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Problem 104

A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN.

Determine the outside diameter of the tube if the stress is limited to 120 MN/m2.

Solution 104

P= A

P= A

where:

P=400kN=400000N=120MPa

A=41 D2−41 (1002)

A=41 (D2−10000)

thus,

400000=120[41 (D2−10000)]

400000=30 D2−300000

D2=30 400000+300000

D=119 35mm  answer

Strength of Materials 4th Edition by Pytel and Singer

Problem 105 page 12

Given:

Weight of bar = 800 kg

Maximum allowable stress for bronze = 90 MPa

Maximum allowable stress for steel = 120 MPa

Required: Smallest area of bronze and steel cables

Solution 105

By symmetry:

Pbr=Pst=21(7848)

Pbr=3924N

Pst=3924N

For bronze cable:

Pbr= brAbr

3924=90Abr

Abr=43 6mm2  answer

For steel cable:

Pst= stAst

3924=120Ast

Ast=32 7mm2  answer

Problem 106 page 12

Given:

Diameter of cable = 0.6 inch

Weight of bar = 6000 lb

Required: Stress in the cable

Solution 106

MC=0

5T+10 3 34T =5(6000) 

T=2957 13lb

T= A

2957 13= 41 (0 62)  =10458 72psi  answer

Problem 107 page 12

Given:

Axial load P = 3000 lb

Cross-sectional area of the rod = 0.5 in2

Required: Stress in steel, aluminum, and bronze sections

Solution 107

For steel:stAst=Pst

st(0 5)=12

st=24ksi  answer

For aluminum:alAal=Pal

al(0 5)=12

al=24ksi  answer

For bronze:brAbr=Pbr

br(0 5)=9

br=18ksi  answer

Problem 108 page 12

Given:Maximum allowable stress for steel = 140 MPaMaximum allowable stress for aluminum = 90 MPaMaximum allowable stress for bronze = 100 MPa

Required: Maximum safe value of axial load P

Solution 108

For bronze:brAbr=2P

100(200)=2PP=10000NFor aluminum:alAal=P

90(400)=PP=36000NFor Steel:stAst=5P

P=14000NFor safe P, use P=10000N=10kN  answer

Problem 109 page 13

Given:Maximum allowable stress of the wire = 30 ksiCross-sectional area of wire AB = 0.4 in2

Cross-sectional area of wire AC = 0.5 in2

Required: Largest weight W

 

Solution 109

 

For wire AB: By sine law (from the force polygon):TABsin40 =Wsin80TAB=0 6527WABAAB=0 6527W

30(0 4)=0 6527WW=18 4kips

 

For wire AC:TACsin60 =Wsin80TAC=0 8794WTAC= ACAAC

0 8794W=30(0 5)W=17 1kips

 

Safe load W=17 1kips  answer

Problem 110 page 13

Given:Size of steel bearing plate = 12-inches squareSize of concrete footing = 12-inches squareSize of wooden post = 8-inches diameterMaximum allowable stress for wood = 1800 psiMaximum allowable stress for concrete = 650 psi

Required: Maximum safe value of load P

Solution 110For wood:

Pw= wAw

Pw=1800[41 (82)]Pw=90477 9lb

From FBD of Wood:

P=Pw=90477 9lbFor concrete:Pc= cAc

Pc=650(122)Pc=93600lb

From FBD of Concrete:

P=Pc=93600lbSafe load P=90478lb  answer

Problem 111 page 14

Given:Cross-sectional area of each member = 1.8 in2

Required: Stresses in members CE, DE, and DF

Solution 111From the FBD of the truss:

MA=024RF=16(30)RF=20k

At joint F:

FV=053DF=20DF=3331k(Compression)

At joint D: (by symmetry)

BD=DF=3331k(Compression)ΣFV=0DE=53BD+53DFDE=53(3331)+53(3331)DE=40k(Tension)

At joint E:

FV=053CE+30=40CE=1632k(Tension)Stresses:Stress = Force/AreaCE=1 81632=9 26ksi (Tension)  answer

DE=401 8=22 22ksi (Tension)  answer

DF=1 83331=18 52ksi (Compression)  answer

Problem 112 page 14

Given:Maximum allowable stress in tension = 20 ksiMaximum allowable stress in compression = 14 ksi

Required: Cross-sectional areas of members AG, BC, and CE

FV=0RAV=40+25=65k

 

AV=018RD=8(25)+4(40)RD=20k

 

FH=0RAH=RD=20k

 

Check:MD=0

12RAV=18(RAH)+4(25)+8(40)12(65)=18(20)+4(25)+8(40)780 ft kip=780 ft kip (OK!)

 

For member AG (At joint A):

 

FV=03 13AB=65 AB=78 12k

 

FH=0AG+20=2 13AB AG=20 33kTension

 

AG= tensionAAG

20 33=20AAG

AAG=1 17in2  answer

 

For member BC (At section through MN):

 

MF=06(2 13BC)=12(20) BC=72 11k Compression

 

BC= compressionABC

72 11=14ABC

ABC=5 15in2  answer

 

For member CE (At joint D):

 

FH=02 13CD=20 CD=36 06k

 

FV=0DE=3 13CD=3 13(36 06)=30k 

 

At joint E:

 

FV=03 13EF=30 EF=36 06k

 

FH=0CE=2 13EF=2 13(36 06)=20k  Compression

 

CF= compressionACE

20=14ACE

ACE=1 43in2  answer

 

Problem 113 page 15

Given:Cross sectional area of each member = 1600 mm2.

Required: Stresses in members BC, BD, and CF

Solution 113

For member BD: (See FBD 01)MC=0

3(54BD)=3(60)BD=75kN Tension

BD= BDA75(1000)= BD(1600)BD=46 875MPa (Tension)  answer

For member CF: (See FBD 01)MD=0

4(1 2CF)=4(90)+7(60) CF=275 77kN Compression

CF= CFA275 77(1000)= CF(1600)CF=172 357MPa (Compression)  answer

For member BC: (See FBD 02)

MD=04BC=7(60)BC=105kN Compression

BC= BCA105(1000)= BC(1600)BC=65 625MPa (Compression)  answer

Problem 114 page 15

Given:Maximum allowable stress in each cable = 100 MPaArea of cable AB = 250 mm2

Area of cable at C = 300 mm2

Required: Mass of the heaviest bar that can be supported

Solution 114

FH=0TABcos30 =RDsin50RD=1 1305TAB

FV=0TABsin30 +TAB+TC+RDcos50 =WTABsin30 +TAB+TC+(1 1305TAB)cos50 =W2 2267TAB+TC=WTC=W−2 2267TAB

MD=06(TABsin30 )+4TAB+2TC=3W7TAB+2(W−2 2267TAB)=3W2 5466TAB=WTAB=0 3927WTC=W−2 2267TAB

TC=W−2 2267(0 3927W)TC=0 1256WBased on cable AB:TAB= ABAAB

0 3927W=100(250)W=63661 83NBased on cable at C:T2= CAC

0 1256W=100(300)W=238853 50NSafe weight W=63669 92NW=mg63669 92=m(9 81)m=6490kgm=6 49Mg  answer

SHEAR STRESS

Problem 115 page 16

Given:Required diameter of hole = 20 mmThickness of plate = 25 mmShear strength of plate = 350 MN/m2

Required: Force required to punch a 20-mm-diameter hole

Solution 115

The resisting area is the shaded area along the perimeter and the shear force V is

equal to the punching force P.

V= AP=350[ (20)(25)]

P=549778 7NP=549 8kN  answer

Problem 116 page 16

Given:Shear strength of plate = 40 ksiAllowable compressive stress of punch = 50 ksiThe figure below:

Required:

a. Maximum thickness of plate to punch a 2.5 inches diameter hole

b. Diameter of smallest hole if the plate is 0.25 inch thick

Solution 116

a. Maximum thickness of plate:

Based on puncher strength:

P= AP=50[41 (2 52)]P=78 125 kips  Equivalent shear force of the plate

Based on shear strength of plate:V= A V=P78 125 =40[ (2 5t)]t=0 781inch  answer

b. Diameter of smallest hole:

Based on compression of puncher:

P= A

P=50(41 d2)P=12 5 d2  Equivalent shear force for plate

Based on shearing of plate:V= A V=P12 5 d2=40[ d(0 25)]d=0 8in  answer

Problem 117 page 17

Given:Force P = 400 kNShear strength of the bolt = 300 MPaThe figure below:

Required: Diameter of the smallest bolt

Solution 117The bolt is subject to double shear.V= A400(1000)=300[2(41 d2)]d=29 13mm  answer

Problem 118 page 17

Given:Diameter of pulley = 200 mmDiameter of shaft = 60 mmLength of key = 70 mmApplied torque to the shaft = 2.5 kN·mAllowable shearing stress in the key = 60 MPa

Required: Width b of the key

Solution 118

T=0 03F2 2=0 03FF=73 33kNV= A

Where:V=F=73 33kNA=70b=60MPa

73 33(1000)=60(70b)b=17 46mm  answer

Problem 119 page 17

Given:Diameter of pin at B = 20 mm

Required: Shearing stress of the pin at B

Solution 119

From the FBD:MC=0

0 25RBV=0 25(40sin35 )+0 2(40cos35 )RBV=49 156kNFH=0

RBH=40cos35RBH=32 766kN

RB= R2BH+R2BV RB= 32 7662+49 1562 RB=59 076kN  shear force of pin at B

VB= BA  double shear

59 076(1000)= B 2 41 (202)  B=94 02MPa  answer

Problem 120 page 17

Given:Unit weight of each member = 200 lb/ftMaximum shearing stress for pin at A = 5 000 psi

Required: The smallest diameter pin that can be used at A

 

Solution 120For member AB:

 

 

Length, LAB= 42+42=5 66ft Weight, WAB=5 66(200)=1132lb

 

MA=04RBH+4RBV=2WAB

4RBH+4RBV=2(1132)RBH+RBV=566  Equation (1)

 

For member BC:

 

 

Length, LBC= 32+62=6 71ft Weight, WBC=6 71(200)=WBC=1342lb

 

MC=06RBH=1 5WBC+3RBV

6RBH−3RBV=1 5(1342)2RBH−RBV=671  Equation (2)

 

Add equations (1) and (2)RBH RBH 3RBH + âˆ’ + RBV RBV RBV = = = 566 671 1237  Equation (1) Equation (2)  RBH=412 33lb

 

From equation (1):412 33+RBV=566RBV=153 67lb

 

From the FBD of member ABFH=0

RAH=RBH=412 33lb

 

FV=0RAV+RBV=WAB

RAV+153 67=1132RAV=978 33lb

 

RA= R2AH+R2AV RA= 412 332+978 332 RA=1061 67lb  shear force of pin at A

 

V= A1061 67=5000(41 d2)d=0 520in  answer

Problem 121 page 18

Given:Allowable shearing stress in the pin at B = 4000 psiAllowable axial stress in the control rod at C = 5000 psiDiameter of the pin = 0.25 inchDiameter of control rod = 0.5 inchPin at B is at single shear

Required: The maximum force P that can be applied by the operator

Solution 121

MB=06P=2Tsin10°  Equation (1)

FH=0BH=Tcos10°From Equation (1), T=3Psin10°

BH= 3Psin10° cos10° BH=3cot10°PFV=0

BV=Tsin10°+PFrom Equation (1), Tsin10°=3PBV=3P+PBV=4PR2B=B2H+B2V

R2B=(3cot10°P)2+(4P)2

R2B=305 47P2

RB=17 48PP=RB17 48  Equation (2)Based on tension of rod (equation 1):P=31Tsin10°P=31[5000 41 (0 5)2]sin10°P=56 83lbBased on shear of rivet (equation 2):P=17 484000[41 (0 25)2]P=11 23lbSafe load P=11 23lb  answer

Problem 122 page 18

Given:Width of wood = wThickness of wood = tAngle of Inclination of glued joint = Cross sectional area = A

 

Required: Show that shearing stress on glued joint  =Psin2 2A 

 

Solution 122

 

 

Shear area, Ashear=t(wcsc )Shear area, Ashear=twcscShear area, Ashear=AcscShear force, V=Pcos

 

V= AshearPcos = (Acsc )=APsin cos=2AP(2sin cos )=Psin2 2A  (ok!)

Strength of Materials 4th Edition by Pytel and Singer Problem 123 page 18

Given: Cross-section of wood = 50 mm by 100 mm Maximum allowable compressive stress in wood = 20 MN/m2 Maximum allowable shear stress parallel to the grain in wood = 5 MN/m2Inclination of the grain from the horizontal = 20 degree Required: The

axial force P that can be safely applied to the block Solution 123   

Based on maximum compressive stress: Normal force: N=Pcos20  Normal

area: AN=50(100sec20 )AN=5320 89mm2 N= AN Pcos20 =20(532089) P=113247N P=133 25kNBased on maximum shearing stress: Shear

force: V=Psin20  Shear area: AV=ANAV=5320 89mm2 V= AV Psin20 =5(532089) P=77786N P=77 79kN For safe compressive force, use P=77 79kN  answer

BEARING STRESS

Problem 125

In Fig. 1-12, assume that a 20-mm-diameter rivet joins the plates that are each 110 mm wide. The allowable stresses are 120 MPa for bearing in the plate material and 60 MPa for shearing of rivet. Determine (a) the minimum thickness of each plate; and (b) the largest average tensile stress in the plates.

 

 

Solution 125

Part (a): From shearing of rivet:P= Arivets

P=60[41 (202)]P=6000 textN

 

From bearing of plate material:P= bAb

6000 =120(20t)t=7 85mm  answer

 

Part (b): Largest average tensile stress in the plate:P= A6000 = [7 85(110−20)]

=26 67MPa  answer

Problem 126 page 21

Given:Diameter of each rivet = 3/4 inchMaximum allowable shear stress of rivet = 14 ksiMaximum allowable bearing stress of plate = 18 ksi

The figure below:

Required: The maximum safe value of P that can be applied

Solution 126

Based on shearing of rivets:P= AP=14[4(41 )(43)2]P=24 74kipsBased on bearing of plates:P= bAb

P=18[4(43)(87)]P=47 25kipsSafe load P=24 74kips  answer

Problem 127 page 21

Given:Load P = 14 kipsMaximum shearing stress = 12 ksiMaximum bearing stress = 20 ksi

The figure below:

Required: Minimum bolt diameter and minimum thickness of each yoke

Solution 127

For shearing of rivets (double shear)P= A14=12[2(41 d2)]d=0 8618in  diameter of bolt answerFor bearing of yoke:P= bAb

14=20[2(0 8618t)]t=0 4061in  thickness of yoke answer

Problem 128 page 21

Given:Shape of beam = W18 × 86Shape of girder = W24 × 117Shape of angles = 4 × 3-½ × 3/8Diameter of rivets = 7/8 inchAllowable shear stress = 15 ksiAllowable bearing stress = 32 ksi

Required: Allowable load on the connection

Solution 128Relevant data from the table (Appendix B of textbook): Properties of Wide-Flange Sections (W shapes): U.S. Customary Units

Designation Web thicknessW18 × 86 0.480 inW24 × 117 0.550 in

Shearing strength of rivets:There are 8 single-shear rivets in the girder and 4 double-shear (equivalent to 8 single-shear) in the beam, thus, the shear strength of rivets in girder and beam are equal.

V= A=15[41 (87)2(8)]V=72 16kipsBearing strength on the girder:The thickness of girder W24 × 117 is 0.550 inch while that of the angle clip L4 321

83 is 83 or 0.375 inch, thus, the critical in bearing is the clip.P= bAb=32[87(0 375)(8)]P=84kips

Bearing strength on the beam:The thickness of beam W18 × 86 is 0.480 inch while that of the clip angle is 2 × 0.375 = 0.75 inch (clip angles are on both sides of the beam), thus, the critical in bearing is the beam.

P= bAb=32[87(0 480)(4)]P=53 76kipsThe allowable load on the connection is P=53 76kips  answer

Problem 129 page 21

Given:Diameter of bolt = 7/8 inchDiameter at the root of the thread (bolt) = 0.731 inchInside diameter of washer = 9/8 inchTensile stress in the nut = 18 ksiBearing stress = 800 psi

Required:Shearing stress in the head of the boltShearing stress in threads of the boltOutside diameter of the washer

Solution 129

Tensile force on the bolt:P= A=18[41 (87)2]P=10 82kipsShearing stress in the head of the bolt:=AP=10 82 (87)(21)=7 872ksi  answer

Shearing stress in the threads:=AP=10 82 (0 731)(85)=7 538ksi  answer

Outside diameter of washer:P= bAb

10 82(1000)=800 41 [d2−(89)2]  d=4 3inch  answer

Problem 130 page 22

Given:Allowable shear stress = 70 MPaAllowable bearing stress = 140 MPaDiameter of rivets = 19 mmThe truss below:

Required:Number of rivets to fasten member BC to the gusset plateNumber of rivets to fasten member BE to the gusset plateLargest average tensile or compressive stress in members BC and BE

Solution 130At Joint C:

FV=0BC=96kN (Tension)

Consider the section through member BD, BE, and CE:

MA=08(53BE)=4(96)BE=80kN (Compression)For Member BC:Based on shearing of rivets:BC= A Where A = area of 1 rivet × number of rivets, n

96000=70[41 (192)n]n=4 8 say 5 rivetsBased on bearing of member:BC= bAb

Where Ab = diameter of rivet × thickness of BC × number of rivets, n96000=140[19(6)n]n=6 02 say 7 rivets

use 7 rivets for member BC answer

For member BE:Based on shearing of rivets:BE= AWhere A = area of 1 rivet × number of rivets, n80000=70[41 (192)n]n=4 03 say 5 rivetsBased on bearing of member:BE= bAb

Where Ab = diameter of rivet × thickness of BE × number of rivets, n

80000=140[19(13)n]n=2 3 say 3 rivets

use 5 rivets for member BE answer

Relevant data from the table (Appendix B of textbook): Properties of Equal Angle Sections: SI Units

Designation AreaL75 × 75 × 6 864 mm2

L75 × 75 × 13

1780 mm2

Tensile stress of member BC (L75 × 75 × 6):=AP=96(1000)864−19(6)=128Mpa  answer

Compressive stress of member BE (L75 × 75 × 13):=AP=178080(1000)=44 94Mpa  answer

Problem 131Repeat Problem 130 if the rivet diameter is 22 mm and all other data remain unchanged.

 

Solution 131For member BC:P=96kN (Tension)

 

Based on shearing of rivets:P= A96000=70[41 (222)n]n=3 6 say 4 rivets

 

Based on bearing of member:P= bAb

96000=140[22(6)n]n=5 2 say 6 rivets

 

Use 6 rivets for member BC answer

 

Tensile stress:=AP=96(1000)864−22(6)=131 15MPa  answer

 

For member BE:P=80kN (Compression)

 

Based on shearing of rivets:P= A80000=70[41 (222)n]n=3 01 say 4 rivets

 

Based on bearing of member:P= bAb

80000=140[22(13)n]n=1 998 say 2 rivets

 

use 4 rivets for member BE answer

 

Compressive stress:=AP=178080(1000)=44 94MPa  answer

A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist bursting, developed across longitudinal and transverse sections.

TANGENTIAL STRESS (Circumferential Stress)Consider the tank shown being subjected to an internal pressure p. The length of the

tank is Land the wall thickness is t. Isolating the right half of the tank:

The forces acting are the total pressures caused by the internal pressure p and the total

tension in the walls T.

F=pA=pDLT= tAwall= ttLFH=0

F=2TpDL=2( ttL)

t=2tpD

If there exist an external pressure po and an internal pressure pi, the formula may be expressed as:

t=2t(pi−po)D

LONGITUDINAL STRESS,  L

Consider the free body diagram in the transverse section of the tank:

The total force acting at the rear of the tank F must equal to the total longitudinal stress

on the wall PT= LAwall. Since t is so small compared to D, the area of the wall is close

to  DtF=pA=p4 D2

PT= L DtFH=0

PT=FL Dt=p4 D2

t=4tpD

If there exist an external pressure po and an internal pressure pi, the formula may be expressed as:

t=4t(pi−po)D

It can be observed that the tangential stress is twice that of the longitudinal stress.

t=2 L

SPHERICAL SHELL

If a spherical tank of diameter D and thickness t contains gas under a pressure of p, the stress at the wall can be expressed as:

t=4t(pi−po)D

Problem 133 page 28

Given:Diameter of cylindrical pressure vessel = 400 mmWall thickness = 20 mmInternal pressure = 4.5 MN/m2

Allowable stress = 120 MN/m2

Required:Longitudinal stressTangential stressMaximum amount of internal pressure that can be appliedExpected fracture if failure occurs

Solution 133

Part (a)Tangential stress (longitudinal section):

F=2TpDL=2( ttL)t=2tpD=2(20)4 5(400)t=45MPa  answer

Longitudinal Stress (transverse section):

F=P41 D2p= l( Dt)

l=4tpD=4(20)4 5(400)l=22 5MPa  answer

Part (b)From (a),  t=2tpD and  l=4tpD thus,  t=2 l, this shows that tangential stress is the critical.t=2tpD

120=2(20)p(400)p=12MPa  answer

The bursting force will cause a stress on the longitudinal section that is twice to that of the transverse section. Thus, fracture is expected as shown.

Problem 134 page 28

Given:Diameter of spherical tank = 4 ftWall thickness = 5/16 inchMaximum stress = 8000 psi

Required: Allowable internal pressure

Solution 134

Total internal pressure:P=p(41 D2)

Resisting wall:F=PA=p(41 D2)( Dt)=p(41 D2)=4tpD

8000=4(516)p(4 12)p=208 33psi  answer

Problem 135

Calculate the minimum wall thickness for a cylindrical vessel that is to carry a gas at a pressure of 1400 psi. The diameter of the vessel is 2 ft, and the stress is limited to 12 ksi.

Solution 135

The critical stress is the tangential stresst=2tpD

12000=2t1400(2 12)t=1 4in  answer

Problem 136 page 28

Given:Thickness of steel plating = 20 mmDiameter of pressure vessel = 450 mmLength of pressure vessel = 2.0 mMaximum longitudinal stress = 140 MPaMaximum circumferential stress = 60 MPa

Required: The maximum internal pressure that can be applied

Solution 136

Based on circumferential stress (tangential):

FV=0F=2Tp(DL)=2( tLt)t=2tpD

60=2(20)p(450)p=5 33MPa

Based on longitudinal stress:

FH=0F=Pp(41 D2)= l( D)l=4tpD

140=4(20)p(450)p=24 89MPaUse p=5 33MPa  answer

Problem 137 page 28

Given:Diameter of the water tank = 22 ftThickness of steel plate = 1/2 inchMaximum circumferential stress = 6000 psiSpecific weight of water = 62.4 lb/ft3

Required: The maximum height to which the tank may be filled with water.

Solution 137

t=6000psi=6000lb/in2(12in/ft)2

t=864000lb/ft2Assuming pressure distribution to be uniform:p= h=62 4h F=pA=62 4h(Dh)F=62 4(22)h2

F=1372 8h2

T= tAt=864000(th)T=864000(21 112)hT=36000hF=0

F=2T1372 8h2=2(36000h)h=52 45ft  answerCOMMENTGiven a free surface of water, the actual pressure distribution on the vessel is not uniform. It varies linearly from 0 at the free surface to γh at the bottom (see figure below). Using this actual pressure distribution, the total hydrostatic pressure is reduced by 50%. This reduction of force will take our design into critical situation; giving us a maximum height of 200% more than the h above.

Based on actual pressure distribution:

Total hydrostatic force, F:F = volume of pressure diagram

F=21( h2)D=21(62 4h2)(22)F=686 4h2

MA=02T(21h)−F(31h)=0T=31Ft(ht)=31(686 4h2)

h=3 tt686 4=686 43(864000)(21 112)h=157 34ft

Problem 138 page 38

Given:Strength of longitudinal joint = 33 kips/ftStrength of girth joint = 16 kips/ftInternal pressure = 150 psi

Required: Maximum diameter of the cylinder tank

Solution 138

For longitudinal joint (tangential stress):

Consider 1 ft lengthF=2TpD=2 ttt=2tpD

t33000=2t21600DD=3 06ft=36 67in.

For girth joint (longitudinal stress):

F=Pp(41 D2)= l( Dt)l=4tpD

t16000=4t21600DD=2 96ft=35 56in.Use the smaller diameter, D=35 56in. answer

Problem 139 page 28

Given:Allowable stress = 20 ksiWeight of steel = 490 lb/ft3

Mean radius of the ring = 10 inches

Required:The limiting peripheral velocity.The number of revolution per minute for stress to reach 30 ksi.

Solution 139

Centrifugal Force, CF:CF=M 2xwhere:M=gW g V=q RA

=v R x =2R  

CF=q RA vR 2 2R  CF=g2 Av2

2T=CF2 A=g2 Av2

=g v2

From the given data:=20ksi=(20000lb/in2)(12in/ft)a2

=2880000lb/ft2=490lb/ft3 

2880000=32 2490v2

v=435 04ft/sec  answer

When  =30ksi, and R=10in=g v2

30000(122)=32 2490v2

v=532 81ft/sec=v R=10 12532 81=639 37rad/sec=sec639 37rad 1rev2 rad 1min60sec=6 105 54rpm   answer

Problem 140 page 28

Given:Stress in rotating steel ring = 150 MPaMean radius of the ring = 220 mmDensity of steel = 7.85 Mg/m3

Required: Angular velocity of the steel ring

Solution 140

CF=M 2xWhere:M= V= A R x=2R  CF= A R 2(2R ) CF=2 AR2 2 2T=CF2 A=2 AR2 2 

= R2 2 From the given (Note: 1 N = 1 kg·m/sec2):

=150MPa=150000000kg m/sec2 m2

=150000000kg/m sec2

=7 85Mg/m3=7850kg/m3 R=220mm=0 22m150000000=7850(0 22)2 2

=628 33rad/sec  answer

Problem 141 page 28

Given:Wall thickness = 1/8 inchInternal pressure = 125 psiThe figure below:

Required: Maximum longitudinal and circumferential stress

Solution 141

Longitudinal Stress:

F=pA=125[1 5(2)+41 (1 5)2](122)F=85808 62lbsP=Fl[2(2 12)(81)+ (1 5 12)(81)]=85808 62l=6566 02psil=6 57ksi  answer

Circumferential Stress:

F=pA=125[(2 12)L+2(0 75 12)L]F=5250Ltextlbs2T=F2[ t(81)L]=5250Lt=21000psit=21ksi  answer

Problem 142 page 29

Given:Steam pressure = 3.5 MpaOutside diameter of the pipe = 450 mmWall thickness of the pipe = 10 mmDiameter of the bolt = 40 mmAllowable stress of the bolt = 80 MPaInitial stress of the bolt = 50 MPa

Required:Number of boltsCircumferential stress developed in the pipe

Solution 29

F= AF=3 5[41 (4302)]F=508270 42N

P=F( boltA)n=508270 42N(80−55)[41 (402)]n=508270 42n=16 19 say 17 bolts   answer

Circumferential stress (consider 1-m strip):

F=pA=3 5[430(1000)]F=1505000N2T=F2[ t(1000)(10)]=1505000t=75 25MPa  answer

Discussion:

It is necessary to tighten the bolts initially to press the gasket to the flange, to avoid leakage of steam. If the pressure will cause 110 MPa of stress to each bolt causing it to fail, leakage will occur. If this is sudden, the cap may blow.

CHAPTER 2. STRAIN SIMPLE STRAIN

Also known as unit deformation, strain is the ratio of the change in length caused by the applied force, to the original length.

= L

where   is the deformation and L is the original length, thus   is dimensionless.

Suppose that a metal specimen be placed in tension-compression-testing machine. As the axial

load is gradually increased in increments, the total elongation over the gauge length is

measured at each increment of the load and this is continued until failure of the specimen takes

place. Knowing the original cross-sectional area and length of the specimen, the normal

stress  and the strain   can be obtained. The graph of these quantities with the stress   along

the y-axis and the strain   along the x-axis is called the stress-strain diagram. The stress-strain

diagram differs in form for various materials. The diagram shown below is that for a medium-

carbon structural steel. Metallic engineering materials are classified as

either ductile or brittlematerials. A ductile material is one having relatively large tensile strains

up to the point of rupture like structural steel and aluminum, whereas brittle materials has a

relatively small strain up to the point of rupture like cast iron and concrete. An arbitrary strain of

0.05 mm/mm is frequently taken as the dividing line between these two classes.

 

Stress-strain diagram of a medium-carbon structural steel

 

Proportional Limit (Hooke's Law)

From the origin O to the point called proportional limit, the stress-

strain curve is a straight line. This linear relation between elongation

and the axial force causing was first noticed by Sir Robert

Hooke in 1678 and is called Hooke's Law that within the

proportional limit, the stress is directly proportional to strain or

 or  =k

The constant of proportionality k is called the Modulus of Elasticity E or Young's

Modulus and is equal to the slope of the stress-strain diagram from O to P. Then

=E

Elastic Limit

The elastic limit is the limit beyond which the material will no longer go back to its original shape

when the load is removed, or it is the maximum stress that may e developed such that there is

no permanent or residual deformation when the load is entirely removed.

Elastic Limit

The elastic limit is the limit beyond which the material will no longer go back to its original shape

when the load is removed, or it is the maximum stress that may e developed such that there is

no permanent or residual deformation when the load is entirely removed.

Elastic and Plastic Ranges

The region in stress-strain diagram from O to P is called the elastic range. The region from P to

R is called the plastic range.

Yield Point

Yield point is the point at which the material will have an appreciable elongation or yielding

without any increase in load.

Ultimate Strength

The maximum ordinate in the stress-strain diagram is the ultimate strength or tensile strength.

Rapture Strength

Rapture strength is the strength of the material at rupture. This is also known as the breaking

strength.

Modulus of Resilience

Modulus of resilience is the work done on a unit volume of material as the force is gradually

increased from O to P, in N·m/m3. This may be calculated as the area under the stress-strain

curve from the origin O to up to the elastic limit E (the shaded area in the figure). The resilience

of the material is its ability to absorb energy without creating a permanent distortion.

Modulus of Toughness

Modulus of toughness is the work done on a unit volume of material as the force is gradually

increased from O to R, in N·m/m3. This may be calculated as the area under the entire stress-

strain curve (from O to R). The toughness of a material is its ability to absorb energy without

causing it to break.

Working Stress, Allowable Stress, and Factor of Safety

Working stress is defined as the actual stress of a material under a given loading. The

maximum safe stress that a material can carry is termed as the allowable stress. The allowable

stress should be limited to values not exceeding the proportional limit. However, since

proportional limit is difficult to determine accurately, the allowable tress is taken as either the

yield point or ultimate strength divided by a factor of safety. The ratio of this strength (ultimate or

yield strength) to allowable strength is called the factor of safety.

 

Axial DeformationIn the linear portion of the stress-strain diagram, the tress is proportional to strain and is given by

=E

since  =P A  and  = L , then AP=E L=PLAE=E L

To use this formula, the load must be axial, the bar must have a uniform cross-sectional area, and the stress must not exceed the proportional limit.

If however, the cross-sectional area is not uniform, the axial deformation can be determined by considering a differential length and applying integration.

=EP 0LLdx 

where A=ty and y and t, if variable, must be expressed in terms of x.

For a rod of unit mass    suspended vertically from one end, the total elongation due to its own weight is

=2E gL2=2AEMgL

where    is in kg/m3, L is the length of the rod in mm, M is the total mass of the rod in

kg, A is the cross-sectional area of the rod in mm2, and g=9 81m/s2.

Stiffness, k

Stiffness is the ratio of the steady force acting on an elastic body to the resulting displacement. It has the unit of N/mm.

k= P

Stress-strain DiagramStrength of Materials 4th Edition by Pytel and Singer Problem 203 page 39

Given:

Material: 14-mm-diameter mild steel rod

Gage length = 50 mm

Test Result:

Load (N) Elongation (mm)Load(N)

Elongation (mm)

0 0 46 200 1.25

6 310 0.010 52 400 2.50

12 600 0.020 58 500 4.50

18 800 0.030 68 000 7.50

25 100 0.040 59 000 12.5

31 300 0.050 67 800 15.5

37 900 0.060 65 000 20.0

40 100 0.163 65 500 Fracture

41 600 0.433

Required: Stress-strain diagram, Proportional limit, modulus of elasticity, yield point, ultimate

strength, and rupture strength

 

Solution 203

Area, A=41 (14)2=49 mm2; Length, L=50mmStrain = Elongation/Length; Stress = Load/Area

Load (N) Elongation (mm) Strain (mm/mm) Stress (MPa)

0 0 0 0

6 310 0.010 0.0002 40.99

12 600 0.020 0.0004 81.85

18 800 0.030 0.0006 122.13

25 100 0.040 0.0008 163.05

31 300 0.050 0.001 203.33

37 900 0.060 0.0012 246.20

40 100 0.163 0.0033 260.49

41 600 0.433 0.0087 270.24

46 200 1.250 0.025 300.12

52 400 2.500 0.05 340.40

58 500 4.500 0.09 380.02

68 000 7.500 0.15 441.74

59 000 12.500 0.25 383.27

67 800 15.500 0.31 440.44

65 000 20.000 0.4 422.25

61 500 Failure   399.51

 

 

From stress-strain diagram:

a. Proportional Limit = 246.20 MPa

b. Modulus of Elasticity

E = slope of stress-strain diagram within proportional limit

E=0 0012246 20=205166 67MPaE=205 2GPa

c. Yield Point = 270.24 MPa

d. Ultimate Strength = 441.74 MPa

e. Rupture Strength = 399.51 MPa

Solution to Problem 204 Stress-strain Diagram

Problem 204 page 39

Given:Material: Aluminum alloyInitial diameter = 0.505 inchGage length = 2.0 inchesThe result of the test tabulated below:

Load (lb)

Elongation (in.)Load (lb)

Elongation (in.)

0 0 14 000 0.0202 310 0.00220 14 400 0.0254 640 0.00440 14 500 0.0606 950 0.00660 14 600 0.0809 290 0.00880 14 800 0.10011 600 0.0110 14 600 0.12012 600 0.0150 13 600 Fracture

Required:Plot of stress-strain diagram(a) Proportional Limit(b) Modulus of Elasticity(c) Yield Point(d) Yield strength at 0.2% offset(e) Ultimate Strength and(f) Rupture Strength

Solution 204

Area, A=41 (0 505)2=0 0638 in2; Length, L=2inStrain = Elongation/Length; Stress = Load/AreaLoad (lb)

Elongation (in.)Strain (in/in)

Stress (psi)

0 0 0 02 310 0.0022 0.0011 11 532.924 640 0.0044 0.0022 23 165.706 950 0.0066 0.0033 34 698.629 290 0.0088 0.0044 46 381.3211 600 0.011 0.0055 57 914.2412 600 0.015 0.0075 62 906.8514 000 0.02 0.01 69 896.4914 400 0.025 0.0125 71 893.5414 500 0.06 0.03 72 392.8014 600 0.08 0.04 72 892.0614 800 0.1 0.05 73 890.5814 600 0.12 0.06 72 892.0613 600 Fracture   67 899.45

From stress-strain diagram:

a. Proportional Limit = 57,914.24 psib. Modulus of Elasticity:

E=0 005557914 24=10 529 861 82psiE = 10,529.86 ksi

c. Yield Point = 69,896.49 psid. Yield Strength at 0.2% Offset:

Strain of Elastic Limit = ε at PL + 0.002Strain of Elastic Limit = 0.0055 + 0.002Strain of Elastic Limit = 0.0075 in/in

The offset line will pass through Q(See figure):

Slope of 0.2% offset = E = 10,529,861.82 psi

Test for location:slope = rise / run

10 529 861 82=run6989 64+4992 61run = 0.00113793 < 0.0025, therefore, the required point is just before YP.Slope of EL to YP

1 1=0 00256989 641 1=27958561  12795856

For the required point:E= 14992 61+ 1

10529861 82= 127958564992 61+ 1

3 7662 1=4992 61+ 1

1=1804 84psi

Yield Strength at 0.2% Offset= EL+ σ1

= 62906.85 + 1804.84= 64,711.69 psi

e. Ultimate Strength = 73,890.58 psif. Rupture Strength = 67,899.45 psi

Axial DeformationProblem 205 page 39

Given:Length of bar = LCross-sectional area = AUnit mass = ρThe bar is suspended vertically from one end

Required:Show that the total elongation δ = ρgL2 / 2E.If total mass is M, show that δ = MgL/2AE

Solution 205

=PLAE

From the figure:δ = dδP = Wy = (ρAy)gL = dy

d =AE( Ay)gdy

=E g 0Lydy=E g 2y2 0L = g2E[L2−02]= gL2 2E  ok!

Given the total mass M=M V=M AL =2E gL2=2EMAL gL2

=2AEMgL ok!

Another Solution:

=PLAE

Where:P = W = (ρAL)gL = L/2

=AE[( AL)g](L 2)= gL2 2E  ok!

For you to feel the situation, position yourself in pull-up exercise with your hands on the bar and your body hang freely above the ground. Notice that your arms suffer all your weight and your lower body fells no stress (center of weight is approximately just below the chest). If your body is the bar, the elongation will occur at the upper half of it.

Problem 206 page 39

Given:Cross-sectional area = 300 mm2

Length = 150 mtensile load at the lower end = 20 kN

Unit mass of steel = 7850 kg/m3

E = 200 × 103 MN/m2

Required: Total elongation of the rod

Solution 206

Elongation due to its own weight:1=PLAE

Where:P = W = 7850(1/1000)3(9.81)[300(150)(1000)]P = 3465.3825 NL = 75(1000) = 75 000 mmA = 300 mm2

E = 200 000 MPa1=300(200000)3465 3825(75000)1 = 4.33 mm

Elongation due to applied load:2=PLAE

Where:P = 20 kN = 20 000 NL = 150 m = 150 000 mmA = 300 mm2

E = 200 000 MPa2=300(200000)20000(150000)2 = 50 mm

Total elongation:= 1+ 2

=4 33+50=54 33mm  answer

Problem 207 page 39

Given:Length of steel wire = 30 ftLoad = 500 lbMaximum allowable stress = 20 ksiMaximum allowable elongation = 0.20 inchE = 29 × 106 psi

Required: Diameter of the wire

Solution 207

Based on maximum allowable stress:=AP

20000=50041 d2

d=0 1784in

 

Based on maximum allowable deformation:=PLAE

0 20=500(30 12)41 d2(29 106)d=0 1988in

 

Use the bigger diameter, d = 0.1988 inch

 

Problem 208 page 40

Given:Thickness of steel tire = 100 mmWidth of steel tire = 80 mmInside diameter of steel tire = 1500.0 mmDiameter of steel wheel = 1500.5 mmCoefficient of static friction = 0.30E = 200 GPa

Required: Torque to twist the tire relative to the wheel

Solution 208

=PLAE

Where:δ = π (1500.5 - 1500) = 0.5π mmP = T

L = 1500π mmA = 10(80) = 800 mm2

E = 200 000 MPa

0 5 =T(1500 )800(200000)T=53333 33N

F=2Tp(1500)(80)=2(53333 33)p=0 8889MPa  internal pressure

Total normal force, N:N = p × contact area between tire and wheelN = 0.8889 × π(1500.5)(80)N = 335 214.92 N

Friction resistance, f:f = μN = 0.30(335 214.92)f = 100 564.48 N = 100.56 kNTorque = f × ½(diameter of wheel)Torque = 100.56 × 0.75025Torque = 75.44 kN · m

Problem 209 page 40

Given:Cross-section area = 0.5 in2

E = 10 × 106 psiThe figure below:

Required: Total change in length

Solution 209

P1 = 6000 lb tensionP2 = 1000 lb compressionP3 = 4000 lb tension

=PLAE= 1− 2+ 3

=6000(3 12)0 5(10 106)−1000(5 12)0 5(10 106)+4000(4 12)0 5(10 106)=0 0696in (lengthening)  answer

Problem 210

Solve Prob. 209 if the points of application of the 6000-lb and the 4000-lb forces are interchanged.

Solution 210

P1 = 4000 lb compressionP2 = 11000 lb compressionP3 = 6000 lb compression

=PLAE=− 1− 2− 3

=−0 5(10 106)4000(3 12)−0 5(10 106)11000(5 12)−0 5(10 106)6000(412)=−0 19248in=0 19248in (shortening)  answer

Problem 211 page 40

Given:Maximum overall deformation = 3.0 mmMaximum allowable stress for steel = 140 MPaMaximum allowable stress for bronze = 120 MPaMaximum allowable stress for aluminum = 80 MPaEst = 200 GPaEal = 70 GPaEbr = 83 GPaThe figure below:

Required: The largest value of P

Solution 211

Based on allowable stresses:

Steel:

Pst= stAst

P=140(480)=67200NP=67 2kN

Bronze:Pbr= brAbr

2P=120(650)=78000NP=39000N=39kN

Aluminum:Pal= alAal

2P=80(320)=25600NP=12800N=12 8kN

Based on allowable deformation:(steel and aluminum lengthens, bronze shortens)= st− br+ al

3=P(1000)480(200000)−2P(2000)650(70000)+2P(1500)320(83000)3= 196000−111375+326560 P P=84610 99N=84 61kN

Use the smallest value of P, P = 12.8 kN

Problem 212 page 40

Given:Maximum stress in steel rod = 30 ksiMaximum vertical movement at C = 0.10 inchThe figure below:

Required: The largest load P that can be applied at C

Solution 212

Based on maximum stress of steel rod:MA=0

5P=2Pst

P=0 4Pst

P=0 4 atAst

P=0 4[30(0 50)]P=6kipsBased on movement at C:2 st=50 1st=0 04inAEPstL=0 04Pst(4 12)0 50(29 106)=0 04Pst=12083 33lbMA=0

5P=2Pst

P=0 4Pst

P=0 4(12083 33)P=4833 33lb=4 83kips

Use the smaller value, P = 4.83 kips

Problem 213 page 41

Given:Rigid bar is horizontal before P = 50 kN is applied

The figure below:

Required: Vertical movement of P

Solution 213

Free body diagram:

For aluminum:MB=0

6Pal=2 5(50)Pal=20 83kN=PLAEal=500(70000)20 83(3)10002

al=1 78mmFor steel:MA=0

6Pst=3 5(50)Pst=29 17kN=PLAEst=300(200000)29 17(4)10002

st=1 94mm

Movement diagram:

y3 5=61 94−1 78y=0 09mmB=vertical movement of PB=1 78+y=1 78+0 09B=1 87mm  answer

Problem 214 page 41

Given:Maximum vertical movement of P = 5 mmThe figure below:

Required: The maximum force P that can be applied neglecting the weight of all members.

Solution 41

Member AB:

MA=03Pal=6Pst

Pal=2Pst

By ratio and proportion:6 B=3 al

B=2 al=2 PLAE al 

B=2 Pal(2000)500(70000)  B=18750Pal=18750(2Pst)B=14375Pst  movement of B

Member CD:

Movement of D:

D= st+ B= PLAE st+14375Pst D=Pst(2000)300(200000)+14375Pst

D=1142000Pst

MC=06Pst=3PPst=21P

By ratio and proportion:3 P=6 D

P=21 D=21(1142000Pst)P=1184000Pst

5=1184000(21P)P=76363 64N=76 4kN  answer

Problem 215 page 41

Given:The figure below:

Required: Ratio of the areas of the rods

Solution 215

Mal=06Pst=2WPst=31WMst=0

6Pal=4WPal=32Wst= al

PLAE st= PLAE al 

31W(6 12)Ast(29 106)=32W(4 12)Aal(10 106)AstAal=31W(6 12)(10 106)32W(4 12)(29 106)AstAal=3 867  answer

Problem 216 page 42

Given:Vertical load P = 6000 lbCross-sectional area of each rod = 0.60 in2

E = 10 × 106 psiα = 30°θ = 30°The figure below:

Required: Elongation of each rod and the horizontal and vertical displacements of point B

Solution 216

FH=0PABcos30 =PBCcos30PAB=PBC

FV=0PABsin30 +PBCsin30 =6000PAB(0 5)+PAB(0 5)=6000

PAB=6000lb tension

PBC=6000lb compression

=PLAEAB=0 6(10 106)6000(10 12)=0 12inch lengthening  answer

BC=6000(6 12)0 6(10 106)=0 072inch shortening  answer

DB= AB=0 12inchBE= BE=0 072inchB=BB = displacement of BB = final position of B after elongation

Triangle BDB':cos = B0 12B=0 12cos

Triangle BEB':cos(120 − )= B0 072B=0 072cos(120 − )B= B

0 12cos =0 072cos(120 − )cos cos120 cos +sin120 sin =0 6−0 5+sin120 tan =0 6 tan =1 1sin120

=51 79  =90−(30 + )=90 −(30 +51 79 ) =8 21  B=0 12cos51 79B=0 194inch

Triangle BFB':h=BF= Bsin =0 194sin8 21

h=0 0277inchh=0 0023ft  horizontal displacement of B

v=BF= Bcos =0 194cos8 21v=0 192inchv=0 016ft  vertical displacement of B

Problem 217

Solve Prob. 216 if rod AB is of steel, with E = 29 × 106 psi. Assume α = 45° and θ = 30°; all other data remain unchanged.

Solution 217

By Sine LawPABsin60 =6000sin75PAB=5379 45lb (Tension)

PBCsin45 =6000sin75PBC=4392 30lb (Compression)

=PLAEAB=0 6(29 106)5379 45(10 12)==0 0371inch (lengthening)

BC=0 6(10 106)4392 30(6 12)=0 0527inch (shortening)

DB= AB=0 0371inchBE= BE=0 0527inchB=BB = displacement of BB  = final position of B after deformation

Triangle BDB':cos = B0 0371B=cos 0 0371

Triangle BEB':cos(105 − )= B0 0527B=0 0527cos(105 − )B= B

cos 0 0371=0 0527cos(105 − )cos cos105 cos +sin105 sin =1 4205−0 2588+0 9659tan =1 4205 tan =0 96591 4205+0 2588tan =1 7386 

=60 1  B=0 0371cos60 1B=0 0744inch=(45 + )−90  =(45 +60 1 )−90  =15 1  

Triangle BFB':h=FB = Bsin =0 0744sin15 1h=0 0194inchh=0 00162ft  horizontal displacement of Bv=BF= Bcos =0 0744cos15 1v=0 07183inchv=0 00598ft  vertical displacement of B

Problem 218

A uniform slender rod of length L and cross sectional area A is rotating in a horizontal plane about a vertical axis through one end. If the unit mass of the rod is ρ, and it is rotating at a constant angular velocity of ω rad/sec, show that the total elongation of the rod is ρω2 L3/3E.

Solution 218

=PLAE

from the frigure:d =AEdPx

Where:dP = centrifugal force of differential massdP = dM ω2 x = (ρA dx)ω2 xdP = ρAω2 x dx

d =AE( A 2xdx)x

=E 2 0Lx2dx=E 2 3x3 0L =E 2[L3−03]= 2L3 3E  ok!

Problem 219

A round bar of length L, which tapers uniformly from a diameter D at one end to a smaller diameter d at the other, is suspended vertically from the large end. If w is the weight per unit volume, find the elongation of ω the rod caused by its own weight. Use this result to determine the elongation of a cone suspended from its base.

Solution 219

=PLAE

For the differential strip shown:δ = dδP = weight carried by the strip = weight of segment yL = dyA = area of the strip

For weight of segment y (Frustum of a cone):

P=wVy

From section along the axis:

yx=LD−dx=LD−dy

 

Volume for frustum of coneV=31 h(R2+r2+Rr)Vy=31 h[41(x+d)2+41d2+21(x+d)(21d)]Vy=112 y[(x+d)2+d2+(x+d)d]

 

P=112 w[(x+d)2+d2+(x+d)d]yP=112 w[x2+2xd+d2+d2+xd+d2]yP=112 w[x2+3xd+3d2]y

P=12 w L2(D−d)2y2+L3d(D−d)y+3d3 y 

 

Area of the strip:

A=41 (x+d)2=4 LD−dy+d 2 

 

Thus,=PLAE

d =4 LD−dy+d 2E12 w L2(D−d)2y2+L3d(D−d)y+3d3 ydy 

d =4w12E         L2(D−d)2y2+L2d(D−d)y+d2L2(D−d)2y2+L3d(D−d)y+3d2      

  ydy

d =w3E         L2(D−d)2y2+2Ld(D−d)y+L2d2L2(D−d)2y2+3Ld(D−d)y+3L2d2  

      ydy

d =w3E (D−d)2y2+2Ld(D−d)y+L2d2(D−d)2y2+3Ld(D−d)y+3L2d2 ydy 

 

Let: a=D−d and b=Ld

d =w3E a2y2+2aby+b2a2y2+3aby+3b2 ydy 

d =w3E a2y2+3aby+3b2(ay)2+2(ay)b+b2 aa ydy 

d =w3aE (ay+b)2a3y3+3(a2y2)b+3(ay)b2 dy 

d =w3aE (ay+b)2[(ay)3+3(ay)2b+3(ay)b2+b3]−b3 dy 

 

The quantity (ay)3+3(ay)2b+3(ay)b2+b3 is the expansion of (ay+b)3

 

d =w3aE (ay+b)2(ay+b)3−b3 dy 

d =w3aE (ay+b)2(ay+b)3−b3(ay+b)2 dy d =w3aE[(ay+b)−b3(ay+b)−2]dy

 

=w3aE 0L[(ay+b)−b3(ay+b)−2]dy 

=w3aE 2a(ay+b)2−−ab3(ay+b)−1 0L 

=w3a2E 2(ay+b)2+b3ay+b 0L 

=w3a2E 21(aL+b)2+b3aL+b − 21b2+bb3  

=w3a2E 21(aL+b)2+b3aL+b−23b2  

=w3a2E 2(aL+b)(aL+b)3+2b3−3b2(aL+b)  

=w6a2E aL+b(aL)3+3(aL)2b+3(aL)b2+b3+2b3−3ab2L−3b3  

=w6a2E aL+ba3L3+3a2bL2  

 

Note that we let a=D−d and b=Ld

 

=w6(D−d)2E (D−d)L+Ld(D−d)3L3+3(D−d)2(Ld)L2  

=w6(D−d)2E LD−Ld+Ld(D−d)L3[(D−d)2+3d(D−d)]  

=wL36(D−d)E LD(D−d)2+3d(D−d)  

=wL36(D−d)E LDD2−2Dd+d2+3Dd−3d2  

=wL36(D−d)E LDD2+Dd−2d2  

=wL36(D−d)E LDD(D+d)−2d2  

=wL36(D−d)E LDD(D+d) −wL36(D−d)E LD2d2  =6E(D−d)wL2(D+d)−wL2d23ED(D−d)  answer

 

For a cone: D=D and d=0=6E(D−0)wL2(D+0)−wL2(02)3ED(D−0)=6EwL2  answer