS (EQS. 7-39) x y z 1 (s Ex 1 (sy E E (sx sy) sx ) sy) 1 30 * 10 1 30 * 106 0.3
6
STRAINS FROM EQS. (1), (2), AND (3) x 0.3sy) (1) (2) xy plane: sy) (3) xz plane: sin 2u yz plane: z 576 * 10
6 6
y
100 * 10
6
(18,000 (sy
204 * 10
MAXIMUM SHEAR STRAINS (EQ. 7-75) 5400) (gmax) xy 2 gxy 2 gxz 2 gyz 0 0 (gmax) yz 0 (gmax)xz a x 2 y b +a
2 2
(18,000 30 * 106 30 (EQ. 7-71a) y 2 cos 2u + gxy 2
(gmax) xy a x 2
STRAINS AT ANGLE f x1 x + y 2 + x
A
A
gxy 2
6
b
2
676 * 10 z b +a 2
; b
2
gxz
1 1 407 * 10 6 a ba b(12,600 + 0.7sy) 2 30 * 106 1 1 + a ba b (23,400 1.3sy) cos 60 2 30 * 106 Solve for sy : sy 2400 psi
A
(gmax) xz a y 2
780 * 10 z 2 b +a 2
6
;
gyz 2 b
6
(gmax) yz
104 * 10
;
(4)
07Ch07.qxd
9/27/08
1:24 PM
Page 631
SECTION 7.7
Plane Strain
631
Problem 7.7-12 Solve the preceding problem if the plate is made of aluminum with E sx is 86.4 MPa, the angle f is 21 , and the strain P is 946 * 10 6.
72 GPa and
1/3, the stress
Solution 7.7-12 sx 86.4 MPa
E 72 GPa Strain gage: f
Aluminum plate in biaxial stress gxy 0 sy ?
1/3 21 946 * 10
6
MAXIMUM IN-PLANE SHEAR STRESS (tmax) xy sx 2 sy 32.4 MPa ;
UNITS: All stresses in MPa. STRAIN IN BIAXIAL STRESS (EQS. 7-39) x y z 1 (s Ex 1 (sy E E (sx sy) sx) sy) 1 (86.4 72,000 1 (sy 72,000 1 s) 3y 28.8) sy) (1) (2) (3)
STRAINS FROM EQS. (1), (2), AND (3) x z 1100 * 10 500 * 10
6 6
y
101 * 10
6
MAXIMUM SHEAR STRAINS (EQ. 7-75) xy plane: (gmax) xy 2 gxy xz plane: 2 gxz 0 2 gyz 0 (gmax) yz 0 (gmax)xz a x 2 y b +a
2 2
1/ 3 (86.4 72,000 21 (EQ. 7-71a) y 2 cos 2u + gxy 2
(gmax) xy a x 2
STRAINS AT ANGLE f x1 x + y 2
6
A
A
gxy 2
6
b
2
1200 * 10 z b +a gxz 2
;
2
x +
sin 2u yz plane:
946 * 10
1 1 2 a ba b a57.6 + sy b 2 72,000 3 4 s b cos 42 3y (4)
A
b
(gmax) xz a y 2
1600 * 10 z 2 b +a 2 b
6
6
;
gyz 2 ;
(gmax) yz
399 * 10
1 1 + a ba b a115.2 2 72,000 Solve for sy: sy
21.55 MPa
07Ch07.qxd
9/27/08
1:24 PM
Page 632
632
CHAPTER 7
Analysis of Stress and Strain
Problem 7.7-13 An element in plane stress is subjected to stresses sx 8400 psi, 1700 psi (see figure). The material is aluminum with modulus sy 1100 psi, and txy 0.33. of elasticity E 10,000 ksi and Poissons ratio Determine the following quantities: (a) the strains for an element oriented at an angle u 30, (b) the principal strains, and (c) the maximum shear strains. Show the results on sketches of properly oriented elements.
sy txy x sx
y O
Probs. 7.7-13 and 7.7-14
Solution 7.7-13
sx txy 8400 psi 1700 psi
Element in plane strain
sy E 1100 psi 10,000 ksi 0.33 PRINCIPAL STRAINS 1,2 x + y 2 ;
HOOKES LAW (EQS. 7-34 AND 7-35) x y gxy FOR u x1 1 (s Ex 1 (s Ey txy G 30: x + y 2 x 2 434 * 10 gx1y1 y1 868 * 10 x + y +
6
A
a
x 2 2
y
b +a
6
2
gxy 2
b
2
245 * 10 sy) sx) 876.3 * 10 387.2 * 10
6 6
6
; 671 * 10
1
426 * 10 gxy x
6
961 * 10
6
tan 2up
6
y
0.3579
2txy(1 + ) E x 2 y
452.2 * 10
2up up FOR up
19.7 and 199.7 9.8 and 99.8 9.8: x +
6
cos 2u +
gxy 2
sin 2u x1
x + y 2
y 2
cos 2u +
gxy 2 ; ;
sin 2u
756 * 10 gx1y1 2 y
sin 2u +
6 6
gxy 2
916 * 10 cos 2u up 1 up2 99.8 1 9.8 2
426 * 10 916 * 10
6
6
x1
267 * 10
6
07Ch07.qxd
9/27/08
1:24 PM
Page 633
SECTION 7.7
Plane Strain
633
MAXIMUM SHEAR STRAINS x gxy 2 y 2 gmax a b +a b 2 A 2 2 671 * 10 gmax us1 gmax us2 gmin aver 1342 * 10 up 1 45 1342 * 10 us1 + 90
6 6
54.8
6
;
6
144.8 ;
6
1342 * 10 x + y 2
245 * 10
txy
Problem 7.7-14 Solve the preceding problem for the following data: sx 16 MPa, and u 50. The material is brass with E 100 GPa and
150 MPa, sy 0.34.
210 MPa,
Solution 7.7-14
sx txy
Element in plane strain
210 MPa 100 GPa 0.34
150 MPa sy 16 MPa E
HOOKES LAW (EQS. 7-34 AND 7-35) x y gxy FOR u x1 1 (s Ex 1 (s Ey txy G 50: x + y 2 x 2 + x 2
6
sy) sx)
786 * 10 1590 * 10
6
6
2txy(1 + ) E
429 * 10
6
y
cos 2u +
gxy 2
sin 2u
PRINCIPAL STRAINS x + y x gxy 2 y 2 ; b +a b 1,2 a 2 A 2 2 1188 * 10 1 732 * 10 gxy x y
6 6
; 456 * 10 2 0.5333
6 6
1469 * 10 gx1y1 2 gx1y1 y1 y
1644 * 10
sin 2u +
6
gxy 2
cos 2u
tan 2up 2up
358.5 * 10 717 * 10 x + y x1
6
151.9 and 331.9 76.0 and 166.0
907 * 10
6
up
07Ch07.qxd
9/27/08
1:24 PM
Page 634
634
CHAPTER 7
Analysis of Stress and Strain
FOR up x1
76.0: x + 2
6
MAXIMUM SHEAR STRAINS y cos 2u + gxy 2
6 6
x + y 2
sin 2u
gmax 2 gmax us1 gmax us2 gmin aver
A
a
x 2
y
6 6
b +a
2
gxy 2
b
2
1644 * 10 up 1 up2 76.0 2
456 * 10 732 * 10 1644 * 10 ; ; 911 * 10 up 1 us1 45 911 * 10 90
166.0 1
121.0
6
; 31.0
6
911 * 10 x + y 2
;
6
1190 * 10
Problem 7.7-15 During a test of an airplane wing, the strain gage readings from a 45 rosette (see figure) are as follows: gage A, 520 * 10 6; gage B, 360 * 10 6; and gage C, 80 * 10 6 . Determine the principal strains and maximum shear strains, and show them on sketches of properly oriented elements.
C
y 45 B 45 A O x
Probs. 7.7-15 and 7.7-16
07Ch07.qxd
9/27/08
1:24 PM
Page 635
SECTION 7.7
Plane Strain
635
Solution 7.7-15
A C 520 * 10 80 * 10
6
45 strain rosette
B
6
360 * 10
6
up1 up2
12.5 1
551 * 10 111 * 10
6 6
; ;
102.5 2
FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8: x gxy A 2B 520 * 10 A
6
MAXIMUM SHEAR STRAINS
6
y
C
6
80 * 10
C
280 * 10
gmax 2 gmax us1 gmax us2 gmin aver
A
a
x 2
y
6 6
b +a
2
gxy 2
b
2
PRINCIPAL STRAINS 1,2 x + y 2 ; 220 * 10 1 551 * 10
A
331 * 10 a x 2 2 y b +a
6 2
gxy 2
b
2
662 * 10 up 1 45 662 * 10 us1 + 90 x + y 2
32.5or 147.5
6
6 6
;
6
; 331 * 10
111 * 10
6
57.5 ;
6
662 * 10
220 * 10
tan 2up 2up up
gxy x y
0.4667
25.0 and 205.0 12.5 and 102.5 12.5: x + y 2 +
6
For up x1
x 2
y
cos 2u +
gxy 2
sin 2u
551 * 10
Problem 7.7-16 A 45strain rosette (see figure) mounted on the surface of an automobile frame gives the following readings: gage A, 310 * 10 6; gage B, 180 * 10 6; and gage C, 160 * 10 6. Determine the principal strains and maximum shear strains, and show them on sketches of properly oriented elements.
07Ch07.qxd
9/27/08
1:24 PM
Page 636
636
CHAPTER 7
Analysis of Stress and Strain
Solution 7.7-16
A C 310 * 10
6
45 strain rosette
B
6
180 * 10
6
MAXIMUM SHEAR STRAINS gmax 2 a x 2
6 6
160 * 10
A
y
b +a
2
gxy 2
b
2
FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8: x gxy A 2B 310 * 10 A
6
y
C
6
160 * 10
6
257 * 10 gmax us1 gmax us2 gmin aver 515 * 10 up 1 45 515 * 10 us1 + 90 x + y 2
C
210 * 10
PRINCIPAL STRAINS 1,2 x + y 2 75 * 10 1 tan 2up 2up up 332 * 10 gxy x y ;
6
A
33.0 or 147.0
6
;
6
a
x 2
y
b +a
6
2
gxy 2
6
b
2
57.0 ;
6
515 * 10
; 257 * 10
6
75 * 10
2 0.4468
182 * 10
24.1 and 204.1 12.0 and 102.0
FOR up 12.0: x + y x gxy y x1 + cos 2u + sin 2u 2 2 2 332 * 10 up1 up2
6
12.0 1 102.0 2
332 * 10
6 6
; ;
182 * 10
07Ch07.qxd
9/27/08
1:24 PM
Page 637
SECTION 7.7
Plane Strain
637
Problem 7.7-17 A solid circular bar of diameter d 1.5 in. is subjected to an axial force P and a torque T (see figure). Strain gages A and B mounted on the surface of the bar give reading Pa 100 * 10 6 and Pb 55 * 10 6. The bar is made of steel 0.29. having E 30 * 106 psi and
(a) Determine the axial force P and the torque T. (b) Determine the maximum shear strain gmax and the maximum shear stress tmax in the bar.
d C
T P
B 45 A C
Solution 7.7-17
6
Circular bar (plane stress)
STRAIN AT u x1 x1 A x 100 * 10
6 6
Bar is subjected to a torque T and an axial force P. E 30 * 10 psi 1.5 in. 0.29 Diameter d STRAIN GAGES At u At u 0: 45: B 55 * 10
45 x + 2
6
x + y 2 B
6
y
cos 2u + 2u
6
gxy 2 90
sin 2u
(1)
55 * 10
Substitute numerical values into Eq. (1): 55 * 10 Solve for T: 16T pd 3 29 * 10
6
35.5 * 10 T
(0.0649 * 10 ;
6
)T
1390 lb-in
ELEMENT IN PLANE STRESS sx x P A 4P pd 2
6
MAXIMUM SHEAR STRAIN AND MAXIMUM SHEAR STRESS gxy (0.1298 * 10 gmax 2 gmax tmax a
sy y
0
txy x
100 * 10
Eq. (7-75):
A
6
)T y 2
180.4 * 10 b +a
6 6 2
6 2
rad
x
gxy 2
b
AXIAL FORCE P x sx E 4P pd 2E P pd 2Ex 4 5300 lb ;
111 * 10 222 * 10 Ggmax
rad rad ; ;
2580 psi
SHEAR STRAIN txy 2txy(1 + ) gxy G E (0.1298 * 10
6
32T(1 + ) pd 3E (T lb-in.)
)T
07Ch07.qxd
9/27/08
1:24 PM
Page 638
638
CHAPTER 7
Analysis of Stress and Strain
Problem 7.7-18 A cantilever beam of rectangular cross section (width b 25 mm, height h 100 mm) is loaded by a force P that acts at the midheight of the beam and is inclined at an angle a to the vertical (see figure). Two strain gages are placed at point C, which also is at the midheight of the beam. Gage A measures the strain in the horizontal direction and gage B measures the strain at an angle b = 60 to the horizontal. The measured strains are Pa 125 * 10 6 375 * 10 6. and Pb Determine the force P and the angle a, assuming the material is steel with E 200 GPa and 1/3.
b C
h h a P b
B A C
b
Probs. 7.7-18 and 7.7-19
Solution 7.7-18 Cantilever beam (plane stress) Beam loaded by a force P acting at an angle a.
E h 200 GPa 100 mm P sin a P cos a 1/3 b 25 mm
HOOKES LAW x sx E P sin a bhE (1)
Axial force F Shear force V
(At the neutral axis, the bending moment produces no stresses.) STRAIN GAGES At u At u 0: 60: A B x 125 * 10 375 * 10
6 6
P sin a bhEx 62,500 N txy 3(1 + ) P cos a 3P cos a gxy G 2bhG bhE (8.0 * 10 FOR u x1 x1 60: x + y 2 B
6 9
) P cos a
(2)
x + 2
y
6
cos 2u + 2u
6
gxy 2
sin 2u
(3)
ELEMENT IN PLANE STRESS sx txy x F A P sin a bh 3V 2A 125 * 10
6
375 * 10
120 41.67 * 10
9 6
Substitute into Eq. (3): 0 375 * 10 or P cos a x 41.67 * 10
6
sy
41.67 * 10
3P cos a 2bh y
(3.464 * 10 108,260 N a ; 30 SOLVE EQS. (1) AND (4): tan a P 0.5773
)P cos a (4)
;
125 kN
Problem 7.7-19 Solve the preceding problem if the cross-sectional dimensions are b 1.0 in. and h 3.0 in., the gage angle is b = 75, the measure strains are Pa 171 * 10 6 and Pb 266 * 10 6, and the material is a magnesium alloy 6 with modulus E 6.0 * 10 psi and Poissons ratio 0.35.
07Ch07.qxd
9/27/08
1:24 PM
Page 639
SECTION 7.7
Plane Strain
639
Solution 7.7-19
E h 6.0 * 10 psi 3.0 in.
6
Cantilever beam (plane stress)
HOOKES LAW x sx E P sin a bhE (1) 3(1 + v)P cos a bhE (2) 0.35 b 1.0 in.
Beam loaded by a force P acting at an angle a.
Axial force F P sin a Shear foce V P cos a (At the neutral axis, the bending moment produces no stresses.) STRAIN GAGES At u At u 0: 75: A B x 171 * 10 266 * 10
6 6
P sin a bhEx 3078 lb txy 3P cos a gxy G 2bhG (225.0 * 10 FOR u x1 75: x + y 2 B
6 9
)P cos a
x + 2
y
6
cos 2u + 2u
6 9
gxy 2
sin 2u
(3)
ELEMENT IN PLANE STRESS sx txy x F A P sin a bh 3V 2A 171 * 10 x1 sy 0
266 * 10
150 99.961 * 10 )P cos a (4)
6
Substitute into Eq. (3): 266 * 10
6
3P cos a 2bh
6
55.575 * 10
(56.25 * 10 x 59.85 * 10 or P cos a 3939.8 lb
y
SOLVE EQS. (1) AND (4): tan a P 0.7813 a ;
y
38
;
5000 lb
Problem 7.7-20 A 60 strain rosette, or delta rosette, consists of three electrical-resistance strain gages arranged as shown in the figure. Gage A measures the normal strain Pa in the direction of the x axis. Gages B and C measure the strains Pb and Pc in the inclined directions shown. Obtain the equations for the strains Px, Py, and gxy associated with the xy axis.
B
60
C
60 O
A
60 x
Solution 7.7-20 Delta rosette (60 strain rosette) STRAIN GAGES
Gage A at u Gage B at u Gage C at u FOR u 0: 0 60 120 x A Strain Strain Strain ; A B C
FOR u x1 B B
60: x + y 2 A + y 2 +
x 2 A
y y
3y gxy 13 A + + 4 4 4 + 2
cos 2u +
gxy 2
sin 2u gxy
2
(cos 120) +
(sin 120) (1)
07Ch07.qxd
9/27/08
1:24 PM
Page 640
640
CHAPTER 7
Analysis of Stress and Strain
FOR u x1 C C
120: x + y 2 A + y 2 +
SOLVE EQS. (1) AND (2): x 2 A + y y gxy 13 2 4 cos 2u + gxy 2 gxy 2 sin 2u (sin 240) (2) y gxy 1 (2B + 2C 3 2 (B 13 C) A) ; ;
(cos 240)
3y A + 4 4
Problem 7.7-21 On the surface of a structural component in a space vehicle, the strainsare monitored by means of three strain gages arranged as shown in the figure. During a certain maneuver, the following strains were recorded: Pa 1100 * 10 6, Pb 200 * 10 6, and Pc 200 * 10 6. Determine the principal strains and principal stresses in the material, which is a magnesium alloy for which E 6000 ksi and 0.35. (Show the principal strains and principal stresses on sketches of properly oriented element.)
y
B
C
30 O A x
Solution 7.7-21 30-60-90 strain rosette Magnesium alloy: E 6000 ksi 0.35
STRAIN GAGES Gage A at u Gage B at u Gage C at u FOR u FOR u FOR u x1 C 0: 90: 0 90 150 x y PA B C A B x +
6 6 6 6 6 6
tan 2up 2up
gxy x 60 y up
13 30
1.7321
1100 * 10 200 * 10 200 * 10 1100 * 10 200 * 10 y 2
FOR up 30: x + y x gxy y x1 + cos 2u + sin 2u 2 2 2 1550 * 10 up1 30 120 up2 sin 2u
6
150: x + y 2
6
1 2
1550 * 10 250 * 10
6 6
; ;
cos 2u +
gxy 2
6
200 * 10
650 * 10
+ 225 * 10
6
0.43301gxy Solve for gxy: gxy PRINCIPAL STRAINS x + y x gxy 2 y 2 1,2 ; a b +a b 2 A 2 2 650 * 10 1 1550 * 10
6
1558.9 * 10
; 900 * 10
6
6
2
250 * 10
6
07Ch07.qxd
9/27/08
1:24 PM
Page 641
SECTION 7.7
Plane Strain
641
PRINCIPAL STRESSES (see Eqs. 7-36) s1 E 1
2
(1 + 2)
s2
E 1
2
(2 + 1)
Substitute numerical values: s1 10,000 psi s2 2,000 psi ;
y
Problem 7.7-22 The strains on the surface of an experimental device made of pure 0.33) and tested in a space shuttle were measured by aluminum (E 70 Gpa, means of strain gages. The gages were oriented as shown in the figure, and the measured 39.44 * 10 * 6. strains were Pa 1100 * 10 6, Pb 1496 * 10 6, and Pc What is the stress sx in the x direction?
O
B
C
40
A
40
x
Solution 7.7-22 40-40-100 strain rosette 0.33 Pure aluminum: E 70 GPa
STRAIN GAGES Gage A at u Gage B at u Gage C at u FOR u FOR u x1 0: 40: x + y 2 + B
6
FOR u x1
140: x + y 2 +
x 2
6
y
cos 2u +
gxy 2
6
sin 2u
0 40 140 x
A B C A
1100 * 10 1496 * 10
6 6 6
Substitute x1 x 1100 * 10 0.41318y
c 0.49240gxy
39.44 * 10
and
6
; then simplify and rearrange: 684.95 * 10 (2)
39.44 * 10 1100 * 10
6
SOLVE EQS. (1) AND (2): y 200.3 * 10
6
gxy
1559.2 * 10
6
x 2
y
cos 2u +
6
gxy 2 and
sin 2u
HOOKES LAW sx E 1
2
Substitute x1 x 1100 * 10
1496 * 10
(x + y)
91.6 MPa
;
; then simplify and rearrange: 850.49 * 10
6
0.41318y + 0.49240gxy
(1)
07Ch07.qxd
9/27/08
1:24 PM
Page 642
642
CHAPTER 7
Analysis of Stress and Strain
Problem 7.7-23 Solve Problem 7.7-5 by using Mohrs circle for plane strain.
Solution 7.7-23
Element in plane strain
x gxy R
2(130 * 10 180 * 10 158.11 * 10 90 130 a
220 * 10
6 6
y gxy 2
62 6
480 * 10 90 * 10
6
6
POINT C: x1 POINT D (u u 50 x1 350 * 10 gx1y1 R sin b 2 gx1y1
350 * 10 50 ):
6
6
+ R cos b 112.4 * 10
6 6
461 * 10
6
) + (90 * 10
62
)
a b
arctan 180
34.70 2u 45.30
225 * 10
POINT D (u
140): R cos b 239 * 10
6 6
x1 350 * 10 6 gx1y1 R sin b 2 gx1y1 225 * 10
112.4 * 10
6
07Ch07.qxd
9/27/08
1:24 PM
Page 643
SECTION 7.7
Plane Strain
643
Problem 7.7-24 Solve Problem 7.7-6 by using Mohrs circle for plane strain.
Solution 7.7-24
x gxy 420 * 10 310 * 10
6
Element in plane strain
y gxy 2 170 * 10 155 * 10
6 6
POINT C: x1 u 37.5 POINT D (u
125 * 10 37.5):
6
6
x1 125 * 10 6 + R cos b 351 * 10 gx1y1 R sin b 244.8 * 10 6 2 gx1y1 490 * 10
6
6
POINT D (u x1 125 * 10 gx1y1 R sin b 2 gx1y1 2(295 * 10 333.24 * 10 a b arctan 2u 155 295 a
127.5):
6
R cos b 244.8 * 10
6 6
101 * 10
6
490 * 10
R
62 6
) + (155 * 10
62
)
27.72 47.28
07Ch07.qxd
9/27/08
1:24 PM
Page 644
644
CHAPTER 7
Analysis of Stress and Strain
Problem 7.7-25 Solve Problem 7.7-7 by using Mohrs circle for plane strain.
Solution 7.7-25
x gxy 480 * 10
6
Element in plane strain
y
6
350 * 10
140 * 10 6 gxy 175 * 10 2
6
MAXIMUM SHEAR STRAINS 2us2 2us1 gmax gmin
62 6
90
a
44.17 310 * 10
6
us2
22.1 us1 112.1
2us2 + 180 2R
224.17
6 6
Point S1: aver 2(175 * 10 243.98 * 10 a arctan 175 170 x1 Point S2: aver ) + (170 * 10
62
488 * 10
6
310 * 10
488 * 10
R
)
45.83 310 * 10
6
POINT C:
PRINCIPAL STRAINS 2up2 2up1 180 a 134.2
6 6
up2 +R R
67.1 157.1 554 * 10 66 * 10
6 6
2up2 + 180
314.2 up1
Point P1: 1 Point P2: 2
310 * 10 310 * 10
07Ch07.qxd
9/27/08
1:24 PM
Page 645
SECTION 7.7
Plane Strain
645
Problem 7.7-26 Solve Problem 7.7-8 by using Mohrs circle for plane strain.
Solution 7.7-26
x gxy 120 * 10
6
Element in plane strain
y
6
450 * 10 gxy 2
6 6
360 * 10
180 * 10
MAXIMUM SHEAR STRAINS 2us2 2us1 gmax gmin
62 6
90
a
57.72 237.72 165 * 10
us2
6
28.9 us1 118.9
2us2 + 180 2R
Point S1: aver 2(285 * 10 337.08 * 10 a arctan 180 285 Point S2: aver 674 * 10 ) + (180 * 10
62
674 * 10
6
6 6
165 * 10
R
)
32.28 165 * 10
6
Point C: x1 2up2 2up1 180 a R
PRINCIPAL STRAINS 147.72 327.72 165 * 10 165 * 10
6 6
up 2 up1 R
73.9 163.9
6 6
2up2 + 180
Point P1: 1 Point P2: 2
172 * 10
502 * 10
07Ch07.qxd
9/27/08
1:24 PM
Page 646
646
CHAPTER 7
Analysis of Stress and Strain
Problem 7.7-27 Solve Problem 7.7-9 by using Mohrs circle for plane strain.
Solution 7.7-27
x gxy 480 * 10 420 * 10
Element in plane strain
6 6
y gxy 2
70 * 10 210 * 10
6 6
u
75
PRINCIPAL STRAINS 2up1 2up2 a 45.69 up1
6 6
22.8 up2 112.8
6 6
2up1 + 180
225.69 +R R
Point P1: 1 2(205 * 10 293.47 * 10 a b arctan 210 205 Point P2: 2
275 * 10 275 * 10
568 * 10 18 * 10
R
62 6
) + (210 * 10
62
)
45.69 2u 75.69
6
a + 180
Point C: x1 Point D (u
275 * 10 75):
MAXIMUM SHEAR STRAINS 2us2 90 + a 135.69 315.69
6 6 6
us2
67.8 us1 157.8
x1 275 * 10 6 gx1y1 R sin b 2 gx1y1 Point D 569 * 10 (u 165):
6
R cos b
202 * 10
6
6
2us1 gmax gmin
2us2 + 180 2R
Point S1: aver 284.36 * 10
6
275 * 10 587 * 10
6
Point S2: aver
275 * 10
587 * 10
x1 275 * 10 gx1y1 R sin b 2 gx1y1
+ R cos b 284.36 * 10
6
348 * 10
6
6
569 * 10
07Ch07.qxd
9/27/08
1:24 PM
Page 647
SECTION 7.7
Plane Strain
647
Problem 7.7-28 Solve Problem 7.7-10 by using Mohrs circle for plane strain.
Solution 7.7-28
x gxy
Element in plane strain
6
1120 * 10 780 * 10
6
y gxy 2
430 * 10 390 * 10
6
6
u
45
PRINCIPAL STRAINS 2up1 2up2 180 a 131.50 311.50
6 6
up 1 up2 R +R
65.7 155.7 254 * 10 1296 * 10
6 6
2up1 + 180
Point P1: 1 2(345 * 10 520.70 * 10 a b arctan 180 390 345 a Point P2: 2
775 * 10 775 * 10
R
62 6
) + (390 * 10
62
)
48.50 2u 41.50
6
Point C: x1 Point D: (u x1 gx1y1 2
775 * 10 45):
6
MAXIMUM SHEAR STRAINS 2us1 90 a 41.50 221.50 775 * 10 1041 * 10
6 6 6 6
us1
20.7 us2 110.7
775 * 10 R sin b
+ R cos b
6
385 * 10 gx1y1
6 6
2us2 gmax gmin
2us1 + 180 2R
Point S1: aver Point S2: aver
345 * 10 135)
6
690 * 10
775 * 10
Point D : (u x1 gx1y1 2 gx1y1
1041 * 10
775 * 10 R sin b
R cos b 345 * 10
6 6
1165 * 10
6
690 * 10
07Ch07.qxd
9/27/08
1:24 PM
Page 648