3–1. A tension test was performed on a steel specimenhaving an original diameter of 0.503 in. and gauge length of2.00 in. The data is listed in the table. Plot the stress–strain diagram and determine approximately themodulus of elasticity, the yield stress, the ultimate stress, andthe rupture stress. Use a scale of 1 in. 20 ksi and 1 in. = 0.05 in. in. Redraw the elastic region, using the samestress scale but a strain scale of 1 in. 0.001 in. in.>=
3–2. Data taken from a stress–strain test for a ceramic aregiven in the table.The curve is linear between the origin andthe first point. Plot the diagram, and determine the modulusof elasticity and the modulus of resilience.
033.245.549.451.553.4
00.00060.00100.00140.00180.0022
S (ksi) P (in./in.)
Modulus of Elasticity: From the stress–strain diagram
Ans.
Modulus of Resilience: The modulus of resilience is equal to the area under thelinear portion of the stress–strain diagram (shown shaded).
Modulus of Toughness: The modulus of toughness is equal to the area under thestress-strain diagram (shown shaded).
Ans. = 85.0 in # lb
in3
+
12
(12.3) A103 B ¢ lbin2 ≤(0.0004)¢ in.
in.≤
+
12
(7.90) A103 B ¢ lbin2 ≤(0.0012)¢ in.
in.≤
+ 45.5 A103 B ¢ lbin2 ≤(0.0012)¢ in.
in.≤
(ut)approx =
12
(33.2) A103 B ¢ lbin2 ≤(0.0004 + 0.0010)¢ in.
in.≤
3–3. Data taken from a stress–strain test for a ceramic aregiven in the table.The curve is linear between the origin andthe first point. Plot the diagram, and determineapproximately the modulus of toughness.The rupture stressis sr = 53.4 ksi.
*3–4. A tension test was performed on a steel specimenhaving an original diameter of 0.503 in. and a gauge lengthof 2.00 in. The data is listed in the table. Plot thestress–strain diagram and determine approximately themodulus of elasticity, the ultimate stress, and the rupturestress. Use a scale of 1 in. 15 ksi and 1 in. 0.05 in. in.Redraw the linear-elastic region, using the same stress scalebut a strain scale of 1 in. 0.001 in.=
3–5. A tension test was performed on a steel specimenhaving an original diameter of 0.503 in. and gauge length of2.00 in. Using the data listed in the table, plot thestress–strain diagram and determine approximately themodulus of toughness.
Modulus of toughness (approx)
total area under the curve
(1)
Ans.
In Eq.(1), 87 is the number of squares under the curve.
3–6. A specimen is originally 1 ft long, has a diameter of0.5 in., and is subjected to a force of 500 lb. When the forceis increased from 500 lb to 1800 lb, the specimen elongates0.009 in. Determine the modulus of elasticity for thematerial if it remains linear elastic.
3–7. A structural member in a nuclear reactor is made of azirconium alloy. If an axial load of 4 kip is to be supportedby the member, determine its required cross-sectional area.Use a factor of safety of 3 relative to yielding. What is theload on the member if it is 3 ft long and its elongation is 0.02 in.? The material haselastic behavior.
Ezr = 14(103) ksi, sY = 57.5 ksi.
Ans:
Allowable Normal Stress:
Ans.
Stress–Strain Relationship: Applying Hooke’s law with
*3–8. The strut is supported by a pin at C and an A-36 steelguy wire AB. If the wire has a diameter of 0.2 in., determinehow much it stretches when the distributed load acts on the strut.
3–9. The diagram for elastic fibers that make uphuman skin and muscle is shown. Determine the modulus ofelasticity of the fibers and estimate their modulus oftoughness and modulus of resilience.
Ans. Pult = sult A = 100 Cp4 (0.52) D = 19.63 kip = 19.6 kip
PY = sYA = 60 Cp4 (0.52) D = 11.78 kip = 11.8 kip
sy = 60 ksi sult = 100 ksi
E
1=
60 ksi - 00.002 - 0
; E = 30.0(103) ksi
3–10. The stress–strain diagram for a metal alloy havingan original diameter of 0.5 in. and a gauge length of 2 in. isgiven in the figure. Determine approximately the modulusof elasticity for the material, the load on the specimen thatcauses yielding, and the ultimate load the specimen willsupport.
From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is
when the specimen is unloaded, its normal strain recovers along line AB, Fig. a,which has a slope of E. Thus
Ans.
Thus, the permanent set is
.
Then, the increase in gauge length is
Ans.¢L = PPL = 0.047(2) = 0.094 in.
PP = 0.05 - 0.003 = 0.047 in>in
Elastic Recovery =
90E
=
90 ksi30.0(103) ksi
= 0.003 in>in.
E
1=
60 ksi - 00.002 - 0
; E = 30.0(103) ksi
3–11. The stress–strain diagram for a steel alloy having anoriginal diameter of 0.5 in. and a gauge length of 2 in. isgiven in the figure. If the specimen is loaded until it isstressed to 90 ksi, determine the approximate amount ofelastic recovery and the increase in the gauge length after itis unloaded.
The Modulus of resilience is equal to the area under the stress–strain diagram up tothe proportional limit.
Thus,
Ans.
The modulus of toughness is equal to the area under the entire stress–straindiagram. This area can be approximated by counting the number of squares. Thetotal number is 38. Thus,
Ans.C(ui)t Dapprox = 38 c15(103) lbin2 d a0.05
in.in.b = 28.5(103)
in. # lbin3
(ui)r =
12
sPLPPL =
12
C60(103) D(0.002) = 60.0 in. # lb
in3
sPL = 60 ksi PPL = 0.002 in.>in.
*3–12. The stress–strain diagram for a steel alloy havingan original diameter of 0.5 in. and a gauge length of 2 in. isgiven in the figure. Determine approximately the modulusof resilience and the modulus of toughness for the material.
3–13. A bar having a length of 5 in. and cross-sectionalarea of 0.7 in.2 is subjected to an axial force of 8000 lb. If thebar stretches 0.002 in., determine the modulus of elasticityof the material. The material has linear-elastic behavior.
3–14. The rigid pipe is supported by a pin at A and an A-36 steel guy wire BD. If the wire has a diameter of 0.25 in., determine how much it stretches when a load ofP 600 lb acts on the pipe.=
3–15. The rigid pipe is supported by a pin at A and an A-36 guy wire BD. If the wire has a diameter of 0.25 in.,determine the load P if the end C is displaced 0.15 in.downward.
Ans:P = 570 lb
Here, we are only interested in determining the force in wire BD. Referring to theFBD in Fig. a
a
The unstretched length for wire BD is . From thegeometry shown in Fig. b, the stretched length of wire BD is
Thus, the normal strain is
Then, the normal stress can be obtain by applying Hooke’s Law.
Equations of Equilibrium: The force developed in wire DE can be determined bywriting the moment equation of equilibrium about A with reference to the free-body diagram shown in Fig. a,
a
Normal Stress and Strain:
Since , Hooke’s Law can be applied
The unstretched length of wire DE is Thus, theelongation of this wire is given by
3–17. A tension test was performed on a magnesium alloyspecimen having a diameter 0.5 in. and gauge length 2 in.The resulting stress–strain diagram is shown in the figure.Determine the approximate modulus of elasticity and theyield strength of the alloy using the 0.2% strain offsetmethod.
Modulus of Elasticity: From the stress–strain diagram, when , itscorresponding stress is Thus,
Ans.
Yield Strength: The intersection point between the stress–strain diagram and thestraight line drawn parallel to the initial straight portion of the stress–strain diagramfrom the offset strain of is the yield strength of the alloy. From thestress–strain diagram,
3–18. A tension test was performed on a magnesium alloyspecimen having a diameter 0.5 in. and gauge length of 2 in.The resulting stress–strain diagram is shown in the figure. Ifthe specimen is stressed to 30 ksi and unloaded, determinethe permanent elongation of the specimen.
Permanent Elongation: From the stress–strain diagram, the strain recovered isalong the straight line BC which is parallel to the straight line OA. Since
*3–20. The stress–strain diagram for a bone is shown and can be described by the equation �
where is in kPa. Determine the modulus oftoughness and the amount of elongation of a 200-mm-long regionjust before it fractures if failure occurs at P = 0.12 mm>mm.
3–21. The two bars are made of polystyrene, which has thestress–strain diagram shown. If the cross-sectional area ofbar AB is 1.5 in2 and BC is 4 in2, determine the largest forceP that can be supported before any member ruptures.Assume that buckling does not occur.
3–22. The two bars are made of polystyrene, which has thestress–strain diagram shown. Determine the cross-sectionalarea of each bar so that the bars rupture simultaneouslywhen the load P 3 kip. Assume that buckling does notoccur.
3–23. The stress–strain diagram for many metal alloys canbe described analytically using the Ramberg-Osgood threeparameter equation , where E, k, and n aredetermined from measurements taken from the diagram.Using the stress–strain diagram shown in the figure, take
ksi and determine the other two parameters kand n and thereby obtain an analytical expression for thecurve.
3–24. The wires AB and BC have original lengths of 2 ftand 3 ft, and diameters of in. and in., respectively. Ifthese wires are made of a material that has the approximatestress–strain diagram shown, determine the elongations ofthe wires after the 1500-lb load is placed on the platform.
316
18
Equations of Equilibrium: The forces developed in wires AB and BC can bedetermined by analyzing the equilibrium of joint B, Fig. a,
(1)
(2)
Solving Eqs. (1) and (2),
Normal Stress and Strain:
The corresponding normal strain can be determined from the stress–strain diagram,Fig. b.
3–25. The acrylic plastic rod is 200 mm long and 15 mm indiameter. If an axial load of 300 N is applied to it, determinethe change in its length and the change in its diameter.
3–26. The thin-walled tube is subjected to an axial force of40 kN. If the tube elongates 3 mm and its circumferencedecreases 0.09 mm, determine the modulus of elasticity,Poisson’s ratio, and the shear modulus of the tube’smaterial. The material behaves elastically.
Normal Stress and Strain:
Applying Hooke’s law,
Ans.
Poisson’s Ratio: The circumference of the loaded tube is Thus, the outer radius of the tube is
3–27. When the two forces are placed on the beam, thediameter of the A-36 steel rod BC decreases from 40 mm to39.99 mm. Determine the magnitude of each force P.
Ans:P = 157 kN
Equations of Equilibrium: The force developed in rod BC can be determined bywriting the moment equation of equilibrium about A with reference to the free-body diagram of the beam shown in Fig. a.
a
Normal Stress and Strain: The lateral strain of rod BC is
*3–28. If P 150 kN, determine the elastic elongation ofrod BC and the decrease in its diameter. Rod BC is made of A-36 steel and has a diameter of 40 mm.
=
Equations of Equilibrium: The force developed in rod BC can be determined bywriting the moment equation of equilibrium about A with reference to the free-body diagram of the beam shown in Fig. a.
a
Normal Stress and Strain: The lateral strain of rod BC is
Since , Hooke’s Law can be applied. Thus,
The unstretched length of rod BC is Thus theelongation of this rod is given by
Ans.
We obtain,
;
Thus,
Ans.dd = Plat dBC = -0.2387(10-3)(40) = -9.55(10-3) mm
3–29. The friction pad A is used to support the member,which is subjected to an axial force of P 2 kN. The pad ismade from a material having a modulus of elasticity of E 4 MPa and Poisson’s ratio . If slipping does notoccur, determine the normal and shear strains in the pad.The width is 50 mm. Assume that the material is linearlyelastic. Also, neglect the effect of the moment acting on the pad.
n = 0.4=
=
Internal Loading: The normal force and shear force acting on the friction pad can bedetermined by considering the equilibrium of the pin shown in Fig. a.
Normal and Shear Stress:
Normal and Shear Strain: The shear modulus of the friction pad is
3–30. The lap joint is connected together using a 1.25 in.diameter bolt. If the bolt is made from a material having ashear stress–strain diagram that is approximated as shown,determine the shear strain developed in the shear plane ofthe bolt when P 75 kip.=
Internal Loadings: The shear force developed in the shear planes of the bolt can bedetermined by considering the equilibrium of the free-body diagram shown in Fig. a.
Shear Stress and Strain:
Using this result, the corresponding shear strain can be obtained from the shearstress–strain diagram, Fig. b.
3–31. The lap joint is connected together using a 1.25 in.diameter bolt. If the bolt is made from a material having ashear stress–strain diagram that is approximated as shown,determine the permanent shear strain in the shear plane ofthe bolt when the applied force P 150 kip is removed.=
Internal Loadings: The shear force developed in the shear planes of the bolt can bedetermined by considering the equilibrium of the free-body diagram shown in Fig. a.
Shear Stress and Strain:
Using this result, the corresponding shear strain can be obtained from the shearstress–strain diagram, Fig. b.
When force P is removed, the shear strain recovers linearly along line BC, Fig. b,with a slope that is the same as line OA. This slope represents the shear modulus.
Thus, the elastic recovery of shear strain is
And the permanent shear strain is
Ans.gP = g - gr = 0.02501 - 6.112(10-3) = 0.0189 rad
Shear Stress–Strain Relationship: Applying Hooke’s law with .
(Q.E.D)
If is small, then tan . Therefore,
At
Then,
At
Ans.d =
P
2p h G ln
ro
ri
r = ri, y = d
y =
P
2p h G ln
ro
r
C =
P
2p h G ln ro
0 = - P
2p h G ln ro + C
r = ro, y = 0
y = - P
2p h G ln r + C
y = - P
2p h G L
drr
dy
dr= -
P
2p h G r
g = gg
dy
dr= -tan g = -tan a P
2p h G rb
g =
tA
G=
P
2p h G r
tA =
P
2p r h
*3–32. A shear spring is made by bonding the rubberannulus to a rigid fixed ring and a plug. When an axial loadP is placed on the plug, show that the slope at point y inthe rubber is For smallangles we can write Integrate thisexpression and evaluate the constant of integration usingthe condition that at From the result computethe deflection of the plug.y = d
r = ro.y = 0
dy>dr = -P>12phGr2.-tan1P>12phGr22.dy>dr = -tan g =
3–33. The aluminum block has a rectangular cross sectionand is subjected to an axial compressive force of 8 kip. If the1.5-in. side changed its length to 1.500132 in., determinePoisson’s ratio and the new length of the 2-in. side.
Average Shear Stress: The rubber block is subjected to a shear force of .
Shear Strain: Applying Hooke’s law for shear
Thus,
Ans.d = a g = =
P a2 b h G
g =
t
G=
P2 b h
G=
P
2 b h G
t =
V
A=
P2
b h=
P
2 b h
V =
P
2
3–34. A shear spring is made from two blocks of rubber,each having a height h, width b, and thickness a. Theblocks are bonded to three plates as shown. If the platesare rigid and the shear modulus of the rubber is G,determine the displacement of plate A if a vertical load P isapplied to this plate. Assume that the displacement is smallso that d = a tan g L ag.
3–35. The elastic portion of the tension stress–straindiagram for an aluminum alloy is shown in the figure. Thespecimen used for the test has a gauge length of 2 in. and adiameter of 0.5 in. When the applied load is 9 kip, the newdiameter of the specimen is 0.49935 in. Compute the shearmodulus for the aluminum.Gal
*3–36. The elastic portion of the tension stress–straindiagram for an aluminum alloy is shown in the figure. Thespecimen used for the test has a gauge length of 2 in. and adiameter of 0.5 in. If the applied load is 10 kip, determinethe new diameter of the specimen. The shear modulus isGal = 3.811032 ksi.
3–37. The rigid beam rests in the horizontal position ontwo 2014-T6 aluminum cylinders having the unloadedlengths shown. If each cylinder has a diameter of 30 mm.determine the placement x of the applied 80-kN load sothat the beam remains horizontal.What is the new diameterof cylinder A after the load is applied? .nal = 0.35
a (1)
a (2)
Since the beam is held horizontally,
Ans.
From Eq. (2),
Ans.dA¿ = dA + d Plat = 30 + 30(0.0002646) = 30.008 mm
3–38. The wires each have a diameter of in., length of 2 ft, and are made from 304 stainless steel. If P 6 kip,determine the angle of tilt of the rigid beam AB.
=
12
Equations of Equilibrium: Referring to the free-body diagram of beam AB shownin Fig. a,
a
Normal Stress and Strain:
Since and , Hooke’s Law can be applied.
Thus, the elongation of cables BC and AD are given by
Referring to the geometry shown in Fig. b and using small angle analysis,
u =
dBC - dAD
36=
0.017462 - 0.00873136
= 0.2425(10-3) rada 180°prad
b = 0.0139°
dAD = PADLAD = 0.3638(10-3)(24) = 0.008731 in.
dBC = PBCLBC = 0.7276(10-3)(24) = 0.017462 in.
PAD = 0.3638(10-3) in.>in.10.19 = 28.0(103)PADsAD = EPAD;
3–39. The wires each have a diameter of in., length of 2 ft, and are made from 304 stainless steel. Determine themagnitude of force P so that the rigid beam tilts 0.015°.
12
Equations of Equilibrium: Referring to the free-body diagram of beam AB shownin Fig. a,
a
Normal Stress and Strain:
Assuming that and and applying Hooke’s Law,
Thus, the elongation of cables BC and AD are given by
Here, the angle of the tile is Using small
angle analysis,
Ans.
Since and
, the assumption is correct.11.00 ksi 6 sY
sAD = 1.6977(6476.93) =sBC = 3.3953(6476.93) = 21.99 ksi 6 sY
P = 6476.93 lb = 6.48 kip
0.2618(10-3) =
2.9103(10-6)P - 1.4551(10-6)P
36u =
dBC - dAD
36;
u = 0.015°aprad180°
b = 0.2618(10-3) rad.
dAD = PADLAD = 60.6305(10-9)P(24) = 1.4551(10-6)P
dBC = PBCLBC = 0.12126(10-6)P(24) = 2.9103(10-6)P
PAD = 60.6305(10-9)P1.6977P = 28.0(106)PADsAD = EPAD;
Normal Strain: Since , Hooke’s law is still valid.
Ans.
If the nut is unscrewed, the load is zero. Therefore, the strain Ans.P = 0
P =
s
E=
28.9729(103)
= 0.000999 in.>in.
s 6 sg
s =
P
A=
800p4 A 3
16 B2 = 28.97 ksi 6 sg = 40 ksi
*3–40. The head H is connected to the cylinder of acompressor using six steel bolts. If the clamping force ineach bolt is 800 lb, determine the normal strain in thebolts. Each bolt has a diameter of If and
what is the strain in each bolt when thenut is unscrewed so that the clamping force is released?Est = 2911032 ksi,
3–41. The stress–strain diagram for polyethylene, which isused to sheath coaxial cables, is determined from testing aspecimen that has a gauge length of 10 in. If a load P on thespecimen develops a strain of determinethe approximate length of the specimen, measured betweenthe gauge points, when the load is removed. Assume thespecimen recovers elastically.
P = 0.024 in.>in.,
Modulus of Elasticity: From the stress–strain diagram, when
Elastic Recovery: From the stress–strain diagram, when
3–42. The pipe with two rigid caps attached to its ends issubjected to an axial force P. If the pipe is made from amaterial having a modulus of elasticity E and Poisson’sratio n, determine the change in volume of the material.
Normal Stress: The rod is subjected to uniaxial loading.Thus, and .
3–43. The 8-mm-diameter bolt is made of an aluminumalloy. It fits through a magnesium sleeve that has an innerdiameter of 12 mm and an outer diameter of 20 mm. If theoriginal lengths of the bolt and sleeve are 80 mm and50 mm, respectively, determine the strains in the sleeve andthe bolt if the nut on the bolt is tightened so that the tensionin the bolt is 8 kN. Assume the material at A is rigid.
*3–44. An acetal polymer block is fixed to the rigid platesat its top and bottom surfaces. If the top plate displaces 2 mm horizontally when it is subjected to a horizontal force P 2 kN, determine the shear modulus of the polymer.The width of the block is 100 mm. Assume that the polymeris linearly elastic and use small angle analysis.
=
400 mm
200 mm
P � 2 kN
Normal and Shear Stress:
Referring to the geometry of the undeformed and deformed shape of the blockshown in Fig. a,
Applying Hooke’s Law,
Ans.G = 5 MPa
50(103) = G(0.01)t = Gg;
g =
2200
= 0.01 rad
t =
V
A=
2(103)
0.4(0.1)= 50 kPa
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