Design of Simple Connections

53 Allowable Stress allow fail P P S F = . allow fail S F σ σ = . allow fail S F τ τ = .

6. Design of Simple Connections

allow P A σ = allow V A τ = • Cross-Sectional Area of a Tension Member P P a a P τ allow allow P A σ = 54 σ b allow P V = P P • Cross-Sectional Area of a Connector Subjected to Shear allow P A σ = P P ττττ allow • Required Area to Resist Bearing allow b P A σ = 55 • Required Area to Resist Shear by Axial Load P d P l allow π τ = P d Uniform shear stress τ allow 56 Example 12a The control arm is subjected to the loading shown. a Determine the shear stress for the steel at all pin b Determine normal stress in rod AB, plate D and E. The thickness of plate D and E is 10 mm. pin C 25 kN 13 kN 3 4 5 75 mm 50 mm 200 mm A B C D E φ = 10 mm φ = 20 mm D E φ = 25 mm 13 kN 50 mm 25 kN 50 mm φ = 25 mm 57 25 kN 13 kN 3 4 5 75 mm 50 mm 200 mm • Reaction C 125 . 87 . 36 sin 25 075 . 13 2 . = − + AB F + ΣM C = 0: F AB = 4.5 kN , ← ΣF y = 0: + C y + 13 - 25sin36.87 o = 0 C y = 2 kN, ↑ ΣF x = 0: + -4.5 - C x + 25cos36.87 o = 0 C x = 15.5 kN, ← kN 63 . 15 2 5 . 15 2 2 = + = C 25 kN 13 kN 75 mm 50 mm 200 mm F AB C y C x A B C C x C y C 36.87 o 58 25 kN 13 kN 75 mm 50 mm 200 mm 4.5 kN a Shear stress 15.63 kN 15.63 kN Pin C Double shear 7.815 kN 7.815 kN Pin D and E Single shear pin C 25 kN 13 kN 3 4 5 75 mm 50 mm 200 mm A B C D E = 35 mm φ = 20 mm φ = 25 mm φ = 25 mm ⇐ = = = MPa 88 . 24 02 . 4 kN 815 . 7 2 π A V τ C C C ⇐ = = = MPa 48 . 26 025 . 4 kN 13 2 π A V τ D D D ⇐ = = = MPa 93 . 50 025 . 4 kN 25 2 π A V τ E E E 59 25 kN 13 kN 75 mm 50 mm 200 mm 4.5 kN b Normal stress 15.63 kN Cale AB Plate D φ rod = 10 mm A C D E σ ⇐ = = = MPa 7 . 56 010 . 4 kN 5 . 4 2 π σ AB AB A P D E φ = 25 mm 13 kN 50 mm 25 kN 50 mm φ = 25 mm 13 kN σ D 13 kN 13 kN t = 0.01 m 0.05 m ⇐ = = = MPa 26 01 . 05 . kN 13 A P σ D 60 25 kN 13 kN 75 mm 50 mm 200 mm 4.5 kN 15.63 kN φ rod = 35 mm A C D E Plate E ⇐ = − = = MPa 100 01 . 025 . 05 . kN 25 A P E σ D E φ = 25 mm 13 kN 50 mm 25 kN 50 mm φ = 25 mm 25 kN 50 mm 25 kN 25 kN 50 mm σ E 0. 05 m 0. 02 5 m t = .0 1 m 61 Example 12b The control arm is subjected to the loading shown. Determine the required diameter of the steel pin at C if the allowable shear stress for the steel is τ allow = 55 MPa. Note in the figure that the pin is subjected to double shear. 22 kN 13 kN 3 4 5 75 mm 50 mm 200 mm A B C 62 • Internal Shear Force 125 . 87 . 36 sin 22 075 . 13 2 . = − − AB F + ΣM C = 0: F AB = 13.125 kN ΣF y = 0: + C y - 13 - 22sin36.87 o = 0 C y = 26.2 kN ΣF x = 0: + -13.125 - C x + 22cos36.87 o = 0 C x = 4.47 kN kN 58 . 26 2 . 26 47 . 4 2 2 = + = C 22 kN 13 kN 75 mm 50 mm 200 mm F AB C y C x 22 kN 13 kN 3 4 5 75 mm 50 mm 200 mm A B C C x C y C 63 • Required Area 2 2 6 6 3 mm 242 m 10 6 . 241 10 55 10 29 . 13 = × = × × = = − allow τ V A 2 2 mm 242 2 = d π d = 17.55 mm Use d = 20 mm 22 kN 13 kN 3 4 5 75 mm 50 mm 200 mm A B C 22 kN 13 kN 75 mm 50 mm 200 mm 13.125 kN 26.58 kN 26.58 kN 13.29 kN 13.29 kN 64 Example 13a The two members are pinned together at B as shown below. Top views of the pin connections at A and B are also given. If the pins have an allowable shear stress of τ allow = 86 MPa , the allowable tensile stress of rod CB is σ t allow = 112 MPa and the allowable bearing stress is σ b allow = 150 MPa, determine to the smallest diameter of pins A and B ,the diameter of rod CB and the thickness of AB necessary to support the load. 13 kN A B C 1 m 0.6 m 3 4 5 Top view A B 65 • Internal Force 13 kN A B F BC 1 m 0.6 m 36.87 o A y A x + Σ M A = 0: 6 . 1 87 . 36 sin 1 13 = + − BC F F BC = 13.54 kN Σ F y = 0: + 87 . 36 sin 54 . 13 13 = + − y A A y = 4.88 kN Σ F x = 0: + 87 . 36 cos 54 . 13 = + − x A A x = 10.83 kN 66 • Diameter of the Pins 13 kN A B F BC = 13.54 kN 1 m 0.6 m 36.87 o A y = 4.88 kN A x = 10.83 kN Top view A B 11.88 kN Pin A 11.88 kN V A = 5.94 kN V A = 5.94 kN 2 2 3 mm 07 . 69 m kN 10 86 kN 94 . 5 = × = = allow A A τ V A 2 2 mm 07 . 69 4 = A d π d A = 9.38 mm, Use d A = 10 mm Pin B F BC = 13.54 kN V B = 13.54 kN 2 2 3 mm 4 . 157 m kN 10 86 kN 54 . 13 = × = = allow B B V A τ 2 2 mm 4 . 157 4 = B d π d B = 14.16 mm, Use d B = 15 mm 67 • Diameter of Rod 13.54 kN 13.54 kN 2 2 3 mm 9 . 120 m kN 10 112 kN 54 . 13 = × = = allow t BC σ P A 2 2 mm 9 . 120 4 = BC d π d BC = 12.4 mm, Use d BC = 15 mm 13 kN A B F BC = 13.54 kN 1 m 0.6 m 36.87 o A y = 4.88 kN A x = 10.83 kN 11.88 kN 68 t AB φ B = 15 mm 20 kN A B 13.54 kN 11.88 kN t AB φ A = 10 mm A B σ b A σ b B m 00792 . 010 . 10 88 . 11 10 150 3 6 = × = × = AB AB allow b t t A P σ • Thickness 13 kN A B F BC = 13.54 kN 1 m 0.6 m 36.87 o A y = 4.88 kN A x = 10.83 kN 11.88 kN m 00602 . 015 . 10 54 . 13 10 150 3 6 = × = × = AB AB allow b t t A P σ mm 8 = AB t use comparison By 69 Example 13b The two members are pinned together at B as shown below. Top views of the pin connections at A and B are also given. If the pins have an allowable shear stress of τ allow = 86 MPa, the allowable tensile stress of rod CB is σ t allow = 112 MPa and the allowable bearing stress is σ b allow = 150 MPa, determine to the maximum load P that the beam can supported. P A B C 1 m 0.6 m 3 4 5 Top view A B φ = 15 mm φ = 10 mm t = 8 mm φ = 15 mm t = 8 mm 70 • Internal Force P A B F BC 1 m 0.6 m 36.87 o A y A x + Σ M A = 0: 6 . 1 87 . 36 sin 1 = + − BC F P F BC = 1.042P T 87 . 36 sin 042 . 1 ; = + − = Σ ↑ + P P A F y y A y = 0.375P 87 . 36 cos 042 . 1 ; = + − = Σ →  + P A F x x A x = 0.834 kN 71 Pin A double shear 0.914P V A = 0.457P V A = 0.457P kN 78 . 14 01 . 4 457 . kPa 10 86 2 3 = = × = P π P A V τ A A allow Pin B single shear F BC = 1.042P V B = 1.042P kN 58 . 14 015 . 4 042 . 1 kPa 10 86 2 3 = = × = P π P A V τ B B allow P A B F BC = 1.042P 1 m 0.6 m 36.87 o A y = 0.375P 0.834P Top view A B 0.914P φ = 10 mm φ = 15 mm 72 1.042P 1.042P Rod AB P A B F BC = 1.042P 1 m 0.6 m 36.87 o A y = 0.375P 0.834P 0.914P kN 19 015 . 4 042 . 1 kPa 10 112 2 3 = = × = P π P A P σ BC allow φ rod = 15 mm 73 t AB = 10 mm φ B = 15 mm 20 kN A B F BC = 1.042P 0.914P t AB = 10 mm φ A = 10 mm A B σ b A σ b B kN 41 . 16 010 . 010 . 914 . 10 150 6 = = × = P P A P σ allow b kN 60 . 21 01 . 015 . 042 . 1 10 150 6 = = × = P P A P σ allow b P A B F BC = 1.042P 1 m 0.6 m 36.87 o A y = 0.375P 0.834P Top view A B 0.914P φ = 10 mm φ = 15 mm t = 10 mm t = 10 mm ⇐ = kN 58 . 14 P all comparison By 74 Example 14 The suspender rod is supported at its end by a fixed-connected circular disk as shown. If the rod passes through a 40-mm diameter hole, determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 20 kN load. The allowable normal stress for the rod is σ allow = 60 MPa, and the allowable shear stress for the disk is τ allow = 35 MPa. d 40 mm t 20 kN 20 kN 40 mm τ allow 75 d 40 mm t 20 kN 20 kN 40 mm τ allow • Diameter of Rod 2 3 m kN 10 60 kN 20 × = = allow σ P A 2 3 2 m kN 10 60 kN 20 4 × = = d π A d = 0.0206 m = 20.6 mm • Thickness of Disk 2 3 m kN 10 35 kN 20 × = = allow τ V A 2 3 m kN 10 35 kN 20 m 02 . 2 × = t π t = 4.55x10 -3 m = 4.55 mm 76 Example 15 An axial load on the shaft shown is resisted by the collar at C, which is attached to the shaft and located on the right side of the bearing at B. Determine the largest value of P for the two axial forces at E and F so that the stress in the collar does not exceed an allowable bearing stress at C of σ b allow = 75 MPa and allowable shearing stress of the adhesive at C of τ allow = 100 MPa , and the average normal stress in the shaft does not exceed an allowable tensile stress of σ t allow = 55 MPa. A B P C E F 2P 60 mm 20 mm 80 mm 77 A B P C E F 2P 60 mm 20 mm 80 mm 3P Position Load 3P 2P • Axial Stress 3P P 2P A P allow t = σ 2 2 3 m 03 . 3 m kN 10 55 π P = × P 1 = 51.8 kN • Bearing Stress A P allow b = σ ] m 03 . m 04 . [ 3 m kN 10 75 2 2 2 3 π π P − = × B C 60 mm 80 mm 3P P 2 = 55 kN 78 A B P C E F 2P 60 mm 20 mm 80 mm 3P Position Load 3P 2P The largest load that can be applied to the shaft is P = 51.8 kN. shear allow A P = τ ] 020 . m 04 . 2 [ 3 m kN 10 100 2 3 π P = × P 3 = 55 kN • Shearing Stress B C 60 mm 80 mm 3P 20 mm • Axial Stress • Shearing Stress • Bearing Stress P 3 = 55 kN P 1 = 51.8 kN P 2 = 55 kN 79 Example 16 The rigid bar AB shown supported by a steel rod AC having a diameter of 20 mm and an aluminum block having a cross-sectional area of 1800 mm 2 . The18-mm- diameter pins at A and C are subjected to single shear. If the failure stress for the steel and aluminum is σ st fail = 680 Mpa and σ al fail = 70 MPa, respectively, and the failure shear stress for each pin is τ fail = 900 MPa, determine the largest load P that can be applied to the bar. Apply a factor of safety of F.S = 2.0. P Steel 0.75 m 2 m Aluminum A B C 80 + ΣM B = 0: P1.25 m - F AC 2 m = 0 --------- F AC = 0.625P 1.25 m P A B F AC F B 0.75 m + ΣM A = 0: P0.75 m - F B 2 m = 0 ---------- F B = 0.375P P Steel 0.75 m 2 m Aluminum A B C • Rod AC AC AC fail st allow st A F S F = = . σ σ 2 3 m π 0.01 0.625 2 kPa 10 680 P = × P 1 = 171 kN • Block B B B fail al allow al A F S F = = . σ σ 2 6 3 m 10 1800 0.375 2 kPa 10 70 − × = × P P 2 = 168 kN 81 P Steel 0.75 m 2 m Aluminum A B C • Pin A or C pin AC fail allow A F S F = = . τ τ 2 3 m π 0.009 0.625 2 kPa 10 900 P = × P 3 = 183 kN Largest load P = 168 kN 1.25 m P A B F AC = 0.625P F B = 0.375P 0.75 m • Pin A or C P 3 = 183 kN Summary • Rod AC P 1 = 171 kN • Block B P 2 = 168 kN 1 STRAIN Deformation Strain Normal Strain Shear Strain Cartesian Strain Components 2 Deformation Deformed body θ A B C Undeformed body uA A´ B´ C´ θ´ 3 Strain x y z x y z x y 1 ε x 1 x y 1 1 ε y • Normal Strain 4 x y x y 2 π x y xy γ xy γ ε y 1 1 ε x • Shear Strain 5 Example 1 The slender rod shown subjected to an increase of temperature along its axis, which creates a normal strain in the rod of ε z = 4010 -3 z 12 , where z is given in meters. Determine a the displacement of the end B of the rod due to the temperature increase, and b the average normal strain in the rod. 200 mm z dz A B 6 200 mm z dz A B a the displacement of the end B of the rod due to the temperature increase dz z dz ] 10 40 [ 2 1 3 − = ∫ − = ∆ m 2 . 2 1 3 ] 10 40 [ dz z B ↓ = = − m z 00239 . 3 2 10 40 m 2 . 2 3 3 b the average normal strain in the rod. mm mm 0119 . mm 200 mm 39 . 2 = = ∆ ∆ − ∆ = s s s avg ε 7 Example 2 A force acting on the grip of the lever arm shown causes the arm to rotate clockwise through an angle of θ = 0.002 rad. Determine the average normal strain developed in the wire BC. B A L 2L C 8 B A L 2L C B´ 2L B A L 2L C Lcos θ Lsin θ θ • Solution 001 . 2 002 . 2 2 = = = = − = θ θ ε L L CB CB CB avg Since θ is small, the stretch in wire CB, is using Eq.2-6. The average normal strain in the wire is then , sin θ θ L L ≈ 9 Example 3 The plate is deformed into the dashed shape shown. If in this deformed shape horizontal lines on the plate remain horizontal and do not change their length, determine a the average normal strain along the side AB, and b the average shear strain in the plate relative to the x and y axes. 250 mm 300 mm 3 mm 2 mm B C x y A 10 250 mm 3 mm 2 mm B y A B´ mm 018 . 248 3 2 250 2 2 = + − = AB a the average normal strain along the side AB 250 250 018 . 248 − = − = AB AB AB avg AB ε mmmm 10 93 . 7 3 − − = b the average shear strain in the plate relative to the x and y axes. 250 mm 300 mm 3 mm 2 mm B C x y A B´ γ xy θ´ rad 0121 . 2 250 3 tan 1 = − = − xy γ 2 θ π γ − = xy 11 Example 4 The plate shown is held in the rigid horizontal guides at its top and bottom, AD and BC. If its right side CD is given a uniform horizontal displacement of 2 mm, determine a the average normal strain along the diagonal AC, and b the shear strain at E relative to the x, y axes. 150 mm 150 mm A B C D E 2 mm x y 12 A B 75 mm C´ D´ E´ 75 mm 76 mm 76 mm a the average normal strain along the diagonal AC m 21213 . 150 . 150 . 2 2 = + = AC m 21355 . 152 . 150 . 2 2 = + = AC 150 mm 150 mm A B C D E 2 mm x y mmmm 00669 . 21213 . 21213 . 21355 . = − = − = AC AC AC avg AC ε θ ´ 13 A B 75 mm C´ D´ E´ 75 mm 76 mm 76 mm θ ´ b the shear strain at E relative to the x, y axes. 75 76 2 tan = θ 150 mm 150 mm A B C D E 2 mm x y rad 58404 . 1 759 . 90 180 759 . 90 = = = o o o π θ rad 0132 . 58404 . 1 2 − = − = π γ xy 1 M ECHANICAL P ROPERTIES OF M ATERIALS Simple Tension Test The Stress-Strain Diagram Stress-Strain Behavior of Ductile and Brittle Materials Hooke’s Law Strain Energy Poisson’s Ratio The Shear Stress-Strain Diagram Failure of Materials Due to Creep and Fatigue 2 400 350 300 250 200 150 100 50 εεεε mmmm σ σ σ σ MPa 0.10 0.20 0.30 Upper scale 0.40 0.00 0.0020 0.0030 0.0010 Lower scale 0.0040 0.0000 Stress Strain Relationship L d P P 3 necking strain hardening yielding elastic region The Stress-Strain Diagram εεεε σ σ σ σ σ pl proportional limit elastic limit σ f σ u σ f ´ fracture stress ultimate stress elastic behavior plastic behavior true fracture stress yield stress σ Y 4 σ Y 0.002 0.2 offset 300 250 200 150 100 50 ε mmmm σ MPa 0.005 0.010 Offset yield strength for material with no yield points Offset Yield Stress 5 σ y u = 230 MPa σ y l = 220 MPa A P = σ L δ ε = 400 350 300 250 200 150 100 50 ε mmmm σ MPa 0.10 0.20 0.30 Upper scale 0.40 0.00 0.0020 0.0030 0.0010 Lower scale 0.0040 0.0000 E = = 200x10 3 MPa= 200 GPa 0.001 σ fail = 295 MPa σ pl = 200 MPa σ u = 390 MPa 200 L d P P 6 400 350 300 250 200 150 100 50 ε mmmm σ MPa 0.10 0.20 0.30 Upper scale 0.40 0.00 0.0020 0.0030 0.0010 Lower scale 0.0040 0.0000 σ pl = 180 MPa σ y = 250 MPa σ u = 360 MPa σ fail = 310 MPa E = = 200x10 3 MPa = 200 GPa 0.0005 100 7 Stress-Strain Behavior of Ductile and Brittle Materials 0.02 0.04 0.06 100 200 300 400 σ MPa ε mmmm σ−ε σ−ε σ−ε σ−ε diagrams for a methacrylate plastic Brittle material Ductile material 8 • Elongation 100 L L L elongation Percent f − = 100 A A A area of reduction Percent f − = 9 • Temperature Effects: 0.02 0.04 0.06 100 200 300 400 σ MPa ε mmmm σ−ε σ−ε σ−ε σ−ε diagrams for a methacrylate plastic 10 o C 40 o C 70 o C Brittle to Ductile Ductile to Brittle 10 σ ε Hooke’s Law σ pl ε pl = = pl pl E ε σ Constant 11 - Apply load to failure • Elastic and Plastic Behavior of Materials 1 σ ε PL EL Failure 12 - Apply and release load σ ε PL EL a Load is less than proportional limit 1 2 σ ε PL EL b Load is more than proportional limit, but less than elastic limit 1 2 13 2 3 1 2 3 4 Mechanical hysteresis 1 σ ε PL EL c Load is more than elastic limit, and reaply σ ε PL EL d Repeated load is more than elastic limit loading 14 2 σ ε PL EL Repeated loading n times 3 4 mechanical hysteresis 1 σ εεεε O 1 Apply load once - Comparison 15 E E elastic region plastic region O´ unload ε σ A O load A´ B permanent set elastic region plastic region elastic recovery ε σ O´ A´ mechanical hysteresis 16 u t u r Strain Energy • Modulus of Resilience σ pl ε pl ε σ Modulus of resilience u r • Modulus of Toughness ε σ Modulus of toughness u t E pl u pl pl r 2 2 1 2 1 σ ε σ = = 17 Modulus of resiliency u r = Area under curve proportional limit u r = 120.001180 MPa = 90 kPa = 90 kNm 2 = 90 kN•mm 3 = 90 kJm 3 Upper scale Lower scale 400 350 300 250 200 150 100 50 ε mmmm σ MPa 0.10 0.20 0.30 0.40 0.00 0.0020 0.0030 0.0010 0.0040 0.0000 Energy per unit volume = 90 kNm 2 1 m 3 = 90 kN•m = 90 kJ σ σ σ σ pl = 180 MPa Modulus of resiliency • Modulus of Resilience 18 400 350 300 250 200 150 100 50 ε mmmm σ MPa 0.10 0.20 0.30 0.40 0.00 0.0020 0.0030 0.0010 Upper scale Lower scale 0.0040 0.0000 Modulus of hyper-resiliency EL 2 1 3 19 400 350 300 250 200 150 100 50 εεεε mmmm σ σ σ σ MPa 0.10 0.20 0.30 Upper scale 0.40 0.00 0.0020 0.0030 0.0010 Lower scale 0.0040 0.0000 Failure Modulus of toughness • Modulus of Toughness 20 Example 1 A tension test for a steel alloy results in the stress-strain diagram shown. Calculate the modulus of elasticity and the yield strength based on a 0.2 offset. Identify on the graph the proportional limit, elastic limit, ultimate stress and the fracture stress. 400 350 300 250 200 150 100 50 ε mmmm σ MPa 0.10 0.20 0.30 Upper scale 0.40 0.00 0.0020 0.0030 0.0010 Lower scale 0.0040 0.0000 21 400 350 300 250 200 150 100 50 ε mmmm σ MPa 0.10 0.20 0.30 Upper scale 0.40 0.00 0.0020 0.0030 0.0010 Lower scale 0.0040 0.0000 σ σ σ σ y E • Modulus of Elasticity GPa mm mm MPa E 200 0005 . 100 = = • Yield Strength σ y = 250 MPa σ σ σ σ pl • Proportional Limit σ pl = 180 MPa σ σ σ σ EL • Elastic Limit σ pl = 220 MPa σ σ σ σ u • Ultimate Stress σ u = 365 MPa • Fracture Stress σ f = 310 MPa σ σ σ σ fail 22 Example 2 An aluminum specimen shown has a diameter of d = 25 mm, a gauge length of L = 250 mm and is subjected to an axial load of 294.5 kN. If a portion of the stress-strain diagram for the material is shown, determine the approximate elongation of the rod when the load is applied. If the load is removed, does the rod return to its original length? Also, compute the modulus of resilience both before and after the load application. 0.01 0.02 0.04 150 300 450 σ MPa ε mmmm 0.03 750 600 L d 294.5 kN 294.5 kN 23 0.01 0.02 0.04 150 300 450 σ MPa ε mmmm 0.03 750 600 F O 294.5 kN 294.5 kN L = 250 mm 294.5 kN 294.5 kN L = 250 mm d = 25 mm • The Load is Applied - Normal Stress MPa m N A P 600 025 . 4 10 5 . 294 2 3 = × = = π σ - The Strain ε = 0.023 mmmm - The Elongation δ = ε L = 0.023 mmmm250 mm = 5.75 mm D 0.023 5.75 mm2 5.75 mm2 A B 24 294.5 kN 294.5 kN • The Load is Removed 0.01 0.02 0.04 150 300 450 σ MPa ε mmmm 0.03 750 600 F O D 0.023 A B L = 250 mm d = 25 mm - Permanent Strain CD MPa GPa mm mm MPa E 600 . 75 006 . 450 = = = CD = 0.008 mmmm The permanent strain, ε OC = 0.023 - CD ε OC = 0.023 - 0.008 = 0.015 mmmm - Normal Stress MPa m N A P 600 025 . 4 10 5 . 294 2 3 = × = = π σ - The Permanent Elongation δ = ε L = 0.015 mmmm250 mm = 3.75 mm ε pl =0.006 ε OC σ pl = E E CD C 294.5 kN 294.5 kN L = 250 mm 3.75 mm2 3.75 mm2 25 u r initial u r final 0.01 0.02 0.04 150 300 450 σ MPa ε mmmm 0.03 750 600 F O D 0.023 A B L = 250 mm d = 25 mm 294.5 kN 294.5 kN - Normal Stress MPa m N A P 600 025 . 4 10 5 . 294 2 3 = × = = π σ ε pl =0.006 σ pl = CD = 0.008 C • Modulus of Resilience - Modulus of Resilience 3 40 . 2 008 . 600 2 1 2 1 m MJ mm mm MPa u pl pl final r = = = ε σ 3 35 . 1 006 . 450 2 1 2 1 m MJ mm mm MPa u pl pl initial r = = = ε σ 26 Example 3 An aluminum rod shown has a circular cross section and is subjected to an axial load of 10 kN. If a portion of the stress-strain diagram for the material is shown, determine the approximate elongation of the rod when the load is applied. If the load is removed, does the rod return to its original length? Take E al = 70 GPa. 600 mm 400 mm 20 mm 15 mm A B C 10 kN 10 kN 50 60 40 O ε mmmm σ MPa 0.02 0.08 0.10 0.04 0.06 0.12 30 20 10 27 50 60 40 O ε mmmm σ MPa 0.02 0.08 0.10 0.04 0.06 0.12 30 20 10 • The Load is Applied 600 mm 400 mm 20 mm 15 mm A B C 10 kN 10 kN MPa m kN A P AB 83 . 31 01 . 10 2 = = = π σ MPa m kN A P BC 6 . 56 0075 . 10 2 = = = π σ mm mm Pa Pa E al AB AB 0004547 . 10 70 10 83 . 31 9 6 = × × = = σ ε ε BC = 0.045 mmmm The elongation of the rod is δ = Σ ε L = ε AB L AB + ε BC L BC = 0.0004547600 mm + 0.045400 mm = 18.3 mm 31.83 56.6 28 50 60 40 O ε mmmm σ MPa 0.02 0.08 0.10 0.04 0.06 0.12 30 20 10 • The Load is Removed 600 mm 400 mm 20 mm 15 mm A B C MPa m kN A P AB 83 . 31 01 . 10 2 = = = π σ MPa m kN A P BC 6 . 56 0075 . 10 2 = = = π σ The elongation of the rod is δ´ = Σ ε L = 0 + ε OG L BC = 0.0442400 mm = 17.7 mm parallel G ε BC = 0.045 mmmm mm mm Pa Pa E al BC rec 000808 . 10 70 10 6 . 56 9 6 = × × = = σ ε ε OG The permanent strain, ε OG = 0.0450 - 0.000808 = 0.0442 mmmm 56.6 10 kN 10 kN σ pl = 31.83 29 L Tension r Poisson’s Ratio L Compression Original Shape r P P Final Shape δ´ δ2 δ2 Original Shape Final Shape δ2 δ2 P P δ´ r and L lat long δ ε δ ε = = long lat ε ε ν − = 30 P z x y t P L b z y z x ε ε ε ε ν − = − = Assumption: • Homogeneous • Isotropic • Elastic 31 Example 4 A bar made of A-36 steel has the dimensions shown. If an axial force of P = 80 kN is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically. Take E = 200 GPa and ν st = 0.32. P = 80 kN z x y 50 mm P = 80 kN 1.5 m 100 mm 32 MPa m m kN A P z 16 05 . 1 . 80 = = = σ z Pa Pa ε 10 16 10 200 6 9 = δ z = ε z L z = [8010 -6 1.5 m] = 120 µm P = 80 kN z x y 50 mm P = 80 kN 1.5 m 100 mm • The change in the bar’s length δ δ δ δ - z direction z z st E ε σ = µ ε 80 10 80 6 = = − mm mm z - x and y direction z y z x long lat st Ratio s Poisson ε ε ε ε ε ε ν − = − = − = : ε x = ε y = - ν st ε z = -0.32[8010 -6 ] = -25.6 µ δ x = -ε x L x = -[25.610 -6 0.1 m = -2.56 µm δ y = -ε y L y = -[25.610 -6 0.05 m = -1.28 µm 33 x y The Shear Stress-Strain Diagram x y τ xy π2 - γ xy γ xy 2 γ xy 2 τ = Gγ τ γ τ pl γ pl τ u γ u τ f γ f G 1 2 ν + = E G 34 Example 5 A specimen of titanium alloy is tested in torsion and the shear stress-strain diagram is shown. a Determine the shear modulus G, the proportional limit, and the ultimate shear stress. b Determine the distance d that the top of a block of this material, shown, could be displaced horizontally by a shear force V of 135 MN . 300 200 O γ rad τ MPa 0.73 100 400 0.54 0.008 50 mm 100 mm 75 mm γ d V 35 300 200 O γ rad τ MPa 0.73 100 400 - Proportional limit ; τ pl = 270 MPa - Ultimate shear stress; τ u = 370 MPa γ pl = 0.008 G a The shear modulus, proportional limit, and the ultimate shear stress. GPa rad MPa G 75 . 33 008 . 270 = = - Shear Modulus ; τ u = 370 τ pl = 270 36 50 mm 100 mm 75 mm

1.35 MN