Russell C. Hibbeler. Chapter 1: Stress

Russell C. Hibbeler Chapter 1: Stress

Introduction Mechanics of materials is a study of the relationship between the external loads on a body and the intensity of the internal loads within the body. This subject also involves the deformations and stability of a body when subjected to external forces. Chapter 1: Stress

Equilibrium of a Deformable Body External Forces 1.Surface Forces - caused by direct contact of other body s surface.body Forces - other body exerts a force without contact Chapter 1: Stress

Equilibrium of a Deformable Body Reactions Surface forces developed at the supports/points of contact between bodies. Chapter 1: Stress

Equilibrium of a Deformable Body Equations of Equilibrium Equilibrium of a body requires a balance of forces and a balance of moments F 0 MO 0 For a body with x, y, z coordinate system with origin O, F x M 0, x 0, F M y 0, 0, Best way to account for these forces is to draw the body s free-body diagram (FBD). y F M z z 0 0 Chapter 1: Stress

Equilibrium of a Deformable Body Internal Resultant Loadings Objective of FBD is to determine the resultant force and moment acting within a body. In general, there are 4 different types of resultant loadings: a) Normal force, N b) Shear force, V c) Torsional moment or torque, T d) Bending moment, M Chapter 1: Stress

Example 1.1 Determine the resultant internal loadings acting on the cross section at C of the beam. Solution: Free body Diagram Distributed loading at C is found by proportion, w 70 w 180 N m 6 9 Magnitude of the resultant of the distributed load, F ( ) m ( 180)( 6) 540N 1 which acts 1 6 from C 3 Chapter 1: Stress

Solution: Equations of Equilibrium Applying the equations of equilibrium we have Chapter 1: Stress + + + F F y M x 0; 0; C 0; V N C 0 0 (Ans) 540 0 M M C N V C C C C 540 (Ans) 540 ( ) 0 1080 N m (Ans)

Example 1.5 Determine the resultant internal loadings acting on the cross section at B of the pipe. The pipe has a mass of kg/m and is subjected to both a vertical force of 50 N and a couple moment of 70 N m at its end A. It is fixed to the wall at C. Chapter 1: Stress

FBD. Chapter 1: Stress

Solution Free-Body Diagram Calculating the weight of each segment of pipe, W W BD AD ( )( 0.5)( 9.81) 9.81 N ( )( 1.5)( 9.81) 4.55 N Chapter 1: Stress Applying the six scalar equations of equilibrium, Fx 0; ( FB ) x Fy 0; ( FB ) y Fz 0; ( FB ) z 0 (Ans) 0 (Ans) 9.81 4.55 50 ( FB ) 84.3 N (Ans) x ( M B ) 0; ( M ) + 70 50( 0.5) 4.55( 0.5) 9.81( 0.5) x B x ( M B ) x 30.3N m (Ans) ( M B ) 0; ( M ) y B y + 4.55( 0.65) + 50( 1.5) 0 ( M B ) y 77.8N m (Ans) ( M ) 0; ( M ) 0 (Ans) B z B z 0 0

Stress Distribution of internal loading is important in mechanics of materials. We will consider the material to be continuous. This intensity of internal force at a point is called stress. Chapter 1: Stress

Stress Normal Stress σ Force per unit area acting normal to ΔA Shear Stress τ σ z lim Δ A 0 Δ ΔA Force per unit area acting tangent to ΔA F z τ τ zx zy lim ΔA 0 lim ΔA 0 ΔFx ΔA ΔF y ΔA Chapter 1: Stress

Average Normal Stress in an Axially Loaded Bar When a cross-sectional area bar is subjected to axial force through the centroid, it is only subjected to normal stress. Stress is assumed to be averaged over the area. Chapter 1: Stress

Average Normal Stress in an Axially Loaded Bar Average Normal Stress Distribution When a bar is subjected to a constant deformation, df σda A P σa σ P A Equilibrium σ average normal stress P resultant normal force A cross sectional area of bar normal stress components that are equal in magnitude but opposite in direction. Chapter 1: Stress

Example 1.6 The bar has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown. Solution: By inspection, different sections have different internal forces. Chapter 1: Stress

Solution: Graphically, the normal force diagram is as shown. By inspection, the largest loading is in region BC, P BC 30 kn Since the cross-sectional area of the bar is constant, the largest average normal stress is σ BC P BC A 3 ( ) 30 10 ( 0.035)( 0.01) 85.7 MPa (Ans) Chapter 1: Stress

Example 1.8 The casting is made of steel that has a specific weight of 3 γ st 80 kn/m. Determine the average compressive stress acting at points A and B. Solution: By drawing a free-body diagram of the top segment, the internal axial force P at the section is + F z 0; P P W st 0 ( 80)( 0.8) π ( 0.) P 8.04 kn 0 The average compressive stress becomes P 8.04 σ π 64.0 kn/m A ( 0.) (Ans) Chapter 1: Stress

Average Shear Stress The average shear stress distributed over each sectioned area that develops a shear force. τ V avg A τ average shear stress P internal resultant shear force A area at that section different types of shear: a) Single Shear b) Double Shear Chapter 1: Stress

Example 1.1 The inclined member is subjected to a compressive force of 3000 N. Determine the average compressive stress along the smooth areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by EDB. Solution: The compressive forces acting on the areas of contact are + + F F y x 0; 0; F F AB BC 3000 3000 3 ( ) 5 0 FAB 4 () 0 F 400 N 5 BC 1800 N Chapter 1: Stress

Solution: The shear force acting on the sectioned horizontal plane EDB is + 0 ; V 1800 N F x Average compressive stresses along the AB and BC planes are σ σ AB BC 1800 5 ( )( 40) 400 50 ( )( 40) 1.80 N/mm 1.0 N/mm (Ans) (Ans) Average shear stress acting on the BD plane is τ avg 1800 0.60 N/mm ( 75)( 40) (Ans) Chapter 1: Stress

Allowable Stress Many unknown factors that influence the actual stress in a member. A factor of safety is needed to obtained allowable load. The factor of safety (F.S.) is a ratio of the failure load divided by the allowable load Chapter 1: Stress F. S F. S F. S F F σ σ τ τ fail allow fail allow fail allow

Example 1.14 The control arm is subjected to the loading. Determine to the nearest 5 mm the required diameter of the steel pin at C if the allowable shear stress for the steel is 55 MPa. Note in the figure that the pin is subjected to double shear. τ allowable + + Solution: For equilibrium we have + M F F y x C 0; 0; 0; F AB 3 ( 0.) 15( 0.075) 5( )( 0.15) 5 4 C + 5() 0 C 5 kn 15 x 5 x 3 C 15 5() 0 C 30 kn y 5 y 0 F AB 15 kn Chapter 1: Stress

Solution: The pin at C resists the resultant force at C. Therefore, F C () 5 ( 30) 30.41kN The pin is subjected to double shear, a shear force of 15.05 kn acts over its crosssectional area between the arm and each supporting leaf for the pin. The required area is A τ V allowable d π d 15.05 3 55 10 46.45 mm 18.8 mm 76.45 10 6 m Use a pin with a diameter of d 0 mm. (Ans) Chapter 1: Stress

Example 1.17 The rigid bar AB supported by a steel rod AC having a diameter of 0 mm and an aluminum block having a cross sectional area of 1800 mm. The 18-mm-diameter pins at A and C are subjected to single shear. If the failure stress for the steel and aluminum is ( σ st ) 680 MPa fail and ( σ al ) 70 MPa fail 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.. Solution: The allowable stresses are ( σ ) ( σ ) τ st al allow allow ( σ ) F. S. ( σ ) fail F. S. τ fail F. S. allow 900 Chapter 1: Stress 680 70 450 MPa st al fail 340 MPa 35 MPa

Solution: There are three unknowns and we apply the equations of equilibrium, + + M M B A 0; 0; P F ( 1.5) FAC ( ) 0 (1) ( ) P( 0.75) 0 () B We will now determine each value of P that creates the allowable stress in the rod, block, and pins, respectively. 6 For rod AC, F AC ( σ ) ( A ) 340( 10 ) π ( 0.01) st allow ( )( ) Using Eq. 1, 106.8 P 1.5 AC 171kN [ ] 106.8 kn 6 6 For block B, F ( σ ) A 35( 10 )[ 1800( 10 )] 63.0 kn B al allow Using Eq., ( 63.0)( ) P 0.75 B 168 kn Chapter 1: Stress

Solution: 6 For pin A or C, V F τ A 450( 10 ) π ( 0.009) Using Eq. 1, P AC ( 114.5)( ) 1.5 allow 183 kn [ ] 114.5 kn When P reaches its smallest value (168 kn), it develops the allowable normal stress in the aluminium block. Hence, P 168 kn (Ans) Chapter 1: Stress

Russell C. Hibbeler Chapter : Strain

Deformation When a force is applied to a body, it will change the body s shape and size. These changes are deformation. Note the before and after positions of 3 line segments where the material is subjected to tension. Chapter : Strain

Strain Normal Strain The elongation / contraction of a line segment per unit of length is referred to as normal strain. Average normal strain is defined as If the normal strain is known, then the approximate final length is Δs' ( 1+ ε ) Δs Chapter : Strain ε avg Δs' Δs Δs +ε line elongate -ε line contracts

Strain Units Normal strain is a dimensionless quantity since it is a ratio of two lengths. Shear Strain Change in angle between line segments that were perpendicular to one another refers to shear strain. Chapter : Strain γ nt π lim B A along n C A along t θ ' θ<90 +shear strain θ>90 -shear strain

Example.1 3 1/ The slender rod creates a normal strain in the rod of ε z 40 10 z where z is in meters. Determine (a) displacement of end B due to the temperature increase, and (b) the average normal strain in the rod. Solution: Part (a) Since the normal strain is reported at each point along the rod, it has a deformed length of 3 1/ dz' [ 1+ 40( 10 ) z ]dz The sum along the axis yields the deformed length of the rod is The displacement of the end of the rod is therefore Δ B Chapter : Strain 0. z' 0 3 [ 1+ 40( 10 ) z ] 1/ dz 0.039 m ( ) 0.039 0. 0.0039m.39mm (Ans)

Solution: Part (b) Assumes the rod has an original length of 00 mm and a change in length of.39 mm. Hence, ε avg Δs' Δs Δs.39 00 0.0119 mm/mm (Ans) Example.3 The plate is deformed into the dashed shape. 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. Chapter : Strain

Solution: Part (a) Line AB, coincident with the y axis, becomes line after deformation, thus the length of this line is AB ' ( 50 ) + 3 48.018 mm The average normal strain for AB is therefore ' AB 3 48.018 50 40 ( ) 7.93( 10 ) mm/mm (Ans) ε AB avg AB AB The negative sign indicates the strain causes a contraction of AB. Chapter : Strain

Solution: Part (b) As noted, the once 90 angle BAC between the sides of the plate, referenced from the x, y axes, changes to θ due to the displacement of B to B. γ xy Since γ π then is the angle shown in the figure. Thus, xy θ ' γ xy tan 3 1 50 0.11 rad (Ans) Chapter : Strain

Russell C. Hibbeler Chapter 3: Mechanical Properties of Materials

The Tension and Compression Test The strength of a material depends on its ability to sustain a load. This property is to perform under the tension or compression test. The following machine is designed to read the load required to maintain specimen stretching. Chapter 3: Mechanical Properties of Materials

The Stress Strain Diagram Conventional Stress Strain Strain Diagram Nominal or engineering stress is obtained by dividing the applied load P by the specimen s original cross-sectional area. Nominal or engineering strain is obtained by dividing the change in the specimen s gauge length by the specimen s original gauge length. Chapter 3: Mechanical Properties of Materials σ ε P A 0 δ L 0

The Stress Strain Diagram Conventional Stress Strain Strain Diagram Stress-Strain Diagram Elastic Behaviour Stress is proportional to the strain. Material is said to be linearly elastic. Yielding Increase in stress above elastic limit will cause material to deform permanently. Chapter 3: Mechanical Properties of Materials

The Stress Strain Diagram Conventional Stress Strain Strain Diagram Stress-Strain Diagram Strain Hardening. After yielding a further load will reaches a ultimate stress. Necking At ultimate stress, cross-sectional area begins to decrease in a localized region of the specimen. Specimen breaks at the fracture stress. Chapter 3: Mechanical Properties of Materials

The Stress Strain Diagram True Stress Strain Strain Diagram The values of stress and strain computed from these measurements are called true stress and true strain. Use this diagram since most engineering design is done within the elastic range. Chapter 3: Mechanical Properties of Materials

Stress Strain Behavior of Ductile and Brittle Materials Ductile Materials Material that can subjected to large strains before it ruptures is called a ductile material. Brittle Materials Materials that exhibit little or no yielding before failure are referred to as brittle materials. Chapter 3: Mechanical Properties of Materials

Hooke s Law Hooke s Law defines the linear relationship between stress and strain within the elastic region. σ Eε σ stress E modulus of elasticity or Young s modulus ε strain E can be used only if a material has linear elastic behaviour. Chapter 3: Mechanical Properties of Materials

Hooke s Law Strain Hardening When ductile material is loaded into the plastic region and then unloaded, elastic strain is recovered. The plastic strain remains and material is subjected to a permanent set. Chapter 3: Mechanical Properties of Materials

Strain Energy When material is deformed by external loading, it will store energy internally throughout its volume. Energy is related to the strains called strain energy. Modulus of Resilience When stress reaches the proportional limit, the strain-energy density is the modulus of resilience, u r. u r 1 1 σ plε σ pl E pl Chapter 3: Mechanical Properties of Materials

Strain Energy Modulus of Toughness Modulus of toughness, u t, represents the entire area under the stress strain diagram. It indicates the strain-energy density of the material just before it fractures. Chapter 3: Mechanical Properties of Materials

Example 3. The stress strain diagram for an aluminum alloy that is used for making aircraft parts is shown. When material is stressed to 600 MPa, find the permanent strain that remains in the specimen when load is released. Also, compute the modulus of resilience both before and after the load application. Solution: When the specimen is subjected to the load, the strain is approximately 0.03 mm/mm. The slope of line OA is the modulus of elasticity, E 450 0.006 75.0 GPa From triangle CBD, Chapter 3: Mechanical Properties of Materials ( 6 10 ) 75.0 ( 10 ) 0.008 mm/mm BD 600 9 E CD CD CD

Solution: This strain represents the amount of recovered elastic strain. The permanent strain is ε OC 0.03 0.008 0.0150 mm/mm (Ans) Computing the modulus of resilience, 1 1 1 1 ( u ) σ ε ( 450)( 0.006) r 1.35 MJ/m (Ans) 3 ( u ) σ ε ( 600)( 0.008).40 MJ/m (Ans) r initial final pl pl pl pl Note that the SI system of units is measured in joules, where 1 J 1 N m. 3 Chapter 3: Mechanical Properties of Materials

Poisson s Ratio Poisson s ratio, v (nu), states that in the elastic range, the ratio of these strains is a constant since the deformations are proportional. v ε ε lat long Poisson s ratio is dimensionless. Typical values are 1/3 or 1/4. Negative sign since longitudinal elongation (positive strain) causes lateral contraction (negative strain), and vice versa. Chapter 3: Mechanical Properties of Materials

Example 3.4 A bar made of A-36 steel has the dimensions shown. If an axial force of 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. Solution: The normal stress in the bar is σ z ε z A P σ z E st 3 ( 10 ) 80 6 16.0( 10 )Pa ( 0.1)( 0.05) From the table for A-36 steel, E st 00 GPa 6 ( ) 6 ( ) 16.0 10 6 80( 10 )mm/mm 00 10 Chapter 3: Mechanical Properties of Materials

Solution: The axial elongation of the bar is therefore 6 [ 80( 10 )( 1.5) ] 10 m (Ans) δ ε z μ z L z The contraction strains in both the x and y directions are ε x ε y v st ε 0.3 80 10 z 6 [ ( )] 5.6 μm/m The changes in the dimensions of the cross section are δ ε L x δ ε L y x y x y 6 [ 5.6( 10 )( 0.1) ] 6 5.6( 10 )( 0. 05).56μm (Ans) [ ] 1.8μm (Ans) Chapter 3: Mechanical Properties of Materials

The Shear Stress Strain Diagram For pure shear, equilibrium requires equal shear stresses on each face of the element. When material is homogeneous and isotropic, shear stress will distort the element uniformly. Chapter 3: Mechanical Properties of Materials

The Shear Stress Strain Diagram For most engineering materials the elastic behaviour is linear, so Hooke s Law for shear applies. τ Gγ G shear modulus of elasticity or the modulus of rigidity 3 material constants, E, and G are actually related by the equation G E 1+ ( v) Chapter 3: Mechanical Properties of Materials

Example 3.5 A specimen of titanium alloy is tested in torsion and the shear stress strain diagram is shown. Find the shear modulus G, the proportional limit, and the ultimate shear stress. Also, find the maximum distance d that the top of a block of this material could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement? Solution: The coordinates of point A are (0.008 rad, 360 MPa). Thus, shear modulus is G 360 45 10 3 0.008 ( ) MPa (Ans) Chapter 3: Mechanical Properties of Materials

Solution: By inspection, the graph ceases to be linear at point A. Thus, the proportional limit is τ pl 360 MPa (Ans) This value represents the maximum shear stress, point B. Thus the ultimate stress is τ u 504 MPa (Ans) Since the angle is small, the top of the will be displaced horizontally by d tan 50 mm ( 0.008 rad) 0.008 d 0.4 mm The shear force V needed to cause the displacement is τ avg V V ; 360 MPa V A ( 75)( 100) 700 kn (Ans) Chapter 3: Mechanical Properties of Materials

*Failure of Materials Due to Creep and Fatigue Creep When material support a load for long period of time, it will deform until a sudden fracture occurs. This time-dependent permanent deformation is known as creep. Both stress and/or temperature play a significant role in the rate of creep. Creep strength will decrease for higher temperatures or higher applied stresses. Chapter 3: Mechanical Properties of Materials

*Failure of Materials Due to Creep and Fatigue Fatigue When metal subjected to repeated cycles of stress or strain, it will ultimately leads to fracture. This behaviour is called fatigue. Endurance or fatigue limit is a limit which no failure can be detected after applying a load for a specified number of cycles. This limit can be determined in S-N diagram. Chapter 3: Mechanical Properties of Materials

Russell C. Hibbeler Chapter 4: Axial Load

Saint-Venant s Principle Saint-Venant s principle states that both localized deformation and stress tend to even out at a distance sufficiently removed from these regions. Chapter 4: Axial Load

Elastic Deformation of an Axially Loaded Member Using Hooke s law and the definitions of stress and strain, we are able to develop the elastic deformation of a member subjected to axial loads. Suppose an element subjected to loads, σ P A ( x) dδ and ε ( x) dx L δ δ 0 P A ( x) ( x) dx E small displacement L original length P(x) internal axial force A(x) cross-sectional area E modulus of elasticity Chapter 4: Axial Load

Elastic Deformation of an Axially Loaded Member Constant Load and Cross-Sectional Area When a constant external force is applied at each end of the member, δ PL AE Sign Convention Force and displacement is positive when tension and elongation and negative will be compression and contraction. Chapter 4: Axial Load

Example 4. The assembly consists of an aluminum tube AB having a cross-sectional area of 400 mm. A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kn is applied to the rod, determine the displacement of the end C of the rod. (E st 00 GPa, E al 70 GPa ) Solution: Find the displacement of end C with respect to end B. 3 PL [ + 80( 10 )]( 0.6) δc / B + 0.003056 m 9 AE π 0.005 00 10 δ B 3 [ 80( 10 )]( 0.4) 6 9 [ 400( 10 )][ 70( 10 )] PL 0.001143 0.001143 m AE Chapter 4: Axial Load ( )[ ( )] Displacement of end B with respect to the fixed end A, Since both displacements are to the right, δ C δ + δ 0.004 m 4.0 mm C C / B (Ans)

Example 4.4 A member is made from a material that has a specific weight and modulus of γ and elasticity E. If it is formed into a cone, find how far its end is displaced due to gravity when it is suspended in the vertical position. Solution: Radius x of the cone as a function of y is determined by proportion, x y r ; L x o ro L y The volume of a cone having a base of radius x and height y is V π yx 3 πr 3L o y 3 Chapter 4: Axial Load

Solution: Since W γv, the internal force at the section becomes + F y γπr 0; P y 3L A o 3 ( y) The area of the cross section is also a function of position y, πr L o ( y) π x y Between the limits of y 0 and L yields L ( y) ( y) L [( γπro 3L )] [( γπr L )] P dy dy δ γl 6 A E E E 0 0 o (Ans) Chapter 4: Axial Load

Principle of Superposition Principle of superposition is to simplify stress and displacement problems by subdividing the loading into components and adding the results. Statically Indeterminate Axially Loaded Member A member is statically indeterminate when equations of equilibrium are not sufficient to determine the reactions on a member. Chapter 4: Axial Load

Example 4.5 The steel rod has a diameter of 5 mm. It is attached to the fixed wall at A, and before it is loaded, there is a gap between the wall at and B and the rod of 1 mm. Find the reactions at A and B if the rod is subjected to an axial force of P 0 kn. Neglect the size of the collar at C. (E st 00 GPa) Solution: Equilibrium of the rod requires + Fx 0; FA FB Chapter 4: Axial Load A + 0 10 ( 3 ) 0 (1) The compatibility condition for the rod is. By using the load displacement relationship, FALAC FB LCB δ B / A 0.001 AE AE F ( 0.4) F ( 0.8) 397.0 N m () B δ B / A 0.001m Solving Eqs. 1 and yields F A 16.6 kn and F B 3.39 kn. (Ans)

Example 4.8 The bolt is made of 014-T6 aluminum alloy and is tightened so it compresses a cylindrical tube made of Am 1004-T61 magnesium alloy. The tube has an outer radius of 10 mm, and both the inner radius of the tube and the radius of the bolt are 5 mm. The washers at the top and bottom of the tube are considered to be rigid and have a negligible thickness. Initially the nut is hand-tightened slightly; then, using a wrench, the nut is further tightened one-half turn. If the bolt has 0 threads per inch, determine the stress in the bolt. Solution: Equilibrium requires + Fy 0 ; Fb Ft 0 When the nut is tightened on the bolt, the tube will shorten. (1) ( + ) δt 0. 5 δb Chapter 4: Axial Load

Solution: Taking the modulus of elasticity, π F ( 60) t 3 [ 10 5 ][ 45( 10 )] 5F 0.5 π t 15π Solving Eqs. 1 and simultaneously, we get F ( 60) b 3 [ 5 ][ 75( 10 )] ( 115) 9F () b F b F t 31556 31.56 kn The stresses in the bolt and tube are therefore σ σ b s F A b b Ft A t 31556 π () 5 31556 π ( 10 5 ) 401.8 N/mm 133.9 N/mm 401.8 MPa (Ans) 133.9 MPa (Ans) Chapter 4: Axial Load

Example 4.9 The A-36 steel rod shown has a diameter of 5 mm. It is attached to the fixed wall at A, and before it is loaded there is a gap between the wall at and the rod of 1 mm. Determine the reactions at A and B. Solution: Consider the support at B as redundant and using principle of superposition, + 0.001 δ p δ B (1) Thus, δ δ P B PL AE AC FB L AE AB π 3 [ 0( 10 )]( 0.4) 9 ( 0.005) [ 00( 10 )] π F ( 1.) B 9 ( 0.005) [ 00( 10 )] 0.00037 m 0.3056 10 6 ( ) F B Chapter 4: Axial Load

Solution: By substituting into Eq. 1, 0.001 0.00037 F B 3.39 10 From the free-body diagram, 6 0.3056( 10 ) 3 ( ) 3.39 kn (Ans) F B + F x 0; F A + 0 3.39 0 F A 16.6 kn (Ans) Chapter 4: Axial Load

Thermal Stress Change in temperature cause a material to change its dimensions. Since the material is homogeneous and isotropic, α ΔT T δ T δt αδtl linear coefficient of thermal expansion, property of the material algebraic change in temperature of the member original length of the member algebraic change in length of the member Chapter 4: Axial Load

Example 4.1 The rigid bar is fixed to the top of the three posts made of A-36 steel and 014-T6 aluminum. The posts each have a length of 50 mm when no load is applied to the bar, and the temperature is T 1 0 C. Determine the force supported by each post if the bar is subjected to a uniform distributed load of 150 kn/m and the temperature is raised to T 0 C. Solution: From free-body diagram we have + Fy 0; Fst + Fal 90 10 ( 3 ) 0 (1) The top of each post is displaced by an equal amount and hence, ( + ) δ () st δ al Chapter 4: Axial Load

Solution: The final position of the top of each post is equal to its displacement caused by the temperature increase and internal axial compressive force. Applying Eq. gives ( + ) δ st ( δ st ) + ( δ ) T st F ( + ) δ al ( δ al ) + ( δ al ) T F ( δ st ) + ( δ st ) ( δ st ) + ( δ al ) T F T F With reference from the material properties, we have 6 [ 1( 10 )]( 80 0)( 0.5) + π F ( 0.5) st 9 ( 0.0) [ 00( 10 )] F st ( 0.5) 6 Fal [ 310 ( )]( 80 0)( 0.5) + 9 π ( 0.03) [ 73.110 ( )] 3 1.16F 165.9( 10 ) (3) al Solving Eqs. 1 and 3 simultaneously yields Chapter 4: Axial Load F st 16.4 kn and F 13 kn al (Ans)

Russell C. Hibbeler Chapter 5: Torsion

Torsional Deformation of a Circular Shaft Torque is a moment that twists a member about its longitudinal axis. If the angle of rotation is small, the length of the shaft and its radius will remain unchanged. Chapter 5: Torsion

The Torsion Formula When material is linear-elastic, Hooke s law applies. A linear variation in shear strain leads to a corresponding linear variation in shear stress along any radial line on the cross section. τ max τ T J c p τ max Tc or τ J Tp J maximum shear stress in the shaft shear stress resultant internal torque polar moment of inertia of cross-sectional area outer radius of the shaft intermediate distance Chapter 5: Torsion

The Torsion Formula If the shaft has a solid circular cross section, J π 4 c If a shaft has a tubular cross section, J 4 4 ( ) π c o c i Chapter 5: Torsion

Example 5. The solid shaft of radius c is subjected to a torque T. Find the fraction of T that is resisted by the material contained within the outer region of the shaft, which has an inner radius of c/ and outer radius c. Solution: Stress in the shaft varies linearly, thus τ The torque on the ring (area) located within the lighter-shaded region is ( τda) ρ( ρ c) τ ( πρdρ ) dt' ρ max For the entire lighter-shaded area the torque is ( ρ c) τ max πτ max 3 T ' ρ dρ c c c / 15π τ 3 max c 3 (1) Chapter 5: Torsion

Solution: Using the torsion formula to determine the maximum stress in the shaft, we have τ τ max max Tc J T 3 πc Tc c ( π ) 4 Substituting this into Eq. 1 yields 15 T ' T 16 (Ans) Chapter 5: Torsion

Example 5.3 The shaft is supported by two bearings and is subjected to three torques. Determine the shear stress developed at points A and B, located at section a a of the shaft. Solution: From the free-body diagram of the left segment, M x 0 ; 450 3000 T 0 T 150 knmm The polar moment of inertia for the shaft is 4 7 J π ( 75) 4.97 10 mm Since point A is at ρ c 75 mm, ( 150)( 75) Tc τ 1.89 MPa (Ans) 7 4.97 10 B J Likewise for point B, at ρ 15 mm, we have Tc ( 150)( 15) 0.377 MPa (Ans) 7 4.97 10 J τ B Chapter 5: Torsion

Power Transmission Power is defined as the work performed per unit of time. For a rotating shaft with a torque, the power is P Tω whereshaft angular velocity, ω dθ / dt Since 1cycle π rad ω πf, the power equation is For shaft design, the design or geometric parameter is Chapter 5: Torsion P πft J c T τ allow

Example 5.5 A solid steel shaft AB is to be used to transmit 3750 W from the motor M to which it is attached. If the shaft rotates at w 175 rpm and the steel has an allowable shear stress of allow τ allow 100 MPa, determine the required diameter of the shaft to the nearest mm. Solution: The torque on the shaft is P Tω 175 π 3750 T T 04.6 Nm 60 4 J π c T Since c c τ T c πτ allow 1/3 Chapter 5: Torsion allow ( 04.6)( 1000) π ( 100) 1/3 10.9 mm As c 1.84 mm, select a shaft having a diameter of mm.

Angle of Twist Integrating over the entire length L of the shaft, we have L φ 0 T J ( x) ( x) dx G Φ angle of twist T(x) internal torque J(x) shaft s polar moment of inertia G shear modulus of elasticity for the material Assume material is homogeneous, G is constant, thus φ TL JG Sign convention is determined by right hand rule, Chapter 5: Torsion

Example 5.8 The two solid steel shafts are coupled together using the meshed gears. Determine the angle of twist of end A of shaft AB when the torque 45 Nm is applied. Take G to be 80 GPa. Shaft AB is free to rotate within bearings E and F, whereas shaft DC is fixed at D. Each shaft has a diameter of 0 mm. Solution: From free body diagram, F 45/ 0.15 300 N ( ) 300( 0.075).5 Nm T D x Angle of twist at C is φ C TL DC JG Chapter 5: Torsion ( +.5)( 1.5) ( π )( 0.001) 80( 10) 9 [ ] + 4 0.069 rad Since the gears at the end of the shaft are in mesh, φ B ( 0.15) ( 0.069)( 0.075) 0.0134 rad

Solution: Since the angle of twist of end A with respect to end B of shaft AB caused by the torque 45 Nm, φ T L JG ( + 45)( ) AB AB A/ B + 4 9 ( π )( 0.010) [ 80( 10 )] 0.0716 rad The rotation of end A is therefore φ A φb B + φa/ 0.0134 + 0.0716 + 0.0850 rad (Ans) Chapter 5: Torsion

Example 5.11 The solid steel shaft has a diameter of 0 mm. If it is subjected to the two torques, determine the reactions at the fixed supports A and B. Solution: By inspection of the free-body diagram, M x 0 ; T + 800 500 T 0 b A (1) Since the ends of the shaft are fixed, φ A / B 0 Using the sign convention, TB JG ( 0.) ( T + 500)( 1.5) T ( 0.3) + A JG + 1.8T A A JG 0.T B 0 750 Solving Eqs. 1 and yields T A -345 Nm and T B 645 Nm. () Chapter 5: Torsion

Russell C. Hibbeler Chapter 6: Bending

Shear and Moment Diagrams Members with support loadings applied perpendicular to their longitudinal axis are called beams. Beams classified according to the way they are supported. Chapter 6: Bending

Shear and Moment Diagrams Shear and moment functions can be plotted in graphs called shear and moment diagrams. Positive directions indicate the distributed load acting downward on the beam and clockwise rotation of the beam segment on which it acts. Chapter 6: Bending

Example 6.1 Draw the shear and moment diagrams for the beam shown. Solution: From the free-body diagram of the left segment, we apply the equilibrium equations, F y F y 0; 0; + + Chapter 6: Bending F y M 0; 0; P V P M x Left segment of the beam extending a distance x within region BC is as follow, P P P V 0 V L P M + P x x M (1) () (3) P ( L-x) (4)

Solution: The shear diagram represents a plot of Eqs. 1 and 3 The moment diagram represents a plot of Eqs. and 4 Chapter 6: Bending

Example 6.4 Draw the shear and moment diagrams for the beam shown. Solution: The distributed load is replaced by its resultant force and the reactions. Intensity of the triangular load at the section is found by proportion, w w0 w or w x L 0 L Resultant of the distributed loading is determined from the area under the diagram, + + M F y 0; 0; w0l w0l 3 1 w0l w0 x x V L ( x) + 1 0 V w0 x x L 1 3 w0 L x + M ( L x ) 0 () (1) Chapter 6: Bending

Solution: The shear diagram represents a plot of Eqs. 1 The moment diagram represents a plot of Eqs. Chapter 6: Bending

Example 6.6 Draw the shear and moment diagrams for the beam shown. Solution: regions of x must be considered in order to describe the shear and moment functions for the entire beam. 0 x < 5 m, + + 1 M 5 m x + + M 1 F F y 0; 0; 0; < 10 m, y 0; 5.75 V 80 5.75x 80 5.75x M Chapter 6: Bending 0 V 1 5.75 15 5 1 + M 5.75 kn 0 M (1) ( 5.75x + 80) knm () ( x 5) V 0 V ( 15.75 5x ) + + 15 + 5 ( x 5) 5 (.5x + 15.75x + 9.5) knm (4) x 1 + M 0 kn (3)

Solution: The shear diagram represents a plot of Eqs. 1 and 3 The moment diagram represents a plot of Eqs. and 4 Chapter 6: Bending

Graphical Method for Constructing Shear and Moment Diagrams Regions of Distributed Load The following equations provide a convenient means for quickly obtaining the shear and moment diagrams for a beam. Slope of the shear diagram at each point dv dx w ( x) -distributed load intensity at each point Slope of moment diagram at each point dm V dx Shear at each point Chapter 6: Bending

Graphical Method for Constructing Shear and Moment Diagrams We can integrate these areas between any two points to get change in shear and moment. Change in shear ΔV w ( x)dx -area under distributed loading Change in moment ΔM V ( x)dx Area under shear diagram Chapter 6: Bending

Graphical Method for Constructing Shear and Moment Diagrams Regions of Concentrated Force and Moment Some of the common loading cases are shown below. Chapter 6: Bending *Note: Not to be memorized

Example 6.7 Draw the shear and moment diagrams for the beam shown. Solution: The reactions are shown on a free-body diagram. For shear diagram according to the sign convention, Chapter 6: Bending at x 0, V + P and at x L, V + P Since w 0, the slope of the shear diagram will be zero, thus dv dx w 0 at all points For moment diagram according to the sign convention, at x 0, M PL and at x L, M 0 The shear diagram indicates that the shear is constant Positive, thus dm dx V + P at all points

Example 6.8 Draw the shear and moment diagrams for the beam shown. Solution: The reaction at the fixed support is shown on the free-body diagram. Since no distributed load exists on the beam the shear diagram will have zero slope, at all points. From the shear diagram the slope of the moment diagram will be zero since V 0. Chapter 6: Bending

Example 6.10 Draw the shear and moment diagrams for the beam shown. Solution: The reaction at the support is calculated and shown on the free-body diagram. The distributed loading on the beam is positive yet Decreasing, thus negative slope. The curve of the moment diagram having this slope behaviour is a cubic function of x. Chapter 6: Bending

Example 6.11 Draw the shear and moment diagrams for the beam shown. Solution: The reaction at the support is calculated and shown on the free-body diagram. The slope of the shear diagram will vary from zero at x 0 to and at x 4.5. The point of zero shear can be found by 1 x + F y 0 ; 1.5 0 4.5 x x.6 m Slope of the moment diagram will begin at 1.5, then becomes decreasingly positive until it reaches zero at.6 m. It then becomes increasingly negative reaching 3 at x 4.5 m. Chapter 6: Bending

Bending Deformation of a Straight Member Cross section of a straight beam remains plane when the beam deforms due to bending. There will be tensile stress on one side and compressive stress on the other side. Chapter 6: Bending

Bending Deformation of a Straight Member Longitudinal strain varies linearly from zero at the neutral axis. Hooke s law applies when material is homogeneous. Neutral axis passes through the centroid of the cross-sectional area for linear-elastic material. Chapter 6: Bending

The Flexure Formula Resultant moment on the cross section is equal to the moment produced by the linear normal stress distribution about the neutral axis. My σ I σ normal stress in the member M resultant internal moment I moment of inertia y perpendicular distance from the neutral axis By the right-hand rule, negative sign is compressive since it acts in the negative x direction. Chapter 6: Bending

Example 6.15 The simply supported beam has the cross-sectional area as shown. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this location. Solution: The maximum internal moment in the beam is M.5 knm Chapter 6: Bending

Solution: By symmetry, the centroid C and thus the neutral axis pass through the midheight of the beam, and the moment of inertia is I 1 1 ( I + ) Ad 301.3 10 3 ( 0.5)( 0.0) + ( 0.5)( 0.00)( 0.16) + ( 0.0)( 0.3) 6 4 ( ) m 1 1 Applying the flexure formula where c 170 mm, 3 σ Mc I ( 0.17).5 301.3 10 max ; σ max 6 ( ) 1.7 MPa (Ans) Chapter 6: Bending

Example 6.17 The beam has a cross-sectional area in the shape of a channel. Determine the maximum bending stress that occurs in the beam at section a a. Solution: The resultant internal moment must be computed about the beam s neutral axis at section a a. Since the neutral axis passes through the centroid, y ya A ( 0.1)( 0.)( 0.015) + ( 0.01)( 0.0)( 0.5) ( 0.)( 0.015) + ( 0.0)( 0.5) 0.05909 m 59.09 mm Chapter 6: Bending

Solution: Applying the moment equation of equilibrium about the neutral axis, we have + M M NA ( ) + 1.0( 0.05909) M 0 4.859 knm 0 ;.4 The moment of inertia about the neutral axis is 1 I 1 1 + 1 4.6 10 3 ( 0.5)( 0.0) + ( 0.5)( 0.0)( 0.05909 0.01) 3 ( 0.015)( 0.) + ( 0.015)( 0.)( 0.1 0.05909) 6 4 ( ) m The maximum bending stress occurs at points farthest away from the neutral axis. Mc 4.859( 0. 0.05909) σ max 16. MPa (Ans) 6 I 4.6 10 Chapter 6: Bending ( )

Russell C. Hibbeler Chapter 6: Bending

Shear and Moment Diagrams Members with support loadings applied perpendicular to their longitudinal axis are called beams. Beams classified according to the way they are supported. Chapter 6: Bending

Solution: The shear diagram represents a plot of Eqs. 1 and 3 The moment diagram represents a plot of Eqs. and 4 Chapter 6: Bending

Solution: The shear diagram represents a plot of Eqs. 1 The moment diagram represents a plot of Eqs. Chapter 6: Bending

Solution: The shear diagram represents a plot of Eqs. 1 and 3 The moment diagram represents a plot of Eqs. and 4 Chapter 6: Bending

Graphical Method for Constructing Shear and Moment Diagrams Regions of Concentrated Force and Moment Some of the common loading cases are shown below. Chapter 6: Bending *Note: Not to be memorized

Russell C. Hibbeler Chapter 7: Transverse Shear

Shear in Straight Members When a shear V is applied, non-uniform shear-strain distribution over the cross section will cause the cross section to warp. The relationship between moment and shear is V dm dx Chapter 7: Transverse Shear

The Shear Formula The shear formula is used to find the transverse shear stress on the beam s cross-sectional area. τ VQ It where Q A' yda y' A' τ the shear stress in the member V internal resultant shear force I moment of inertia of the entire cross-sectional area t width of the member s cross-sectional area Chapter 7: Transverse Shear

Shear Stresses in Beams For rectangular cross section, shear stress varies parabolically with depth and maximum shear stress is along the neutral axis. Chapter 7: Transverse Shear

Example 7.1 The beam is made of wood and is subjected to a resultant internal vertical shear force of V 3 kn. (a) Determine the shear stress in the beam at point P, and (b) compute the maximum shear stress in the beam. Solution: (a) The moment of inertia of the cross sectional area computed about the neutral axis is 1 I bh 1 1 Q ya ' 15 + Applying the shear formula, we have τ p VQ It Chapter 7: Transverse Shear 3 1 1 3 6 4 ( 100)( 15) 16.8 10 mm 3 ( 50) ( 50)( 100) 18.75 10 4 mm 4 ()( 3 18.75 10 ) 6 ( 16.8 10 )( 100) 0.346 MPa (Ans)

Solution: (b) Maximum shear stress occurs at the neutral axis, since t is constant throughout the cross section, Q 65. y' A ' 3 ( 100)( 6.5) 19.53 10 4 mm Applying the shear formula yields τ VQ It 4 ()( 3 19.53 10 ) 6 ( 16.8 10 )( 100) max 0.360 MPa (Ans) Chapter 7: Transverse Shear

Shear Flow in Built-Up Members For fasteners it is necessary to know the shear force by the fastener along the member s length. This loading is referred as the shear flow q, measured as a force per unit length. q VQ I q shear flow V internal resultant shear I moment of inertia of the entire cross-sectional area Chapter 7: Transverse Shear

Example 7.4 The beam is constructed from four boards glued together. If it is subjected to a shear of V 850 kn, determine the shear flow at B and C that must be resisted by the glue. Solution: The neutral axis (centroid) will be located from the bottom of the beam, ~ ya y 0.1968 m A The moment of inertia computed about the neutral axis is thus Since the glue at B and holds the top board to the beam I 87.5 10 4 ( 6 ) m Q B y' B A' B 3 [ 0.305 0.1968]( 0.50)( 0.01) 0.7110 ( 3 ) m Chapter 7: Transverse Shear

Solution: Likewise, the glue at C and C holds the inner board to the beam Q C y' C A' C 3 [ 0.05 0.1968]( 0.15)( 0.01) 0.0106( 10 3 ) m Therefore the shear flow for BB and CC, q' q' B C VQ I VQ I B C 3 ( 850)( 0.71 10 ) 87.5 10 6 3 ( 850)( 0.0106 10 ) 0.0996 MN/m 87.5 10 6.63 MN/m Since two seams are used to secure each board, the glue per meter length of beam at each seam must be strong enough to resist one-half of each calculated value of q. q B 1.31MN/m and q 0.0498 MN/m C (Ans) Chapter 7: Transverse Shear

Example 7.7 The thin-walled box beam is subjected to a shear of 10 kn. Determine the variation of the shear flow throughout the cross section. Solution: The moment of inertia is For point B, the area A' 0 thus q B 0. For point C, VQ I Chapter 7: Transverse Shear q The shear flow at D is q D VQ I D C C ( 30 / ) 10 184 I 1 1 3 Also, QC ya' ( 3.5)( 5)( 1) 17.5 cm 3 Q y ' ( )( 1)( 4) 30 cm D A ( ) 10 17.5/ 184 1 1 3 3 4 ( 6)( 8) ( 4)( 6) 184 mm 1.63 kn/cm 163 N/mm 0.951kN/cm 91.5 N/mm

Russell C. Hibbeler Chapter 8: Combined Loading

Thin-Walled Pressure Vessels Thin wall refers to a vessel having an inner-radius to-wall-thickness ratio of 10 or more. ( r / t 10) For cylindrical vessels under normal loading, there are normal stresses in the circumferential or hoop direction and in the longitudinal or axial direction. σ1 σ pr t pr t normal stress in hoop direction normal stress in longitudinal direction Chapter 8: Combined Loadings

Thin-Walled Pressure Vessels The following is a summary of loadings that can be applied onto a member: a) Normal Force b) Shear Force c) Bending Moment d) Torsional Moment e) Thin-Walled Pressure Vessels f) Superposition Chapter 8: Combined Loadings

Example 8. A force of 15 000 N is applied to the edge of the member. Neglect the weight of the member and determine the state of stress at points B and C. Solution: For equilibrium at the section there must be an axial force of 15 000 N acting through the centroid and a bending moment of 750 000 N mm about the centroidal or principal axis. A P σ 15000 3.75 MPa ( 100)( 40) The maximum stress is σ Mc I 75000 1 1 ( 50) max 3 ( 40)( 100) 11.5 MPa Chapter 8: Combined Loadings

Solution: The location of the line of zero stress can be determined by proportional triangles 75 x 15 100 ( x) x 33.3 mm Elements of material at B and C are subjected only to normal or uniaxial stress. σ σ B C 7.5 MPa 15 MPa (tension) (Ans) (compression) (Ans) Chapter 8: Combined Loadings

Example 8.3 The tank has an inner radius of 600 mm and a thickness of 1 mm. It is filled to the top with water having a specific weight of γ w 10 kn/m 3. If it is made of steel having a specific weight of γ st 78 kn/m 3, determine the state of stress at point A. The tank is open at the top. Solution: The weight of the tank is 61 W st γ stvst 78 π π 1000 st A 61 600 st π 1000 1000 Chapter 8: Combined Loadings 600 1000 () 1 3.56 kn The pressure on the tank at level A is p γ z ( 10 )( 1) 10 kpa For circumferential and longitudinal stress, we have 600 pr 10( 1000 ) σ1 500 kpa (Ans) 1 t W σ ( ) 1000 3.56 [( ) ( ) ] 77.9 kpa (Ans) w

Example 8.6 The rectangular block of negligible weight is subjected to a vertical force of 40 kn, which is applied to its corner. Determine the normal-stress distribution acting on a section through ABCD. Solution: For uniform normal-stress distribution the stress is A P σ Chapter 8: Combined Loadings 40 0.8 ( )( 0.4) 15 kpa For 8 knm, the maximum stress is M xcy 8( 0.) σ I 0.8 0.4 3 [ ( )( ) ] max x 1 1 For 16 knm, the maximum stress is M ycx 16( 0.4) σ I 1 0.4 0.8 3 [ ( )( ) ] max y 1 375 kpa 375 kpa

Solution: Assuming that tensile stress is positive, we have σ σ σ σ A B C D 15 + 375 + 375 65 kpa 15 375 + 375 15 kpa 15 375 375 875 kpa 15 + 375 375 15 kpa The line of zero stress can be located along each side by proportional triangles ( 0.4 e) e ( 0.8 e) 65 e 0.0667 m 15 and 65 h h 0.133 m 15 Chapter 8: Combined Loadings

Russell C. Hibbeler Chapter 9: Stress Transformation

Plane-Stress Transformation General state of stress at a point is characterized by 6 independent normal and shear stress components. It can be analyzed in a single plane of a body, the material can said to be subjected to plane stress. Chapter 9: Stress Transformation

Plane-Stress Transformation Stress components from one orientation of an element can transform to an element having a different orientation. Chapter 9: Stress Transformation

Example 9.1 The state of plane stress at a point on the surface of the airplane fuselage is represented on the element oriented as shown. Represent the state of stress at the point on an element that is oriented 30 clockwise from the position shown. Solution: The element is sectioned by the line a a. The free-body diagram of the segment is as shown. Chapter 9: Stress Transformation

+ σ x + Solution: Applying the equations of force equilibrium in the x and y direction, F 0; x' ' ΔA ( 50ΔAcos30 ) cos30 + ( 5ΔAcos30 ) ( 80ΔAsin30 ) sin30 + ( 5ΔAsin30 ) cos30 σ x' 4.15 MPa (Ans) sin30 0 + F y' 0; τ τ ΔA ( 50ΔAcos30 ) sin 30 ( 5ΔAcos30 ) ( 80ΔAsin 30 ) cos30 + ( 5ΔAcos30 ) sin 30 x'y' x'y' 68.8 MPa (Ans) cos30 0 Chapter 9: Stress Transformation

Solution: Repeat the procedure to obtain the stress on the perpendicular plane b b. + + F F x' y' 0; 0; σ σ -τ τ x' ΔA ( 5ΔAcos30 ) sin 30 + ( 80ΔAcos30 ) ( 5ΔAcos30 ) cos30 ( 50ΔAsin 30 ) sin 30 x' x'y' ΔA + ( 5ΔAcos30 ) cos30 + ( 80ΔAcos30 ) ( 5ΔAsin 30 ) sin 30 + ( 50ΔAsin 30 ) cos30 x'y' 5.8 MPa (Ans) 68.8 MPa (Ans) cos30 0 sin 30 0 The state of stress at the point can be represented by choosing an element oriented. Chapter 9: Stress Transformation

General Equations of Plane-Stress Transformation Positive normal stress acts outward from all faces and positive shear stress acts upward on the right-hand face of the element. σ x + σ y σ x σ y σ x' + cosθ + τ xy σ x σ y τ x' y' sin θ + τ xy cosθ sin θ Chapter 9: Stress Transformation

Example 9. The state of plane stress at a point is represented by the element. Determine the state of stress at the point on another element oriented 30 clockwise from the position shown. Solution: From the sign convention we have, σ 80 MPa σ 50 MPa τ 5 MPa θ 30 x y xy To obtain the stress components on plane CD, σ x + σ y σ x σ y σ x' + cosθ + τ xy sin θ 5.8 MPa σ x σ y τ x' y' sin θ + τ xy cosθ 68.8 MPa (Ans) (Ans) Chapter 9: Stress Transformation

Solution: To obtain the stress components on plane BC, σ x 80 MPa σ 50 MPa τ 5 MPa θ 60 y xy σ τ x' x' y' σ x + σ y σ x σ y + cosθ + τ xy sin θ 4.15 MPa σ x σ y sin θ + τ xy cosθ 68.8 MPa (Ans) (Ans) The results are shown on the element as shown. Chapter 9: Stress Transformation

Principal Stresses and Maximum In-Plane Shear Stress In-Plane Principal Stresses Orientation of the planes will determine the maximum and minimum normal stress. tan θ p ( σ σ )/ x τ xy y Chapter 9: Stress Transformation

. The solution has two roots, thus we obtain the following principle stress. σ x + σ y σ x σ y ± + τ xy σ 1, whereσ1 σ > Chapter 9: Stress Transformation

Principal Stresses and Maximum In-Plane Shear Stress Maximum In-Plane Shear Stress Orientation of an element will determine the maximum and minimum shear stress. tan θ s ( σ σ ) x τ xy y / The solution has two roots, thus we obtain the maximum in-plane shear stress and averaged normal stress. σ x σ y τ max in-plane + τ Chapter 9: Stress Transformation xy σ avg σ x σ y

Example 9.3 When the torsional loading T is applied to the bar it produces a state of pure shear stress in the material. Determine (a) the maximum in-plane shear stress and the associated average normal stress, and (b) the principal stress. Solution: From the sign convention we have, σ x 0 σ y 0 τ xy τ a) Maximum in-plane shear stress is τ σ x σ y σ x + σ y max in-plane + τ xy ± τ σ avg b) For principal stress, τ xy tan θ p σ p σ σ / ( ) x y 45, σ p1 135 0 (Ans) σ 1, σ x + σ y ± σ x σ y + τ xy ± τ (Ans) Chapter 9: Stress Transformation

Example 9.6 The state of plane stress at a point on a body is represented on the element. Represent this stress state in terms of the maximum in-plane shear stress and associated average normal stress. Solution: Since σ 0, σ 90, τ 60 we have x y xy ( ) σ x σ y / tan θ s σ s 1.3, σ s1 111. 3 τ xy The maximum in-plane shear stress and average normal stress is τ σ max in-plane avg σ x σ x σ y + σ y + τ xy 35 MPa (Ans) 81.4 MPa (Ans) Chapter 9: Stress Transformation

Mohr s Circle Plane Stress Plane stress transformation is able to have a graphical solution that is easy to remember. Chapter 9: Stress Transformation

Example 9.8 The torsional loading T produces the state of stress in the shaft as shown. Draw Mohr s circle for this case. Solution: We first construct of the circle, σ τ max in-plane τ, σ 0 avg x 0, σ y 0 and τ xy τ σ x + σ y The center of the circle C is on the axis at σ avg 0 Point A is the average normal stress and maximum in-plane shear stress, Principal stresses are identified as points B and D on the circle. σ1 τ, σ τ Chapter 9: Stress Transformation

Example 9.10 The state of plane stress at a point is shown on the element. Determine the maximum in-plane shear stresses and the orientation of the element upon which they act. Solution: We first construct of the circle, σ 0, σ 90 and τ 60 The center of the circle C is on the axis at x y xy σ avg 0 + 90 35 MPa From point C and the A(-0, 60) are plotted, we have R 60 + 55 81.4 MPa Max in-plane shear stress and average normal stress are τ max in-plane 81.4 MPa, σ 35 MPa avg The counter-clockwise angle is 1 0 + 35 θ s tan 1.3 (Ans) 1 60 Chapter 9: Stress Transformation (Ans)

Example 9.1 An axial force of 900 N and a torque.5 Nm of are applied to the shaft. The shaft diameter is 40 mm, find the principal stresses at a point P on its surface. τ Solution: The stresses produced at point P is Tc J ( 0.0) P 900 198.9 kpa, σ 4 ( 0.0) A π ( 0.0).5 π 716. kpa The principal stresses can be found using Mohr s circle, σ avg 0 + Chapter 9: Stress Transformation 716. 358.1 kpa Principal stresses are represented by points B and D, σ 358.1+ 409.7 767.7 kpa 1 σ 358.1 409.7 51.5 kpa (Ans) (Ans)

Example 9.13 The beam is subjected to the distributed loading of w 10 kn/m. Determine the principal stresses in the beam at point P, which lies at the top of the web. Neglect the size of the fillets and stress concentrations at this point. I 67.4(10-6 ) m 4. Solution: The equilibrium of the sectioned beam is as shown where V At point P, σ τ VQ It 3 ( ) 6 ( ) 30.6 10 0.1 45.4 67.4 10 84 My I Chapter 9: Stress Transformation [( 0.1075)( 0.175)( 0.015) ] 6 67.4( 10 )( 0.01) (Ans) 35. MPa (Ans) 84 kn M 30.6 knm

Solution: Thus the results are as follows, The centre of the circle is 45.4 + 0.7 and point A is (-45.4, -3.5). Thus the radius is calculated as 41.9, thus the principle stresses are σ σ 1 ( 41.9.7) 19. MPa (.7 + 41.9) 64.6 MPa The counter-clockwise angle is p θ p 57. θ 8. 6 Chapter 9: Stress Transformation

Russell C. Hibbeler Chapter 1: Deflection of Beams and Shafts

The Elastic Curve For elastic curve, positive internal moment tends to bend the beam concave upward, and vice versa. There must be an inflection point at point C, where the curve changes from concave up to concave down, since this is a point of zero moment. Chapter 1: Deflection of Beams and Shafts

The Elastic Curve Moment-Curvature Relationship Due to loading, deformation of the beam is caused by both the internal shear force and bending moment. If material is homogeneous and behaves in a linearelastic manner, Hooke s law applies thus, 1 ρ M EI ρ radius of curvature at a specific point M internal moment in the beam at the point E material s modulus of elasticity I beam s moment of inertia computed about the neutral axis EI flexural rigidity Chapter 1: Deflection of Beams and Shafts

Slope and Displacement by Integration For most problems the flexural rigidity will be constant along the length of the beam. The slope and displacement relationship of the beam is 4 3 d v d v d v EI w( x) EI V ( x) EI M ( x) dx 4 3 dx Each integration is used to solve for all the constants to obtain a unique solution for a particular problem. dx Chapter 1: Deflection of Beams and Shafts

Slope and Displacement by Integration Boundary and Continuity Conditions The constants of integration are determined by evaluating the functions for shear, moment, slope, or displacement. These values are called boundary conditions. Chapter 1: Deflection of Beams and Shafts

Example 1. The simply supported beam supports the triangular distributed loading. Determine its maximum deflection. EI is constant. Solution: Due to symmetry only one x coordinate is needed for the solution, 0 x L / w0 The equation for the distributed loading is w x.hence, L + M NA 0; M M w0 x + L w0 x 3L x 3 w0 L 4 w0l + x 4 ( x) 0 Chapter 1: Deflection of Beams and Shafts

Solution: Integrating twice, we have d v w0 3 w0 L EI M x + x dx 3L 4 dv w0 4 w0l EI x + x + C1 dx 1L 8 w0 5 w0l 3 EIv x + x + C1x + C 60L 4 3 5w0 L For boundary condition, v 0, x 0 and dv dx 0, x L we havec1, C 0 19 Hence, 3 w0 5 w0l 3 5w0L EIv x + x x 60L 4 19 For maximum deflection at x L/, v max 4 w0l 10EI (Ans) Chapter 1: Deflection of Beams and Shafts

Example 1.4 The beam is subjected to a load P at its end. Determine the displacement at C. EI is constant. Solution: Due to the loading x coordinates will be considered, 0 x1 a and 0 x P Using the free-body diagrams, M1 x1 M Px d v1 P Thus, for 0 x < a EI x 1 dx EIv Chapter 1: Deflection of Beams and Shafts 1 1 dv EI dx 1 1 P x 4 P x 1 3 1 a 1 + C 1 1 + C x 1 + C

Solution: And for 0 x Chapter 1: Deflection of Beams and Shafts C < a d v EI dx v 0, x 0; v 0, x a; v 0 x a 1 1 1 1, 1 dv EI dx EIv Pa 3 P x 6 Px C 3 P x + C 0 + C 3 x 3 The four constants of integration are determined using three boundary conditions, Solving, we obtain Thus solving the equations, When x 0, we get v c v Pa EI 3 P 6EI x (Ans) C 3 + C 3 4 7 6 7Pa + 6EI Pa x C 4 Pa EI Pa 3 3

Discontinuity Functions When expressing load or internal moment of the beam, we need to use discontinuity functions. 1) Macaulay Functions X is the point along the beam and a is the location on the beam where a discontinuity occurs. General equation can used for distributed loadings: x a n 0 n a ( x a) n for for x x < a a Chapter 1: Deflection of Beams and Shafts

Discontinuity Functions The Macaulay functions below describe both the uniform load and triangular load. Chapter 1: Deflection of Beams and Shafts

Discontinuity Functions Singularity Functions The functions are used to describe the point location forces or couple moments acting on a beam. i) To describe a force, w P x a 1 0 for x a P for x a ii) To describe a couple moment, w M 0 x a 0 M 0 for x a for x a iii) Integration of both equations will give x a dx x a, n 1, Chapter 1: Deflection of Beams and Shafts n n+ 1

Example 1.5 Determine the equation of the elastic curve for the cantilevered beam, EI is constant. Solution: The boundary conditions require zero slope and displacement at A. The support reactions at A have been calculated by statics and are shown on the free-body diagram, 1 w 5 x 0 + 58 x 0 + x 0 50 x 5 8 x 5 0 0 Chapter 1: Deflection of Beams and Shafts