Mechanics of Solids Chapter 2 1/25 AXIALLY LOADED MEMBERS 2.1 INTRODUCTION i.e., SOLID BARS (truss members, jet eng struts) CABLES We will consider structural elements that: COIL SPRINGS 1. have straight longitudinal axes 2. carry only axial forces (Tension or Compression) Will determine: 1. Maximum Stresses 2. Displacements Structural Analysis = the calculation of stresses, strains, displacements and load-carrying capacity. Design = The determination of the geometric configuration and other characteristics of a structure (i.e., material) in order to fulfill all the prescribed functions. Designing a structure is usually a repetitive chore (iterative). Design-analyze, if it does not meet specifications or perform as intended, Redesign – analyze……. Design is more difficult than analyzing. 2.2 DISPLACEMENTS OF AXIALLY LOADED MEMEBERS L Prismatic Bar: δ P CONDITIONS For Eqn ( 1 ) P Prismatic Bar P ________________(a) A ________________(b) L Homogeneous Material E _______________(c) Linearly Elastic From (b) into (c): E From (a) into (d): Sign Convention: L ____________________(d) P E A L + = lengthening - = shorthening or PL _______________(1) EA (usually not needed) Mechanics of Solids Chapter 2 2/25 Analogous to an axially loaded spring: Stiffness (spring constant): k Flexibility (compliance): f From (1): P __________________(e) 1 _______________(f) k P PL EA P EA _________________(g) L FROM PHYSICS: F = - k (Δx) From (f) into (g): k EA __________________(2) L f L EA (used a lot) (seldom used) 2.3 MULTIPLE LOADS and / or MULTIPLE SECTIONS ( each one prismatic bar ) P 1 P 1 A a P 2 B a b C P 2 c b D n i 1 Pi Li Ei Ai _____________________(3) Pi = axial force in section i of the bar n = total No. of sections section = (discontinuity) load change, geometry change, or material change don’t cut section at a point where the discontinuity occurs because won’t know which part to apply load or geometry or material to. Mechanics of Solids Chapter 2 3/25 CONTINUOUSLY VARYING AXIAL FORCE and / or CROSS-SECTIONAL AREA Can not use above equation (eqn 3) A FBD C B A C P(x) dx X X L from eqn. (1): PL EA using a differential element: d P( x) dx E A( x) Integrating: L L 0 0 d P( x) dx _________________________(4) E A( x) Based on the following assumption: - stress distribution is uniform over the entire cross section ( based on σ = P/A ). (not valid for tapered bars) However, eqn (4) gives accurate results for small tapers. < 20° - Due to presence of E, material must be Linearly Elastic Mechanics of Solids Chapter 2 4/25 EXAMPLE No. 1 GIVEN: The composite shown is made from 2 segments, AB and BD having cross-sectional areas AAB = 1 in2 and ABD = 2 in2, respectively. It is subject to the loads shown. Let E = 29 x 103 ksi. 15 kip A FIND: Vertical displacement of end A. 4 kip B 8 kip 8 kip SOLn: 2 ft 4 kip 1.5 ft C D 1 ft Mechanics of Solids Chapter 2 5/25 EXAMPLE No. 2 GIVEN: A member is made from a material that has a specific weight γ and modulus of elasticity E. It is formed into the cone shown and suspended in the vertical position. y FIND: r 0 The displacement of the end due to gravity SOLn: L x Mechanics of Solids Chapter 2 6/25 2.4 STATICALLY INDETERMINATE STRUCTURES Statically determinate = a structure for which the internal forces can be found using equations of equilibrium. Statically Indeterminate = a structure for which the equations of equilibrium are not sufficient to find the internal forces. A A P P B B STATICALLY DETERMINATE STATICALLY INDETERMINATE Fy RB P 0 Fy RB P RA 0 ___( 1 ) 1 eqn / 1 unkwn 1 eqn / 2 unkwns 2 methods for analyzing statically indeterminate members: 1. Flexibility Method 2. Stiffness Method FLEXIBILITY METHOD (Force Method…..forces are unknown quantities) RA FBD: A P a P RA + RB = P L C b B RB Let RA = statically redundant force – remove it. From eqn ( 1 ): (2) Mechanics of Solids Chapter 2 7/25 A This is the RELEASED or PRIMARY structure. P Now, consider the effect of P on the displacement of A. C b p B Pb EA displacement is Next, consider the effect of RA RA A R RA L EA displacement is B Final displacement of A is: δA = δP - δR + ( + is in direction of load) We know: δA = 0, therefore; δR = δP Compatibility Eqn R A L Pb EA EA RA Pb __________ ( 3a ) L Solving for RB in eqn ( 2 ): RB = P – RA Substituting from eqn ( 3 ): RB P Pb L RB PL Pb P ( L b) L L RB Pa _____________ ( 3b ) L NOTE: a L b Mechanics of Solids Chapter 2 8/25 FLEXIBILITY METHOD SUMMARY: 1. select 1 unknown reaction as redundant and release it from the structure. 2. calculate displacement due to the actual load. 3. calculate displacement due to the redundant force. 4. combine the 2 displacements into an equation of compatibility ( δR = δP ). 5. substitute expressions for displacements and solve for redundant force. 6. solve for the remaining unknown force using equilibrium eqn. called Flexibility Method because eqn of compatibility contains: f L EA Also called, FORCE METHOD….forces are unknown quantities. RESTRICTIONS: material must behave in a linearly elastic manner. Flexibility Method is based on: 1. eqns of static equilibrium 2. compatibility eqns…..will always have as many compatibility eqns. as redundants will always have sufficient no. of eqns to solve for all unknowns STIFFNESS METHOD (Displacement Method…displacements are unknown quantities) RA A P a P L C δC is taken as the unknown. C b B RB Assume C is moving downward: AC elongates R a C A EA RA EA C ________ ( 4a ) a BC shortens R b C B EA RB EA C _________ ( 4b ) b Mechanics of Solids Chapter 2 Isolate C as freebody: 9/25 +Fy = RA + RB – P = 0 RA RA + RB = P P C Substituting from eqns 4a & 4b: RB EA EA C C P a b Stiffness Terms k Thus, C Pab Pab EA(a b) EAL NOTE: L = a + b Substituting this back into eqns 4a & 4b: RA EA Pab a EAL RB EA Pab b EAL RA Pb L RB Pa L STIFFNESS METHOD SUMMARY: 1. select suitable displacement as unknown quantity 2. express forces in terms of this displacement 3. substitute into the equilibrium eqn 4. solve for displacement 5. solve for forces 2.5 TEMPERATURE & PRE-STRAIN EFFECTS εT = UNIFORM THERMAL STRAIN = Coefficient of Thermal Expansion - property of material - values for various materials from Table H – 4 (text appendix H, pg 915) T T _________________ ( 5 ) units of : SI 1 (Kelvin) or k 1 C sign convention: + = expansion – = contraction US 1 F EA L Mechanics of Solids Chapter 2 10/25 Changes in dimensions, such as length: T T L T L ___________ ( 6 ) - usually only length of structural members are considered. Transverse direction is usually neglected. OMIT statically determinant – no thermal stress because member is free to expand or contract ( uniform ΔT ) READ TEXT Pg 93 - 100 statically indeterminate – thermal stresses non-uniform ΔT also produces thermal stresses. Consider a statically indeterminate bar: R Remove upper support FBD δT ΔT L T T L R Applying R R R RL EA δT – δR = 0 T L RL 0 EA R = (ΔT)EA STRESS: R T E A (compression) read about PRE-STRAIN {important for pre-loading in bolts} pg 101 – 104 analysis is essentially the same as that of temperature change. Mechanics of Solids Chapter 2 11/25 EXAMPLE No 1 GIVEN: The aluminum post shown is reinforced with a brass core. It supports a resultant compressive axial load of P = 9 kip. EAL = 10x103 ksi EBR = 15 x 103 ksi 9 kip P FBD A A 1.5’ FBR r = 1” r = 2” FAL y FIND: Average normal stress in the aluminum and brass via the Flexibility Method. SOLn: Mechanics of Solids Chapter 2 EXAMPLE No. 2 GIVEN: Same as example No. 1 FIND: Average normal stress in the aluminum and brass via the Stiffness Method. SOLn: 12/25 Mechanics of Solids Chapter 2 13/25 OMIT EXAMPLE No. 3 GIVEN: The rigid bar shown below is fixed to the top of the 3 posts made of steel and aluminum. The posts have a length of 250 mm when no load is applied to the bar and the temperature is 20°C. The bar is subjected to a uniform distributed load of 150 kN/m and the temperature is raised to 80°C. 600 mm STEEL: Est = 200(109) Pa ALUM: Eal = 70(109) Pa αst = 12(10-6)/ °C αal = 23(10-6)/ °C FIND: The force supported by each post. 60 mm 40 mm SOLn: 150 kN/m st st al 40 mm 250 mm Mechanics of Solids Chapter 2 14/25 2.6 STRESSES ON INCLINED SURFACES Axially loaded prismatic bar y m P P x O n z y m P O x x P A ( 1 ) normal stress on m n y n z x x x In 2-D: y z O m P x x O n P A There are other internal forces acting on the bar. y p P P x x P _________ ( 1 ) A q IMPORTANT: is w.r.t. direction of load And outward normal !! // ┴ N P P sin P cos V N = Normal Force (force normal to surface) V = Shearing Force (force parallel to surface) Mechanics of Solids Chapter 2 + 15/25 F┴ = 0 F// = 0 + P sin – V = 0 __________ ( 2b ) N – P cos = 0 ________ ( 2a ) These forces produce stresses: // P A σ = normal stress ( ┴ to surface) ┴ σ τ = shearing stress ( // to cut surface) τ A = cross sectional area of cut surface SIGN CONVENTION: σ = + when in tension τ = + when it produces a CCW rotation of material (matches moment sign) A A = A cos A A N A N A __________ ( 4a ) A ________________( 3 ) cos V A V A __________ ( 4b ) FROM eqns ( 2 ) & ( 4 ): A P cos 0 P sin A 0 A P cos A P sin FROM eqn ( 3 ): A P cos cos P cos 2 A FROM eqn ( 1 ): x cos 2 ________ ( 5a ) A P sin cos P sin cos A x sin cos ASIDE: sin cos 12 sin( 2 ) 12 x sin( 2 ) _________ ( 5b ) Mechanics of Solids Chapter 2 16/25 Eqns 5 gives the normal and shear stress at an angle w.r.t. the axial load in terms of x which is the normal stress in the direction of the axial load. These eqns (5a & b) are only good for axial loading. However, they were derived based only on the equilibrium eqns so they are valid for any material whether it behaves linearly or nonlinearly, elastically or nonelastically. FROM eqn ( 5 ): cos = max when = 0° sin (2) = max when 2 = 90° = 45° max x @ 0 THEREFORE: max x 2 @ 45 Maximum Normal Stress Maximum Shear Stress STRESS ELEMENT 45° P y σx σx P σx σx 2 = 45° = 0° σx 2 σ τmax = x 2 τmax = σx x σx 2 σx 2 y x τmax = σx 2 τmax = σx 2 σx 2 FROM eqn ( 5 ): (cos 0°)2 = 1 σ = σx = σmax sin ( 2 x 0° ) = 0 τ = 0 (cos 45° )2 = ½ sin (2 x 45°) = 1 σ = ½ σ x τ = - ½ σx = τmax Mechanics of Solids Chapter 2 17/25 For 0° or 45° { for angles of other than 0° and 45° } σ2 σ1 τ y x τ τ τ σ2 σ1 Eqns ( 5a ) and ( 5b ): 1 x cos 2 1 12 x sin( 21 ) 1 For angles other than 0° and 45° must consider the complimentary angle, 2 = 90° + 1 to obtain: 2 x cos 2 2 _________________ ( 5c ) ALTERNATTE STRESS ELEMENT (found in some texts) τ x - advantage: gives original stresses , τ - disadvantage: doesn’t give stress on other planes τxy τyx y - homework should include stress element. - stress element is just as important as a FBD > gives stress at a point in the structure Once you know the magnitude and direction of 1 shear stress, you have it for the others on the element because they are equal but opposite…..keeps element in equilibrium. This is not true for the normal stress. Mechanics of Solids Chapter 2 18/25 2.7 STRAIN ENERGY OMIT READ: pg 116 - 122 P is a static load. This Load is applied slowly (gradually) from P = 0 to P = Pmax δ L Thus, δ = 0 to δ = δmax P LOAD DISPLACEMENT DIAGRAM P Work = area under curve dP1 W P1 d 1 0 Pmax P1 δ δ1 dδ1 δ Strain Energy (internal work) = the energy absorbed by the bar during the loading process. U = strain energy U W P1 d 1 _____________ ( 1 ) 0 P1 = any value of load between 0 and Pmax UNITS: SI 1J = 1Nm US ft lbs Mechanics of Solids Chapter 2 19/25 Now, we unload the bar: P Elastic limit (pt A) exceeded B A Permanent deformation OD remains LOADING UNLOADING δ O D C Elastic strain energy – recoverable energy Inelastic strain energy – lost energy Now, assume we stay below the elastic limit load P A Material Conditions from O to A: 1. Elastic 2. Follows Hooke’s Law LOADING P U FROM eqn ( 1 ): UNLOADING B O δ U 12 P _________ ( 2 ) δ RECALL: for a prismatic bar: FROM eqn ( 2 ): PL __________ ( a ) EA 2U _______________ ( b ) P FROM eqns ( a ) & ( b ): 2U PL P EA P2L U ____________ ( 3a ) in terms of load 2 EA Mechanics of Solids Chapter 2 20/25 FROM eqn ( a ): P EA ________________ ( c ) L FROM ( c ) & ( 3 ): EA L L U 2 EA 2 U EA 2 _______________ ( 3b ) in terms of 2L k EA ______________ ( d ) L deformation LET: k = spring stiffness FROM ( d ) & ( 3a ): FROM ( d ) & ( 3b ): U 2 P 2k U k 2 2 For a bar with changes in cross-section: Pa Pb Pi = axial force in part (section) i of the bar. n = number of parts (sections) based on eqn ( 3a ): n U i 1 Pi 2 Li 2 Ei Ai Mechanics of Solids Chapter 2 21/25 Nonprismatic bar with varying axial force: based on eqn ( 3a ): P(x) 2 Px U dx 2 E Ax 0 L x dx L Strain energy density = strain energy per unit volume u volume: V = AL RECALL: P E A U U V AL (strain energy density) L L FROM eqn ( 3a ): u P2 1 P2L P2 2 2 EA( AL) 2EA2 A 2 2 E 2E u u 2 2E FROM eqn ( 3b ): u EA 2 E 2 E (L) 2 E 2 2L( AL) 2L2 2 2 L2 E 2 2 Mechanics of Solids Chapter 2 22/25 10 mm EXAMPLE No. 1 600 N GIVEN: An axial force of 600 N acts on the steel bar. 150 mm 600 N 3200 mm mm FIND: The stresses on all faces of a stress element for = - 30°. SOLn: Mechanics of Solids Chapter 2 EXAMPLE No. 2 23/25 OMIT Pass out SOLn GIVEN: A conical bar of diameter d at the support and length L hangs vertically under its own weight. LET: γ = specific weight d E = modulus of elasticity DERIVE: Formula for the strain energy, U, of the bar. SOLn: L Mechanics of Solids Chapter 2 24/25 2.10 STRESS CONCENTRATIONS Changes in geometry cause stress concentrations: concentrations. P P changes in AREA cause stress A P Which can sustain greater load? VS For a prismatic bar, we know that: P A But what about the following? c P /2 P d c /2 STRESS DISTRIBUTION: σmax P Stress Concentrations = highly localized stresses caused by changes in geometry. The more abrupt the change, the higher the stress concentration. Stress Raisers = the geometric features that cause stress concentrations Stress Concentration Factor = K K max nom nom P P A c 2 t 2 nom for flat bar, above K, stress concentration factor is equivalent to an additional safety factor. P ct (average stress) t = thickness ct = net area RECALL: Factor of Safety n y allow Text pg 141 & 142 gives the stress concentration factor, K, for common geometries. Mark’s Engineering Hndbk and other engineering references books have stress concentration factors Mechanics of Solids Chapter 2 25/25 EXAMPLE No. 1 GIVEN: The steel strap shown below is subject to an axial load of 80kN. σy = 700 x 106 Pa E = 200 x 109 Pa 20 mm 80 kN 80 kN 40 mm 6 mm 300 mm 800 mm 300 mm FIND: a.) Maximum normal stress developed in the strap b.) Displacement of one end of the strap w.r.t. the other end. SOLn: 10 mm