a) Tension stresses tension forces b) Compression stresses compression forces c) Shear stresses shear forces

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1 1.5 Basic loadings: Bending and Torsion External forces and internal stresses: a) Tension stresses tension forces ) Compression stresses compression forces c) Shear stresses shear forces Other asic loading of machine parts: Bending of a eam: d) M : ending moment, e) T t : torsion torque, f) normal and shear stresses comined LECTUE-DT-1_03 Seite 1 von 0

2 The eam deflects y the force applied to the free (cantilever eam): Distance l to fixed end is important for the deflection y at the free end. If the force F is increased the deformation increases on the magnitude of the force and the length l (lever arm) of the force Fl = M (ending moment) the ending moment causes internal forces which causes strains: extensions on upper side ε + σ + tension stresses compressions on lower side ε σ compression stresses Strains (deformations) due to ending: a) undeformed element in the eam, ) deformed element in the eam LECTUE-DT-1_03 Seite von 0

3 Deformed element : the normal strain along a line element s s ε = s with x = s = r i ϑ and s = ( r + y) i ϑ ( r + y) i ϑ r i ϑ ε = r i ϑ y ε = r s is: The strain varies linear with y from the neutral axis. Let c the imum distance e to the outer fier from neutral axis: ε ε y r y = ε = i ε C C r This is a linear relationship for the strain distriution: ε σ y σ = iσ C ; according to Hooke s law View on the cross section of the eam: ll internal moments must e equal to external ending moment M( ex) = M( in) Strain and stress distriution in a ar (eam) ( ) ε y y y = ε( y ) = i ε linear relation of strain ε C C if: ( ) ε ( y ) ( ) ε( ) y = C : ε y = C = y = 0: ε = 0 = 0 y < C : ε y < C = y LECTUE-DT-1_03 Seite 3 von 0

4 σ y from a linear strain distriution follows: ( ) y = i σ acc. to Hooke s law C σ :. ending stress due to external ending moment ll external ending moments must e equal to all internal ending moments dmι = σ idx idy iy = σ idiy For equilirium condition: dm = σ y d = M M I y σ = σ C y = σ yd C σ d = y C This integral is a special parameter in ending: y i d: area moment of inertia = Ι y (section moment nd degree) LECTUE-DT-1_03 Seite 4 von 0

5 For a rectangular ar: h The area moment of inertia is: ( aout y-axis ; the ending axis) ya + x 1 + h + y 1 y e x h I = yd = ydy dx = ydy dx ; h 3 h h + y x 3 h h h h h ih 3 4 = 3 3 ( ) = + = =Ι = y mmimm mm 1 = = For the ending moment is: M σ h h C 1 = h h σ = σ 1 h 6 3 = with C= 3 For ending the imum stress at the outer fier of the section is of interest: ih Ι y Ι y = Wy = 1 C section modulus for y-axis (Widerstandsmoment) Therefore in ending calculations the section module is used: σ M = σ = W y M σ = mmimm W = = y h i 1 6 mm 3 LECTUE-DT-1_03 Seite 5 von 0

6 Stress distriution in cantilever ending: The neutral axis = ending axis gives the neutral plane where no stress are acting or this is the stress free plain in the ending section: For many standard sections: rectangular areas, circular areas, hollow circular areas and others the values for the following data: - area moment of inertia for x-and y- axis - section (ending) modulus are taulated. - LECTUE-DT-1_03 Seite 6 von 0

7 Tale of area moments of inertia and section modulus for solid, hollow or comined sections LECTUE-DT-1_03 Seite 7 von 0

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13 1.6 Basic loadings: Torsion of a circular shaft further asic loading is the torsion of a circular shaft: Deformation at the end of the shaft: r iϑ = γϑ l r the strain comes from internal shear stresses τϑr γϑr as like to Hooke s law for torsion, there is a linear relation for shear stress τϑr = G iγϑr G: shear module N mm The imum shear deformation is at the outer diameter, there is a linear relationship for the shear stresses: ϑ τ ( r) = Giγ = Gi ir l The external torque M t = T is equal to the internal torques in the twisted section: LECTUE-DT-1_03 Seite 13 von 0

14 Torsion of a circular section: a) distortion of two sections with stress element; ) distorted grid on the surface Calculation of internal torques in a circular section: t π π ( ) ( ) M = T = r τ r d = r τ r r dr dα π G ϑ G ϑ π 3 = r rrdrd r drd 0 α = α 0 l l α = 0 r= 0 polar area momentof inertia π π π α α π α = 0 r= r D IP = r dr d = = = with = 4 4 I P π D π D = = 3 ccording to the ending of eams, the imum shear stress is at the surface of the circular section: I Wt = P : section torsion modulus D T T T τt = = = : imum torsion shear stress 4 3 Wt πid i πid 3D 16 Til ϑ = : imum twist angle at the end of the circular section G I i P ; LECTUE-DT-1_03 Seite 14 von 0

15 The G Module (Shear Module) is a elastic material constant: E E E G = N G for E mm N ν Poisson' s ratio ν = = ; ( ) ( ).6 mm Torsion and ending are the most applied technical loadings to machine elements: Example: Torsion of shaft: a motor connected to a machine = rotational power of a torque: iπin πin Nimi1imin Nim P= Tiω = Ti = Ti = sec min sec = W i n = rpm of the motor = revolutions per minute ; I p = r d with d= r dα dr π r π IP = r r dα dr= r dα dr = r dr dα ( α) 0 = π = π = π D π D π D = = = 16 3 ( ) ( ) T= r rdrd r= r r drd τ α τ α with τ τ r ( r) = τ 3 T = r dr d α LECTUE-DT-1_03 Seite 15 von 0

16 In the case of torsion the torque rotates around the central axis of the section: a polar area moment of inertia is defined. For the polar area moment of inertia is: ( ) IP = r d = x + y d ; r² = x² + y² = x dx+ y dy = Ix+ Iy Ix = Iy acc. to Pythagoras for circular sections IP π π d π d d with 4 Ix = Iy = = = = = W x π d π d = = 64 d From the sum of all internal torques the external torque is calculated: 4 G π 3 G r π ( ) l 0 α α α ; 0 l T = r dr d = G πg d T = π = with = l 4 l 4 4 π d G πd T = with = Ip 3 l 3 4 I P G mm N G ϑ G ϑ Ip T l T = I = p = ϑ = l mm mm l l G I Or the twist angle due to the torque T is : Tili3 ϑ = Giπ id 4 In mechanical engineering the. value for twisted ars is ϑ 0, 5 / m length or rad / m is allowed from this equation. The necessary diameter for given torque T with a limit value for the twist angle is: 4 Tili3 Tili3 d = d = 4 Gi i Gi i π ϑ π ϑ p ϑ LECTUE-DT-1_03 Seite 16 von 0

17 Other solution of the prolem: τ ( r) r = τ τ τ π 4 P T = τ ( r) r d = r d = = τ = τ WP IP polar section module 4 3 IP πi πi = = i for circular sections = d there for: I 3 3 p π d π d T 16 π = Wp = = τ 3 = = 3 16 W π d for circular sections is valid: I = I + I and I = I P x y x y IP WP Ix = Iy = therefor : Wx = Wy = πd W π d π d WP = = = = M M 3 σ ( ending) = = P Wx 3 W π d P I LECTUE-DT-1_03 Seite 17 von 0

18 Strength of materials rea moment of inertia and section modulus of sections: Bending and uckling Cross section LECTUE-DT-1_03 rea moment of inertia xial section modulus W Torsion Polar section modulus Wp Seite 18 von 0

19 1.7 Basic loadings: Comined loading (ending and tension): F σ = T T ( tension stress: normal stress) at the section point M σ = W at point σ tension ending stress total stress at point is: σ = σ + σ tot T addition of normal stresses at point B: σ = σ σ tot B T B B addition of normal stresses in the same section area Same condition for shear stresses is valid: Equal types of stresses can e superimposed with respect to their sign (+) for tension stresses (-) and compression stresses. But: a linear comination of shear and normal stresses is not allowed. Comination of a ending moment and a torque: The common loading of a circular shaft of a ending moment and a torque is very often: M = Fil ; T t σ and τ t can not added in a simple way, therefore we need an equivalent σ eq stress: (, t ) σ = f σ τ σ eq all The equivalent stress σ eq is compared with an allowale stress for the material: LECTUE-DT-1_03 Seite 19 von 0

20 There are different theories for equivalent stresses, for ductile materials the imum-distortion-energy theory has een proved as the est one for comined normal and shear stresses: (DET): the Mises criterion ( ) 0 σ = + 3 i t eq σ α τ σ : τ t : α 0 :. ending stress. shear stress due to torsion stress intensity factor α σ ( ) 0,7 0,6 0,8,lim 0 = 3iτt,lim σ : limit ending stress,lim τ t,lim : limit torsion stress σ or σ y then yielding starts if eq P0, LECTUE-DT-1_03 Seite 0 von 0