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Home , Bending, Bending moment, Stress (mechanics)

Failure Theories Why do mechanical components fail? Mechanical components fail because the applied stresses exceeds the material’s strength (Too simple). What kind of stresses cause failure? Under any load combination, there is always a combination of normal and shearing stresses in the material. What is the definition of Failure? Obviously but in some components yielding can also be considered as failure, if yielding distorts the material in such a way that it no longer functions properly

Which causes the material to fail? Usually ductile materials are limited by their strengths. While brittle materials ( < 5%) are limited by their tensile strengths.

Stress at which point?

Stress at which point? Failure Theories

Load type Material Property Application of Stress Uniaxial Ductile Static Biaxial Brittle Dynamic

Static Loading Maximum Normal Stress Dynamic Loading Modified Mohr Goodman strength Gerber Maximum Soderberg Distortion Static Failure Theories The idea behind the various classical failure theories is that whatever is responsible for failure in the standard tensile test will also be responsible for failure under all other conditions of static loading. Ductile Material Brittle Material Characteristic Yield Stress Ultimate Stress Failure Stress

Important 1. Maximum Shear Stress 1. Maximum Normal Theories 2. Maximum Octahedral Stress Shear Stress 2. Modified Mohr. Maximum Ductile Materials Shear Stress Failure occurs when the maximum shear stress in Theory the part exceeds the shear stress in a tensile test specimen (of the same material) at yield. Hence in a tensile test, S τ = y max 2 For a general state of stresses −σσ S τ = 31 = y max 22 This leads to an hexagonal failure envelop. A stress system in the interior of the envelop is considered SAFE

The Maximum Shear Stress Theory for Ductile Materials is also known as the Tresca Theory.

for design purposes, the failure relation can be S y modified to include a factor of safety (n): n = −σσ31 Several cases can be analyzed in problems:

Case 1: σ σ 21 ≥≥ 0

In this case σ3=0 −σσ σ S τ = 31 1 == y max 222

σ1 ≥ S y

Case 2: σ1 0 ≥≥ σ 3 Yielding condition

−σσ S τ = 31 = y max 22

σσ31 ≥− S y Distortion Energy Theory Based on the consideration of angular distortion of stressed elements. The theory states that failure occurs when the distortion energy in the material exceeds the distortion strain energy in a tensile test specimen (of the same material) at yield. Resilience Resilience is the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered.

ε y Modulus of resilience Ur r = dU εσ ∫0 If it is in a linear elastic region, 2 1 1 ⎛ ⎞ σσ ⎜ y ⎟ y U r yy == σεσy ⎜ ⎟ = 2 2 ⎝ ⎠ 2EE 1 For general 3-D stresses: u ()++= εσεσεσ 2 332211 1 Applying Hooke’s Law u ()2 2 2 2 ()++−++= σσσσσσνσσσ 2E 1 2 3 133221 There are two components in ++ σσσ321 σ +σ +σ zyx this energy a mean component σ M = = and deviatoric component. 3 3

σ D = σ1,1 −σ M σ D = σ 2,2 −σ M σ D = σ 3,3 −σ M

The energy due to the mean stress (it gives a volumetric change but not a distortion: 1 u ()222 −++= 2 ()++ σσσσσσνσσσ Mean 2E MMM MMMMMM 1 − 21 ν u = []2 ()213 νσ=− ()2 2 2 +++++ 222 σσσσσσσσσ Mean 2E M 6E 1 2 3 133221 1+ν uuu =−= ( 2 2 2 −−−++ σσσσσσσσσ) D Mean 3E 1 2 3 133221 Compare the distortion energy of a tensile test with the distortion energy of the material. 1+ν 1+ν u = 2 uS == ( 2 2 2 −−−++ σσσσσσσσσ) Tensile 3E Dy 3E 1 2 3 133221

2 2 2 S y 1 2 3 −−−++= σσσσσσσσσ133221

2 2 S y 1 3 −+= σσσσ13 Plane Stress

Von Mises : Defined as the uniaxial tensile stress that creates the same distortion energy as any actual combination of applied stresses. This simplifies the approach since we can use the following failure criterion

σ ≥ S yVM S n = y σVM

()2 ( )2 ()2 6( +++−+−+− τττσσσσσσ222 ) σ = yx zy xz zxyzxy VM 2 22 2 2D VM +−+= 3τσσσσσxyyxyx Case of Pure Shear

VM 3τσ≥= S yxy S τ y == 577.0 S Max 3 y

Brittle Materials Several theories have been developed to describe the failure of brittle materials, such as: Maximum Normal Stress Theory Coulomb-Mohr Theory Modified-Mohr Theory Maximum Normal Stress Theory Failure occurs when one of the three principal stresses reaches a permissible strength (TS). σ > σ 21 Failure is predicted to occur when

σ1=St and σ2<-Sc σ2 Where St and Sc are the tensile and St For a biaxial state of stresses

-Sc St σ1

-Sc Coulomb-Mohr Theory or Internal Theory (IFT) This theory is a modification of the maximum normal stress theory in the which the failure envelope is constructed by connecting the opposite corners of quadrants I and III.

The result is an hexagonal failure envelop. Similar to the maximum shear stress theory but also accounts for the uneven material properties of brittle material Mohr’s Theory The theory predicts that a material will fail if a stress state is on the envelope that is tangent to the three Mohr’s circles corresponding to: a. uni-axial ultimate stress in , b. uni-axial ultimate stress in , and c. pure shear. Modified Mohr’s Theory

σ σ 12 ≤− 1 σ σ TC

This theory is a modification of the Coulomb-Mohr theory and is the preferred theory for brittle materials.

Maximum Normal-Strain Theory Also known as the Saint-Venant’s Theory. Applies only in the elastic range.

Failure is predicted to occur if σ −νσ 21 ±= S y or σ −νσ12 = ±S y Where S is the yield strength. y σ For a biaxial state of stress 2 Sy

-Sy Sy σ1

-Sy Maximum Strain-Energy Theory Yielding is predicted to occur when the total strain energy in a given is greater than or exceeds the strain energy in the same volume corresponding to the yield strength in tension or compression. 2 The strain energy stored per unit volume (u ) S y s u = during uniaxial loading is s 2⋅ E ⋅ ⋅σεσε In a biaxial state of stress u = + 2211 σ 22 1 u ()2 2 2 ⋅⋅⋅−+= σσνσσ σ 2E 1 2 21

This theory is no longer used

Example:

Given the material SY , σx , σv and τxy find the safety factors for all the applicable criteria. a. Pure aluminum

Y = 30MPaS σ x =10 MPa σ y = −10 MPa τ xy = 0 MPa

σ1 =10MPa σ 3 = −10MPa τ Max =10MPa Is Al ductile or brittle? Ductile Use either the Maximum Shear Stress Theory (MSST) or the Distortion Theory (DT) -10 10 MSST Theory S 30 30 n = y = = =1.5 σ1 −σ 3 10 − (−10) 20

2 2 2 DT Theory σVM = σ x +σ y −σ xσ y + 3τ xy = 300 =17.32MPa S 30MPa n = y = =1.73 σVM 17.32MPa b. 0.2%C Carbon Steel

Y = 65KsiS σ x = −5 Ksi σ y = −35 Ksi τ xy =10 Ksi In the plane XY the principal stresses are - 1.973Ksi and -38.03Ksi with a maximum shear stress in the XY plane of 18.03Ksi In any orientation

σ1 = 0Ksi σ 2 = −1.973Ksi σ 3 = −38.03Ksi τ =19.01Ksi Max Ductile S 65 MSST Theory n = y = =1.71 σ1 −σ 3 0 − (−38.03)

σ = σ 2 +σ 2 −σ σ = 38.03Ksi DT Theory VM 1 3 1 3 S 65Ksi n = y = =1.71 σVM 38.03MPa C. Gray Cast Iron

ut = 30 uc =120 KsiSKsiS σ x = −35 Ksi σ y =10 Ksi τ xy = 0 Ksi

σ1 =10Ksi σ 2 = 0Ksi σ 3 = −35Ksi

τ Max = 22.5Ksi

-35 10 Brittle Use Maximum Normal Stress Theory (MNST), Internal Friction Theory (IFT), Modified Mohr Theory (MMT) S 30 MNST Theory (tensile) n = ut = = 3.0 σ1 10 S 120 MNST Theory (compression) n = uc = = 3.4 σ 3 35 σ σ IFT 12 ≤− 1 th σ σ TC 1 0 σσ3 ≤≥ 0 _4 quadrant

Suc line _ equation σ 3 Suc +−= σ1 Sut 1 σ σ 10 − 35 1 3 −=−= = 625.0 SSn ucut 30 120 n = 6.1 MMT th 1 0 σσ3 ≤≥ 0 _4 quadrant

Sut 1 SS utuc σ1 − σ 3 = − utuc nSS − SS utuc 1 = 54.0 30 )30)(120(1 10 − )35( =− n − 30120 n − )30120( n = 84.1 Example 1 The tube shown is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276MPa. We wish to select a stock size tube (according to the table below). Using a design factor of n=4. The load is F=1.75kN, the axial tension is P=9.0kN and the is T=72N.m. What is the realized factor of safety?

Consider the critical area ( top surface). P Mc σ += x A I Maximum bending = 120F ⎛ d ⎞ mm× 75.1120 kNx⎜ ⎟ 9kN 2 σ += ⎝ ⎠ x A I ⎛ d ⎞ 72×⎜ ⎟ Tr 2 36d τ == ⎝ ⎠ = zx J J J 1 22 2 VM ()+= 3τσσzxx S 276.0 σ y =≤ GPa = 0690.0 GPa VM n 4

For the dimensions of that tube S 0.276 n = y = = 4.57 σVM 0.06043 Example 2: A certain F is applied at D near the end of the 15-in lever, which is similar to a socket wrench. The bar OABC is made of AISI 1035 steel, forged and heat treated so that it has a minimum (ASTM) yield strength of 81kpsi. Find the force (F) required to initiate yielding. Assume that the lever DC will not yield and that there is no at A. Solution: 1) Find the critical section The critical sections will be either point A or Point O. As the moment of inertia varies with r4 then point A in the 1in diameter is the weakest section. 2) Determine the stresses at the ⎛ d ⎞ critical section M ⎜ ⎟ My 2 ×× 1432 inF σ == ⎝ ⎠ = = 6.142 F x I πd 4 πd 3 64 3) Chose the failure ⎛ d ⎞ criteria. T⎜ ⎟ Tr 2 ×× 1516 inF τ == ⎝ ⎠ = = 4.76 F The AISI 1035 is a zx J πd 4 π in)1( 3 ductile material. Hence, we need to employ the 32 distortion-energy theory. 22 2 2 2 VM 3 xyyxyx x τστσσσσσzx =+=+−+= 5.1943 F S 81000 F y === 416lbf σVM 5.194 Apply the MSS theory. For a point undergoing plane stress with only one non-zero normal stress and one shear stress, the two non- zero principal stresses (σA and σB) will have opposite signs (Case 2).

2 −σσBA S y ⎛σ x ⎞ 2 τ max = ±== ⎜ ⎟ +τ zx ⎝ 222 ⎠

2 ⎛σ x ⎞ 2 2 2 σσS yBA =≥− 2 ⎜ ⎟ zx x +=+ 4τστzx ⎝ 2 ⎠ 1 81000 = ()()()F 2 ×+ 4.7646.142 F 2 2 = 388lbfF Example 3: A round cantilever bar is subjected to torsion plus a transverse load at the free end. The bar is made of a ductile material having a yield strength of 50000psi. The transverse force (P) is 500lb and the is 1000lb-in applied to the free end. The bar is 5in long (L) and a safety factor of 2 is assumed. Transverse shear can be neglected. Determine the minimum diameter to avoid yielding using both MSS and DET criteria. Solution 1) Determine the critical section

The critical section occurs at the wall. ⎛ d ⎞ ⎛ d ⎞ PL⎜ ⎟ T⎜ ⎟ Mc ⎝ 2 ⎠ 32PL Tc ⎝ 2 ⎠ 16T σ x == 4 = 3 τ == = I πd πd xy J πd 4 πd 3 64 32

2 2 ⎛ + yx ⎞ ⎛ −σσσσyx ⎞ 2 ⎛ ⎞ ⎛σσ⎞ 2 ⎜ ⎟ ⎜ ⎟ x x σ 2,1 = ⎜ ⎟ ± ⎜ ⎟ ()τ xy =+ ⎜ ⎟ ± ⎜ ⎟ + ()τ xy ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ ⎠ ⎝ 22 ⎠

2 2 16PL ⎛16PL ⎞ ⎛ T ⎞ 1616 ⎡ 2 2 ⎤ σ 2,1 ±= ⎜ ⎟ + ⎜ ⎟ ()+±= TPLPL d 3 ⎝ d 3 ⎠ ⎝ 3 ⎠ ππππdd 3 ⎣⎢ ⎦⎥ 16 σ = ⎡ ()2 +×±× 100055005500 2 ⎤ 2,1 πd 3 ⎣⎢ ⎦⎥ 26450 8.980 The stresses are in the wrong σ = σ −= 1 d 3 2 d 3 order.. Rearranged to 26450 8.980 σ = σ −= 1 d 3 3 d 3 σ − σ − − )8.980(26450 4.13715 τ = 31 = = MSS MAX 2 2d 3 d 3 S 50000 2τσσ y ==≤=− 000,25 31 MAX n 2 d ≥ 031.1 in

2 2 2 2 ⎛ 26450 ⎞ ⎛ 8.980 ⎞ ⎛ 26450 ⎞⎛ 8.980 ⎞ σσσσσ=−+= ⎜ ⎟ ⎜−+ ⎟ − ⎜ ⎟⎜− ⎟ VM 1 3 31 d 3 d 3 3 dd 3 DET ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ 26950 S 50000 σ y =≤= VM d 3 n 2 ≥ 025.1 ind Example 4:

In the wheel suspension of a car, the motion is provided by a torsion bar fastened to the arm on which the wheel is mounted. The torque in the torsion bar is created by the 2500N force acting on the wheel from the ground through a 300mm long lever arm. Because of space limitations, the holding the torsion bar is situated 100mm from the wheel shaft. The diameter of the torsion bar is 28mm. Find the stresses in the torsion bar at the bearing by using the DET theory. Solution The stresses acting on a torsion bar are: d × lengtharmF )(_( ) Tc ()()× 014.03.02500 1. Torsion τ == 2 = 32 Pa =174MPa J πd 4 π ()028.0 4 32 ⎛ d ⎞ ()×bearingF _ length ⎜ ⎟ 2. bending Mc 2 ()()× 014.01.02500 σ == ⎝ ⎠ = 64 Pa =116MPa I πd 4 π ()028.0 4 64 The principal stresses are:

2 2 ⎛ + yx ⎞ ⎛ −σσσσyx ⎞ 2 ⎛ ⎞ ⎛σσ⎞ 2 ⎜ ⎟ ⎜ ⎟ x x σ 2,1 = ⎜ ⎟ ± ⎜ ⎟ ()τ xy =+ ⎜ ⎟ ± ⎜ ⎟ + ()τ xy ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ ⎠ ⎝ 22 ⎠

2 0.116 ⎛ 0.116 ⎞ 2 σ 2,1 ±= ⎜ ⎟ + ()0.174 2 ⎝ 2 ⎠

σ1 = 4.241 MPa σ 2 −= 4.125 MPa DET

2 2 2 2 VM 1 3 σσσσσ31 =−+= ()()()()−−−+ 4.1254.2414.1254.241 S σ = 6.322 MPa ≤ y VM n

MSS σ −σ − (− 4.1254.241 ) τ = 31 = = 4.183 MPa Max 2 2 S 2 τ =× 8.366 MPa ≤ Y Max n Example 5: The factor of safety for a machine element depends on the particular point selected for the analysis. Based upon the DET theory, determine the safety factor for points A and B.

This bar is made of AISI 1006 cold-drawn steel (Sy=280MPa) and it is loaded by the F=0.55kN, P=8.0kN and T=30N.m ⎛ d ⎞ Solution: Fl⎜ ⎟ Mc P 2 P Fl 432 P Point A σ =+= ⎝ ⎠ +=+ x I Area d 4 ππd 2 d 3 ππd 2 464

()( 3 )( 1.01055.032 ) ( )(1084 3 ) σ = + = 49.95 MPa x π ()02.0 3 π ()02.0 2 Tr T (301616 ) τ === = 10.19 MPa xy J πd 3 π ()020.0 3 1 22 2 2 2 VM ()τσσxyx [ +=+= ()] = 1.1011.19349.953 MPa S 280 n y === 77.2 σVM 1.101

Point B P ( )(10844 3 ) σ == = 47.25 MPa x πd 2 π ()02.0 2 T 416 V ()3016 ()()1055.04 3 τ xy 3 =+= 3 + = 43.21 MPa πd 3A π ()02.0 ⎛ π ⎞ 2 3⎜ ⎟()02.0 ⎝ 4 ⎠ 1 2 2 2 σVM []+= ()= 02.4543.21347.25 MPa 280 n == 22.6 02.45