Stresses. Normal stress. (Note this is average value) P A. Example: Normal stress in member A B F A AB AB. Normal Stress at a point F AB.

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1 tresses Normal stress (Note this is average value) xample: Normal stress in member Normal tress at a point lim Δ 0 Δ Δ Δ Δ d d

2 tress concentrations at points of load application more uniform stress distribution M R stress not uniform due to M R Y tress oncentration actor, way of estimating the effects of holes and other discontinuities on the local stresses around the discontinuity xample: stresses at edge of a hole, where may range from to 3, depending on size of hole ee eterson s tress oncentration actors, nd d., by W. D. ilkey, John Wiley & ons, 997

3 I units : ascal ilo ascal Mega ascal U.. units : Units of tress N a m 3 N ka 0 m 6 N Ma 0 lb in m ounds per square inch psi onversion : psi ka ksi Mpa hear tress glue joint of area j j hear force on glue joint verage shear stress tress at a point τ τ jo int j j lim ΔΑ 0 Δ Δ 3

4 xample: pin at in truss member imaginary cut imaginary cut pin pin shear force pin I pin pin pin t bearing force d bearing stress b td 4

5 tress on Oblique lane under xial oading Two directions required to specify stress component, one for orientation of face and another for force acting on face erpendicular plane: Y (uniform) arallel plane: bonded joint τ Y Y Y Y Y Y 5

6 Oblique plane: x Y Y θ θ Y Note: cosθ orce Normal to Oblique lane orce arallel to Oblique lane θ cos θ sin Y Normal tress or cos θ ( cos θ ) cos θ x cos θ 6

7 hear tress τ Y Y sinθ cosθ τ Y sinθ cosθ Variation of normal stress and shear stress with angle θ θ 0 ο cos θ τ θ 45 ο θ τ Y sin θ cos θ τ min min θ 90 0 ο,90 ο 0 π / 4 π / θ 7

8 Importance of Normal tresses and hear tresses in Material ailure. rittle materials fail by maximum tensile normal stress. Ductile materials fail by maximum shear stress at 45 0 rittle: Ductile: 45 ctual-cup-cone fracture surface for ductile material 8

9 9 V V lane : ree ody Diagram olution: 0 + V / V ( )( ) a V I /

10 lane : ree ody Diagram olution: + 0 ( b )( ) b lane : ree ody Diagram Y Y 45 olution: 0 On lane : c () c cos θ cos 45 c τ Y sin θ cos θ sin 45 cos 45 c c c 0

11 lane D: ree ody Diagram D olution: D D 0 0 llowable tresses: hear stress along grain τ allow 85 psi Normal stress perpendicular to grain psi allow

12 ind the maximum value of so that allow or are allow not exceeded τ Normal tress: cos θ 365 cos lb ( 4 ) cos θ τ Y hear tress: τ Y sin θ cos θ 85 sin 30 cos 30 4 ( 4 ) sin θ cos 6836 lb and the shear stress governs θ

13 ross section glue joint 4 ind the allowable load if the maximum allowable shear stress along the glue joint is 00 psi and the maximum allowable tensile normal stress across the glue joint is 300 psi Y Y

14 Y Y 3 4 τ Y θ 60 Normal stress: 3 4 cos 60 cos ( )( 4) psi, 800 lb hear stress: 3 τ sin θ cos θ 4 sin 60 cos 60 Y ( )( 4) 00 psi, 463 lb therefore failure by shear governs and,463 lb is the maximum allowable load. 4

15 tress at a point: consider an imaginary cut perpendicular to the - axis as shown below. Y Δ Y Δ Total force Z Δ Z Δ Δ rea of cut face which is perpendicular to -axis (- face) Normal stress on cut : lim Δ 0 Δ Δ hear stresses on cut : τ Y lim Δ 0 Δ Δ Y τ Z lim Δ 0 Δ Δ Z 5

16 imilarly, we can find three components of stress acting on the Y and Z faces YY lim ΔY 0 Δ Δ Y Y ZZ lim Δ Z 0 Δ Δ Z Z τ Y lim ΔY 0 Δ Δ Y τ Z lim Δ Z 0 Δ Δ Z τ YZ lim ΔY 0 Δ Δ Z Y τ ZY lim Δ Z 0 Δ Δ Y Z YY τ YZ τ ZY τ Z τ Y τ Z τ Y ZZ Due to moment equilibrium: τ Y τ Y τ τ Z τ τ YZ Z ZY (will show later) 6

17 train Normal strain under xial oading + Normal strain ε (for constant uniform stress) When normal strain varies from point to point, need to define normal strain at a point: Δ ε lim Δ 0 ΔΧ (for varying stress) Χ + ΔΧ + Δ trains in structural materials are generally very small ε on order of 6 in 0 or mm in mm 6 in mm microstrain 0 or in mm 7

18 more useful to work with and rather than &. & ε apply in any situation, whereas & are specific to a given structural element. ε curves rittle material inear curve to racture (e.g., glass, ceramics) ε U breaking stress ultimate stress U ε ε U ε racture surface to load Ductile Materials up-cone failure surface with 45 hear lips 8

19 xample of ductile material- ow arbon teel U YU Y ε ε ε xample of ductile material- luminum lloy U Y ε 0.% offset ε ε 9

20 ercent elongation 00 ε or linear lastic Region Hooke s law applies ε (-D only!) Young s modulus 3 6 Units: ksi 0 psi, 0 psi, 6 N 9 N Mpa 0, Gpa 0 m m see ppendix, Tables -7, -8 (table of properties) onversion: ksi Ma, 0 6 psi Ga O O 0

21 ample material property data sheet from MatWeb Data heet lclad luminum 606-T6, T65 eywords: al606, lclad 606-T6, T65; lclad 606-T6, T65 ubat: luminum lloy, Nonferrous Metal, 6000 eries luminum lloy, Metal Material Notes: Data points with the note have been provided by the luminum ssociation, Inc. and are NOT OR DIGN. omponent Value Min Max luminum, l hromium, r opper, u Iron, e 0.7 Magnesium, Mg 0.8. Manganese, Mn ilicon, i Titanium, Ti 0.5 Zinc, Zn 0.5 roperties Value Min Max hysical Density, g/cc Mechanical Ultimate Tensile trength, Ma Tensile Yield trength, Ma longation at reak, % Modulus of lasticity, Ga hear trength, Ma rocessing olution Temperature, ging Temperature, ging Temperature, Inelastic ehavior - lip and dislocation movement, plastic deformation lip occurs by passage of dislocations along slip planes and not by simultaneous shearing of atomic planes - rug wrinkle analogy. rug floor arge force is required to pull rug across floor Rug can be moved with less force by creating wrinkle and pushing it along from one end to the other

22 Tensile stress-strain curves train hardening (work hardening) large scale dislocation movement ε True tress-train urve ngineering tress-train curve lastic region inear elastic region ε ngineering vs. True tress and train ngineering True o i ε o based on original dimensions, o o ε f o d ln f o based on instantaneous dimensions

23 hear train Unlike normal strain, which characterizes elongation or contraction under normal stress, shear strain characterizes distortion under shear stress. γ xy τ xy deformed shape due to shear stress γ xy τ xy hear strain (radians) or shear angle (dimensionless) π γ xy y x τ xy γ xy Rotating deformed shape W by s γ xy for convenience, γ xy deformed shape Where for small deformations, the average shear strain is γ tan γ xy s xy s shear deformation When shear strain varies from point to point, need to define shear strain at a point: Δ s d s γ xy lim 0 Δ d 3

24 Deformations under axial load tress: Hooke s aw: ; + ε ubstituting above equations ε train: ε (one dimensional) (for linear elastic region) longation (or contraction) of uniaxially loaded element xample: this equation applies to any -force member in a loaded structure 4

25 Members and are made of steel ( psi) and each member has a cross sectional area of.5in.ind the stresses and the axial deflections of each member. 000 lb 4 0, 3 tatics: 4 + Y 0 5 5, , lb lb Deflections: (,500 )( 5)( ) (.5)( 30 ) in (contraction) 5

26 ( 7,500 )( 3)( ) (.5)( 30 ) in (extension) tresses:,500 5, ,500 3, psi psi () (T) concrete column ( 0 Ga) having a variable cross section is subjected to several loads as shown. ind the tresses and deflections of elements, and D and the total vertical deflection at 6N N 0N 6N N D D rea ( m ) ength (m )

27 : 0N 0 0, 0 N ( T ) Y 0.67 N / m ( T ) 6 3 ( )( 0 ) 9 ( 6)( 0 0 ) (elongation) 8 m : 6N 0N 6N 0 (6) + 0, N ( ) Y 0.5 N / m ( )( 0 ) 9 ( 0 0 ) ( ) (contraction) m D: 6N 0N 6N N N D 0 (6 ) ( ) + D 0, D 6 N ( ) 6 D 0.6 N / m ( ) 0 Y D d 3 DD 6( )( 0 ) 8 D DD 0( 0 0 ) D m( ) m ( ) 0 (contraction) 8 7

28 xample of internal force diagram showing distribution of internal forces along length of structure 0 kn 0 kn 6 kn 6 kn kn kn - kn D -6 kn D (T) () Internal orce (kn) Thermal train and deflection: Isotropic materials expand or contract uniformly in all directions under temperature change ΔΤ. or bar, the thermal strain is ε T α ( ΔΤ ) Where α coefficient of thermal expansion (T) and deflection is T ε T α ( Δ T ) o the total strain in bar along axis is ε + α ( Δ T ) + α ( Δ T ) applied load or Reaction force developed due to Δ T and total deflection is + α ( ΔΤ ) Mechanical part Thermal part 8

29 xample: free expansion with no applied loads and temperature change ΔΤ. train: ε 0 + α ( ΔΤ ), 0 tress: 0 xample: Restrained Rigid wall ΔΤ Rigid wall total strain ε + α ( Δ T ) 0 total deflection + α ( Δ T ) thermal stress is α ( ΔΤ ) 0 9

30 if ΔΤ > 0, if ΔΤ < 0, tress is compressive tress is tensile Thermal stresses only develop when thermal deformation is restrained. If no restraint, we get free expansion and thermal strain ε α ( ΔΤ ) tatically Indeterminate roblems (or, know when to give up with statics!) When the number of unknown reactions is greater than the number of independent, nontrivial static equilibrium equations, the structure is statically indeterminate and the unknown reactions cannot be determined from statics alone. dditional equations must be generated by considering deformations. xample: oncrete post under compressive load D s orce in post Rigid lock oncrete post oncrete post 30

31 or rigid block: + Y + 0 nd since can be determined from statics, the structure is statically determinate. However, if we attach a steel reinforcement bar to the concrete post, we make the structure statically indeterminate, ince there are two unknown reactions tructure: D: Rigid block concrete steel + Y + + Z 0 0 trivial 0 unknown reactions independent, non trivial tatics equation 3

32 3 One additional equation is needed to supplement the statics equation, and that equation can be generated by considering the deformations. Geometric compatibility: where deformation of steel deformation of concrete orce-deformation: ombining these equations, we get and s c ince we know the cross sectional areas and and The moduli of elasticity and, we can solve this equation simultaneously with the statics equation. nd determine the unknown reactions and y ince

33 xample: if 4,000 lb determine stresses in bars 6 teel bar rigid block 30 0 in 6 psi ronze bar in 6 psi unknown reactions & independent non-trivial statics equation 0 y Geometry: at the block orce-deflection: ( 6 )( ) 6 ()( 30 0 ) 6 ; ( )( ) 6 ( 4 )( 5 0 ) quilibrium: / / 33

34 + + Y + 4, , ,000 lb ( T ), 6000 psi ( T ) 4, , 000 lb ( ) 36, psi ( ) uperposition: nother method for solving statically indeterminate problems. or example, if we remove the support at and analyze the structure as shown below, the geometric compatibility equation is / / ( ) ( ) 0 / / ( ) ( ) + 34

35 35 irst consider alone: ( ) / / ( ) + 0 (note that only the bar deforms here) Now consider alone: ( ) ( ) + (note that the bars and both deform now under the same load ) Now, ( ) ( ) 0 since point is fixed 0 +

36 4,000 4 ( ) 6 + ( 5 ) ( 30 ) 4 ( 5 ) 0 6, 000 lb Then from statics, 36, 000 lb, etc. as before The rod of length, modulus of elasticity and diameter D is fixed on one end and free on the other end before the load is applied. efore the load is applied, there is a gap of width between the free end of the rod and the wall. fter the axial load is applied, the free end of the rod contacts the wall as shown. nswer each of the questions below in terms of the given parameters. Δ d diameter efore loading after loading 36

37 37 Draw the free-body diagram of the rod after the load is applied Y Write the static equilibrium equation for the above free body diagram 0 + Write the geometric compatibility equation Δ Write the force-deformation equation ( ) Δ D 4 π D Δ 4 π Determine the stress in the rod after the load is applied. D Δ 4 π Reaction orce: Δ

38 xample: ind the stress in b ( ) R 38

39 39 ( ) + a R b R R ( ) ( ) ( ) 0 + R total + b a R b + b a b R R R R 0 + b a b

40 thin-walled hemispherical pressure vessel is attached to a flat place with a flange and eight (8) bolts, as shown below. The allowable stress in each bolt is max, the bolt diameter is d, the stresses length of each bolt is, the mean diameter of the pressure vessel is D, and the modulus of elasticity of each bolt is. The bolts are initially unstressed and the vessel is then pressurized to pressure. nswer each of the questions below in terms of the given parameters. olts (8) ressure vessel D d Draw the free-body diagram of the pressure vessel gas 4 4 Write the static equilibrium equation for the above free body diagram D + y 0 π

41 Write the geometric compatibility equation: ssuming that bolts are loaded in parallel: (all bolts have same deflection) Write the force-deformation equation max max π d 4 max Determine the allowable pressure, max, in the vessel π D max 8 4 d max 8 D π d 8 4 max max Determine the allowable elongation, max, of each bolt max max 4

42 ind forces in members and deflection of pin : 3 etc 3 ( symmetry ) quilibrium: 3 sin θ 3 sin θ 0 3 Y cos θ cos θ 0 () + cos θ tatically indeterminate 4

43 orce -Deflection: 3 ( ) Geometric ompatibility: deformed undeformed θ θ D D θ D D for small deflections cos θ ( 3 ) cos θ ( ) + ( 3 ) from this + cos substituting () θ cos + deflection at D cos θ θ then + cos θ cos θ 43

44 ind the equivalent stiffness for the two spring arrangements shown below eries arrangement arallel arrangement or the parallel arrangement, the free body diagrams are as shown below quilibrium equation: + + Y + Geometric ompatibility: assuming the spring supports remain parallel: orce-deformation: 0 44

45 ombining the above equations, we get ( + ) o the equivalent stiffness of two parallel springs is eq + or a system of n springs in parallel, n eq i i or the series arrangement, the free body diagrams are as shown below It is clear that static equilibrium requires that Geometric ompatibility: the total deformation is + orce-deformation:, 45

46 ombining the above equations, we get + + o the equivalent stiffness of two series springs is then eq + or a system of n springs in series, eq n i Note that for a uniform bar loaded with an axial load, the stiffness is found as shown below, quivalent spring 46

47 eries and parallel arrangement of axially loaded members + quivalent spring pring stiffness, eries arrangement total total total total total total orces: total Deformations: total + quivalent stiffness: or eq eq total total + total

48 arallel arrangement total total orces: total + Deformations: total quivalent stiffness: eq eq total total + + total + 48

49 Designing gainst ailure The design of structural and machine elements against failure involves the selection of materials, shapes and sizes for the elements so that they will withstand the imposed loading conditions without failure. There are many possible modes of failure, so the designer must examine the design carefully to make sure that all of the likely failure modes are covered in the analysis. ome of the most common causes of structural failure:. xcessive stress - leads to fracture or permanent deformation (yielding, slip). xcessive deformation may or may not lead to fracture or permanent deformation. 3. atigue under repetitive or cyclic loading, initial defects or flaws grow to critical size, causing failure. 4. Instability, or buckling. 5. reep - time-dependent deformation. 6. Others to be covered in advanced courses. 49

50 xample: uto ront uspension pring wheel D D D x pring s y s y x x. xample of excessive stress: D D D D D reaches level required for fracture or deformation. xample of excessive deformation: s 50

51 3. atigue: Due to wheel movement and bumps in road, all loads are repetitive with time allowable stresses are reduced 4. uckling: D D reaches buckling load member becomes unstable Design roblem round steel rod ( 00Gpa) 6 meters long must carry a tensile load of 7000 N. The two design constraints are, () the allowable stress in the steel is 5 Ma, and () The allowable elongation of the rod is.5 mm. What is the minimum required diameter of the rod? d? 7000N 6 m 5

52 tress: π 4 d ( 7000 ) 4 4 d π π ( 5 0 ) m 8.4 mm longation: π d 4 d 4 π π 4 ( 7000 )( 6 ) 9 ( )( 00 0 ) d m 0. 3mm Only the diameter d 0.3mm will satisfy both constraints, so the elongation governs the design 5

53 actor of afety (uncertainty).. avg. ultimate stress avg. allowable stress UT allow where ultimate stress stress at failure (material property) allowable stress working stress under normal use (actual stress) obviously, we want.. >, but if.. is too large, the design may not be practical. election of.. somewhat arbitrary, but depends on such factors as variability in material properties, number of loadings, expected failure mode, uncertainty of analysis, degradation of properties during operation, importance of the member to integrity of structure. xample: primary & secondary structure in aircraft No substitute for experience in choosing.! Typical...0 If uncertain about materials, loads, etc., choose higher.. If good material data base, loading well understood, use lower. Dynamic load (suddenly applied) stress is twice the stress under static load of same value. 53

54 or design against yielding, use elastic strength in Tables -7 or -8. yield allow or design against Ultimate ailure, use Ultimate trength in Tables -7 or -8. Ultimate allow tress 0.% offset yield roportional limit Ultimate strength Two different definitions of yield strength llowable stress, 0.00 train ε 54

55 robabilistic Design - takes into account the statistical scatter in the loads, geometrical dimensions, and material strengths in order to estimate the probability of failure allowable (working)stress allow ultimate strength of material probability UT probability of failure.. UT allow 55