05 Practice Problems
Problem 1:
A pin-connected truss composed of members AB and BC is subjected to a vertical force P = 40 kN at joint B (see figure). Each member is of constant crosssectional area: AAB = 0.004 m 2 and ABC = 0.002 m 2 . The diameter d of all pins is 20 mm, clevis thickness t is 10 mm, and the thickness t1 of the bracket is 15 mm. Determine the normal stress acting in each member and the shearing and bearing stresses at joint C.
Solution 1:
A free-body diagram of the truss is shown in Fig. b. The magnitudes of the axially directed end forces of members AB and BC, which are equal to the support reactions at A and C, are labeled FA and FC , respectively. For computational convenience the x and y components of the inclined forces are used rather than the forces themselves. Hence force FC is resolved into FCx and FC y , as shown.
(1) Calculation of support reactions. Relative dimensions are shown by a small triangle on the member BC in Fig. b. From the similarity of force and relative-dimension triangles, 3 4 FC x = FC , FC y = FC . (*) 5 5 4 It follows then that FCx = FCy . Application of equilibrium conditions to 3 the free-body diagram in Fig. b leads to 3 ∑ M c = 0 : P (1.5) − FA ( 2 ) = 0; FA = 4 P = 30kN (right directed), ∑ Fy = 0 : FCy − P = 0; FCy = P = 40kN (up directed),
∑ Fx = 0 :
− FCx + FA = 0; FCx = FA = 30kN (left directed). We thus have 5 FC = P = 50kN . 4 Note. The positive sign of FA and FC means that the sense of each of the forces was assumed correctly in the free-body diagram. (2) Calculation of internal forces. If imaginary cutting planes are passed perpendicular to the axes of the members AB and BC, separating each into two parts, it is observed that each portion is a two-force member. Therefore the internal forces in each member are the axial forces FA = 30 kN and FC = 50 kN. (3) Calculation of stresses. The normal stresses in each member are
FA 30 × 103 =− = −7.5MPa , σ AB = − AAB 0.004
FC 50 × 103 = = 25MPa , ABC 0.002 where the minus sign indicates compression. Referring to Fig. c, we see that the double shear in the pin C is 1 Fc 25 × 103 2 τc = 2 = = 79.6MPa . π d / 4 π ( 0.02 )2 / 4
σ BC =
For the bearing stress in the bracket at joint C, we have Fc 50 × 103 = = 166.7MPa , σb = t1d ( 0.015 ) ( 0.02 ) while the bearing stress in the clevis at joint C is given by Fc 50 ×103 = = 125MPa . σb = 2td 2 ( 0.01)( 0.02 ) The shear and bearing stresses in the other joints may be determined in a like manner.
Problem 2:
A force P of magnitude 200 N is applied to the handles of the bolt cutter shown in Fig. a. Compute (1) the force exerted on the bolt and rivets at joints A, B, and C and (2) the normal stress in member AD, which has a uniform crosssectional area of 2 × 10−4 m 2 . Dimensions are given in millimeters.
P
75
25
A
B
A
C
Q
25
(b) B
FA FBy
FBx
200 N 480
D 25
75
FBx
480
12
P=200 N
25
C FCx
12 FCy
(a)
(c)
Solution 2: Solution The conditions of equilibrium must be satisfied by the entire cutter. To determine the unknown forces, we consider component parts. Let the force between the bolt and the jaw be Q. The free-body diagrams for the jaw and the handle are shown in Figs. b and c. Since AD is a two-force member, the orientation of force FA is known. Note, that the force components on the two members at joint B must be equal and opposite, as indicated in the diagrams.
(1) Referring to the free-body diagram in Fig. b, we have
∑ Fx = 0 : ∑ Fy = 0 : ∑MB = 0:
FBx = 0, Q − FA + FBy = 0, Q ( 0.1) − FA ( 0.075 ) = 0,
FA = Q + FBy , FA =
Q , 0.75
from which Q = 3FBy . Using the free-body diagram in Fig. c, we obtain
∑ Fx = 0 :
− FBx + FCx = 0,
FCx = 0,
Q + 0.2, 3 ∑ M C = 0 : FBx ( 0.025) − FBy ( 0.012 ) + 0.2 ( 0.48) , FBx = 8kN. It follows that Q = 3(8) = 24 kN. Therefore the shear forces on the rivet at the joints A, B, and C are FA = 32 kN, FB = FBy = 8 kN, and FC = FCy = 8.2 kN, respectively.
∑ Fy = 0 :
− FBy + FCy − 0.2 = 0,
FCy =
(2) The normal stress in the member AD is given by
FA 32 × 103 = = 160MPa . σ= A 2 × 10−4 The shear stress in the pins of the cutter is investigated as described in Example 1.1. Note that the handles and jaws are subject to combined flexural and shearing stresses. Problem 3:
250 kN
100 cm
A short post constructed from a hollow circular tube of aluminum supports a compressive load of 250 kN (see figure). The inner and outer diameters of the tube are d1 = 9 cm and d 2 = 13 cm, respectively, and its length is 100 cm. The shortening of the post due to the load is measured as 0.5 mm. Determine the compressive stress and strain in the post. (Disregard the weight of the post itself, and assume that the post does not buckle under the load).
Hollow aluminum post in compression
Solution 3:
Assuming that the compressive load acts at the center of the hollow tube, we can use the equation σ = P / A to calculate the normal stress. The force P equals 250 kN, and the cross-sectional area A is A=
π 2 2 d 22 − d12 ) = ⎡(13 cm ) − ( 9 cm ) ⎤ = 69.08 cm 2 . ( ⎦⎥ 4 4 ⎣⎢
π
Therefore, the compressive stress in the post is P − 250 × 103 N σ= = = −36.19 MPa . A 69.08 × 10 − 4 m 2 The compressive strain is
ε=
δ L
=
0.5 × 10−3 100 × 10
−2
= 0.5 × 10−3 .
Problem 4:
A circular steel rod of length L and diameter d hangs in a mine shaft and holds an ore bucket of weight W at its lower end (see figure). (1) Obtain a formula for the maximum stress σ max in the rod, taking into account the weight of the rod itself. (2) Calculate the maximum stress if L = 40 m, d = 8 mm, and W = 1.5 kN.
Steel supporting weight W
rod a
Solution 4:
(1) The maximum axial force Fmax in the rod occurs at the upper end and is equal to the weight W plus the weight W0 of the rod itself. The latter is equal to the weight density γ of the steel times the volume V of the rod, or W0 = γ V = γ AL , in which A is the cross-sectional area of the rod. Therefore, the formula for the maximum stress becomes F W + γ AL W = +γL. σ max = max = A A A (2) To calculate the maximum stress, we substitute numerical values into the preceding equation. The cross-
πd2
, where d = 8 mm, and the weight 4 density γ of steel is 77.0 kN/ m3 . Thus, 1.5 kN σ max = + 77.0 kN/m3 ( 40 m ) = 2 π ( 8 mm ) / 4 = 29.84 MPa + 3.11 MPa = 33.0 MPa . Note, the weight of the rod contributes noticeably to the maximum stress and should not be disregarded. sectional area A equals
(
)
Problem 5:
A thin, triangular plate ABC is uniformly deformed into a shape ABC', as shown by the dashed lines in the figure. Calculate (1) the normal strain along the centerline OC; (2) the normal strain along the edge AC; and (3) the shearing strain between the edges AC and BC.
Solution 5:
Referring to the figure, we have
LOC = b and
LAC = LBC = b 2 = 1.41421b . (1), (2) Normal strains. As the change in length OC is Δb = 0.001b , Eq. (1.5) yields 0.001b ε oc = = 0.001 = 1.0 × 10−3 . b The lengths of the deformed edges are 1/2
2 LAC ' = LBC ' = ⎡b 2 + (1.001b ) ⎤ = 1.41492b . ⎢⎣ ⎥⎦ Thus 1.41492 − 1.41421 ε AC = ε BC = = 0.502 × 10−3 . 1.41421 (3) Shearing strain. Subsequent to deformation, angle ACB becomes ⎛ b ⎞ D AC ' B = 2 tan −1 ⎜ ⎟ = 89.943 . ⎝ 1.001b ⎠
The change in the right angle is then 90 − 89.943 = 0.057D . The corresponding shearing strain (in radians) is ⎛ π ⎞ −3 γ = 0.057 ⎜ ⎟ = 0.995 × 10 . ⎝ 180 ⎠ Note. Since the angle ACB is decreased, the shear strain is positive.
Problem 7:
A steel strut S serving as a brace for a boat hoist transmits a compressive force P = 54 kN to the deck of a pier (see figure (a)). The strut has a hollow square cross section with a wall thickness t = 12 mm (see figure (b)), and the angle θ between the strut and the horizontal is 40°. A pin through the strut transmits the compressive force from the strut to two gussets G that are welded to the base plate B. Four anchor bolts fasten the base plate to the deck. The diameter of the pin is d pin = 18 mm, the thickness of the gussets is tG = 15 mm, the thickness of the base plate is t B = 8 mm, and the diameter of the anchor bolts is dbolt = 12 mm. Determine the following stresses: (1) the bearing stress between the strut and the pin; (2) the shear stress in the pin; (3) the bearing stress between the pin and the gussets; (4) the bearing stress between the anchor bolts and the base plate, and (5) the shear stress in the anchor bolts. In solution, disregard any friction between the base plate and the deck. Solution 7:
(1) Bearing stress between strut and pin. The average value of the bearing stress between the strut and the pin is found by dividing the force in the strut by the total bearing area of the strut against the pin. The latter is equal to twice the thickness of the strut (because bearing occurs at two locations) times the diameter of the pin (see figure (b)). Thus, the bearing stress is P 54 kN σ b1 = = = 125 MPa . 2td pin 2 (12 mm )(18 mm )
This, stress is not excessive for a strut made of structural steel, since the yield stress is probably near 200 MPa (see Appendix A). Assuming the factor of safety f s = 1.5 allowable stress σ allow = 133 MPa. It means that the strut will be strong in bearing. (2) Shear stress in pin. As can be seen from figure (b), the pin tends to shear on two planes, namely, the planes between the strut and the gussets.
P 40
S
Pin G
B
S
G
G
(a)
t (b)
(a) Pin connection between strut S and base plate B. (b) Cross section through the strut S
Therefore, the average shear stress in the pin (which is in double shear) is equal to the total load applied to the pin divided by twice its cross-sectional area: 54 kN P τ pin = = = 106 MPa . 2π d pin 2 / 4 2π (18 mm )2 / 4 The pin would normally be made of high-strength steel (tensile yield stress greater than 340 MPa) and could easily withstand this shear stress (the yield stress in shear is usually at least 50% of the yield stress in tension). (3) Bearing stress between pin and gussets. The pin bears against the gussets at two locations, so the bearing area is twice the thickness of the gussets times the pin diameter; thus, P 54 kN σ b2 = = = 100 MPa , 2tG d pin 2 (15 mm )(18 mm ) which is less than the bearing stress against the strut. (4) Bearing stress between anchor bolts and base plate. The vertical component of the force P (see figure (a)) is transmitted to the pier by direct bearing between the base plate and the pier. The horizontal component, however, is transmitted through the anchor bolts. The average bearing stress between the base plate and the anchor bolts is equal to the horizontal component of the force P divided by the bearing area of four bolts. The bearing area for one bolt is equal to the thickness of the plate times the bolt diameter. Consequently, the bearing stress is P cos 40° ( 54 kN )( cos 40° ) σ b3 = = = 108 MPa . 4t B d bolt 4 ( 8 mm )(12 mm ) (5) Shear stress in anchor bolts. The average shear stress in the anchor bolts is equal to the horizontal component of the force P divided by the total cross-sectional area of four bolts (note that each bolt is in single shear). Therefore, ( 54 kN )( cos 40° ) = 119 MPa . P cos 40° τ bolt = = 4π d bolt 2 / 4 4π (12 mm )2 4 Note. Any friction between the base plate and the pier would reduce the load on the anchor bolts.
Problem 8:
A punch for making holes in steel plates is shown in Fig. a. Assume that a punch having a diameter of 19 mm is used to punch a hole in a 6-mm plate, as shown in the cross-sectional view (see figure (b)). If a force P = 125 kN is required, what is the average shear stress in the plate and the average compressive stress in the punch? P P d t
(b)
(a) Punching a hole in a steel plate
Solution 8:
The average shear stress in the plate is obtained by dividing the force P by the shear area of the plate. The shear area as is equal to the circumference of the hole times the thickness of the plate, or As = π dt = π (19 mm )( 6 mm ) = 358 mm 2 , in which d is the diameter of the punch and t is the thickness of the plate. Therefore, the average shear stress in the plate is P 125,000 N τ aver = = = 349 MPa . As 358 × 10−6 m 2 The average compressive stress in the punch is P P 125,000 N σc = = 2 = = 441 MPa , A punch πd 4 π 19 × 10 − 3 2 4
(
)
in which Apunch is the cross-sectional area of the punch. Note. This analysis is highly idealized because we are disregarding impact effects that occur when a punch is rammed through a plate.
Problem 9:
A bearing pad of the kind used to support machines and bridge girders consists of a linearly elastic material (usually an elastomer, such as rubber) capped by a steel plate (see figure (a)). Assume that the thickness of the elastomer is h, the dimensions of the plate are a × b , and the pad is subjected to a horizontal shear force Q. Obtain formulas for the average shear stress τ aver in the elastomer and the horizontal displacement d of the plate (see figure (b)).
Bearing pad in shear
Solution 9:
Assume that the shear stresses in the elastomer are uniformly distributed throughout its entire volume. Then the shear stress on any horizontal plane through the elastomer equals the shear force Q divided by the Q area of the plane (see figure (a)): τ . aver = ab The corresponding shear strain (from Hooke's law in shear, which will be considered below) is τ Q γ = aver = , Ge abGe in which Ge is the shear modulus of the elastomeric material. Finally, the horizontal displacement d is equal to h tan γ : ⎛ Q ⎞ d = h tan γ = h tan ⎜ ⎟. abG e⎠ ⎝ In most practical situations the shear strain γ is a small angle, and in such cases we may replace tan γ by γ and obtain hQ d = hγ = . abGe Equations mentioned above give approximate results for the horizontal displacement of the plate because they are based upon the assumption that the
shear stress and strain are constant throughout the volume of the elastomeric material. In reality the shear stress is zero at the edges of the material (because there are no shear stresses on the free vertical faces), and therefore the deformation of the material is more complex than pictured in the figure. However, if the length a of the plate is large compared with the thickness h of the elastomer, the results are satisfactory for design purposes.
A pin-connected truss composed of members AB and BC is subjected to a vertical force P = 40 kN at joint B (see figure). Each member is of constant crosssectional area: AAB = 0.004 m 2 and ABC = 0.002 m 2 . The diameter d of all pins is 20 mm, clevis thickness t is 10 mm, and the thickness t1 of the bracket is 15 mm. Determine the normal stress acting in each member and the shearing and bearing stresses at joint C.
Solution 1:
A free-body diagram of the truss is shown in Fig. b. The magnitudes of the axially directed end forces of members AB and BC, which are equal to the support reactions at A and C, are labeled FA and FC , respectively. For computational convenience the x and y components of the inclined forces are used rather than the forces themselves. Hence force FC is resolved into FCx and FC y , as shown.
(1) Calculation of support reactions. Relative dimensions are shown by a small triangle on the member BC in Fig. b. From the similarity of force and relative-dimension triangles, 3 4 FC x = FC , FC y = FC . (*) 5 5 4 It follows then that FCx = FCy . Application of equilibrium conditions to 3 the free-body diagram in Fig. b leads to 3 ∑ M c = 0 : P (1.5) − FA ( 2 ) = 0; FA = 4 P = 30kN (right directed), ∑ Fy = 0 : FCy − P = 0; FCy = P = 40kN (up directed),
∑ Fx = 0 :
− FCx + FA = 0; FCx = FA = 30kN (left directed). We thus have 5 FC = P = 50kN . 4 Note. The positive sign of FA and FC means that the sense of each of the forces was assumed correctly in the free-body diagram. (2) Calculation of internal forces. If imaginary cutting planes are passed perpendicular to the axes of the members AB and BC, separating each into two parts, it is observed that each portion is a two-force member. Therefore the internal forces in each member are the axial forces FA = 30 kN and FC = 50 kN. (3) Calculation of stresses. The normal stresses in each member are
FA 30 × 103 =− = −7.5MPa , σ AB = − AAB 0.004
FC 50 × 103 = = 25MPa , ABC 0.002 where the minus sign indicates compression. Referring to Fig. c, we see that the double shear in the pin C is 1 Fc 25 × 103 2 τc = 2 = = 79.6MPa . π d / 4 π ( 0.02 )2 / 4
σ BC =
For the bearing stress in the bracket at joint C, we have Fc 50 × 103 = = 166.7MPa , σb = t1d ( 0.015 ) ( 0.02 ) while the bearing stress in the clevis at joint C is given by Fc 50 ×103 = = 125MPa . σb = 2td 2 ( 0.01)( 0.02 ) The shear and bearing stresses in the other joints may be determined in a like manner.
Problem 2:
A force P of magnitude 200 N is applied to the handles of the bolt cutter shown in Fig. a. Compute (1) the force exerted on the bolt and rivets at joints A, B, and C and (2) the normal stress in member AD, which has a uniform crosssectional area of 2 × 10−4 m 2 . Dimensions are given in millimeters.
P
75
25
A
B
A
C
Q
25
(b) B
FA FBy
FBx
200 N 480
D 25
75
FBx
480
12
P=200 N
25
C FCx
12 FCy
(a)
(c)
Solution 2: Solution The conditions of equilibrium must be satisfied by the entire cutter. To determine the unknown forces, we consider component parts. Let the force between the bolt and the jaw be Q. The free-body diagrams for the jaw and the handle are shown in Figs. b and c. Since AD is a two-force member, the orientation of force FA is known. Note, that the force components on the two members at joint B must be equal and opposite, as indicated in the diagrams.
(1) Referring to the free-body diagram in Fig. b, we have
∑ Fx = 0 : ∑ Fy = 0 : ∑MB = 0:
FBx = 0, Q − FA + FBy = 0, Q ( 0.1) − FA ( 0.075 ) = 0,
FA = Q + FBy , FA =
Q , 0.75
from which Q = 3FBy . Using the free-body diagram in Fig. c, we obtain
∑ Fx = 0 :
− FBx + FCx = 0,
FCx = 0,
Q + 0.2, 3 ∑ M C = 0 : FBx ( 0.025) − FBy ( 0.012 ) + 0.2 ( 0.48) , FBx = 8kN. It follows that Q = 3(8) = 24 kN. Therefore the shear forces on the rivet at the joints A, B, and C are FA = 32 kN, FB = FBy = 8 kN, and FC = FCy = 8.2 kN, respectively.
∑ Fy = 0 :
− FBy + FCy − 0.2 = 0,
FCy =
(2) The normal stress in the member AD is given by
FA 32 × 103 = = 160MPa . σ= A 2 × 10−4 The shear stress in the pins of the cutter is investigated as described in Example 1.1. Note that the handles and jaws are subject to combined flexural and shearing stresses. Problem 3:
250 kN
100 cm
A short post constructed from a hollow circular tube of aluminum supports a compressive load of 250 kN (see figure). The inner and outer diameters of the tube are d1 = 9 cm and d 2 = 13 cm, respectively, and its length is 100 cm. The shortening of the post due to the load is measured as 0.5 mm. Determine the compressive stress and strain in the post. (Disregard the weight of the post itself, and assume that the post does not buckle under the load).
Hollow aluminum post in compression
Solution 3:
Assuming that the compressive load acts at the center of the hollow tube, we can use the equation σ = P / A to calculate the normal stress. The force P equals 250 kN, and the cross-sectional area A is A=
π 2 2 d 22 − d12 ) = ⎡(13 cm ) − ( 9 cm ) ⎤ = 69.08 cm 2 . ( ⎦⎥ 4 4 ⎣⎢
π
Therefore, the compressive stress in the post is P − 250 × 103 N σ= = = −36.19 MPa . A 69.08 × 10 − 4 m 2 The compressive strain is
ε=
δ L
=
0.5 × 10−3 100 × 10
−2
= 0.5 × 10−3 .
Problem 4:
A circular steel rod of length L and diameter d hangs in a mine shaft and holds an ore bucket of weight W at its lower end (see figure). (1) Obtain a formula for the maximum stress σ max in the rod, taking into account the weight of the rod itself. (2) Calculate the maximum stress if L = 40 m, d = 8 mm, and W = 1.5 kN.
Steel supporting weight W
rod a
Solution 4:
(1) The maximum axial force Fmax in the rod occurs at the upper end and is equal to the weight W plus the weight W0 of the rod itself. The latter is equal to the weight density γ of the steel times the volume V of the rod, or W0 = γ V = γ AL , in which A is the cross-sectional area of the rod. Therefore, the formula for the maximum stress becomes F W + γ AL W = +γL. σ max = max = A A A (2) To calculate the maximum stress, we substitute numerical values into the preceding equation. The cross-
πd2
, where d = 8 mm, and the weight 4 density γ of steel is 77.0 kN/ m3 . Thus, 1.5 kN σ max = + 77.0 kN/m3 ( 40 m ) = 2 π ( 8 mm ) / 4 = 29.84 MPa + 3.11 MPa = 33.0 MPa . Note, the weight of the rod contributes noticeably to the maximum stress and should not be disregarded. sectional area A equals
(
)
Problem 5:
A thin, triangular plate ABC is uniformly deformed into a shape ABC', as shown by the dashed lines in the figure. Calculate (1) the normal strain along the centerline OC; (2) the normal strain along the edge AC; and (3) the shearing strain between the edges AC and BC.
Solution 5:
Referring to the figure, we have
LOC = b and
LAC = LBC = b 2 = 1.41421b . (1), (2) Normal strains. As the change in length OC is Δb = 0.001b , Eq. (1.5) yields 0.001b ε oc = = 0.001 = 1.0 × 10−3 . b The lengths of the deformed edges are 1/2
2 LAC ' = LBC ' = ⎡b 2 + (1.001b ) ⎤ = 1.41492b . ⎢⎣ ⎥⎦ Thus 1.41492 − 1.41421 ε AC = ε BC = = 0.502 × 10−3 . 1.41421 (3) Shearing strain. Subsequent to deformation, angle ACB becomes ⎛ b ⎞ D AC ' B = 2 tan −1 ⎜ ⎟ = 89.943 . ⎝ 1.001b ⎠
The change in the right angle is then 90 − 89.943 = 0.057D . The corresponding shearing strain (in radians) is ⎛ π ⎞ −3 γ = 0.057 ⎜ ⎟ = 0.995 × 10 . ⎝ 180 ⎠ Note. Since the angle ACB is decreased, the shear strain is positive.
Problem 7:
A steel strut S serving as a brace for a boat hoist transmits a compressive force P = 54 kN to the deck of a pier (see figure (a)). The strut has a hollow square cross section with a wall thickness t = 12 mm (see figure (b)), and the angle θ between the strut and the horizontal is 40°. A pin through the strut transmits the compressive force from the strut to two gussets G that are welded to the base plate B. Four anchor bolts fasten the base plate to the deck. The diameter of the pin is d pin = 18 mm, the thickness of the gussets is tG = 15 mm, the thickness of the base plate is t B = 8 mm, and the diameter of the anchor bolts is dbolt = 12 mm. Determine the following stresses: (1) the bearing stress between the strut and the pin; (2) the shear stress in the pin; (3) the bearing stress between the pin and the gussets; (4) the bearing stress between the anchor bolts and the base plate, and (5) the shear stress in the anchor bolts. In solution, disregard any friction between the base plate and the deck. Solution 7:
(1) Bearing stress between strut and pin. The average value of the bearing stress between the strut and the pin is found by dividing the force in the strut by the total bearing area of the strut against the pin. The latter is equal to twice the thickness of the strut (because bearing occurs at two locations) times the diameter of the pin (see figure (b)). Thus, the bearing stress is P 54 kN σ b1 = = = 125 MPa . 2td pin 2 (12 mm )(18 mm )
This, stress is not excessive for a strut made of structural steel, since the yield stress is probably near 200 MPa (see Appendix A). Assuming the factor of safety f s = 1.5 allowable stress σ allow = 133 MPa. It means that the strut will be strong in bearing. (2) Shear stress in pin. As can be seen from figure (b), the pin tends to shear on two planes, namely, the planes between the strut and the gussets.
P 40
S
Pin G
B
S
G
G
(a)
t (b)
(a) Pin connection between strut S and base plate B. (b) Cross section through the strut S
Therefore, the average shear stress in the pin (which is in double shear) is equal to the total load applied to the pin divided by twice its cross-sectional area: 54 kN P τ pin = = = 106 MPa . 2π d pin 2 / 4 2π (18 mm )2 / 4 The pin would normally be made of high-strength steel (tensile yield stress greater than 340 MPa) and could easily withstand this shear stress (the yield stress in shear is usually at least 50% of the yield stress in tension). (3) Bearing stress between pin and gussets. The pin bears against the gussets at two locations, so the bearing area is twice the thickness of the gussets times the pin diameter; thus, P 54 kN σ b2 = = = 100 MPa , 2tG d pin 2 (15 mm )(18 mm ) which is less than the bearing stress against the strut. (4) Bearing stress between anchor bolts and base plate. The vertical component of the force P (see figure (a)) is transmitted to the pier by direct bearing between the base plate and the pier. The horizontal component, however, is transmitted through the anchor bolts. The average bearing stress between the base plate and the anchor bolts is equal to the horizontal component of the force P divided by the bearing area of four bolts. The bearing area for one bolt is equal to the thickness of the plate times the bolt diameter. Consequently, the bearing stress is P cos 40° ( 54 kN )( cos 40° ) σ b3 = = = 108 MPa . 4t B d bolt 4 ( 8 mm )(12 mm ) (5) Shear stress in anchor bolts. The average shear stress in the anchor bolts is equal to the horizontal component of the force P divided by the total cross-sectional area of four bolts (note that each bolt is in single shear). Therefore, ( 54 kN )( cos 40° ) = 119 MPa . P cos 40° τ bolt = = 4π d bolt 2 / 4 4π (12 mm )2 4 Note. Any friction between the base plate and the pier would reduce the load on the anchor bolts.
Problem 8:
A punch for making holes in steel plates is shown in Fig. a. Assume that a punch having a diameter of 19 mm is used to punch a hole in a 6-mm plate, as shown in the cross-sectional view (see figure (b)). If a force P = 125 kN is required, what is the average shear stress in the plate and the average compressive stress in the punch? P P d t
(b)
(a) Punching a hole in a steel plate
Solution 8:
The average shear stress in the plate is obtained by dividing the force P by the shear area of the plate. The shear area as is equal to the circumference of the hole times the thickness of the plate, or As = π dt = π (19 mm )( 6 mm ) = 358 mm 2 , in which d is the diameter of the punch and t is the thickness of the plate. Therefore, the average shear stress in the plate is P 125,000 N τ aver = = = 349 MPa . As 358 × 10−6 m 2 The average compressive stress in the punch is P P 125,000 N σc = = 2 = = 441 MPa , A punch πd 4 π 19 × 10 − 3 2 4
(
)
in which Apunch is the cross-sectional area of the punch. Note. This analysis is highly idealized because we are disregarding impact effects that occur when a punch is rammed through a plate.
Problem 9:
A bearing pad of the kind used to support machines and bridge girders consists of a linearly elastic material (usually an elastomer, such as rubber) capped by a steel plate (see figure (a)). Assume that the thickness of the elastomer is h, the dimensions of the plate are a × b , and the pad is subjected to a horizontal shear force Q. Obtain formulas for the average shear stress τ aver in the elastomer and the horizontal displacement d of the plate (see figure (b)).
Bearing pad in shear
Solution 9:
Assume that the shear stresses in the elastomer are uniformly distributed throughout its entire volume. Then the shear stress on any horizontal plane through the elastomer equals the shear force Q divided by the Q area of the plane (see figure (a)): τ . aver = ab The corresponding shear strain (from Hooke's law in shear, which will be considered below) is τ Q γ = aver = , Ge abGe in which Ge is the shear modulus of the elastomeric material. Finally, the horizontal displacement d is equal to h tan γ : ⎛ Q ⎞ d = h tan γ = h tan ⎜ ⎟. abG e⎠ ⎝ In most practical situations the shear strain γ is a small angle, and in such cases we may replace tan γ by γ and obtain hQ d = hγ = . abGe Equations mentioned above give approximate results for the horizontal displacement of the plate because they are based upon the assumption that the
shear stress and strain are constant throughout the volume of the elastomeric material. In reality the shear stress is zero at the edges of the material (because there are no shear stresses on the free vertical faces), and therefore the deformation of the material is more complex than pictured in the figure. However, if the length a of the plate is large compared with the thickness h of the elastomer, the results are satisfactory for design purposes.
