121_pdfsam_1Mechanics of Materials(3 Ed)[Team Nanban]Tmrg

8/18/2019 121_pdfsam_1Mechanics of Materials(3 Ed)[Team Nanban]Tmrg

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SHEAR STRAIN

Prob. 2.7-11. A rectangular plate (dashed lines show origi-nal configuration) is uniformly deformed into the shape of aparallelogram (shaded figure) as shown in Fig. P2.7-11. (a)

Determine the average shear strain, call it xy( A), betweenlines in the directions x and y shown in the figure. (b)Determine the average shear strain, call it (B), betweenlines in the directions of x and y shown in the figure.(Hint:Don’t forget that shear strain is a signed quantity, that is, itcan be either positive or negative.)

g x¿ y¿

g

shear stress and average shear strain.) (b) If the shear ulus of elasticity of the rubber is Gr 0.6 MPa, what average shear strain, , related to the average shear strcomputed in Part (a)? (c) Based on the average shear sdetermined in Part (b), what is the relative displacemebetween the rectangular bar and the C-shaped bracket wthe load P 250 N is applied?

g

End view

Ball

Solid steel hitch bar

Steel pin

Square tubular

steel hitch

receptacle

Side view

Top view

P (Towing load from trailer)

P

V V

P2.7-10

▼

10 in.

8 in.

y

A B x

x ′

y′

0.10 in. 0.10 in.

Prob. 2.7-12. Shear stress produces a shear strain xy (be-tween lines in the x direction and lines in the y direction) of

xy 1200 (i.e., 0.0012 ). (a) Determine the horizon-tal displacement A of point A. (b) Determine the shearstrain x y between the lines in the x direction and the y

direction, as shown on Fig. P2.7-12.Prob. 2.7-13. Two identical symmetrically placed rubberpads transmit load from a rectangular bar to a C-shapedbracket, as shown in Fig. P2.7-13. (a) Determine the averageshear stress, , in the rubber pads on planes parallel to thetop and bottom surfaces of the pads if P 250 N and the di-mensions of the rubber pads are: b 50 mm, w 80 mm,and h 25 mm. (Although the load is transmitted predom-inately by shearing deformation, the pads are not undergo-ing pure shear. However, you can still calculate the average

t

g

mmgg

g

P2.7-11

τ

τ

b

(b) Deformed ru pad.

δ

P

w

h

b

(a) Configuration of rubber load-transfer pads.

P2.7-13 and P2.7-14

DProb. 2.7-14. Two identical, symmetrically placed rupads transmit load from a rectangular bar to a C-shbracket, as shown in Fig. 2.7-14. The dimensions of theber pads are: b 3 in., w 4 in., and h 2 in. The modulus of elasticity of the rubber is Gr 100 psi. Imaximum relative displacement between the bar andbracket is max 0.25 in., what is the maximum valload P that may be applied? (Use average shear strainaverage shear stress in solving this problem.)

*Prob. 2.7-15. Vibration isolators like the one shown inP2.7-15 are used to support sensitive instruments.Each tor consists of a hollow rubber cylinder of outer diametinner diameter d, and height h.A steel center post of di

ter d is bonded to the inner surface of the rubber cylindethe outer surface of the rubber cylinder is bonded to the surface of a steel-tube base. (a) Derive an expression foaverage shear stress in the rubber as a function of the disr from the center of the isolator. (b) Derive an expressiolating the load P to the downward displacement of the cpost, using G as the shear modulus of the rubber, and asing that the steel post and steel tube are rigid (comparedthe rubber). (Hint: Since the shear strain varies with thtance r from the center, an integral is required.)

150 mm

120 mm

y

x

A

x ′

y′

δ Aτ

τ

P2.7-12


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D

P

d

h

Steel tube

Steelpost

Rubber

Instrument

Vibration

isolator

r

(b)(a)

P2.7-15

DESIGN FOR AXIAL LOADS AND DIRECT SHEAR

DProb. 2.8-1. A bolted lap joint is used to connect a rectan-

gular bar to a hanger bracket, as shown in Fig. P2.8-1. If theallowable shear stress in the bolt is 15 ksi, and the allowabletensile load on the rectangular bar is to beP allow. 2 kips, whatis the required minimum diameter of the bolt shank in inches?

MDS 2.12–2.15

▼

PP

P2.8-1 and P2.8-2

DProb. 2.8-2. A bolted lap joint is used to connect a rectan-

gular bar to a hanger bracket, as shown in Fig. P2.8-2. If the

allowable shear stress in the bolt is 80 MPa, and the diame-ter of the bolt shank 15 mm, what is the allowable tensileload on the rectangular bar, P allow., in kN?D

Prob. 2.8-3. The pin that holds the two halves of a pair of pliers together at B has a diameter d 6.35 mm and is madeof steel for which allow. 75 MPa. What is the allowableforce (P C )allow. (not shown) that can be exerted on the roundrod at C by each jaw, assuming that the corresponding forceP A is applied to the handles at each of the two places marked A in Fig. P2.8-3?

t

P A

40 mm120 mm

A

P A

A

a a

d

View a – a

B C

B

P2.8-3

P

P

A

d p

d r

(a)

(b)

d p

T

Cable stay

U-bracket

Tee-bracket

Boat deck

P2.8-4

P2.8-5

DProb. 2.8-5. The forestay (cable) on a sailboat is attache

to a tee-bracket on the deck of the boat by a (removablstainless steel pin. If the allowable shear stress in the pin allow. 11 ksi, and the diameter of the pin is d p 0.25 inwhat is the allowable tensile force, T allow., in the stay?

DProb. 2.8-4. The brass eye-bar in Fig. P2.8-4a has a diam

ter dr 0.500 in. and is attached to a support bracket bybrass pin of diameter d p 0.375 in. If the allowable shestress in the pin is 12 ksi and the allowable tensile stress the bar is 18 ksi, what is the allowable tensile load P allow.?

102


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DProb. 2.8-6. A compressor of weight W is suspended from

a sloping ceiling beam by long rods AB and CD of diametersd1 and d2, respectively, as shown in Fig. P2.8-6a. A typicalbracket is shown in Fig. P2.8-6b.Using the data given below,determine the allowable compressor weight,W allow.. (Neglectthe weight of the platform between A and C , and neglect theweight of the two rods. Also, assume that rod AB and thepins at A and B are large enough that they do not need to beconsidered.)

a 0.75 m, b 0.50 m

Pins at C and D: d p 7 mm, tallow. 100 MPa

Rod CD: d2 10 mm, sallow. 85 MPa

a high-strength steel pin. Assume that the pin at B isquate to sustain the loading applied to it, and that the decritical components are the bar BD and the pin at Cfactor of safety against failure of BD by yielding is and the factor of safety against ultimate shear failure opin at C is 3.3. (a) Determine the required thickt , of the rectangular bar BD, whose width is b.(b) Deterthe required diameter, d, of the pin at C.

Bar BD: b 1 in., sY 36 ksi, Pin C : tU 60 kP

2400 lb,

L

6 ft,

h

4 ft

FSt

FSs

DProb. 2.8-7. An angle bracket ABC is restrained by a high-

strength steel wire CD, and it supports a load P at A, asshown in Fig. P2.8-7. The strength properties of the wire and

the shear pin at B are y 350 MPa (wire), and U 300MPa (pin at B). If the wire and pin are to be sized to providea factor of safety against yielding of the wire of 3.3and a factor of safety against ultimate shear failure of the pinof 3.5, what are the required diameters of the wire (tothe nearest mm) and the pin (to the nearest mm)?FS

t

FSs

t

A

B

C

D

(1)

L2

L1

a b

(2)

W

(a) (b) Typ. bracket

P2.8-6

h

d

P

C

D

b

a

a

B

t = thickness

View a-a

A

L/2 L/2

P2.8-8 and P2.8-9

90 mm

d p

P = 1 kN

C

d w

B

View a-a of B

A

100 mm

15°

a

a

D

P2.8-7

L

A B

Rigid

C

δC

D

b /2b /2

W

P2.8-10 and P2.8-11

DProb. 2.8-9. Solve Prob. 2.8-8 using the following data

D

Prob. 2.8-10. A load W is to be suspended from a cabendC of a rigid beam AC , whose length is b 3 m.Beamin turn, is supported by a steel rod of diameter d 25and length L 2.5 m. Rod BD is made of steel with a point Y 250 MPa, and modulus of elasticity E 200 If the maximum displacement at C is ( C )max 10 mmthere is to be a factor of safety with respect to yielding oof FS 3.3 and with respect to displacement of FS what is the allowable weight that can be suspended frombeam at C ?

d

Pin C : tU 400 MPa

Bar BD: b 25 mm, sY 250 MPa,

P 10 kN, L 3 m, h 2 m

DProb. 2.8-8. Boom AC in Fig. P2.8-8 is supported by a rec-

tangular steel bar BD, and it is attached to a bracket at C by


4/20

DProb. 2.8-11. A load of W 2 kips is suspended from acable at end C of a rigid beam AC , whose length is b 8 ft.Beam AC , in turn, is supported by a steel rodBD (E 30 103 ksi) of diameter d and length L 12 ft.The rodBD is tothe sized so that there will be a factor of safety with respectto yielding of FS 4.0 and a factor of safety with respect todeflection of FS 3.0. The yield strength of rodBD is Y 50 ksi, and the maximum displacement at C is limited to

(C )

max 0.25 in. . Determine the

required diameter, d, of rod BD to the nearest in.DProb. 2.8-12. A tension rod is spliced together by a pin-and-yoke type connector, as shown in Fig.P2.8-12.The tensionrod is to be designed for an allowable load of P allow. 3 kips.If the allowable tensile stress in the rods is allow. 15 ksi,and the allowable shear stress in the pin is allow. = 12 ksi, de-termine (to the nearest in.) (a) the smallest diameter,dr , of rod that can be used, and (b) the smallest diameter,d p, of pinthat can be used.

116

18

ai.e., (dC )allow

(dC )max

FSd bd

and by a tie-rod AB.Both the tie rod and the pin are to bsized with a factor of safety of FS 3.0, the tie-rod with rspect to tensile yielding and the pin with respect to shefailure. The strength properties of the rod and pin are: Y 340 MPa and 340 MPa; the respective lengths are:L11.5 m and L2 L3 2.0 m. (a) If the loading platform is be able to handle loads W up to W 8 kN, what is the rquired diameter, dr , of the tie-rod (to the nearest millimter)? (b) What is the required diameter, d p, of the shear p

at C (to the nearest millimeter)?DProb. 2.8-15. A three-bar, pin-jointed, planar trusupports a single horizontal load P at joint B.Joint C is free move horizontally.The allowable stress in tension is ( )allow.140 MPa, and the allowable stress in compression is ( C )allow.85 MPa. If the truss is to support a maximum loP allow. 50 kN, what are the required cross-sectional area Ai, of the three truss members?

sT

tU

P P

d r d r

d p

P2.8-12

A

B P

C

(1)

(2) (3)

1200 mm

450 mm

600 mm

P2.8-15

COMPUTER-BASED DESIGN FOR AXIAL LOADS

L1

L2 L3

A

B

C

a a

d r

W

d pView a-a of C

P2.8-13 and P2.8-14

DProb. 2.8-14. The L-shaped loading-platform frame in Fig.P2.8-14 is supported by a high-strength steel shear pin at C

DProb. 2.8-13. The L-shaped loading frame in Fig. P2.8-13 issupported by a high-strength shear pin (d p 0.5 in, U 50 ksi)and by a tie-rod AB (dr 0.625 in., Y 50 ksi). Both thetie-rod and the pin are to be sized with a factor of safety of FS 3.0, the tie-rod with respect to tensile yielding, and the

shear pin with respect to ultimate shear failure. Determine theallowable platform load,W allow.. Let L1 3 ft,L2 L3 4 ft.

CProb. 2.8-16. The pin-jointed planar truss shown in FiP2.8-16a is to be made of two steel two-force members ansupport a single vertical load P 10 kN at joint B. For thsteel truss members,the allowable stress in tension is ( T )allow150 MPa, the allowable stress in compression is ( C )allow 100 MPa, and the weight density is 77.0 kN/m3. You are consider truss designs for which joint B can be located at anpoint along the vertical line that is 1 m to the right of Awith yB varying from yB 0 to yB 2m. (a) Show that,each member has the minimum cross-sectional area thmeets the strength criteria stated above, the weight W of thtruss can be expressed as a function of yB, the position

joint B, by the function that is plotted in Fig. P2.8-16b. (What value of yB gives the minimum-weight truss, and whis the weight of that truss?

For Problems 2.8-16 through 2.8-18 you are to develop acomputer program to generate the required graph(s) tha

will enable you to choose the “optimum design.” You

may use a spreadsheet program or other mathematica

application program (e.g., TK Solver or Mathcad), o

you may write a program in a computer language (e.g.

BASIC or FORTRAN).

▼

104


5/20

C*Prob. 2.8-17. The pin-jointed planar truss shown in Fig.P2.8-17 is to be made of three steel two-force members andis to support vertical loads P

B 2 kips at joint B and P

C

3 kips at joint C.The lengths of members AB and BC are L1 30 in. and L2 24 in., respectively. Joint C is free to movevertically. For the steel truss members, the allowable stress intension is ( T )allow 20 ksi, the allowable stress in compression

is ( C )allow 12 ksi, and the weight density is 0.284 lYou are to consider truss designs for which the vertical mber AC has lengths varying from L3 18 in. to L3 50 iShow that, if each member has the minimum cross-sectarea that meets the strength criteria stated above, the wW of the truss can be expressed as a function of the leL3 of member AC by a function that is similar to theplotted in Fig. P2.8-16b. (Hint: Use the law of cosines ttain expressions for the angle at joint A and the angle at

C.) (b) What value of L3 gives the minimum-weight and what is the weight of that truss?C*Prob. 2.8-18. The pin-jointed planar truss shown inP2.8-18 is to be made of two aluminum two-force memand support a single horizontal load P 50 kN at joiFor the aluminum truss members, the allowable stress insion is ( T )allow 200 MPa, the allowable stress in comsion is ( C )allow 130 MPa, and the weight density iskN/m3.You are to consider truss designs for which suppcan be located at any point along the x axis, with xC vafrom xC 1 m to xC 2.4 m. (a) Show that, if each mehas the minimum cross-sectional area that meetsstrength criteria stated above, the weight W of the trus

be expressed as a function of xC , the position of suppoby a function that is similar to the one plotted in Fig.2.8(b) What value of xC gives the minimum-weight trusswhat is the weight of that truss?

2 m

A

C

B

(1)

y B

1 m

(2)

P

25

20

15

10

5

00.00 0.40 0.80 1.20 1.60 2.000.20 0.60 1.00 1.40 1.80

W ( N )

y B (m)

W min = 12.58 N @ y B = 0.80 m

(a) A two-bar planar truss.

(b) Minimum-weight design for the two-bar planar truss.

y

x

P2.8-16

A

B

C

P B = 2 kips

PC = 3 kips

L1 = 30 in.

L2 = 24 in.

L3

P2.8-17

x C

x

y

1 m

1 m

A

C

B P = 50 kN

P2.8-18

STRESSES ON INCLINED PLANES▼

In Problems 2.9-1 through 2.9-13, use free-body diagr

and equilibrium equations to solve for the required stres

Prob. 2.9-1. The plane NN makes an angle 30

respect to the cross section of the prismatic bar showFig. P2.9-1. The dimensions of the rectangular cross se

P2.9-1

N ′

P

P N

θ


6/20

of the bar are 1 in. 2 in. Under the action of an axialtensile load P , the normal stress on the NN plane is n 8 ksi. (a) Determine the value of the axial load P ; (b) de-termine the shear stress nt on the NN plane; and (c) deter-mine the maximum shear stress in the bar. Use free-bodydiagrams and equilibrium equations to solve for the re-quired stresses.

Prob. 2.9-2. The prismatic bar in Fig. P2.9-2 is subjected toan axial compressive load P 70 kips. The cross-sectional

area of the bar is 2.0 in2

. Determine the normal stress andthe shear stress on the n face and on the t face of an elementoriented at angle 40. Use free-body diagrams and equi-librium equations to solve for the required stresses.

DProb. 2.9-6. A brass bar with a square cross section of d

mension b is subjected to a compressive load P 10 kips,shown in Fig. P2.9-6. If the allowable compressive stress fthe brass is allow. 12 ksi, and the allowable shear stress allow. 7 ksi, what is the minimum value of the dimensionto the nearest in.?116

P2.9-2 and P2.9-3

θPP

t n P2.9-6

P

P b

b

P2.9-4

P

P

P2.9-5

P

N ′

P

N

θ

Prob. 2.9-3. A prismatic bar in tension has a cross section

that measures 20 mm 50 mm and supports a tensile loadP 200 kN, as illustrated in Fig.P2.9-3. Determine the normaland shear stresses on the n and t faces of an element orientedat angle 30. Use free-body diagrams and equilibriumequations to solve for the required stresses.D

Prob. 2.9-4. Determine the allowable tensile load P for theprismatic bar shown in Fig. P2.9-4 if the allowable tensilestress is allow 135 MPa and the allowable shear stress is allow 100 MPa. The cross-sectional dimensions of the barare 12.7 mm 50.8 mm.

Prob. 2.9-5. A bar with rectangular cross section is subjectedto an axial tensile load P , as shown in Fig.P2.9-5.(a) Determinethe angle, call it na, of the plane NN on which nt 2 n, thatis, the plane on which the magnitude of the shear stress istwice the magnitude of the normal stress. (b) Determine theangle, call it nb, of the plane on which n 2 nt . (Hint:Youcan get approximate answers from Fig. 2.34.)

Prob. 2.9-7. A 6-in.-diameter concrete test cylinder is su jected to a compressive load P 110 kips, as shown in FP2.9-7. The cylinder fails along a plane that makes an angof 62 to the horizontal. (a) Determine the (compressivaxial stress in the cylinder when it reaches the failure loa(b) Determine the normal stress, , and shear stress, , on th

failure plane at failure.

P2.9-7

P = 110 kips

62°

Prob. 2.9-8. A wood cube that has dimension b on eaedge is tested in compression, as illustrated in Fig. P2.9The direction of the grain of the wood is shown in the figurDetermine the normal stress n and shear stress nt on planthat are parallel to the grain of the wood.

P 5 kN, b 150 mm, a 55°

P2.9-8

P

α

106


7/20

D*Prob. 2.9-9. Either a finger-joint splice, Fig. P2.9-9a, or a

diagonal lap-joint splice, Fig. P2.9-9b, may be used to gluetwo wood strips together to form a longer tension member.Determine the ratio of allowable loads, (P f )allow./(P d)allow. forthe following two glue-strength cases: (a) the glue is twice asstrong in tension as it is in shear, that is allow. 0.5 allow., and(b) the glue is twice as strong in shear as it is in tension, thatis allow. 2 allow.. (Hint:For each of the above cases, determineP allow. in terms of allow., using the given glue strength ratios.)

Prob. 2.12-2 The normal stress on the rectangular crostion ABCD in Fig. P2.12-2 varies linearly with respect t y coordinate.That is, x has the form x a by, varyinearly from xb at the bottom edge of the cross section tat the top edge of the cross section. (a) Show thatM y this symmetrical normal-stress distribution. (b) Deteran expression for the axial force F x in terms of the str xb and xt and the dimensions of the cross section, wi

and height h. (c) Determine an expression for the corresding value of the bending moment M z.P2.9-9

(b)

P f P f 8°

(a)

Pd Pd

30°

Prob. 2.9-10. At room temperature (70F) and with no axialload (P 0) the extensional strain of the prismatic bar (Fig.P2.9-10) in the axial direction is zero, that is, x 0.Subsequently, the bar is heated to 120F and a tensile loadP is applied. The material properties for the bar are: E 10 103 ksi and 13 106/F, and the cross-sectionalarea of the bar is 1.8 in2. For the latter load-temperature condi-tion, the extensional strain is found to be .(a) Determine the value of the axial tensile load P . (b) De-

termine the normal stress and the shear stress on the obliqueplane NN . (Note: The total strain is the sum of strain asso-ciated with normal stress x (Eq. 2.14) and the strain due tochange of temperature T (Eq. 2.8).)

x

900 106 in.in.

P2.9-10

P

N ′

P

x

N

25°

Gage measuresextensionalstrain x

STRESS RESULTANTS

Prob. 2.12-1. The normal stress, x, over the top half of thecross section of the rectangular bar in Fig.P2.12-1 is 0, whilethe normal stress acting on the bottom half of the cross sec-tion is 2 0. (a) Determine the value of the resultant axialforce,F x. (b) Locate the point R in the cross section throughwhich the resultant axial force, F x, acts.

▼

P2.12-1

F x

y

R

b

b

b /2 b /2

b

(b)(a)

z

σO

2σO

x y R

y b

P2.12-2

A A

x F x

B B

C

y

z

h2

h2

b2

(b)(a)

y

z

M z

M y

σ xb

σ xt

Prob. 2.12-3. The normal stress on the rectangular section ABCD in Fig. P2.12-3 varies linearly with respe

P2.12-3

A A

x F x

B B

C

dA

y

z

2 in.

4 in.

4 in.

2

(b)(a)

y

z

M z

M y

σ xA = 12 ksi

σ xC = 12 ksi

σ xB = 8 ksi


8/20

position ( y, z) in the cross section. That is, x has the form x a by cz.The values of x at corners A, B, and C are: xA 12 ksi, xB 8 ksi,and xC 12 ksi. (a) Determine thevalue of xD, the normal stress at corner D. (b) Determinethe axial force, F x. (c) Determine the bending moment M y.

Prob. 2.12-4. The stress distribution on the cross sectionshown in Fig. P2.12-4a is given by

Determine expressions for the resultant forces F x and V y andthe bending moment M z in terms of stress-related quantitiesa, b, and c and the dimensions d and h of the cross section.(See Example 2.13.)

s x a by ; t xy c

c ah

2b

2

y2

d ; t xz 0

Prob. 2.12-6. On the cross section of a circular rod, the shestress at a point acts in the circumferential direction at thpoint, as illustrated in Fig. P2.12-6. The shear stress magntude varies linearly with distance from the center of the cro

section, that is, .Using the ring-shaped area in Fi

P2.12-6b, determine the formula that relates max and thresultant torque, T.

t tmaxr

r

P2.12-4

y

h /2

h /2

d /2

d /2

z

F

T

M y

V y

V z

M z

x σ

xyτ

xzτ

x

(a) The stresses on cross section x .

(b) The stress resultants at x .

y

z

DProb. 2.12-5.

On a particular cross section of the rectangu-lar bar shown in Fig.2.12-5 there is shear stress whose distri-bution has the form

where y is measured in mm from the centroid of the crosssection (Fig. P2.12-5b). If the shear stress xy may not exceed allow. 50 MPa, what is the maximum shear force V y thatmay be applied to the bar at this cross section?

t xy tmax c 1 a y50

b2 d

P2.12-5

x

V y

y

z

50 mm

50 mm

20 mm20 mm

(b)(a)

y

C

z

τmax = τ xy ( y = 0)

P2.12-6

T

τ(ρ)

d ρ

r

ρ

dA = 2πρ

(a) (b)

Prob. 2.12-7. Determine the relationship between max anT if, instead of acting on a solid circular bar, as in Fig. P2.12

the shear stress distribution acts on a tubul

cylinder with outer radius r o and inner radius r i. (The crossectional dimensions are shown in Fig. P2.12-7. See Pro2.12-6 for an illustration of the shear stress distribution oncircular cross section and for the definitions of T and .)

t tmaxr

r o

P2.12-7

r or i

108


9/20

*Prob.2.12-8. If the magnitude of the shear stress on a solid,circular rod of radius r varies with radial position as shownin Fig. P2.12-8, determine the formula that relates the result-ant torque T to the maximum shear stress Y . (See Prob.2.12-6for an illustration of the shear stress distribution on the crosssection and for the definitions of T and , and use the areashown in Fig. 2.12-6b.)

so that it is elongated by an amount .Determine an exsion for the (uniform) extensional strain n of the diag AC. Express your answer in terms of , L, and the angAssume that and see Appendix A.2 for releapproximations.

Prob. 2.12-11. A rectangular plate ABCD with base bheight h is uniformly stretched an amount x in the x direand y in the y direction to become the enlarged rect AB*C*D* shown in Fig. P2.12-11. Determine an expre

for the (uniform) extensional strain n of the diagonaExpress your answer in terms of x, y, L, and , w

and tan h/b.Base your calculations osmall-displacement assumptions, that is, assume thatand . (See Appendix A.2 for relevant approximatid y V L

d x

L 2 b2 h2

d V L

P2.12-8

ρ

τ

τY

r r /2

1.2 τY

STRAIN-DEFORMATION EQUATIONS

Prob. 2.12-9. (a) Using Figs. P2.12-9 and the definition of

extensional strain given in Eq. 2.35, show that the changein length, L, of a thin wire whose original length is L isgiven by , where x( x) is the extensionalstrain of the wire at x. (b) Determine the elongation of a2-m-long wire if it has a coefficient of thermal expansion 20 106/C, and if the change in temperature alongthe wire is given by T 10 x2 (C).

¢L L

0 x( x) dx

▼

P2.12-9

dx x

L

(a) Before deformation.

(b) After deformation.

dx (1 + x )

L∆ L

Prob. 2.12-10. The thin, rectangular plate ABCD shown inFig. P2.12-10a undergoes uniform stretching in the x direction,

P2.12-10

(a) Undeformed plate. (b) Deformed plate.

B C

n δ

A

B*

A*

C *

D* D

b

L

θ

P2.12-11

δ x

B*

A

C *

n

B

L

θ

C

D*

D x

y

b

δ y

h

Prob. 2.12-12. A thin, square plate ABCD undergoes dmation in which no point in the plate moves in the y dtion. Every horizontal line (except the bottom edguniformly stretched as edge CD remains straight and roclockwise aboutD. Using the definition of extensional str

Eq. 2.35, determine an expression for the extensional sin the x direction, x( x, y).

P2.12-12 and P2.12-13

(b) Deformed plate.(a) Undeformed plate.

A D

B C

P Q

R

P*

B*

A* D

a

a

a

0.1a y

x

y

*Prob. 2.12-13. Using the definition of shear strain, Eq.and using the “undeformed plate” and “deformed psketches in Fig.P2.12-13, determine an expression for the strain xy as a function of position in the plate, that is, xy(


10/20

Prob. 2.12-14. A thin, square plate ABCD undergoes defor-mation such that a typical point P with coordinates ( x, y)moves horizontally an amount

The undeformed and deformed plates are shown in Figs.P2.12-14a and P2.12-14b, respectively. Using the definitionextensional strain in Eq. 2.35, determine an expression for

x( x, y), the extensional strain in the x direction.

u( x, y) PP * 1

100 (a x) a y

ab2

HOOKE’S LAW FOR ISOTROPIC MATERIALS;

DILATATION

Prob. 2.13-1. When thin sheets of material, like the to“skin” of the airplane wing in Fig. P2.13-1, are subjected stress, they are said to be in a state of plane stress, with z xz yz 0. Starting with Eqs. 2.38, with T 0, show th

for the case of plane stress Hooke’s Law can be written as

s x E1 n2

( x v y),

s y E1 n2

( y v x)

t

P2.12-14 and P2.12-15

(b) After deformation.(a) Before deformation.

A D

P

Q R

A*

B* C *

D*

P

u x

P*

x

a

a

y

y

B C

∆ x

*Prob. 2.12-15. Using the definition of shear strain, Eq. 2.36,and using the “before deformation” and “after deformation”sketches in Fig. P2.12-15, determine an expression for theshear strain xy as a function of position in the plate, that is,determine xy( x, y).

*Prob. 2.12-16. A typical point P at coordinates ( x, y) in a

flat plate moves through small displacements u( x, y) and v( x, y)in the x direction and the y direction, respectively. Using thedefinition of extensional strain, Eq. 2.35, and using the “be-fore deformation” and “after deformation” sketches in Fig.P2.12-16, show that the formula for the extensional strain inthe x direction, x( x, y), is the partial differential equation

x 0u

0 x

P2.12-16

(a) Before deformation. (b) After deformation.

PP*

Q*

QP Q

u( x , y)

v( x , y)

u( x + ∆ x , y)

v( x + ∆ x , y)

▼

P2.13-1

x

y

z

Prob. 2.13-2. Figure P2.13-2 shows a small portion of a thaluminum-alloy plate in plane stress ( z yz 0At a particular point in the plate x 600, y 200, an xy 200. For the aluminum alloy, E 10 10

3 ksi and0.33.Determine the stresses x, y, and xy at this point in thplate. (Note: Start with Eqs. 2.38, not with Eqs. 2.40.)

n

t xz

P2.13-2

x

y

z

Prob. 2.13-3. Determine the state of strain that correspon

to the following three-dimensional state of stress at a certapoint in a steel machine component:

Use E 210 GPa and 0.30 for the steel.

Prob. 2.13-4. The flat-bar plastic test specimen shown Fig. P2.13-4 has a reduced-area “test section” that measur

n

t yz 10 MPat xz 15 MPa,t xy 20 MPa,

sz 30 MPas y 20 MPa,s x 60 MPa,

110


11/20

0.5 in. 1.0 in. Within the test section a strain gage orientedin the axial direction measures , while a strain gagemounted in the transverse direction measures ,when the load on the specimen is P 300 lb. (a) Determinethe values of the modulus of elasticity,E, and Poisson’s ratio,. (b) Determine the value of the dilatation, V , within the

test section.n

y 0.0008in.in.

x 0.002in.in.

Prob. 2.13-7. A block of linearly elastic material (E,compressed between two rigid, perfectly smooth surfacan applied stress x 0, as depicted in Fig.P2.13-7.Theother nonzero stress is the stress y induced by the restrasurfaces at y 0 and y b. (a) Determine the value of thstraining stress y . (b) Determine a, the change in themension of the block. (c) Determine the change t inthickness t in the z direction.P2.13-4

x

y

PP

y

(1)1 in.

0.5′′

(1) x gage

(2) y gage

(2)

Electricalleads

Prob. 2.13-5. A titanium-alloy bar has the following original

dimensions: a 10 in., b 4 in.,and c 2 in.The bar is subjected to stresses x 14 ksi and y 6 ksi, as indicatedin Fig. P2.13-5. The remaining stresses— z, xy, , and —are all zero. Let E 16 103 ksi and 0.33 for thetitanium alloy. (a) Determine the changes in the lengths: a,b, and c, where a* a a, etc. (b) Determine thedilatation, V .

n

t yzt xz

P2.13-5

z

x

y

c*

b*

a*14 ksi

6 ksi

Prob. 2.13-6. An aluminum-alloy plate is subjected to a bi-axial state of stress, as illustrated in Fig. P2.13-6 ( z xz yz xy 0). For the aluminum alloy, E 72 GPa and

0.33. Determine the stresses x and y if x 200, and y 140. (Note: Start with Eqs. 2.38, not with Eqs. 2.40.)n

P2.13-6

x σ x σ y

y

z

P2.13-7

x

σ0σ0

∆a

y

b

a

Prob. 2.13-8. A thin, rectangular plate is subjected to aform biaxial state of stress ( x, y). All other componenstress are zero.The initial dimensions of the plate are L x and L y 2 in., but after the loading is applied, the disions are , and If known that x 10 ksi and E 10 10

3 ksi, (a) what value of Poisson’s ratio? (b) What is the value of y?

L* y 2.00344 in.L* x 4.00176 in.

P2.13-8

x

σ x = 10 ksi

σ y = ?

y

L y*

L x *


12/20

*Prob. 2.13-10. A block of linearly elastic material (E, ) isplaced under hydrostatic pressure: x y z p; xy xz yz 0, as shown in Fig. P2.13-10. (a) Determinean expression for the extensional strain x ( y z). (b)Determine an expression for the dilatation, V . (c) The bulk

n

modulus, kb, of a material is defined as the ratio of the hdrostatic pressure, p, to the magnitude of the volume chanper unit volume, V , that is,

Determine an expression for the bulk modulus of thblock of linearly elastic material. Express your answer terms of E and .

Prob. 2.14-1. You are to evaluate a new concept for an envronmentally friendly building product, a laminated composifloor panel. This composite panel will use a new material cosisting of a recycled polymer filled with recycled wood parcles. This recycled material has an elastic modulus of 6 GPand is produced in sheets 2 mm thick. These are laminatwith thin, 0.5-mm-thick sheets of aluminum, EA1 70 GPThe two different materials are firmly bonded by a strong ahesive to create the laminated composite panel.The final laminated composite panel contains 10 sheets of aluminum an11 sheets of the recycled material in alternating layers.

Use the techniques discussed in Section 2.14 to calculaapproximate values of elastic modulus in the plane of the laminated panel and through the thickness of the laminated paneBefore you begin your calculations, be sure to draw a simpschematic of the laminated composite structure,and use this help you determine the volume fractions of each material.

Prob. 2.14-2. Consider a polymer matrix having Em 2.8 GPwhich is reinforced with V f 0.2 volume fraction of radomly oriented, short glass fibers having E f 72 GP(a) Calculate an approximate elastic modulus, Ec, for thcomposite material. (b) Would you expect the actual elastmodulus to be higher, or lower, than your approximation?

n

kb p

V

Prob. 2.13-9. At a point in a thin steel plate in plane stress( z xz yz 0), x 800, y 400, and xy 200.For the steel plate,E 200 GPa and 0.30. (a) Determinethe extensional strain z at this point. (b) Determine thestresses x, y and xy at this point. (c) Determine the dilata-tion, V , at this point.

n

P2.13-9

x

y

z

P2.13-10

x

p

(Stresses on hidden faces not shown.)

p

p

y

z

112


13/20

SectionSuggesReviewProblem

Section 2.1 points out the need for defini-tions of stress and strain in order to explainhow force is distributed throughout a de-formable body under load and how thebody deforms point by point.

C H A P T E R 2 R E V I E W — S T R E S S A N D S T R A I N;

I N T R O D U C T I O N T O

D E S I G N

You should familiarize yourself with this material.2.1

Section 2.2 defines stress as “Force dividedby Area.” Normal stress is normal i.e., per-pendicular to the plane on which it acts,and it is denoted by the Greek symbolsigma ( ). Figure 2.2 and Eqs. 2.1 and 2.2define:

• Normal stress at a point ( y, z) on cross

section x.• Average normal stress on cross section x.

Normal Stress

(2.1)

(2.2)savg F

A

s( x, y, z) lim¢ AS0

a ¢F ¢ A

b2.2

Normal force on a cross section (Fig. 2.2).

Axial Stress

(2.5)s

i

F i

Ai

Normal force through centroid (Fig. 2.4b).

The sign convention for normal stress is:

• Positive normal stress is called tensilestress.

• Negative normal stress is called compres-sive stress.

The normal stress on cross sections of anaxially loaded member is called axial stress.The resultant normal force on the crosssection must act through the centroid.

2.2-32.2-92.2-15

Extensional Strain

(2.7) ¢L

L

2.3Section 2.3 defines extensional strain, thestrain that goes with normal stress.

Extensional strain (Fig. 2.5).

A

BC

D

E

F

L

(a) The undeformed bar.

A*

B*C*

D* E*

F*

L L*

∆ L

(b) The deformed bar.

2.3-72.3-11

z

y

z z

z R

y

x

x

x

R

(a) Distributed normal stress on a cross section.

(b) Resultant of distributed normal stress in (a).

∆F ∆ A

F ( x )

y y R

x

F i

F ( x ) = F i

σ


14/20

SectionSuggestedReviewProblems

2.4

A typical tension-test specimen (Fig. 2.10).

Section 2.4 discusses stress–strain diagramsand the mechanical properties of materialsthat are obtained from testing tension spec-imens and compression specimens.

114

where is the coefficient of thermal expansion and

T is the change in temperature from the referencetemperature.

a¢T Section 2.3 also defines thermal strain,the strain that is produced by a changein temperature. 2.3-15

Stress-Strain Diagrams

Stress-strain diagrams for structural steel intension (Fig. 2.11).

A stress-strain diagram is a graph of the re-sults of a tension test (or compression test):stress ( P/A versus strain ( (L)/L.

From the stress-strain diagram you should

be able to determine directly (or calculate)the following mechanical properties:

• the proportional limit of the material,

• the modulus of elasticity of the material,

• the yield point of the material, and

• the ultimate strength of the material.

2.4-3

60

50

40

30

20

10

0.100.001

0.200.002

0.300.003

0.400.004

σU = 63

σF = 47

(σYP)u = 38

(σYP)l = 36

σPL = 35

(in./in.)

σ (ksi)

Yielding Strainhardening

Necking

Elasticbehavior

Elasticregion

True fracture stress

True stress–true strain

∆

∆σ

σ

σF t

σU

σF (σYP)u

(σYP)l

σPL

A

B

C D E

Gt

G

F

Plastic behavior

(a)

(b)

L*

(b) Deformed specimen.

P P

L0

(a) Undeformed specimen.


15/20


Define each of these design-related properties; dis-cuss how each is determined from stress-strain dia-grams; and discuss how each design propertydiffers from the others.

The principal design properties of materialsare the following:

• Strength

• Stiffness, and

• Ductility.

Define the following terms and indicate how eachis determined:

• Elastic behavior of a material,

• Plastic behavior of a material,

• Linearly elastic behavior of a material, and

• Permanent set.

2.5

Section 2.5 discusses the differences be-tween elastic behavior and plastic behaviorof materials.

Hooke’s Law

(2.14)

where E is the modulus of elasticity, also calledYoung’s modulus.

Poisson’s Ratio

(2.15)

where is called Poisson’s ratio.n

y z n x

s x E x

2.6

Linearly elastic behavior (Fig. 2.25).

Section 2.6 discusses linearly elastic behav-ior. The discussion is restricted to the caseof uniaxial stress applied to homogeneous,isotropic materials.

Define these two terms.

2.6-12.6-7

z

y

Original specimen

x σ x

σ x


16/20

SectionSuggestedReviewProblems

116

Shear Stress

(2.17)

Average Shear Stress

(2.19)

where A s is the area on which the shear force V acts.

tavg V

A s

t lim¢ AS0

a¢V ¢ A

b

2.7

Until now, Chapter 2 has discussed onlynormal stress, the stress that results in aforce perpendicular to the surface on whichthe normal stress acts. Section 2.7 intro-duces the second form of stress, shearstress, whose resultant is parallel to the sur-face on which the shear stress acts.

Shear stress and its resultant shear force(Fig. 2.27).

(a) The distribution of shear force on a sectioning plane.

∆ A∆V

(b) The resultant shear force on the sectioning plane.

V

2.7-12.7-7

2.7

Shear Strain

(2.21)

where and * are defined in the figure below.

g p

2 u*

Definition of shear strain (Fig. 2.31b).

In Section 2.7 an equilibrium argumentshows that the shear stress on perpendicu-

lar faces is required to be equal. Section 2.7also gives the definition of shear strain.

You should be able to prove that the shearstresses on perpendicular faces must beequal to each other, as shown in Fig. 2.31b.

2.7-11

Hooke’s Law for Shear

(2.23)

where G is the shear modulus of elasticity, which isdiscussed further in Section 2.11.

t GgSection 2.7 concludes with the materialproperties in shear.

2.7-13

γ

δs

A*

Ls

(b) Pure shear

deformation.

τ

τ

τ

τθ*


17/20


Transformation of Stresses

(2.30)sn (s x/2)(1 cos 2u)

tnt (s x/2) sin 2uf

State Saint-Venant’s Principle, and discuss why itis an important principle in structural analysis(i.e., the analysis of components of machines andstructures).

(2.24)G E

2(1 n)

There is no Chapter Review Problem for theseoptional sections.

2.9

2.10

2.11

2.122.132.14

Inclined plane of axial-deformation member.(Fig. 2.33a)

Section 2.10 introduces you to Saint-Venant’s Principle.

Section 2.11 shows how Young’s Modulus,E, and the Shear Modulus of Elasticity, G,are related.

Sections 2.12 through 2.14 discuss moreadvanced topics in stress and strain, andmore advanced topics in the mechanicalbehavior of materials. These optionalsections discuss: General Definitions of Stress and Strain Section 2.12, CartesianComponents of Stress: Generalized

Hooke’s law for Isotropic Materials

Section 2.13, Mechanical Properties of Composite Materials Section 2.14.

DeriveEqs. 2.3

2.9-12.9-7

DeriveEq. 2.24

Section 2.9 introduces you to the fact thatnormal stress and shear stress depend onthe orientation of the plane on which thestresses act. This topic is greatly expandedin Chapter 8.

2.8

Factor of Safety

(2.26)

Allowable Stress

(2.28)sallow. sY

FS, or tallow.

tY

FS

FS Failure Load

Allowable Load

Section 2.8 is an Introduction to Design.There are two ways in which design infor-mation is used in engineering practice:

• To evaluate an existing structure, or toevaluate a proposed design.

• To design a new structure.

Although Allowable-Stress Design isemphasized in this textbook, Load andResistance Design is mentioned briefly.

For allowable-stress design, the allowablestress in axial deformation is based on thetensile, or compressive, yield strength of the material; in direct shear, the allowablestress is based on the shear yield strength.

Review Example 2.11, which illustrates the

process of optimal design of a minimum-weight structure.

2.8-12.8-72.8-13

n

P

(a)

P


18/20

118

AXIAL DEFORMATION3

In Chapter 2 the topic of uniform axial deformation was used to introduce the concepof normal stress and extensional strain and to describe the experiments required to dtermine the stress-strain behavior of materials. In this chapter we will pursue the topof axial deformation in greater detail.We begin with a definition of axial deformatio

A structural member having a straight longitudinal axis is said to undergo axial

deformation if, when loads are applied to the member or it is subjected to temperature

change: (1) the axis of the member remains straight, and (2) cross sections of the mem-

ber remain plane, remain perpendicular to the axis, and do not rotate about the axis as

the member deforms.

There are many examples of axial-deformation members: columns in buildinghoist cables, and truss members in space structures, to name just a few. The pictuin Fig. 3.1 illustrates several stages in the construction of columns (piers) for a higway interchange. On the left is an example of the steel reinforcement for a columand on the right is a completed column with reinforcement protruding from the toof the column.The columns of bridges like the one in the background in Fig. 3.1 aprimarily as axial-deformation members.

3.1INTRODUCTION

Let us now develop the theory of axial deformation by applying the three types equations that are fundamental to all of deformable-body mechanics: equilibrium

geometry of deformation, and material behavior .We begin by considering the geometry of deformation.

Geometry of Deformation; Strain-Displacement Analysis. The theoof axial deformation applies to a straight, slender member with cross section thateither constant or that changes slowly along the length of the member. Figure 3shows such a member before and after it has undergone axial deformation causeby axial loading or temperature change.

3.2 BASIC THEORY OF AXIAL DEFORMATION


19/20

Axial deformation, as defined in Section 3.1, is characterized by two fundamen-

tal kinematic assumptions:

1. The axis of the member remains straight .

2. Cross sections,which are plane and are perpendicular to the axis before defor-

mation, remain plane and remain perpendicular to the axis after deformation.And, the cross sections do not rotate about the axis.

These assumptions are illustrated in Fig. 3.2, where A and B designate cross sectionsat x and ( x x) prior to deformation, and where A* and B* designate these samecross sections after deformation.

The distance that a cross section moves in the axial direction is called its axialdisplacement. The displacement of cross section A is labeled u( x), while the neigh-boring section B displaces an amount u( x x). The displacement u( x) is taken tobe positive in the x direction. We can derive a strain-displacement expression thatrelates the axial strain to this axial displacement u by considering the fundamen-tal definition of extensional strain:

Final length Initial length

Initial length

Basic Theory of

Deform

FIGURE 3.1 Some reiforced concrete columns fhighway interchange brid(Courtesy Roy Craig)

FIGURE 3.2 The geometry of axial deformation.

(a) Before Deformation

x

u( x + ∆ x )∆ x

L

x

u( x )

(b) After Deformation

∆ x *

A* B*

A B

x


20/20

The axial strain of any fiber1 of infinitesimal length x that is parallel to the x axand extends from section A to section B of the undeformed member may be detemined from the fundamental definition of extensional strain at a point. By letting thinitial length of a typical fiber be x, and then letting x approach zero, we can writhe following expression for the axial strain (Eq. 2.35):

Therefore, the axial strain at section x is the derivative (with respect to x) of thaxial displacement, or

(3.

This equation relating axial strain to axial displacement is called the straidisplacement equation for axial deformation. The two fundamental kinematic asumptions stated above imply that the axial strain may be a function of x, but thit is not a function of position in the cross section, that is, of y or z.To emphasize thpoint, a plot of the strain distribution at an arbitrary cross section at x is shown

Fig. 3.3 superimposed on a sketch of a portion of the member. To reiterate, axideformation is characterized by extensional strain that is not a function of positio

in the cross section.As indicated in Fig. 3.4, the total elongation of the member is the differenc

between the displacements of its two ends, that is,

(3.

By summing up the changes in length of increments dx over the entire length of thmember, we get the following equation for the elongation of an axial-deformatiomember of initial length L:

(3.Elongation

Formulae

L

0

( x)dx

e u(L) u(0)

Strain-

Displacement

Equation

( x) du( x)

dx

x( x) lim¢ xS0

a ¢ x* ¢ x¢ x

b lim¢ xS0

c u( x ¢ x) u( x)¢ x

d dudx

120

Axial Deformation

1The word fiber is used to signify a line of material particles

Before Deformation

x

L

u( x ) u( L)

u(0)

After Deformation

B* C * A*

A B C

( L + e)

e = u( L) – u(0)

y

x z

( x )

x

FIGURE 3.3 Extensionalstrain distribution for amember undergoing axialdeformation.

FIGURE 3.4 Definition of the total elongation e.

121_pdfsam_1Mechanics of Materials(3 Ed)[Team Nanban]Tmrg

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