(a) The normal stress in the hanger rod is
2 2
rod
rod 2
(0.5 in.) 0.196350 in.
4
2,000 lb
10,185.917 psi
0.196350 in. 10,190 psi
A
Ans.
(b) The cross-sectional area of the bolt is:
2 2
bolt (0.625 in.) 0.306796 in.
A 4
The bolt acts in double shear; therefore, its average shear stress is
bolt 2
2,000 lb
3, 259.493 psi
2(0.306796 in. 3, 260 i
) ps
Ans.
(c) The bearing stress in the bottom strap is based on the projected area of the bolt in contact with the strap. Also, keep in mind that there are two ends of the strap that contact the bolt. The bearing stress is thus
2,000 lb
8,533.334 psi
2(0.625 in.)(3/16 8,530 psi
b in.)
Ans.
P1.30 Rigid bar ABC shown in Figure P1.30 is supported by a pin at bracket A and by tie rod (1).
Tie rod (1) has a diameter of 5 mm, and it is
supported by double-shear pin connections at B and D. The pin at bracket A is a single-shear
connection. All pins are 7 mm in diameter.
Assume a = 600 mm, b = 300 mm, h = 450 mm, P
Equilibrium: Using the FBD shown, calculate the reaction forces that act on rigid bar ABC.
1
(1,843.092 N)cos(36.87 ) (900 N)cos(55 ) 0 958.255 N
The resultant force at A is
2 2
(958.255 N) ( 368.618 N) 1,026.709 N
A
19.635 mm 93.9 MPa
A
Ans.
(b) Shear stress in pin B. The cross-sectional area of a 7-mm-diameter pin is:
2 2
pin (7 mm) 38.485 mm A 4
Pin B is a double shear connection; therefore, its average shear stress is
pin 2
P1.31 The bell crank shown in Figure P1.31 is in equilibrium for the forces acting in rods (1) and (2).
The bell crank is supported by a 10-mm-diameter pin at B that acts in single shear. The thickness of the bell crank is 5 mm. Assume a = 65 mm, b =
Equilibrium: Using the FBD shown, calculate the reaction forces that act on the bell crank.
2 2
(1,100 N)sin(50 )(65 mm) (150 mm) 0
The resultant force at B is
2 2
(341.919 N) ( 842.649 N) 909.376 N
B
(a) Shear stress in pin B. The cross-sectional area of the 10-mm-diameter pin is:
2 2
pin (10 mm) 78.540 mm A 4
Pin B is a single shear connection; therefore, its average shear stress is
pin 2
909.376 N
78.540 mm 11.58 MPa
B Ans.
P1.32 The beam shown in Figure P1.32 is
Equilibrium: Using the FBD shown, calculate the reaction forces that act on the beam.
(47.0731 kN)cos(35 ) 0 38.5600 kN
The resultant force at C is
2 2
(38.5600 kN) (27.0000 kN) 47.0731 kN
C
Shear stress in pin A. The cross-sectional area of a 25-mm-diameter pin is:
2 2
pin (25 mm) 490.8739 mm A 4
Pin A is a single shear connection; therefore, its average shear stress is
pin 2
47,073.1 N
490.8739 m 95.9 MPa
A m
Ans.
Shear stress in pin C.
Pin C is a double shear connection; therefore, its average shear stress is
pin 2
P1.33 The bell-crank mechanism shown in Figure P1.33 is in equilibrium for an applied load of P = 7 kN applied at A.
Assume a = 200 mm, b = 150 mm, and = 65°. Determine the minimum diameter d required for pin B for each of the following conditions:
(a) The average shear stress in the pin may not exceed 40 MPa.
(b) The bearing stress in the bell crank may not exceed 100 MPa.
(c) The bearing stress in the support bracket may not exceed 165 MPa.
FIGURE P1.33 Solution
Equilibrium: Using the FBD shown, calculate the reaction forces that act on the bell crank.
2 2
(7,000 N)sin(65 )(200 mm) (150 mm) 0
The resultant force at B is
2 2
( 11,417.201 N) (6,344.155 N) 13,061.423 N
B
(a) The average shear stress in the pin may not exceed 40 MPa. The shear area required for the pin at B is
Since the pin at B is supported in a double shear connection, the required cross-sectional area for the pin
(b) The bearing stress in the bell crank may not exceed 100 MPa. The projected area of pin B on the bell crank must equal or exceed
2 2
13,061.423 N
130.614 mm 100 N/mm
Ab
The bell crank thickness is 8 mm; therefore, the projected area of the pin is Ab = (8 mm)d. Calculate the required pin diameter d:
130.614 mm2
8 16.3
mm 3 mm
d Ans.
(c) The bearing stress in the support bracket may not exceed 165 MPa. The pin at B bears on two 6-mm-thick support brackets. Thus, the minimum pin diameter required to satisfy the bearing stress limit on the support bracket is
2 2
13,061.423 N
79.160 mm 165 N/mm
Ab
79.160 mm2
2(6 mm) 6.60 mm
d Ans.
P1.34 A structural steel bar with a 25 mm × 75 mm rectangular cross section is subjected to an axial load of 150 kN. Determine the maximum normal and shear stresses in the bar.
Solution
The maximum normal stress in the steel bar is
max
(150 kN)(1,000 N/kN)
(25 mm)(75 mm) 80 MPa F
A Ans.
The maximum shear stress is one-half of the maximum normal stress
max
max 2 40 MPa
Ans.
P1.35 A steel rod of circular cross section will be used to carry an axial load of 92 kips. The maximum stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear. Determine the required diameter for the rod.
Solution
Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is
2 min
max
92 kips
3.0667 in.
30 ksi A F
For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be
2 min
max
92 kips
3.8333 in.
2 2(12 ksi) A F
Therefore, the rod must have a cross-sectional area of at least 3.8333 in.2 in order to satisfy both the normal and shear stress limits.
The minimum rod diameter D is therefore
2 2
min 3.8333 in. min 2.2092 in. 2.21 in.
4d d
Ans.
P1.36 An axial load P is applied to the rectangular bar shown in Figure P1.36. The cross-sectional area of the bar is 400 mm2. Determine the normal stress perpendicular to plane AB and the shear stress parallel to normal force N perpendicular to plane AB is found from
cos (40 kN)cos35 57.3406 kN
NP
and the shear force V parallel to plane AB is sin (70 kN)sin 35 40.1504 kN
V P
The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is
2
The normal stress n perpendicular to plane AB is
2
P1.37 An axial load P is applied to the 1.75 in.
by 0.75 in. rectangular bar shown in Figure P1.37. Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB if the bar is subjected to an axial load of P = 18 kips.
FIGURE P1.37 Solution
The angle for the inclined plane is 60°. The normal force N perpendicular to plane AB is found from
cos (18 kips)cos60 9.0 kips
NP
and the shear force V parallel to plane AB is sin (18 kips)sin 60 15.5885 kips
V P
The cross-sectional area of the bar is (1.75 in.)(0.75 in.) = 1.3125 in.2, but the area along inclined plane AB is
The normal stress n perpendicular to plane AB is
2
9.0 kips
3.4286 ksi
2.6250 in 3.43 ks
. i
P1.38 A compression load of P = 80 kips is applied to a 4 in. by 4 in. square post, as shown in Figure P1.38/39. Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB.
FIGURE P1.38/39 Solution
The angle for the inclined plane is 55°. The normal force N perpendicular to plane AB is found from
cos (80 kips)cos55 45.8861 kips
N P
and the shear force V parallel to plane AB is sin (80 kips)sin 55 65.5322 kips
V P
The cross-sectional area of the post is (4 in.)(4 in.) = 16 in.2, but the area along inclined plane AB is
2
The normal stress n perpendicular to plane AB is
2
45.8861 kips
1.6449 ksi
27.8951 1.645 ksi
n in.
P1.39 Specifications for the 50 mm × 50 mm square bar shown in Figure P1.38/39 require that the normal and shear stresses on plane AB not exceed 120 MPa and 90 MPa, respectively. Determine the maximum load P that can be applied without exceeding the specifications.
FIGURE P1.38/39 Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are (1 cos 2 )
The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can be supported by the square bar is found from Eq. (a):
2 2
The shear stress on plane AB is limited to 90 MPa. From Eq. (b), the maximum load P based the shear stress limit is
2 2
Thus, the maximum load that can be supported by the bar is
max 479 kN
P Ans.
P1.40 Specifications for the 6 in. × 6 in. square post shown in Figure P1.40 require that the normal and shear stresses on plane AB not exceed 800 psi and 400 psi, respectively.
Determine the maximum load P that can be applied without exceeding the specifications.
FIGURE P1.40 Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are (1 cos 2 )
The normal stress on plane AB is limited to 800 psi; therefore, the maximum load P that can be supported by the square post is found from Eq. (a):
2 2(36 in. )(800 psi)2
The shear stress on plane AB is limited to 400 psi. From Eq. (b), the maximum load P based the shear stress limit is
2 2(36 in. )(400 psi)2
Thus, the maximum load that can be supported by the post is
max 29, 200 lb 29.2 ipk s
P Ans.
P1.41 A 90 mm wide bar will be used to carry an axial tension load of 280 kN as shown in Figure P1.41. The normal and shear stresses on plane AB must be limited to 150 MPa and 100 MPa, respectively. Determine the minimum thickness t required for the bar.
FIGURE P1.41 Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are (1 cos 2 )
The normal stress on plane AB is limited to 150 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq. (a):
The shear stress on plane AB is limited to 100 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq. (b):
To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 1,379.7309 mm2. Since the bar width is 90 mm, the minimum bar thickness t must be
2
P1.42 A rectangular bar having width w = 6.00 in. and thickness t = 1.50 in. is subjected to a tension load P as shown in Figure P1.42/43. The normal and shear stresses on plane AB must not exceed 16 ksi and 8 ksi, respectively. Determine the maximum load P that can be applied without exceeding either stress limit.
FIGURE P1.42/43 Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are (1 cos 2 )
The angle for inclined plane AB is calculated from
tan 3 3 71.5651
1
The cross-sectional area of the bar is A = w×t = (6.00 in.)(1.50 in.) = 9.0 in.2.
The normal stress on plane AB is limited to 16 ksi; therefore, the maximum load P can be found from Eq. (a):
The shear stress on plane AB is limited to 8 ksi. From Eq. (b), the maximum load P based the shear stress limit is
2 2(9.0 in. )(8 ksi)2
Thus, the maximum load that can be supported by the bar is
max 240 kips
P Ans.
P1.43 In Figure P1.42/43, a rectangular bar having width w = 1.25 in. and thickness t is subjected to a tension load of P = 30 kips. The normal and shear stresses on plane AB must not exceed 12 ksi and 8 ksi, respectively. Determine the minimum bar thickness t required for the bar.
FIGURE P1.42/43 Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are (1 cos 2 )
The angle for inclined plane AB is calculated from
tan 3 3 71.5651
1
The normal stress on plane AB is limited to 12 ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq. (a):
The shear stress on plane AB is limited to 8 ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq. (b):
30 kips 2
sin 2 sin 2(71.5651 ) 1.1250 in.
2 nt 2(8 ksi)
A P
To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 1.1250 in.2. Since the bar width is 1.25 in., the minimum bar thickness t must be
2
P1.44 The rectangular bar has a width of w = 3.00 in. and a thickness of t = 2.00 in. The normal stress on plane AB of the rectangular block shown in Figure P1.44/45 is 6 ksi (C) when the load P is applied. Determine:
(a) the magnitude of load P.
(b) the shear stress on plane AB.
(c) the maximum normal and shear stresses in the block at any possible orientation.
FIGURE P1.44/45 Solution
The general equation for normal stress on an inclined plane in terms of the angle is (1 cos 2 )
n 2 P
A (a)
and the angle for inclined plane AB is
tan 3 0.75 36.8699
4
The cross-sectional area of the rectangular bar is A = (3.00 in.)(2.00 in.) = 6.00 in.2.
(a) Since the normal stress on plane AB is given as 6 ksi, the magnitude of load P can be calculated from Eq. (a):
(b) The general equation for shear stress on an inclined plane in terms of the angle is sin 2
nt 2 P
A
therefore, the shear stress on plane AB is
2
56.25 kips
sin 2(36.8699 )
2(6.00 in. ) 4.50 ksi
nt Ans.
(c) The maximum normal stress at any possible orientation is
max 2
and the maximum shear stress at any possible orientation in the block is 56.25 kips
4.6875 ksi 4.69 ksi
P Ans.
P1.45 The rectangular bar has a width of w = 100 mm and a thickness of t = 75 mm. The shear stress on plane AB of the rectangular block shown in Figure P1.44/45 is 12 MPa when the load P is applied. Determine:
(a) the magnitude of load P.
(b) the normal stress on plane AB.
(c) the maximum normal and shear stresses in the block at any possible orientation.
FIGURE P1.44/45 Solution
The general equation for shear stress on an inclined plane in terms of the angle is sin 2
nt 2 P
A (a)
and the angle for inclined plane AB is
tan 3 0.75 36.8699
4
The cross-sectional area of the rectangular bar is A = (100 mm)(75 mm) = 7,500 mm2.
(a) Since the shear stress on plane AB is given as 12 MPa, the magnitude of load P can be calculated from Eq. (a):
2 2
2 2(7,500 mm )(12 N/mm )
187,500 N
sin 2 sin 2(36.869 187.5 k
9 N
(b) The general equation for normal stress on an inclined plane in terms of the angle is (1 cos 2 )
n 2 P
A
therefore, the normal stress on plane AB is
2
187,500 N
(1 cos 2(36.8699 ))
2(7,500 mm 16.00 MPa
n )
Ans.
(c) The maximum normal stress at any possible orientation is
max 2
187,500 N
7,500 mm 25.0 MPa P
A Ans.
and the maximum shear stress at any possible orientation in the block is
max 2
187,500 N
2 2(7,500 mm 12.50 MPa )
P
A Ans.
P2.1 When an axial load is applied to the ends of the bar shown in Figure P2.1, the total elongation of the bar between joints A and C is 0.15 in. In segment (2), the normal strain is measured as 1,300 in./in.
Determine:
(a) the elongation of segment (2).
(b) the normal strain in segment (1) of the bar.
FIGURE P2.1
Solution
(a) From the definition of normal strain, the elongation in segment (2) can be computed as
6
2 2L2 (1,300 10 )(90 in.) 0.1170 in. Ans.
(b) The combined elongations of segments (1) and (2) is given as 0.15 in. Therefore, the elongation that occurs in segment (1) must be
1 total 2 0.15 in. 0.1170 in. 0.0330 in.
The strain in segment (1) can now be computed:
1 1
1
0.0330 in.
0.000825 in./in.
40 in. 825 μin./in.
L Ans.
P2.2 The two bars shown in Figure P2.2 are used to support a load P. When unloaded, joint B has coordinates (0, 0). After load P is applied, joint B moves to the coordinate position (0.35 in., –0.60 in.).
Assume a = 11 ft, b = 6 ft, and h = 8 ft. Determine the normal strain in each bar.
FIGURE P2.2 Solution
Given
(11 ft)(12 in./ft) 132 in.
(6 ft)(12 in./ft) 72 in.
(8 ft)(12 in./ft) 96 in.
a b h
The initial length of bar AB is
2 2
(132 in.) (96 in.) 163.2176 in.
LAB
and its length after deformation is
2 2
(132.35 in.) (96.60 in.) 163.8538 in.
LAB
The strain in bar AB is thus
163.8538 in. 163.2176 in. 0.6362 in. 6
3,897.9 10 in./in.
163.2176 in. 163.2176 in 3,900 ε
. μ
AB Ans.
The initial length of bar BC is
2 2
(72 in.) (96 in.) 120.0000 in.
LBC
and its length after deformation is
2 2
(71.65 in.) (96.60 in.) 120.2717 in.
LBC
The strain in bar BC is thus
120.2717 in. 120.0000 in. 0.2717 in. 6
2, 264.2 10 in./in.
120.0000 in. 120.0000 in 2, 260 ε
. μ
BC Ans.
P2.3 A rigid steel bar is supported by three rods, as shown in Figure P2.3. There is no strain in the rods before the load P is applied. After load P is 2-mm gap in the connections between the rigid bar and rods (1) at joints A and C before the load is applied.
(c) the normal strain in rod (2) if there is a 2-mm gap in the connection between the rigid bar and rod (2) at joint B before the load is applied.
FIGURE P2.3 Solution
(a) From the normal strain in rod (1) and its length, the deformation of rod (1) can be calculated:
6
1 1 1L (860 10 )(2, 400 mm) 2.064 mm
Since rod (1) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must move downward by an amount equal to the deformation of rod (1); therefore,
1 2.064 mm (downward)
A C
v v
By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rod (2) is connected to the rigid bar at B, and again, a perfect connection is assumed. The deformation of rod (2) must be equal to the deflection of joint B; thus, 2 = 2.064 mm. The normal strain in rod (2) can now be calculated as:
2 2
2
2.064 mm
0.0011467 mm/mm 1,147 μmm/m
1,800 m m
m
L Ans.
(b) We know the strain in rod (1); hence, we know its deformation must be 2.064 mm. However, the joints at A and C are not perfect connections. The first 2 mm of downward movement by the rigid bar does not elongate rod (1)—it simply closes the gap. To elongate rod (1) by 2.064 mm, joint A must move down:
1 2 mm 2.064 mm 2 mm 4.064 mm vA
2 2
2
4.064 mm
0.0022578 mm/mm
1,800 mm 2, 260 με
L Ans.
(c) We now assume that the bolted connection at A (or C) is perfect; therefore, the rigid bar deflection as calculated in part (a) must be vA = 2.064 mm (downward). Further, the rigid bar must remain horizontal as it deflects downward by virtue of symmetry; thus, vB = vA = vC. Therefore, the deflection downward of joint B is vB = 2.064 mm What effect is caused by the gap at B? When joint B moves downward by 2.064 mm, the first 2 mm of this downward movement does not stretch rod (2)—it just closes the gap. Therefore, rod (2) only gets elongated by the amount
2 2 mm
2.064 mm 2 mm 0.064 mm
vB
This deformation creates a strain in rod (2) of:
2 2
2
0.064 mm
0.0000356 mm/mm
1,800 mm 35.6 με
L Ans.
P2.4 A rigid bar ABCD is supported by two bars as shown in Figure P2.4. There is no strain in the vertical bars before load P is applied. After load P is applied, the normal strain in rod (1) is −570
m/m. Determine:
(a) the normal strain in rod (2).
(b) the normal strain in rod (2) if there is a 1-mm gap in the connection at pin C before the load is applied.
(c) the normal strain in rod (2) if there is a 1-mm gap in the connection at pin B before the load is applied.
From the deformation diagram of rigid bar ABCD 240 mm (240 mm 360 mm)
600 mm
(0.5130 mm) 1.2825 mm 240 mm
B C
C
v v
v
Therefore, 2 = 1.2825 mm (elongation), and thus, from the definition of strain:
6 deformation diagram is unaffected; thus, vB = 0.5130 mm (downward) and vC = 1.2825 mm (downward).
The rigid bar must move downward 1 mm at C
and so the strain in member (2) is
2 6 2
2
0.2825 mm
188.3 10 mm/mm
1,500 mm 188.3 με
L Ans.
(c) From the strain given for rod (1), 1 =
−0.5130 mm. In order to contract rod (1) by this amount, the rigid bar must move downward at B by
0.5130 mm 1 mm 1.5130 mm vB
From deformation diagram of rigid bar ABCD 240 mm (240 mm 360 mm)
600 mm
(1.5130 mm) 3.7825 mm 240 mm
B C
C
v v
v and so
2 6 2
2
3.7825 mm
2,521.7 10 mm/mm
1,500 2,520 ε μ
mm
L Ans.
P2.5 In Figure P2.5, rigid bar ABC is supported by a pin connection at B and two axial members. A slot in member (1) allows the pin at A to slide 0.25-in. before it contacts the axial member. If the load P produces a compression normal strain in member (1) of −1,300 in./in., determine the normal strain in member (2).
FIGURE P2.5 Solution
From the strain given for member (1)
1 1 1
( 1,300 10 )(32 in.)6
0.0416 in.
L
Pin A has to move 0.25 in. before it contacts member (1); therefore, vA = 1.080 mm (downward).
0.0416 in. 0.25 in.
0.2916 in.
0.2916 in. (to the left) vA
From the deformation diagram of rigid bar ABC
12 in. 20 in.
20 in.
(0.2916 in.) 0.4860 in. (downward) 12 in.
A C
C
v v
v
Therefore, 2 = 0.4860 in. (elongation). From the definition of strain,
6 2 0.4860 in.
3,037.5 10 in./in. 3,040 με Ans.
P2.6 The sanding-drum mandrel shown in Figure P2.6 is made for use with a hand drill. The mandrel is made from a rubber-like material that expands when the nut is tightened to secure the sanding sleeve placed over the outside surface. If the diameter D of the mandrel increases from 2.00 in. to 2.15 in. as the nut is tightened, determine (a) the average normal strain along a diameter of
the mandrel.
(b) the circumferential strain at the outside surface
of the mandrel. FIGURE P2.6
Solution
(a) The change in strain along a diameter is found from 2.15 in. 2.00 in.
2.00 in. 0.075 in./in.
D
D
D Ans.
(b) Note that the circumference of a circle is given by D. The change in strain around the circumference of the mandrel is found from
(2.15 in.) (2.00 in.)
(2.00 in.) 0.075 in./in.
C
C
C Ans.
P2.7 The normal strain in a suspended bar of material of varying cross section due to its own weight is given by the expression y/3E where is the specific weight of the material, y is the distance from the free (i.e., bottom) end of the bar, and E is a material constant. Determine, in terms of , L, and E, (a) the change in length of the bar due to its own weight.
(b) the average normal strain over the length L of the bar.
(c) the maximum normal strain in the bar.
Solution
(a) The strain of the suspended bar due to its own weight is given as 3
y E
Consider a slice of the bar having length dy. In general, = L.
Consider a slice of the bar having length dy. In general, = L.