Mechanics of Materials 2nd Edition Philpot Solution Manual PDF

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2.1 When an axial load is applied to the ends of the bar shown in Fig. 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. Fig. P2.1

Solution (a) From the definition of normal strain, the elongation in segment (2) can be computed as 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

The strain in segment (1) can now be computed: 30 in. L1 40 in.

Ans.

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2.2 A rigid steel bar is supported by three rods, as shown in Fig. P2.2. There is no strain in the rods before the load P is applied. After load P is applied, the normal strain in rod (2) is 1,080 in./in. Assume initia rod lengths of L1 = 130 in. and L2 = 75 in. Determine: (a) the normal strain in rods (1). (b) the normal strain in rods (1) if there is a 1/32-in. 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 rods (1) if there is a 1/32-in. gap in the connection between the rigid bar and rod (2) at joint B before the load is applied. Fig. P2.2

Solution (a) From the normal strain in rod (2) and its length, the deformation of rod (2) can be calculated: Since rod (2) 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 (2); therefore, vB in. (downward) By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rods (1) are connected to the rigid bar at A and C, and again, perfect connections are assumed. The deformation of rod (1) must be equal to the deflection of joint A (or C); thus, 1 = 0.0810 in. The normal strain in rods (1) can now be calculated as: 10 in. Ans. L1 130 in. (b) We can assume that the bolted connection at B is perfect; therefore, vB = 0.0810 in. (downward). Further, the rigid bar must remain horizontal as it deflects downward by virtue of symmetry. Therefore, the deflection downward of joints A and C is still equal to 0.0810 in. What effect is caused by the gap at A and C? When joint A (or C) moves downward by 0.0810 in., the first 1/32-in. of this downward movement does not stretch rod (1)—it just closes the gap. Therefore, rod (1) only gets elongated by the amount in.

5 in.

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This deformation creates a strain in rod (1) of: 975 in. L1 130 in.

Ans.

(c) The gap is now at joint B. We know the strain in rod (2); hence, we know its deformation must be 0.0810 in. However, the first 1/32-in. of downward movement by the rigid bar does not elongate the rod—it simply closes the gap. To elongate rod (2) by 0.0810 in., joint B must move down: 5 in. vB

5 in. Again, since the rigid bar remains horizontal, vB = vA = vC. The joints at A and C are assumed to be perfect; thus, 5 in. vA and the normal strain in rod (1) is: 225 in. L1 130 in.

Ans.

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2.3 A rigid steel bar is supported by three rods, as shown in Fig. P2.3. There is no strain in the rods before the load P is applied. After load P is applied, the normal strain in rods (1) is 860 m/m Assume initial rod lengths of L1 = 2,400 mm and L2 = 1,800 mm. Determine: (a) the normal strain in rod (2). (b) the normal strain in rod (2) if there is a 2mm 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 2mm gap in the connection between the rigid bar and rod (2) at joint B before the load is applied. Fig. P2.3

Solution (a) From the normal strain in rod (1) and its length, the deformation of rod (1) can be calculated: 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, vA mm (downward) 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: 4 mm Ans. m L2 1,800 mm (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: vA By symmetry, the rigid bar remains horizontal; therefore, vB = vA = vC. The joint at B is assumed to be perfect; thus, any downward movement of the rigid bar also elongates rod (2) by the same amount: vB mm

and the normal strain in rod (2) is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

L2

4 mm 1,800 mm

mm

Ans.

m

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

mm This deformation creates a strain in rod (2) of: 4 mm L2 1,800 mm

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2.4 A rigid bar ABCD is supported by two bars as shown in Fig. 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. Fig. P2.4

Solution (a) From the strain given for rod (1)

00 mm) mm Therefore, vB = 0.5130 mm (downward). From the deformation diagram of rigid bar ABCD vB vC 240 mm (240 mm ) 600 mm vC 240 mm Therefore, L2

2

= 1.2825 mm (elongation), and thus, from the definition of strain: 25 mm 1,500 mm

Ans.

(b) The 1-mm gap at C doesn’t affect rod (1); therefore, 1 = −0.5130 mm. The rigid bar 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 before it begins to elongate member (2). Therefore, the elongation of member (2) is

mm

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and so the strain in member (2) is 25 mm L2 1,500 mm

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 vB From deformation diagram of rigid bar ABCD vB vC 240 mm (240 mm ) 600 mm vC 240 mm and so 25 mm L2 1,500 mm

Ans.

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2.5 In Fig. 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).

Fig. P2.5

Solution From the strain given for member (1)

in.) in. Pin A has to move 0.25 in. before it contacts member (1); therefore, vA = 1.080 mm (downward). vA in. in. (to the left)

From the deformation diagram of rigid bar ABC vA v 12 in. 20 in. 20 in. vC wnward) 12 in. Therefore, 2 = 0.4860 in. (elongation). From the definition of strain, 60 in. L2 160 in.

Ans.

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2.6 The sanding-drum mandrel shown in Fig. P2.6 is made for use with a hand drill. The mandrel is made from a rubber-like material tha 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.

Fig. P2.6

Solution (a) The change in strain along a diameter is found from D

Ans. D

2.00 in.

(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 C

Ans. .)

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2.7 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 3E Consider a slice of the bar having length dy. In general, = L. Applying this definition to the bar slice, the deformation of slice dy is given by d

3E Since this strain expression varies with y, the total deformation of the bar must be found by integrating d over the bar length: L

L

Ans. 0

3E

3E

(b) The average normal strain is found by dividing the expression above for

by the bar length L Ans.

6E

L

(c) Since the given strain expression varies with y, the maximum normal strain occurs at the maximum value of y, that is, at y = L: y

Ans.

3E

y

3E

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2.8 A steel cable is used to support an elevator cage at the bottom of a 2,000-ft deep mineshaft. A uniform normal strain of 250 in./in. is produced in the cable by the weight of the cage. At each point, the weight of the cable produces an additional normal strain that is proportional to the length of the cable below the point. If the total normal strain in the cable at the cable drum (upper end of the cable) is 700 in./in., determine (a) the strain in the cable at a depth of 500 ft. (b) the total elongation of the cable.

Solution Call the vertical coordinate y and establish the origin of the y axis at the lower end of the steel cable. The strain in the cable has a constant term (i.e., = 250 ) and a term (we will call it k) that varies with the vertical coordinate y. The problem states that the normal strain at the cable drum (i.e., y = 2,000 ft) is 700 in./in. Knowing this value, the constant k can be determined 700 ft) 700 2, 000 ft Substituting this value for k in the strain expression gives

(a) At a depth of 500 ft, the y coordinate is y = 1,500 ft. Therefore, the cable strain at a depth of 500 ft is Ans. (b) The total elongation is found by integrating the strain expression over the cable length; thus, 2000

0 2000

y t

.

Ans.

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2.9 The 16 × 22 × 25-mm rubber blocks shown in Fig. P2.9 are used in a double U shear mount to isolate the vibration of a machine from its supports. An applied load of P = 690 N causes the upper frame to be deflected downward by 7 mm. Determine the average shear strain and the shear stress in the rubber blocks.

Fig. P2.9

Solution Consider the deformation of one block. After a downward deflection of 7 mm: 7 mm tan 16 mm and thus, the shear strain in the block is Ans. ad 0 μrad Note that the small angle approximation most definitely does not apply here! The applied load of 690 N is carried by two blocks; therefore, the shear force applied to one block is V = 345 N. The area subjected to shear stress is the area that is parallel to the direction of the shear force; that is, the 22 mm by 25 mm surface of the block. The shear stress is 345 N (22 mm)(25 mm) MPa a

Ans.

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2.10 A thin polymer plate PQR is deformed such that corner Q is displaced downward 1/16-in. to new position Q’ as shown in Fig. P2.10. Determine the shear strain at Q’ associated with the two edges (PQ and QR).

Fig. P2.10

Solution Before deformation, the angle of the plate at Q was 90° or /2 radians. We must now determine the plate angle at Q′ after deformation. The difference between these angles is the shear strain. After point Q displaces downward by 1/16in., the angle ′ is 25 in. 25 in. tan (10 in. .) 10.0625 in. and the angle ′ is

4 in.

tan

(10 in.

.)

4 in. 10.0625 in.

After deformation, the angle of the plate at Q′ is rad The difference in the plate angle at Q before and after deformation is the shear strain:

Ans. 2

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2.11 A thin polymer plate PQR is deformed so that corner Q is displaced downward 1.0 mm to new position Q’ as shown in Fig. P2.11. Determine the shear strain at Q’ associated with the two edges (PQ and QR).

Fig. P2.11

Solution Before deformation, the angle of the plate at Q was 90° or /2 radians. We must now determine the plate angle at Q′ after deformation. The difference between these angles is the shear strain. After deformation, the angle ′ is 120 mm 120 mm tan (300 mm 301 mm and the angle ′ is 750 mm tan (300 mm

750 mm 301 mm

Therefore, after deformation, the angle of the plate at Q′ is rad The difference in the angle at Q before and after deformation is the shear strain:

Ans. 2

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2.12 A thin rectangular plate is uniformly deformed as shown in Fig. P2.12. Determine the shear strain xy at P.

Fig. P2.12

Solution Before deformation, the angle of the plate at P was 90° or /2 radians. We must now determine the plate angle at P after deformation. The difference between these angles is the shear strain. After deformation, the angle that side PQ makes with the horizo...