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Lesson 03, simple stress and strain

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Simple Stress and Strain 2

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Lesson 03, simple stress and strain

  1. 1. Simple Stress and Strain Lecturer; Dr. Dawood S. Atrushi PP' P' r D d 1 2 d 1 2 ␴max ␴ave November 2014
  2. 2. Content ¢  Elastic deformations of axially loaded members ¢  Statically determinate and indeterminate members loaded axially ¢  Average shear Stress ¢  Stress concentrations ¢  Allowable stress ¢  IS Units November, 142 Strenght of Materials I - DAT
  3. 3. Elastic Deformation of Axially Loaded Members Consider a bar with ¢  gradually varying cross section along its length, and ¢  subjected to concentrated loads at its right end and also a variable external load distributed along its length November, 14Strenght of Materials I - DAT3
  4. 4. Mechanical deformation November, 14Strenght of Materials I - DAT4 (SI&4th p.120-127;3rd p.122-129) Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional are along its length L. For a more general case, the bar is subjected to concentrated loads at its right end and also a variable external load distributed along its length (such as a distributed load could be for example, to represent the weight of a vertical bar or friction forces acting on bar surface). Here we wish to find the relative displacement of one end with respect to the other. x P(x) dx dx (original length) P(x)+dP d (elongation of dx) P FBD a a View a-a A(x) L Fig. 2.7 Thermal and mechanical deformation We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The load P(x) will deform the element into the shape indicated by the dashed outline. The average stress in the cross-sectional area would be )( )( xA xP x d Generalized bar with gradually varying cross-sectional along its length (SI&4th p.120-127;3rd p.122-129) Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional are along its length L. For a more general case, the bar is subjected to concentrated loads at its right end and also a variable external load distributed along its length (such as a distributed load could be for example, to represent the weight of a vertical bar or friction forces acting on bar surface). Here we wish to find the relative displacement of one end with respect to the other. x P(x) dx dx (original length) P(x)+dP d (elongation of dx) P FBD a a View a-a A(x) L Fig. 2.7 Thermal and mechanical deformation We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The load P(x) will deform the element into the shape indicated by the dashed outline. The average stress in the cross-sectional area would be )( )( xA xP x d
  5. 5. ¢  The average stress in the cross- sectional area; November, 14Strenght of Materials I - DAT5 dx dx (original length) Fig. 2.7 Thermal and mechanical deformation element of length dx and cross-sectional area A(x). FBD can b Assume that resultant internal axial force is represented as P he element into the shape indicated by the dashed outline. he cross-sectional area would be )( )( xA xP x he cross-sectional area would be dx d x ties do not exceed the proportional limit, we can relate the dx dx (original length) Fig. 2.7 Thermal and mechanical deformation lement of length dx and cross-sectional area A(x). FBD ca Assume that resultant internal axial force is represented a e element into the shape indicated by the dashed outline. e cross-sectional area would be )( )( xA xP x cross-sectional area would be dx d x es do not exceed the proportional limit, we can relate ¢  The average strain in the cross- sectional area;
  6. 6. Using Hook’s law, i.e. σ = Eε; November, 14Strenght of Materials I - DAT6 Provided these quantities do not exceed the proportional limit, we Hook’s law, i.e. E Therefore dx d xE xA xP )( )( Re-organize the equation, we have dx xExA xP d )( )( For the entire length L of the bar, we must integrate this expression displacement L dx xExA xP 0 )( )( Where: = displacement between two points L = distance between the points P(x) = Internal axial force distribution A(x) = Cross-sectional area E(x) = Young’s modulus dx d xE xA xP )( )( Re-organize the equation, we have dx xExA xP d )( )( For the entire length L of the bar, we m displacement L dx xExA xP 0 )( )( Where: = displacement be L = distance between P(x) = Internal axial for A(x) = Cross-sectional E(x) = Young’s modulu dx d xE xA xP )( )( Re-organize the equation, we have dx xExA xP d )( )( For the entire length L of the bar, we must displacement L dx xExA xP 0 )( )( Where: = displacement betwe L = distance between th P(x) = Internal axial force A(x) = Cross-sectional area E(x) = Young’s modulus Where δ = displacement between two points L = distance between the points P(x) = Internal axial force distribution A(x) = Cross-sectional area E(x) = Young’s modulus For the entire length L of the bar;
  7. 7. Constant Load and Cross- Sectional Area For P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material) the displacement is; November, 14Strenght of Materials I - DAT7 Constant Load and Cross-Sectional Area In many engineering cases, the structural constant cross-sectional area and made of on P(x) = P = constant (no axially d A(x) = A = constant (uniform are E(x) = E = constant (homogeneo From Eq. (2.16), we have EA PL Multi-Segment Bar If the bar is subjected to several different
  8. 8. Multi-Segment Bar ¢  If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the equation can be used for each segment. The total displacement can be computed from algebraic addition as; November, 14Strenght of Materials I - DAT8 constant cross-sectional area and made of one homogeno P(x) = P = constant (no axially distributed loa A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material) From Eq. (2.16), we have EA PL Multi-Segment Bar If the bar is subjected to several different axial forces moduli, the above equation can be used for each segm computed from algebraic addition as i ii ii EA LP Example 2.4: The composite bar shown in the figure is having cross-sectional areas of AAB=200mm2 and ABC= EAB=100GPa and EBC=210GPa respectively. Find the tota
  9. 9. Example 4 The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200 mm2 and ABC=100 mm2. Their Young’s moduli are EAB=100 GPa and EBC=210 GPa respectively. Find the total displacement at the right end. November, 14Strenght of Materials I - DAT9 Multi-Segment Bar If the bar is subjected to several different axial forces or cross- moduli, the above equation can be used for each segment. The computed from algebraic addition as i ii ii EA LP Example 2.4: The composite bar shown in the figure is made of having cross-sectional areas of AAB=200mm2 and ABC=100mm2 . EAB=100GPa and EBC=210GPa respectively. Find the total displace Step 1 FBD Assume th tension. Step 2 Equil Internal forc F (Opposite to 4m 4.2m F1=10kN F2=40kN A B C E1A1 E2A2 F2=40kN
  10. 10. November, 14Strenght of Materials I - DAT10 having cross-sectional areas of AAB=200mm2 and ABC=100mm EAB=100GPa and EBC=210GPa respectively. Find the total displa Step 1 FB Assume tension. Step 2 Eq Internal fo (Opposite so Segme Internal fo PBC Step 3 Compute the total deformation by using Eq. (2.18) 3 4m 4.2m F1=10kN F2=40kN A B C E1A1 E2A2 F1=10kN F2=40kN C E1A1 E2A2 B PAB F1=10kN C PBC FBD1 FBD2 Solution Example 4 Step 1, Free body diagram
  11. 11. Step 2, Equilibriums Internal force in the whole bar, ABC; November, 14Strenght of Materials I - DAT11 Solution Example 4 =200mm and ABC=100mm . Their Young’s moduli are ctively. Find the total displacement at the right end. Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB 00 12 FFPF ABx kNPAB 30 (Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC 00 1FPF BCx kNFP 10 (in tension) F1=10kN C F1=10kN C F1=10kN Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB 00 12 FFPF ABx kNPAB 30 (Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC 00 1FPF BCx kNFPBC 101 (in tension) F1=10kN C F1=10kN C F1=10kN C Internal force in the segment BC;
  12. 12. November, 14Strenght of Materials I - DAT12 Step 3, Deformation The total deformation;Step 3 Compute the total deformation by using Eq. (2. 9 3 2010100 1030 BCBC BCBC ABAB ABAB BCABAC AE LP AE LP mAC 004.0002.0006.0 F1=10kN C E1A1 E2A2 B PAB F1=10kN C PBC FBD1 FBD2 00 12 FFPF ABx kNPAB 30 (Opposite to our assumption of tensio so Segment AB is in compression) Internal force in BC 00 1FPF BCx kNFPBC 101 (in tension) he total deformation by using Eq. (2.18) 69 3 69 3 1010010210 2.41010 1020010100 41030 BCBC BCBC ABAB ABAB AE LP AE LP mmmAC 4004.0002.0006.0 (towards left) F1=10kN F2=40kN C A1 E2A2 B F1=10kN C PBC 0 PF ABx kNPAB 30 (Opposite to our assumptio so Segment AB is in compr Internal force in BC 0 FPF BCx kNFPBC 101 (in ompute the total deformation by using Eq. (2.18) 9 3 69 3 10010210 2.41010 1020010100 41030 BCBC BCBC ABAB ABAB BCB AE LP AE LP mmmAC 4004.0002.0006.0 (towards left) 4m 4.2m F1=10kN F2=40kN C E1A1 E2A2 B B F1=10kN C PBC
  13. 13. Statically Determinate Members Loaded Axially Statically determinate ¢  System with the same number of unknown reactions as equations of statics. ¢  Known reactions can be determined strictly from equilibrium equations. November, 14Strenght of Materials I - DAT13 (SI&4th Ed p. 134-139; 3rd Ed p. 137- Statically Determinate and Indetermin When a bar is supported at one end and s in Fig. 2.8(a), there is only one unknown the unknown reaction can easily be dete unknown reactions as equations of sta reactions can be determined strictly from FA P (a) Statically determinate A B Fig. 2.8 Statically deter
  14. 14. Statically Indeterminate Members Loaded Axially Statically indeterminate The system that has more unknown forces than equations of statics. The reactions can not be determined only from equilibrium equations. November, 14Strenght of Materials I - DAT14 (SI&4th Ed p. 134-139; 3rd Ed p. 137-142) Statically Determinate and Indeterminate When a bar is supported at one end and subjected to an axial force P at the other end as in Fig. 2.8(a), there is only one unknown reaction force FA. By using the equations of the unknown reaction can easily be determined. So such a system with the same num unknown reactions as equations of statics is called statically determinate. – i.e. reactions can be determined strictly from equilibrium equations. FA P (a) Statically determinate P (b) Statically indeterminate FA FB C B AA B LAC LCB L Fig. 2.8 Statically determinate and indeterminate structures
  15. 15. Compatibility Conditions ¢  What we need is an additional equation that specifies how the structure is displaced due to the applied loading. Such an equation is usually termed the compatibility equation. November, 14Strenght of Materials I - DAT15 (SI&4th Ed p. 134-139; 3rd Ed p. 137-142) Statically Determinate and Indeterminate When a bar is supported at one end and subjected to an axial force P at the other end as in Fig. 2.8(a), there is only one unknown reaction force FA. By using the equations of the unknown reaction can easily be determined. So such a system with the same num unknown reactions as equations of statics is called statically determinate. – i.e. reactions can be determined strictly from equilibrium equations. FA P (a) Statically determinate P (b) Statically indeterminate FA FB C B AA B LAC LCB L Fig. 2.8 Statically determinate and indeterminate structures
  16. 16. Compatibility Condition November, 14Strenght of Materials I - DAT16 FA C B A =0 C’ = AC + CB=0 FA AC C C’ PAC FA FBD B C C’ Elongated Contracted CB FB PCB FBD Fig. 2.9 Compatibility condition Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC, The amount that length AC elongates CB contracts
  17. 17. ¢  Now, since both ends of the bar are fully fixed, then the total change in length between A and B must be zero. ¢  The amount that length AC elongates CB contracts; so the equation can be written as: δAC +δCB =0 November, 14Strenght of Materials I - DAT17
  18. 18. From the free body diagram for segment AC and CB; November, 14Strenght of Materials I - DAT18 B = AC + CB=0 B Fig. 2.9 Compatibility condition Let’s respectively look at the free body diagram for segment AC and CB as i (indeed FBD can be in any level of structural system or structural members). For segment AC, 00 ACAy PFF AAC FP Tension (+) ACAC ACA ACAC ACAC AC EA LF EA LP elongation (+) For segment CB, 00 CBBy PFF BCB FP Compression ( ) CBCB CBB CBCB CBCB CB EA LF EA LP contraction ( ) Compatibility condition: LFLF B = AC + CB=0 B Elongated Fig. 2.9 Compatibility condition Let’s respectively look at the free body diagram for segment AC and CB as in Fig. (indeed FBD can be in any level of structural system or structural members). For segment AC, 00 ACAy PFF AAC FP Tension (+) ACAC ACA ACAC ACAC AC EA LF EA LP elongation (+) ( For segment CB, 00 CBBy PFF BCB FP Compression ( ) CBCB CBB CBCB CBCB CB EA LF EA LP contraction ( ) ( Compatibility condition: 0CBBACA CBAC LFLF (
  19. 19. November, 14Strenght of Materials I - DAT19 CBCB CBB CBCB CBCB CB EA LF EA LP contraction ( ) Compatibility condition: 0 CBCB CBB ACAC ACA CBAC EA LF EA LF Combining Compatibility equation (2.23) with the equation of solve for the two unknowns FA and FB as, PFF EA LF EA LF BA CBCB CBB ACAC ACA 0 i.e. P EA L EA L EA L F CBCB CB ACAC AC ACAC AC B A L F A A If ConstEAEA CBCBACAC , we have P L L F AC B and P L L F CB A 00 CBBy PFF BCB FP Compressio CBCB CBB CBCB CBCB CB EA LF EA LP contraction ( ) Compatibility condition: 0 CBCB CBB ACAC ACA CBAC EA LF EA LF Combining Compatibility equation (2.23) with the equation of static solve for the two unknowns FA and FB as, PFF EA LF EA LF BA CBCB CBB ACAC ACA 0 i.e. P EA L EA L EA L F CBCB CB ACAC AC ACAC AC B EA L A L F ACAC AC CB A If ConstEAEA CBCBACAC , we have For segment CB, 00 CBBy PFF BCB FP Compression ( ) CBCB CBB CBCB CBCB CB EA LF EA LP contraction ( ) Compatibility condition: 0 CBCB CBB ACAC ACA CBAC EA LF EA LF Combining Compatibility equation (2.23) with the equation of statics (2.19 solve for the two unknowns FA and FB as, PFF EA LF EA LF BA CBCB CBB ACAC ACA 0 i.e. P EA L EA L EA L F CBCB CB ACAC AC ACAC AC B EA L EA L EA L F CB CB ACAC AC CBCB CB A If ConstEAEA CBCBACAC , we have LAC LCB or segment CB, 00 CBBy PFF BCB FP Compression ( ) CBCB CBB CBCB CBCB CB EA LF EA LP contraction ( ) (2.22) ompatibility condition: 0 CBCB CBB ACAC ACA CBAC EA LF EA LF (2.23) ombining Compatibility equation (2.23) with the equation of statics (2.19), we now olve for the two unknowns FA and FB as, PFF EA LF EA LF BA CBCB CBB ACAC ACA 0 (2.24) e. P EA L EA L EA L F CBCB CB ACAC AC ACAC AC B P EA L EA L EA L F CBCB CB ACAC AC CBCB CB A ConstEAEA CBCBACAC , we have LAC LCB Combining compatibility equation with the equation of statics, we now can solve for the two unknowns FA and FB as;
  20. 20. November, 14Strenght of Materials I - DAT20 Lecture N PFF BA i.e. P EA L EA L EA L F CBCB CB ACAC AC ACAC AC B FA If ConstEAEA CBCBACAC , we have P L L F AC B and P L L F CB A Lecture PFF BA i.e. P EA L EA L EA L F CBCB CB ACAC AC ACAC AC B F If ConstEAEA CBCBACAC , we have P L L F AC B and P L L F CB A
  21. 21. Example 5 Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average normal stress in both bars if increase the temperature from 10°C to 210°C. November, 14Strenght of Materials I - DAT21 Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutmen Originally, there is a gap of 5mm between the ends as shown in the figure. Determine avera normal stress in both bars if increase the temperature from 10 C to 210 C. Copper Aluminumcu=17 × 10-6 Ecu=110GPa al=23×10-6 Eal=69GPa 0.4m 0.8m 0.005m d = 0.01m T,cu F,cu F Thermally expanded T,cu due to T Mechanical force push it back by F,cu Copper Copper d = 0.01m T Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutmen Originally, there is a gap of 5mm between the ends as shown in the figure. Determine avera normal stress in both bars if increase the temperature from 10 C to 210 C. Copper Aluminumcu=17 × 10-6 Ecu=110GPa al=23×10-6 Eal=69GPa 0.4m 0.8m 0.005m d = 0.01m T,cu F,cu F Thermally expanded T,cu due to T Mechanical force push it back by F,cu Copper Copper d = 0.01m T Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutmen Originally, there is a gap of 5mm between the ends as shown in the figure. Determine avera normal stress in both bars if increase the temperature from 10 C to 210 C. Copper Aluminumcu=17 × 10-6 Ecu=110GPa al=23×10-6 Eal=69GPa 0.4m 0.8m 0.005m d = 0.01m T,cu F,cu F Thermally expanded T,cu due to T Mechanical force push it back by F,cu Copper Copper d = 0.01m T
  22. 22. November, 14Strenght of Materials I - DAT22 Originally, there is a gap of 5mm between the ends as shown in the figure. Determine avera normal stress in both bars if increase the temperature from 10 C to 210 C. Copper Aluminumcu=17 × 10-6 Ecu=110GPa al=23×10-6 Eal=69GPa T0=10oC T=210oC T= 200oC 0.4m 0.8m 0.005m d = 0.01m T,cu F,cu T,Al F,Al F F Thermally expanded T,cu due to T T Mechanical force push it back by F,cu Copper Copper Aluminum Aluminum d = 0.01m T 522 10857010 4 143 4 .. . dA m2 and CT o 20010210 Originally, there is a gap of 5mm between the ends as shown in the figure. Determine aver normal stress in both bars if increase the temperature from 10 C to 210 C. Copper Aluminumcu=17 × 10-6 Ecu=110GPa al=23×10-6 Eal=69GPa T0=10oC T=210oC T= 200oC 0.4m 0.8m 0.005m d = 0.01m T,cu F,cu T,Al F,Al F F Thermally expanded T,cu due to T T Mechanical force push it back by F,cu Copper Copper Aluminum Aluminum d = 0.01m T 522 10857010 4 143 4 .. . dA m2 and CT o 20010210 Originally, there is a gap of 5mm between the ends as shown in the figure. Determine aver normal stress in both bars if increase the temperature from 10 C to 210 C. Copper Aluminumcu=17 × 10-6 Ecu=110GPa al=23×10-6 Eal=69GPa T0=10oC T=210oC T= 200oC 0.4m 0.8m 0.005m d = 0.01m T,cu F,cu T,Al F,Al F F Thermally expanded T,cu due to T T Mechanical force push it back by F,cu Copper Copper Aluminum Aluminum d = 0.01m T 522 10857010 4 143 4 .. . dA m2 and CT o 20010210 Originally, there is a gap of 5mm between the ends as shown in the figure. Determine aver normal stress in both bars if increase the temperature from 10 C to 210 C. Copper Aluminumcu=17 × 10-6 Ecu=110GPa al=23×10-6 Eal=69GPa T0=10oC T=210oC T= 200oC 0.4m 0.8m 0.005m d = 0.01m T,cu F,cu T,Al F,Al F F Thermally expanded T,cu due to T T Mechanical force push it back by F,cu Copper Copper Aluminum Aluminum d = 0.01m T 522 10857010 4 143 4 .. . dA m2 and CT o 20010210 Originally, there is a gap of 5mm between the ends as shown in the figure. Determine aver normal stress in both bars if increase the temperature from 10 C to 210 C. Copper Aluminumcu=17 × 10-6 Ecu=110GPa al=23×10-6 Eal=69GPa T0=10oC T=210oC T= 200oC 0.4m 0.8m 0.005m d = 0.01m T,cu F,cu T,Al F,Al F F Thermally expanded T,cu due to T T Mechanical force push it back by F,cu Copper Copper Aluminum Aluminum d = 0.01m T 522 10857010 4 143 4 .. . dA m2 and CT o 20010210 Copper Aluminumcu=17 × 10-6 Ecu=110GPa al=23×1 Eal=69GP T0=10oC T=210oC T= 200oC 0.4m 0.8m 0.005m d = 0.01m T,cu F,cu T,Al F,Al F F Thermally expanded T,cu due to T T Mechanical force push it back by F,cu Copper Copper Aluminum Aluminum d = 0.01 T 522 10857010 4 143 4 .. . dA m2 and CT o 20010210
  23. 23. November, 14Strenght of Materials I - DAT 23 T,Al 522 10857010 4 143 4 .. . dA m2 and Let’s firstly look at the copper bar. When the bar system i copper bar expand towards right by Cu,T . After the copp mechanical force F will develop, which will prevent the We assume that due to such a mechanical force, the copp real total deformation of copper bar will be computed as CuFCuTCu ,, (elongation +, Contractio Similarly, we have Al,FAl,TAl (elongation +, Contraction Because these two expanding bars should fill the gap, we p 0050.CuAl T,Al Aluminum 5 1085 m2 and CT o 20010210 hen the bar system is heated up from 10 C to 210 C, the Cu,T . After the copper bar touch to the aluminum bar, a ch will prevent the copper bar from expanding further. ical force, the copper bar is pressed back by Cu,F . The ll be computed as ion +, Contraction ) on +, Contraction ) Total deformation of copper; T,Al 522 10857010 4 143 4 .. . dA m2 and T 210 Let’s firstly look at the copper bar. When the bar system is heated u copper bar expand towards right by Cu,T . After the copper bar tou mechanical force F will develop, which will prevent the copper ba We assume that due to such a mechanical force, the copper bar is p real total deformation of copper bar will be computed as CuFCuTCu ,, (elongation +, Contraction ) Similarly, we have Al,FAl,TAl (elongation +, Contraction ) Because these two expanding bars should fill the gap, we prescribe a c 0050.CuAl From these two equations, we have 0050.Al,FAl,TCu,FCu,T T= 200oC T,Al T Aluminum 522 10857010 4 143 4 .. . dA m2 and T 210 Let’s firstly look at the copper bar. When the bar system is heated u copper bar expand towards right by Cu,T . After the copper bar tou mechanical force F will develop, which will prevent the copper ba We assume that due to such a mechanical force, the copper bar is p real total deformation of copper bar will be computed as CuFCuTCu ,, (elongation +, Contraction ) Similarly, we have Al,FAl,TAl (elongation +, Contraction ) Because these two expanding bars should fill the gap, we prescribe a 0050.CuAl From these two equations, we have Similarly, we have;
  24. 24. November, 14Strenght of Materials I - DAT24 real total deformation of copper bar will be computed CuFCuTCu ,, (elongation +, Contr Similarly, we have Al,FAl,TAl (elongation +, Contra Because these two expanding bars should fill the gap, 0050.CuAl From these two equations, we have 0050.Al,FAl,TCu,FCu,T i.e. E F TL AE LF TL A AlAl CuCu Cu CuCu LL TLTL F AlCu AlAlCuCu 1017005.0 6 real total deformation of copper bar will be computed as CuFCuTCu ,, (elongation +, Contraction Similarly, we have Al,FAl,TAl (elongation +, Contraction ) Because these two expanding bars should fill the gap, we presc 0050.CuAl From these two equations, we have 0050.Al,FAl,TCu,FCu,T i.e. AE LF TL AE LF TL AlAl Al AlAl CuCu Cu CuCu AE L AE L TLTL F AlAl Al CuCu Cu AlAlCuCu 85.710110 4.0 02001017005.0 9 6 These two expanding bars should fill the gap, we prescribe a compatibility condition as; copper bar expand towards right by Cu,T . After the copper bar touch to mechanical force F will develop, which will prevent the copper bar from We assume that due to such a mechanical force, the copper bar is pressed real total deformation of copper bar will be computed as CuFCuTCu ,, (elongation +, Contraction ) Similarly, we have Al,FAl,TAl (elongation +, Contraction ) Because these two expanding bars should fill the gap, we prescribe a compa 0050.CuAl From these two equations, we have 0050.Al,FAl,TCu,FCu,T i.e. 0050. AE LF TL AE LF TL AlAl Al AlAl CuCu Cu CuCu N AE L AE L TLTL F AlAl Al CuCu Cu AlAlCuCu 2.206 691085.710110 4.0 10234.02001017005.0 59 6 Solve for F; F = 206.2 N
  25. 25. November, 14Strenght of Materials I - DAT25 The average normal stress can be computed as; Lectu i.e. AE LF TL AE LF TL AAl AlAl CuCu Cu CuCu N AE L AE L TLTL F AlAl Al CuCu Cu AlAlCuCu 2.206 10110 0 21017005.0 9 6 The average normal stress can be computed as MPa. . . A F 632 10857 2206 5
  26. 26. Average Shear Stress Shear Stress The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau), and it is expressed as; November, 14Strenght of Materials I - DAT26 m AE LF TL AlAl Al AlAlAlFAlTAl 1065.31005.31068.3 363 ,, 2.9 AVERAGE SHEAR STRESS (S The intensity or force per unit area acting ta is expressed as in Eq. (2.3) as: A F lim t A 0 In order to show how the shear stress can de an example. The block is supported by two vertically as shown in Fig. 2.10. If the force
  27. 27. Let’s take a block supported by two rigid bodies and an external force F as an example; A lim A 0 (2.3) In order to show how the shear stress can develop in a structural member, let’s take a block as an example. The block is supported by two rigid bodies and an external force F is applied vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the block to deform and fail along the vertical planes as shown. A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at each section to hold the segment in equilibrium. F F F V V A B C D Sectioned area A Block Rigid Rigid Fig. 2.10 Average shear stress 020 FVF y November, 14Strenght of Materials I - DAT27 A lim A 0 (2.3) In order to show how the shear stress can develop in a structural member, let’s take a block as an example. The block is supported by two rigid bodies and an external force F is applied vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the block to deform and fail along the vertical planes as shown. A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at each section to hold the segment in equilibrium. F F F V V A B C D Sectioned area A Block Rigid Rigid Fig. 2.10 Average shear stress 020 FVF y A lim A 0 (2.3) In order to show how the shear stress can develop in a structural member, let’s take a block as an example. The block is supported by two rigid bodies and an external force F is applied vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the block to deform and fail along the vertical planes as shown. A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at each section to hold the segment in equilibrium. F F F V V A B C D Sectioned area A Block Rigid Rigid Fig. 2.10 Average shear stress 020 FVF y
  28. 28. Vertical equilibrium; November, 14Strenght of Materials I - DAT28 V Fig. 2.10 Average shear str 020 FVF y 2/FV The average shear stress distributed over each sectioned are defined by A V avg avg = assume to be the same at each point over the sect V = Internal shear force A = Area at the section Average shear stress distributed over each section; Fig. 2.10 Averag 020 FVF y 2/FV The average shear stress distributed over each sec defined by A V avg avg = assume to be the same at each point ov V = Internal shear force A = Area at the section Assume to be uniform over the section Internal shear force Area at the section V Fig. 2.10 Average shear s 020 FVF y 2/FV The average shear stress distributed over each sectioned a defined by A V avg avg = assume to be the same at each point over the sec V = Internal shear force A = Area at the section
  29. 29. Stress concentrations ¢  For a uniform cross- sectional bar that is applied an axial force, both experiment and theory of elasticity find that the normal stress will be uniformly distributed over the cross- section. November, 14Strenght of Materials I - DAT29 intensity of the forces distributed the stress on that section and is igma). The stress in a member of to an axial load P (Fig. 1.8) is he magnitude P of the load by the 5 P A (1.5) dicate a tensile stress (member in dicate a compressive stress (mem- used in this discussion, with P ex- square meters (m2 ), the stress s unit is called a pascal (Pa). How- an exceedingly small quantity and s unit must be used, namely, the (MPa), and the gigapascal (GPa). Fig. 1.8 Member with an axial load. (a) (b) A P A P' P' P ␴ ϭ
  30. 30. Stress Concentration l  If we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements. November, 14Strenght of Materials I - DAT30 avg K max in which avg=P/A’ is the assumed average stress from the figures or tables (as in Fig. 2.11(c)), and from avg=P/A’, where A’ is the smallest cross-sec maximum stress at the cross section can be comput 'A P KK avgmax (a) (b) P P A’ w 2r l  In engineering practice, the actual stress distribution does not need. Instead, only the maximum stress at these sections must be known.
  31. 31. November, 14Strenght of Materials I - DAT31 e points of larger than a structural den change ear the dis- stresses in Figure 2.58 ss distribu- Figure 2.59 widths con- rowest part the use of who has to h an analy- PP' P' r D d 1 2 d 1 2 ␴max ␴ave Fig. 2.58 Stress distribution near circular hole in flat bar under axial loading.
  32. 32. Stress Concentration If we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements. November, 14Strenght of Materials I - DAT32 In engineering practice, though, the actual stress dis Instead, only the maximum stress at these sections designed to resist this highest stress when the axial maximum normal stress at the critical section can be advanced mathematical techniques using the the investigations are usually reported in graphical for Concentration Factor K. avg K max in which avg=P/A’ is the assumed average stress as from the figures or tables (as in Fig. 2.11(c)), and the from avg=P/A’, where A’ is the smallest cross-sectio maximum stress at the cross section can be computed a 'A P KK avgmax from avg=P/A’, where A’ is the smallest cross-sec maximum stress at the cross section can be comput 'A P KK avgmax (a) (b) P P A’ w 2r Fig. 2.11 Stress c Stress concentration occurs in the case that there By observing Fig. 2.11(c), it is interesting to no sectional area, the higher the stress concentration A´ is the smallest cross section
  33. 33. Stress Concentration Factor If we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements. November, 14Strenght of Materials I - DAT33 Stress Concentration Factor In engineering practice, though, the actual stress distribution Instead, only the maximum stress at these sections must be designed to resist this highest stress when the axial load is a maximum normal stress at the critical section can be determin advanced mathematical techniques using the theory of investigations are usually reported in graphical form (as in Concentration Factor K. avg K max in which avg=P/A’ is the assumed average stress as in Fig. 2 from the figures or tables (as in Fig. 2.11(c)), and the average from avg=P/A’, where A’ is the smallest cross-sectional area. maximum stress at the cross section can be computed as: P KK avg K max (2.28) which avg=P/A’ is the assumed average stress as in Fig. 2.11(b). Provided K has been known om the figures or tables (as in Fig. 2.11(c)), and the average normal stress has been calculated om avg=P/A’, where A’ is the smallest cross-sectional area. Then from the above equation the aximum stress at the cross section can be computed as: 'A P KK avgmax (2.29) (a) (b) (c) P P A’ StressConcentrationFactorK w 2r r/w
  34. 34. Allowable Stress ¢  When the stress (intensity of force) of an element exceeds some level, the structure will fail. ¢  We usually adopt allowable force or allowable stress to measure the threshold of safety in engineering. November, 14Strenght of Materials I - DAT34
  35. 35. Uncertainties that we must take into account in engineering: ¢  The load for design may be different from the actual load. ¢  Size of structural member may not be very precise due to manufacturing and assembly. ¢  Various defects in material due to manufacturing processing. November, 14Strenght of Materials I - DAT35
  36. 36. To consider such uncertainties a Factor of Safety, F.S., is usually introduced; November, 14Strenght of Materials I - DAT36 Various defects in material due to manufacturing proce One simple method to consider such uncertainties is to us Safety, F.S., which is a ratio of failure load Ffail (found from the allowable one Fallow allow fail F F .S.F If the applied load is linearly related to the stress developed using A/F , then we can define the factor of safety as the allowable stress allow allow fail SF .. Usually, the factor of safety is chosen to be greater than failure. This is dependent on the specific design case. For safety for some of its components may be as high as 3. For allow fail F F .S.F If the applied load is linearly rela using A/F , then we can defi the allowable stress allow allow fail SF .. Usually, the factor of safety is ch failure. This is dependent on the safety for some of its components F.S. (safer), the heavier the structu to balance the safety and cost. The value of F.S. can be found in use Eq. (2.6) to compute the allow allow Usually, the factor of safety i failure. This is dependent on t safety for some of its compon F.S. (safer), the heavier the str to balance the safety and cost. The value of F.S. can be foun use Eq. (2.6) to compute the al ..SF fail allow Example 2.2: In Example Cu,allow=50MPa, please deter strength point of view. F Fallow : Allowable load Ffail : Failure load σallow : Allowable stress The factor of safety is chosen to be greater than 1
  37. 37. Example 6 ¢  If the maximum allowable stress for copper in Example 1 is σCu,allow=50 MPa, please determine the minimum size of the wire/cable from the material strength point of view. November, 14Strenght of Materials I - DAT37 f Stress its in the SI system is the Newton per square meter or Pascal, i.e. : Pa = N/m2 . neering, Pa seems too small, so we usually use: Kilo Pascal KPa (=Pa 103 ) e.g. 20,000Pa=20kPa Mega Pascal MPa (=Pa 106 ) e.g. 20,000,000Pa=20MPa Giga Pascal GPa (=Pa 109 ) e.g. 20,000,000,000Pa=20GPa le 2.1: An 80 kg lamp is supported by a single electrical cable of diameter d = 3.15 mm. What is the stress carried cable. ermine the stress in the wire/cable as Eq. (2.4), we need ss sectional area A of the cable and the applied internal : 26 22 10793.7 4 00315.0 4 m d A N.mgF 7848980 MPa. .A F 6100 107937 784 6 ble Stress (SI&4th p. 48-49, 3rd p. 51-52) Example 2.1, we may concern whether or not 80kg would be too heavy, or say MPa stress would be too high for the wire/cable, from the safety point of view. Indeed, 80kg F a a Section a-a A d FBD F
  38. 38. Solution November, 14 Strenght of Materials I - DAT 38 Example 2.2: In Example 2.1, if the max Cu,allow=50MPa, please determine the minimum strength point of view. Mathematically, allowCu d mg A F ,2 4 Therefore: mg d allowCu 10469.4 4 , Obviously, the lower the allowable stress, the bigg structural strength and elemental size. In engineering, there are two significant problems Problem (1) Stress Analysis: for a specific str ..SF fail allow mple 2.2: In Example 2.1, if the maximum allowable allow=50MPa, please determine the minimum size of the wire ngth point of view. hematically, allowCu d mg A F ,2 4 refore: mm mg d allowCu 469.410469.4 4 3 , iously, the lower the allowable stress, the bigger the cable size. ctural strength and elemental size. ngineering, there are two significant problems associated with st ..SF fail allow Example 2.2: In Example 2.1, if the maxi Cu,allow=50MPa, please determine the minimum strength point of view. Mathematically, allowCu d mg A F ,2 4 Therefore: mg d allowCu 10469.4 4 , Obviously, the lower the allowable stress, the bigge structural strength and elemental size. In engineering, there are two significant problems as value of F.S. can be found in design codes and engineering hand Eq. (2.6) to compute the allowable stress: ..SF fail allow ample 2.2: In Example 2.1, if the maximum allowable s allow=50MPa, please determine the minimum size of the wire/ca ngth point of view. hematically, allowCu d mg A F ,2 4 refore: mm mg d allowCu 469.410469.4 4 3 , viously, the lower the allowable stress, the bigger the cable size. St ctural strength and elemental size.
  39. 39. IS UNITS from MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. Dewolf David F. Mazurek November, 14Strenght of Materials I - DAT39
  40. 40. SI Prefixes November, 14Strenght of Materials I - DAT40 SI Prefixes Multiplication Factor Prefix† Symbol 1 000 000 000 000 5 1012 tera T 1 000 000 000 5 109 giga G 1 000 000 5 106 mega M 1 000 5 103 kilo k 100 5 102 hecto‡ h 10 5 101 deka‡ da 0.1 5 1021 deci‡ d 0.01 5 1022 centi‡ c 0.001 5 1023 milli m 0.000 001 5 1026 micro m 0.000 000 001 5 1029 nano n 0.000 000 000 001 5 10212 pico p 0.000 000 000 000 001 5 10215 femto f 0.000 000 000 000 000 001 5 10218 atto a † The first syllable of every prefix is accented so that the prefix will retain its identity. Thus, the preferred pronunciation of kilometer places the accent on the first syllable, not the second. ‡ The use of these prefixes should be avoided, except for the measurement of areas and vol-
  41. 41. Principal SI Units Used in Mechanics November, 14Strenght of Materials I - DAT41 Principal SI Units Used in Mechanics Quantity Unit Symbol Formula Acceleration Meter per second squared p m/s2 Angle Radian rad † Angular acceleration Radian per second squared p rad/s2 Angular velocity Radian per second p rad/s Area Square meter p m2 Density Kilogram per cubic meter p kg/m3 Energy Joule J N ? m Force Newton N kg ? m/s2 Frequency Hertz Hz s21 Impulse Newton-second p kg ? m/s Length Meter m ‡ Mass Kilogram kg ‡ Moment of a force Newton-meter p N ? m Power Watt W J/s Pressure Pascal Pa N/m2 Stress Pascal Pa N/m2 Time Second s ‡ Velocity Meter per second p m/s Volume, solids Cubic meter p m3 Liquids Liter L 1023 m3 Work Joule J N ? m † Supplementary unit (1 revolution 5 2p rad 5 3608). ‡ Base unit.
  42. 42. U.S.CustomaryUnitsandTheirSI Equivalents November, 14Strenght of Materials I - DAT42 Quantity U.S. Customary Units SI Equivalent Acceleration ft/s2 0.3048 m/s2 in./s2 0.0254 m/s2 Area ft2 0.0929 m2 in2 645.2 mm2 Energy ft ? lb 1.356 J Force kip 4.448 kN lb 4.448 N oz 0.2780 N Impulse lb ? s 4.448 N ? s Length ft 0.3048 m in. 25.40 mm mi 1.609 km Mass oz mass 28.35 g lb mass 0.4536 kg slug 14.59 kg ton 907.2 kg Moment of a force lb ? ft 1.356 N ? m lb ? in. 0.1130 N ? m Moment of inertia Of an area in4 0.4162 3 106 mm4 Of a mass lb ? ft ? s2 1.356 kg ? m2 Power ft ? lb/s 1.356 W hp 745.7 W Pressure or stress lb/ft2 47.88 Pa lb/in2 (psi) 6.895 kPa Velocity ft/s 0.3048 m/s in./s 0.0254 m/s mi/h (mph) 0.4470 m/s mi/h (mph) 1.609 km/h Volume, solids ft3 0.02832 m3 in3 16.39 cm3 Liquids gal 3.785 L qt 0.9464 L Work ft ? lb 1.356 J
  43. 43. Thank You November, 14Strenght of Materials I - DAT43

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