AAU Strength of Materials Lecture Notes 09-05-21

Department of

Civil Engineering ADDIS ABABA UNIVERSITY Faculty of Technology

Lecture Notes

Strength of Materials

ceng1002

beta version, 29th April 2009 Karsten Schlesier (Dipl.-Ing.)

for use at Addis Abeba University only!

Strength of Materials 2009

Department of Civil Engineering

ADDIS ABABA UNIVERSITY Faculty of Technology

Calendar Semester II 2009

week Date Chapter Content

1 13.04. – 17.04. Introduction to SOM 2 20.04. – 24.04. 1 Stress - Axial Load 3 27.04. – 01.05. 1/2 Stress / Strain - Axial Load 4 04.05. – 08.05. 2 Strain - Axial Load 5 11.05. – 15.05. 3 Bending of Beams 6 18.05. – 22.05. 3 Bending of Beams 7 25.05. – 29.05. Lab Sessions / videos 8 01.06. – 05.06. 9 08.06. – 12.06. Mid-Semester Exam 10 15.06. – 19.06. 4 Shear in Beams 11 22.06. – 26.06. 5 Torsion 12 29.06. – 03.07. 6 Analysis of plane Stress 13 06.07. – 10.07. 7 Deflection of Beams 14 13.07. – 17.07. 8 Stability of Compression Members 15 20.07. – 24.07. 16 27.07. – 31.07. Final Exam

Course Outline

1 Stress – Axial Loads Normal Stress, Shearing Stress, Transformation of Stress, Concept of Design 2 Strain – Axial Loads Strain, Stress-Strain Diagram, Hooke’s Law, Deflection under Axial Load, Material Properties 3 Bending of Beams Stress due to pure Bending, Moment of Inertia, Stress Distribution, Design of Beams 4 Shear in Beams

Shearing Stress in members due to Bending 5 Torsion Moment of Torsion, Shearing Stresses and Deformations of Circular Shafts 6 Analysis of plane Stress Compound Stresses, Combined Stresses, Transformation of Stress, Mohr’s Circle 7 Deflection of Beams

Deflection of members due to Bending 8 Stability of Compression Members Euler Formula, Buckling Load, Buckling Analysis

Assessment / Requirements Attendance is compulsory during lecture hours, tutorials and practical work sessions (except for unpredicted mishaps). Quota of Total Course Credit: 40% mid-semester examination 60% final-semester examination Literature / Teaching Material Popov, E.P., Mechanics of Materials; Beer and Johnson, Mechanics of Materials, 2001; Gere and Timoshenko, Mechanics of Materials, 1990 Course Information: www.elboon.net e-learning board online

Strength of Materials Lecture Notes


ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier

Index

1 Stress – Axial Loads Normal Stress, Shearing Stress, Transformation of Stress, Concept

of Design 2 Strain – Axial Loads Strain, Stress-Strain Diagram, Hooke’s Law, Deflection under Axial

Load, Material Properties 3 Bending of Beams Stress due to pure Bending, Moment of Inertia, Stress Distribution,

Design of Beams 4 Shear in Beams

Shearing Stress in members due to Bending

5 Torsion Moment of Torsion, Shearing Stresses and Deformations of Circular

Shafts 6 Analysis of plane Stress Compound Stresses, Combined Stresses, Transformation of Stress,

Mohr’s Circle 7 Deflection of Beams

Deflection of members due to Bending

8 Stability of Compression Members Euler Formula, Buckling Load, Buckling Analysis

2

11

22

31

38

46

56

60

1

Strength of Materials 1 Stress 1/9



1 Stress By setting up the equilibrium conditions, the inner forces of a member subjected to an external load situation can be determined. So far neither the material nor the type of cross section applied for the member are being taken into account. But both material and type of cross section obviously have an impact on the behaviour of the member subjected to load. To design the member therefore a closer look on how the internal forces act along its cross section needs to be taken.

1.1 Normal Stress – Axial Loading Within this part of the chapter the internal forces are limited to only axial forces (normal forces) acting along the centroidal axis of a member. A suspended rod is subjected to an axial load. The free body diagram in external equilibrium is shown in fig. 1.11a. The rod is cut perpendicular to its axis at any arbitrary distance from its ends and the equations of equilibrium are applied on the part. Thus the internal force found acting normal to the cut surface (area A) is of equal amount but opposite direction of the applied external force (fig. 1.11b). Consider the normal force to equally act on any particle ΔA of the cut surface A (fig. 1.11c).

ΔAΔF

AF=

fig 1.11: axially loaded rod

F

A

F

plane of cut

F

F

A

F

σ A

ΔF ΔA

a) b) c) d)

2




The intensity of a normal force acting on a surface at a certain point is described as the normal stress, denoted by the Greek letter σ (fig.1.1d).

ΔAΔFlimσ

0ΔA→=

Considering a uniform distribution the normal stress is defined as:

AFσ = and dAσF

A

⋅= ∫ (1.1), (1.2)

conclusion: the normal stress acting along a section of a member only depends on

the external load applied (e.g. a normal force F) and the geometry of its cross section A (true for statically determinant systems).

example 1.1 - stress Fig 1.12 shows a typical specimen used for uniaxial tensile testing for materials like timber or plastic. question: At which position will the specimen break if the applied force F is

increased up to failure?

11 A

Fσ = ; 2

2 ΑFσ = A2 < A1, hence σ2 > σ1 linear correlation!

answer: the specimen breaks at the maximum normal stress σ2 along the plane

with the minimum cross sectional area A2.

A1 F F

A2

fig 1.12: specimen for tensile test subjected to axial load

amount of internal force =σ unit area

kN cm2

3




1.2 Average Shearing Stress – Transverse Loading So far the discussion focussed on normal stress, oriented perpendicular to the cutting plane or in direction of the main axis of the member. Stress can also act in the cutting plane thus perpendicular to the main axis of the member. This occurs if the member is subjected to a situation of transverse loads (fig. 1.21). A situation like this is very common in a bolt or rivet connection (fig 1.22). Here the forces acting in the direction of the steel plates are transmitted by the bolt. In fig 1.23 the bolt is cut along the upper two connecting surfaces of the steel plates. To meet the equilibrium conditions, the force being transported along the cutting plane through the bolt is equal to the force being applied on the upper steel plate (F). Dividing the force by the cut area of the bolt, the stress in the plane of cut is determined (fig. 1.24). Assuming the stress is uniformly distributed, the stress is defined as the average shearing stress, denoted by the Greek letter τ:

AFτ =

F

F

fig 1.21: transverse load situation

2 F F

F

fig 1.22: bolt connection fig 1.23: plane of cut

F F

fig 1.24: shearing stress in the plane of cut - cross section through bolt

4




1.3 Stress Analysis and Concept of Design Every material has its individual properties. It can be ductile, flexible or brittle. It deforms under the influence of a temperature change. It may plastically deform at a certain stress (load) and break at another. Its properties according to perpendicular directions may be equal (isotropic) or different (orthotropic). To ensure a safe design, these specific material properties have to be taken into account. The essential information is collected by conducting different tests in a material testing laboratory. At the failure of the material its ultimate stress is reached. The point of plastic deformation of the material is indicated as the yield point, corresponding to the yield stress. Taking this into account, an allowable stress can be defined for each individual material to be used within the design analysis. These stresses such as further indications concerning the maximum allowable deformation (serviceability of a structure) can be found in the respective national codes. A secure design requires a certain safety clearance towards the failure of the employed material. This is ensured by applying a safety factor (in national codes usually denoted by the Greek letter γ). In the design analysis the existing stress due to the existing load increased by the factor of safety (the design stress) has to be proofed less or equal to the allowable stress. Since the applied material might be orthotropic (different properties in different directions, e.g. timber) different allowable stresses are defined for normal and shearing stresses depending on their orientation (parallel or perpendicular, σ║ or σ ┴, see example 1.4). ratio of safety: design analysis: γFFd ⋅= design load = existing load · factor of safety

ΑFσ d

d = design normal stress, axial loaded

allowedd σσ ≤ design stress ≤ allowable stress

ΑFτ d

d = design average shear stress

allowedd ττ ≤ design stress ≤ allowable stress

ultimate load allowable load

5




example 1.2 - design of axially loaded members The lattice truss displayed in fig 1.3 is subjected to a vertical load of 100 kN at its lower chord.

a) determine the normal forces of members S1, S2 and S3 b) carry out the design analysis for diagonal member S2, considering a solid

square cross section 24 mm x 24 mm such as the given safety factor and the allowable stress

c) design lower chord member S3 by choosing the appropriate diameter of a solid circular cross section

a) external equilibrium:

∑ = 0M A 50kNkN10012.0m6.0mFB ==⇒

∑ = 0V 50kNFF BA ==⇒ cutting plane - internal equilibrium at left part:

∑ = 0M3 100kN3.0m

6.0m50kNS1 −=⋅−

=⇒

∑ = 0M 2 50kN3.0m

3.0m50kNS3 =⋅

=⇒

∑ = 0V 50kNsin45S2 =⋅⇒ o

70.71kNS2 =⇒ b) 106.06kN1.570.71kNγSS 2d2, =⋅=⋅=

222

d2,d 18.41kN/cm

cm(2.4)106.06kN

ΑS

σ === 2allow kN/cm22σ =≤ OK

c) 75kN1.550kNγSS 3d3, =⋅=⋅=

2allow

d3,d 22kN/cmσ

ΑS

σ =≤= 22

allow

d3,required 3.41cm

22kN/cm75kN

σS

A ==≥⇒

2.08cmπ

3.41cm4d4dπA

2

required

2

=⋅

=⇒⋅

= chosen: d = 22 mm

given: load safety factor γ = 1.5 material steel: allowable stress σallow = 22 kN/cm2

6.0 m 6.0 m

3.0 m

3.0 m

100 kN

fig 1.31: lattice girder

S1

S2

S3 A B

6.0

S1

S2

S3

3.0

1FA

2

3

4

6




example 1.3 - design of a pin bolt connection A hinge steel connection consisting of three butt straps and a bolt (fig. 1.32) is subjected to a tensile force Fd = 100 kN (design load). Design the steel bolt by choosing the appropriate diameter considering an allowable shearing stress of τallow = 33.6 kN/cm2 (steel 8.8). average shear stress per shear plane in the bolt:

A2F

τ ddaverage, ⋅=

condition from design analysis:

allowd

daverage, τA2

Fτ ≤

⋅=

2

2allow

d 1.49cm33.6kN/cm2

100kNτ2FA =

⋅=

⋅≥⇒

4dπA

2⋅= 1.38cmd ≥⇒

14mmd:chosen =⇒

F

fig 1.32: bolt connection

F

A

F/2

F/2

F

7




example 1.4 - orthotropic properties, design of a timber connection A diagonal member made of timber is connected to a support beam as shown in fig. 1.33. The orientation of the grain is indicated by the hooked lines. The diagonal element is subjected to a negative normal force along its axis of symmetry (centroidal axis). Carry out the design stress analysis for the timber beam (bottom element). Consider all relevant stresses and the load safety factor. resolution of force F: F┴,d = (sin 30˚ F) 1.5 = (0.5 500 N) 1.5 = 375.0 N

F║,d = (cos 30˚ F) 1.5 = (0.87 500 N) 1.5 = 652.5 N areas of bearing planes for respective force components:

A1 = 600 mm2 plane for vertical force (F┴)

A2 = 300 mm2 plane for horizontal force (F║)

A3 = 900 mm2 shear plane (F║) design analysis: σ⊥,d = F┴,d / A1 = 375.0 N / 600 mm2 = 0.625 N/mm² < σ⊥ allow = 2.0 N/mm² OK σ ||,d = F║,d / A2 = 652.5 N / 300 mm2 = 2.175 N/mm² < σ || allow = 6.0 N/mm² OK τd = F║,d / A3 = 652.5 N / 900 mm2 = 0.725 N/mm² < τallow = 0.9 N/mm² OK

given: applied load F = 500 N load safety factor γ = 1.5 material timber: allowable stresses σ║,allow = 6.0 N/mm2

σ┴,allow = 2.0 N/mm2

τallow = 0.9 N/mm2

A3 A2

A1

F

30 mm 20 mm

10 mm

30 mm

fig 1.33: timber connection

30˚

F┴ F

F║

8




1.4 Transformation of Stress – Oblique Plane under Axial Loading Normal and shearing stresses, thus stress acting perpendicular and parallel to the axis of symmetry of the member have been analysed in the previous parts of this chapter. The plane of cut being used to isolate a part of the member was perpendicularly oriented in all the situations regarded so far. How about the situation of stress on an oblique plane of cut? Fig. 1.41 shows a member subjected to an axial load. A part of the member is isolated by a plane of cut, inclined by the angle φ towards the axis of the member. Setting up the free body diagram and the conditions of equilibrium the stress components acting normal to or within the plane of cut are determined.

AFσ x = initial situation, normal stress

N = cos φ · F resolution of force F normal force N

V = sin φ · F shear force V

ϕϕ cosAA = area of oblique plane

A x

fig 1.41: axially loaded member, oblique plane of cut

plane of cut

F F

F F

N

V

σN

τ

a)

b) c)

φ

φ

9




ϕϕ

2N cos

AF

ANσ ==

ϕ2xN cosσσ = normal stress acting on oblique plane

ϕϕϕϕϕ

cossinσcossinAF

AVτ x ⋅=⋅==

using angle function (2 sinφ cosφ = 2 sinφ):

ϕsin22στ x= shear stress acting on oblique plane

conclusion: xN σσmax = o0=ϕ

2σmax τ x= o45=ϕ [ ]4π

10

Strength of Materials 2 Strain 1/11



2 Strain Any object being subjected to load is deformed, changing its initial shape. This is true for any load and any material. It is easily visualised on objects consisting of soft and flexible materials like rubber or foam but also applies for hard materials like steel or rock. Within a certain load limit the object will return to its initial shape again after the load is released. This is called the elastic behaviour of a material. Exceeding the load above a certain limit, the object will not fully return to its initial shape. Some residual deformation is left, being called the plastic behaviour of a material. A further increase of load leads to the break of the object at a certain point. This is also indicated as the rupture or the failure of the material.

2.1 Strain – Axial Loading A suspended rod of an elastic material and of length L is subjected to axial loading situations. The load is not exceeding the elastic limit of the material. The deformations shown in fig. 2.11 can be proven by uniaxial tests. The rod of cross section A is subjected to an axial load F, fig. 2.11 a). Due to the load, the rod is extended by ΔL in its axial direction. Increasing the load F by factor 2, the elongation of the rod amounts to 2·ΔL, fig. 2.11 b). Fig. 2.11 c) shows a situation of a rod of twice the cross section of system a) and b), being equal to two rods of cross section A. The rod again is subjected to the axial load F. The elongation due to the load found on this system amounts to ΔL/2.

fig. 2.11: axially loaded rod

a) b) c)

F

A

F

ΔL

2F

A

2F

2·ΔLΔL/2

F

2A

F

11




The results are summarised in tab. 2.11.

situation a) b) c) load F 2·F F cross section A A 2·A stress σa) * 2·σa) σa)/2 elongation ΔL 2·ΔL ΔL/2

Like the stress for a member of a certain cross section subjected to a certain load also the elongation of the member can be expressed as a generalised term. The strain is equal to the amount of elongation ΔL under the applied load divided by the initial length L of the member. It is denoted by the Greek letter ε.

strain: LΔLε = = deformation of member per unit length

The results of the stress and strain analysis are plotted in a coordinate system of abscissa strain and ordinate stress, the so called stress-strain diagram (fig. 2.12). The graph connecting the origin and the points of results is a straight line. The correlation between stress and strain is linear within the elastic limit of the material. The slope (gradient) of the stress-strain-graph represents the correlation between stress and strain. It is a specific property of a material, indicating its elastic behaviour. It is called the Modulus of Elasticity or the Young’s Modulus of a material denoted by E.

Elastic Modulus: εσE = ⎥⎦

⎤⎢⎣⎡

2cmkN

A high Modulus of Elasticity therefore represents a hard, rigid material like steel, a low Modulus of Elasticity a soft, deformable material like rubber.

*

AFσ a) =

ε

σ

a)

b)

c)

σa)

εa) = ΔL/L

0

fig. 2.12: stress-strain diagram, Hooke’s Law

tab. 2.11

(2.11) [unitless]

(2.12)

12




Precondition to determine a constant Elastic Modulus is the proportional correlation between stress and strain, the linear elastic range of a material. It is represented by a straight curve on the stress-strain diagram (fig.2.12). This is known as Hooke’s Law (Robert Hooke, English Scientist). Transforming equation (2.12), it can also be expressed as:

εEσ ⋅= or AEF

Eσε ==

Hooke’s Law is not only applicable for members consisting of a constant cross section. Fig. 2.13 is showing a general situation of a member consisting of a variable cross section. Generalized equations for Hooke’s Law (see finite element fig 2.13):

dxδxε = strain, factor of elongation of finite element

dxEA

FdxEσdxεΔL

x

0 xx

xx

0 x

xx

0x ⋅=⋅=⋅= ∫∫∫

AEFLΔL =

∑ ⋅⋅

=ii

ii

EALF

ΔL

ε(x)

x

F F dx

fig 2.13: specimen for tensile test subjected to axial load

x

L dx

dx+δx a)

c)

b)

total elongation of a member consisting of a constant section

total elongation of a member consisting of multiple constant sections

13




example 2.1 - strain An aluminium bar is consisting of two different square cross sections is subjected to an axial load situation fig.2.14. question: Determine the total amount of displacement of the member. given: AI,II = 20 cm2; AIII = 10 cm2 Ealuminium = 70 000 N/mm2 answer: horizontal equilibrium: NIII = 100 kN for part III NII = 80 kN for part II NI = 20 kN for part I

∑ ⋅⋅

=ii

iitotal EA

LFΔL

cm286.0cm/kN0007cm01

cm002kN001ΔL 22III =⋅

⋅=

cm057.0cm/kN0007cm02

cm001kN80ΔL 22II =⋅

⋅=

cmcmkNcm

cmkN 014.0/000702

00102ΔL 22I =⋅

⋅=

cm357.0014.0057.0286.0ΔLtotal =++=

100 cm

200 cm 200 cm

I II III

F1=60kN F2=20kN F3=100kN

fig. 2.14

14




2.2 Stress-Strain Diagram Looking at a member of a certain cross section deforming to a certain extent under a certain load reflects an individual situation. Using stress and strain (σ and ε), the situation is generalised. The correlation between stress and strain depends on the applied material and is represented by the Elastic Modulus, E. The established method to determine the Elastic Modulus of a material is to conduct tensile tests in a laboratory. The results of this test are plotted on the stress-strain diagram. Fig. 2.21 shows a typical stress-strain curve of mild steel, being a ductile material. A ductile material shows considerably large deformation before it fails. Up to the yield point the correlation between stress and strain is proportional, the curve is represented by a straight line. It is the linear elastic range of the material. At the yield point, the proportional limit is reached and plastic deformation occurs. Without an increase of stress, a certain amount of deflection takes places (the curve develops parallel to the abscissa). The ultimate stress (highest stress) lies beyond the yield point, correlating to relatively large deflections. Finally the material breaks at the point of rupture. Fig. 2.22 shows a typical stress-strain curve of a brittle material. Relatively small deformation takes place up to its point of rupture. Also it does not possess a well defined yield point indicating the proportional limit. Here the offset method is used to determine an artificial yield point, thus the linear elastic range for the material. Typical Elastic Moduli of materials being common in the field of construction: Steel: 210 000 N/mm2 Aluminium: 70 000 N/mm2 Timber║: 10 000 N/mm2

ε

σ yield point ultimate

stress

rupture

0

fig. 2.21: stress-strain diagram for mild steel

fig. 2.22: stress-strain diagram for a brittle material

ε

σ 0.2% offset

rupture

0

yield point

linear elastic range

plastic range

15




2.3 Thermal Strain Any change of temperature has an impact on the shape of an object. It shrinks at a thermal decline and expands at a thermal increase. The amount of strain is a property being specific to a certain material. It is represented by the coefficient of thermal expansion denoted by αT. coefficient of thermal expansion: αT [1/˚C] thermal strain: εT = αT · ΔT thermal expansion: ΔL = αT · ΔT · L steel: αT = 12 · 10-6 · 1/˚C concrete: αT = 12 · 10-6 · 1/˚C aluminium: αT = 23 · 10-6 · 1/˚C example 2.2 – thermal strain A rod with fixed ends and no external loads, fig 2.31, is subjected to a thermal increase of ΔT. question: Determine the compressive stress in the rod caused by that impact. given: cross section A material E, αT answer:

ΔLT = αT · ΔT · L free expansion of rod

AEPLΔLP −= compression of rod caused by a negative support force

ΔLT = ΔLP

αT · ΔT · L = -AEPL

P = - αT · ΔT·AE σ = - αT · ΔT·E

ΔL

PB (-) PA (-)

L fig. 2.31

16




example 2.3 A steel cable (Ø 8 mm, A = 45 mm2, E = 170 000 N/mm2) is to be tensioned by a pretensile force of 10.0 kN to a length of 10.0 m at a temperature of 20 ˚C.

a) Determine the initial fabrication length of the cable. b) Determine the remaining pretension in the cable if the temperature is

increased to 60 ˚C.

a) Li + ΔL = 10.0 m

AEPLΔL i=

m0.10AEP1Li =⎟

⎠⎞

⎜⎝⎛ +⇒ mm10000

mm/N17000045mm10000N1L 22i =⎟

⎠⎞

⎜⎝⎛

⋅+⇒

mm95.9986Li =⇒ fabrication length

b) αT,steel = 12 · 10-6 · 1/˚C ΔT = 60 – 20 = 40 ˚C PT = - αT · ΔT·AE = 12 · 10-6 · 1/˚C · 40 ˚C · 45 mm2 · 170 000 N/mm2

PT = - 3672 N P60° = 10.0 – 3.67 = 6.33 kN remaining pretension

10.0 m fig. 2.32

17




2.4 Generalised Hooke’s Law – Poisson’s Ratio An object being subjected to an axial force not only deforms in axial direction. It also deforms in its lateral direction, the directions acting perpendicular to the applied load (fig. 2.41). This phenomena is another property of a specific material. It is known as the Poisson’s Ratio, denoted by the Greek letter ν.

ν = (ratio) lateral expansion and compression of a solid body subjected to an axial force precondition: material is homogenous, isotropic and remains elastic

Generalised Hooke’s Law of Strain: General strain of an element in an multiaxial state of stress: precondition: material is homogenous, isotropic and remains elastic,

strain is independent of small shearing deformations

Eσν

Eσ

νEσε zyx

x −−=

Eσν

Eσ

Eσνε zyx

y −+−=

Eσ

Eσ

νEσνε zyx

z +−−=

lateral strain axial strain

fig. 2.41: element subjected to an axial force

initial shape

initial shape deformed

shape

F F

fig. 2.42: orientation of stress on a 3D element

σy

σx σx

σy

σz

σz

y

x z

18




example 2.4 A solid rectangular steel block is subjected to uniform pressure acting along its surface. If edge AB is compressed by 1.2 · 10-2 mm determine

a) the deformation of all other edges b) the pressure p acting on the block

given: steel: E = 210 000 N/mm2; ν = 0.29 answer:

a) σx = σy = σz = – p uniform pressure

εx = εy = εz = ( )2ν1Ep

Epν

Epν

Ep

−−=++− uniform strain

εx = 4-2

10340mm

mm101.2AB

x −⋅−=⋅

−=Δ

− = εy = εz strain in x-, y- and z-direction

mm10620mm103BCεΔy 34

y−− ⋅−=⋅⋅−=⋅= total compression in y-direction

mm10930mm103BDεΔz 34z

−− ⋅−=⋅⋅−=⋅= total compression in z-direction

b) εx = ( )2ν1Ep

−−

0.2921210000N/mm)103(

2ν1Eεp

24x

⋅−⋅⋅−

−=−⋅

−=⇒−

2150N/mmp −= (= -150 MPa) pressure acting on the block

fig. 2.43: rectangular steel block

y

x z

B

A

20 mm

30 mm

40 mm

C D

19




2.5 Shearing Strain 2.5.1 General situation for Shearing Stress To analyse the strain caused by shearing stress, a closer look is to be taken at the general situation for shearing stress first. precondition: material is homogenous, isotropic and remains elastic Fig. 2.51 shows the shearing stresses and their directions acting on mutually perpendicular planes. Equal assumptions can be made for the xz- and yz- directions. Equations of equilibrium: ∑ = 0F : xyxy ττ = ∑ = 0M0 : yxxy ττ = ( ) ( ) dydzdxτdxdzdyτ yxxy ⋅⋅=⋅⋅ area · lever arm conclusion: All shearing stresses on mutually perpendicular planes of an

infinitesimal element are numerically equal this is also shown on Mohr’s Circle – see chapter 6

fig. 2.51: shearing stress acting on a finite element

dx

y

x z

dz dy

τxy

τyx

τxy

τyx

20




2.5.1 Distortion of an element in pure shear Fig. 2.52 shows an element being distorted by pure shear stress acting along its edges. As shearing stresses on mutually perpendicular planes are equal, the indication of the shearing stresses is simplified to τ. The total angle of distortion of the element is denoted by the Greek letter γ. Like stress and strain, σ and ε, also shearing stress and shearing strain, τ and γ, are in linear relationship (proportional correlation). This can be proven experimentally. Hence the same rules can be applied and another material property, the Shearing Modulus is found, denoted by G. Hooke’s Law for shearing strain: τ = G · γ

Shearing Modulus: γτ

=G ⎥⎦⎤

⎢⎣⎡

2cmkN

The material properties E, G and ν are not independent. At this stage the correlation shall be given without derivation as:

( )ν+=12EG

0

fig. 2.52: element in pure shear

τ

τ τ

τ γ/2

γ/2

(2.52)

(2.51)

(2.53)

21

Strength of Materials 3 Pure Bending of Beams 1/9



3 Pure Bending of Beams A beam consisting of a certain cross section and material will react in a certain way (deflection) and have a certain resistance towards applied bending loads. A beam can be subjected to bending in various ways. In many of these situations also shear forces will be present. Shear forces coexist with shear stresses causing additional deflection. To analyse the impact of bending loads on a beam element we therefore focus on a situation being free of shear forces. Lets take into account the statical correlation of the shear force function along a beam being a derivation function of the bending moment. Hence it is concluded that a part of a beam showing a linear constant moment diagram (horizontal line) is free of shearing forces. This internal load situation is called pure bending (M = constant, V = N = 0). Fig. 3.11 a) is illustrating such a situation.

3.1 Normal Stress Fig. 3.11 a) shows a beam subjected to pure bending. To investigate the stresses being caused along the plane of cut by the internal moment, a closer look is to be taken at an isolated element, shown in fig. 3.11 b). All possible stresses acting along the plane of cut are indicated in the figure since so far there is no evidence of the plane being totally free of shearing stresses. To find the isolated element of fig. 3.11 b) in equilibrium, the stresses acting along the plane of cut on the right face have to equal the internal moment Mz on the left face. Any bending moment can be expressed by a couple of forces acting at a certain distance from another. Looking at the orientation of the stress vectors it becomes obvious that both τxy and τxz are irrelevant for the solution. Since both of them act in the plane of cut neither of them acts in a lever arm towards the applied moment. Therefore they are neglected in the following derivation.

b)

Mz

y

x z

τxy·dA

σx·dA

τxz·dA

τxy·dA

σx·dA τxz·dA

+

fig 3.11: beam subjected to pure bending

a)

Mz Mz

-

+

22




Equations of equilibrium concerning σx: ∑ = 0Fx : 0dAσ x =⋅∫

∑ = 0M y : 0dAσz x =⋅⋅∫

∑ = 0M z : zMdAσy- x =⋅⋅∫

3.2 Normal Strain The following geometric derivations are based on the assumption that any cross section of a beam remains plane after being subjected to bending. This is known as the theory of elasticity for slender members undergoing small deflections established by Jacob Bernoulli (1645-1705). A beam member possessing a plane of symmetry is subjected to a situation of pure bending, fig 3.21. The member will deflect uniformly since the internal moment along the member is constant. By deflection the edges of the element (line AB) are transformed into a circular curve. The upper edge AB of the element is decreased in length whereas the lower edge A1B1 is increased. The middle plane, representing the plane of symmetry, remains original in length and is therefore known as the neutral plane (or neutral axis). Fig 3.21 b) shows the situation on an isolated element. ρ represents the radius of curvature of the neutral axis. dx is the original length of the free upper and lower edges. Both decrease and increase in length of the upper and the lower edge are denoted by δx. Since the member is considered to undertake small deflections only, the curvature of the upper and lower edge is neglected. Furthermore the inclined left and right edges of the deformed element are considered to remain original in length.

By geometry: ρc

dxδxε maxx, == (similar triangles)

maxx,x εcy

ρyε ==

(3.11)

A

A1 B1

initial shape

M M

deflected shape

v

B

initial shape

deflected shape

neutral axis

dx+δx dx

c y y

ρ

x fig 3.21: deflection of beam subjected to pure bending

a) b)

(3.12)

23




3.3 Normal Stress Using the proportional correlation of stress and strain:

maxx,x σcyσ =

(3.13) in (3.11):

∑ = 0Fx : 0dAyc

σdAσcydAσ max

maxx ∫∫∫ =⋅=⋅=⋅

∫ dAy : first moment of cross section (statical moment) → about the neutral axis =0 → neutral axis = centroidal axis

∑ = 0M z : z2max

x MdAyc

σdAσy =⋅=⋅⋅ ∫∫

z2max MdAy

cσ

=⋅∫

dAyI 2 ⋅= ∫ : second moment of cross section (moment of inertia)

Transformation of (3.14): z

zmax I

cMσ =

(3.13) in (3.15): z

zx I

yMσ =

Introducing: cIS = elastic section modulus

(3.15) becomes: z

zmax S

Mσ =

since I

yMεEσ ⋅=⋅= →

IEyM⋅⋅

=ε

recalling (3.12): ρyε =

in (3.18): κ=⋅

=ρ IE

M1

(continued in chapter 7, deflection of beams)

(3.13)

(3.14)

(3.15)

(3.16)

flexual stress (linear elastic) elastic flecture formulas

(3.17)

(3.18)

(3.19) curvature of neutral axis EI = bending or flexual stiffness

fig 3.22: stress distribution along section of beam

c

x y

neutral axis

-σmax

+σmax

Mz

24




example 3.1 Discuss the maximum stresses σ and deflections y that will occur on beams subjected to an equal bending moment consisting of the cross sections given in the table below. note: all cross sections have an equal consumption of material (almost equal areas)

cross section [mm2]

A [cm2] 100 100 97 96 97

I [cm4] 833 3333 ... ... ...

S [cm3] 167 333 ... ... ...

σ factor 100% 50% ... ... ...

y factor 100% 25% ... ... ...

tab. 3.31

100

100 200

50

200

200 t=13 t=10

100

400 I360

360

143

t=13 t=19.5

25




example 3.2 A beam consisting of a rectangular cross section is subjected to pure bending. Replace the linear stress distribution along the cross section by its respective force couple. Set up the equation for the maximum stress.

( ) h32h21322a =⋅⋅⋅=

maxmax σ4

bhσb2h

21R =⎟

⎠⎞

⎜⎝⎛=

∑ = 0M : Mσ6

bhaR max

2

==⋅

SM

6bhMσ 2max == flexure formula, linear elastic

allowsafetymax σγσ ≤⋅ e.g. design of beam

→ excursion on inelastic bending, rect. cross section ( elpl S5.1S ⋅= )

3.3 Unsymmetrical bending

3.3.1 Superposition of stresses Principle of superposition (also see chapter 6): Superposition of normal stresses design analysis

y

y

z

zx

IzM

IyM

AFσ ⋅±

+⋅±

−±= allowmax σ≤σ

For algebraic sign convention, see chapter 6.

a

-σmax

+σmax

M R

R

y

y

z z

b

h

26




example 3.3 A beam is being subjected to a load situation of two concentrated loads (see below). Given loads are design loads. Determine the maximum and minimum internal forces. Carry out the design analysis for

a) A rectangular timber cross section 100 x 200 mm, σ║,allow = 0.85 kN/cm2 b) A standard T100 steel section (oriented flange down), A = 20.0 cm2, Iz = 179

cm4, position of centroidal axis see sketch below Internal force diagrams:

a) rect. timber cross section 100 x 200 mm

σ║,allow = 8.5 kN/cm2 A = 200 cm2

43

z cm6666.712bhI ==

3z2

z cm666.7cI

6bhS ===

2Nx, cmkN0.05

ANσ ==

23

z

zMx, cmkN0.75

cm666.7kNcm500

SMσ ±=±=±=

Superposition:

allow

22max

σkN/cm0.85kN/cm0.80.750.05σ =≤=+= OK

M [kNm] N [kN] V [kN]

10.0

-5.0 -2.5

5

1.0m

5 kN

2.0m

10 kN

200

100

y

z

N

+σM

-σM

M

+σN +0.8

-0.7

27




b) steel cross section T100

σ║,allow = 21.8 kN/cm2 A = 20.9 cm2 4

z cm791I =

2Nx, cmkN0.48

ANσ ==

24

z

ztopM,x, cmkN20.27cm7.26

cm179kNcm500-

IyMσ =−== (see sign convention)

24

z

zbottomM,x, cmkN65.7-cm2.74-

cm179kNcm500-

IyMσ =−=−=

Superposition:

allow

22max

σkN/cm.812kN/cm75.0227.0248.0σ =≤=+= OK

3.3.2 Position of neutral surface Position of neutral axis demands: 0σ x =

→ 0I

zMI

yM

y

y

z

z =+−

→ y

z

yz

zy

IIztanθ

IMIzM

y ⋅⋅=⋅⋅

=

→ tanθII

zy

y

z=

→ tanθIItan

y

z=ϕ

(3.20)

y

y

z z 100

72.6

27.4

100

50

fig 3.31: position of neutral axis

θ

φ

P

y

y

z z Mz

=cosθ MP

My =sinθ MP

MP

28




3.4 Cross sections of different materials In the construction industry many structural members consist of more than only one material. Due to their properties some materials cope better with tensile stress whereas others deal well with compression (or are more cost effective). The most commonly composite material being implemented in the construction sector is steel reinforced concrete. To design a cross section consisting of more than one material it is necessary to develop a procedure to determine the stresses in each of the applied materials due to the given load situation. In this part of this chapter cross sections of two different materials are being investigated. The procedure developed can be projected on composite members consisting of even more than two materials. A cross section consisting of two different materials is shown in fig. 3.41a). Both of the applied materials have different elastic moduli (E1 and E2). At their surface of contact both materials are tightly fixed to one another, thus along this surface both materials develop an equal strain under the given load (ε1 = ε2), see fig. 3.41c). Since the elastic moduli are different, a break along the stress distribution is found at the surface of contact (σ = ε E), see fig. 3.41d). To determine the stresses existing in both materials a virtual cross section of one homogenous material is being constructed. For this the ratio n = E2/E1 is determined. The transformed cross section consists of an equal area of material 1 (unchanged). The area of material 2 is extruded parallel to the neutral axis by factor n, see fig. 3.41b). On the next step the centroidal axis of the transformed cross section is computed and the moment of inertia is determined. Now the stresses occurring along the cross section of the homogenous material (material 1) are calculated (e.g. σ1 = ε1 E1 = M/S). To determine the stresses existing along material 2, the respective results for material one are multiplied by factor n (σ2 = ε2 E2 = σ1 n). The same procedure can be used to determine the curvature of a composite cross section. In equation (3.19) the moment of inertia of the transformed cross section is applied.

bI y

y

z z

mat. 1

mat. 2

bII = bI n

n = E2/E1

y

y

z z

ε1 (E1)

σ1

σ2

σ2 σ2 = n σ1

fig 3.41a) – d): determination of stress for composite cross sections

29




example 3.4 A beam consists of a composite cross section of timber and steel. Determine the maximum stresses that develop in each of the materials under the given internal bending moment.

given:

Mz = 30 kNm ET║ = 10000 N/mm2; ES = 200000 N/mm2

choice: transformation into equal section of timber

ratio 02EEn

T

s ==

transformed dimension: cm3001520bnb III === new centroidal axis (from top):

∑∑ ⋅

=i

ii

AyA

y : cm18.330012515300)25.5(125)12.5(15y =

⋅+⋅⋅+⋅

=

moment of inertia:

)yA(II 2iii∑ ⋅+= : 2

32

3

z 2.7)3001(12

13008.5)2515(12

2515I ⋅+⋅

+⋅+⋅

=

4z cm47723I =

stress timber (top):

24maxt, kN/cm1.15

cm47723cm18.3kNcm30000

IcM

σ −=⋅

−=−=

stress steel (bottom):

24Imaxs, kN/cm9.68

cm47723cm7.7kNcm30000

20σnσ =⋅

=⋅=

timber

steel

150

250

10

250

10

150

3000

183

77

58

72

30

Strength of Materials 4 Shear in Beams 1/7



4 Shear Stress in Beams Situation: 3 vertically allocated beams, not fixed along the joint surfaces Pure bending, no shear internal load – force diagram: Bending with shear internal load – force diagram:

M

+M

V 0

M

Mmax=Pl²/4

V +P/2 -P/2

P

σ

M M σ

31




Derivation of Shearing Stresses in a Beam Shear Flow (along the horizontal plane y in the direction of x)

∑Fx = 0 : ΔH + 0dA)( B

A

A =⋅σ−σ∫Δ

I

My=σ

ΔH = dAyIMM CB ∫ ⋅−

Q = yAdAy ⋅Δ=⋅∫

Statical Moment

ΔH = QIMΔ [kN]

q = dxHΔ

= IQ

xMΔΔ [kN/cm] V

dxdM =

lim Δx→0 q =I

VQ [kN/cm] Shear Flow per unit length

equilibrium at particle

ΔH

= MB = MA

σA σB

p(x)

VA VB

segment of beam subjected to bending

y

y

z ΔA

¯ y

x

Δx

32




example 4.1 A beam consists of 3 wooden planks being fixed by nails. Determine the shear force per nail. given: nail spacing e = 25 mm Vd = 500 N Iz = 1620 cm4 Q = ( ) ³cm1206102yA =⋅⋅=⋅Δ

q = cm/N04.37cm1620

³cm120N500I

VQ4 =

⋅=

F = Nail/N6.92cm5.2cm/N04.37 =⋅ Shearing Stress Formula

τ =tq

ItVQ = [kN/cm²] Shear Stress per area

τyx = τxy equal shearing stresses on mutually

perpendicular planes design: τmax,d ≤ τallowed

Δx

t τ

ΔH

at particle

τyx

τxy

y

20 100

100

20

20

y

z ¯

ΔA

33




approximation: τave = webAV

Distribution of Shear Stress example: Rectangular Cross Section

τ (y1) =⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛==⋅=⋅= ∫∫

Δ

²y2h

I2V

2²y

IVdyyb

IbVdAy

IbV

ItVQ

1

22/h

1y

2/h

1yA

distribution parabolically

VdAA

=⋅τ∫ → τave = AV

min τ at y1 = h/2 → τmin = 0

max τ at y1 = 0 → τmax = A2V3

I8²Vh =

for narrow rectangular cross sections example: Shear stress distribution in an I-beam standard section

y1 h

b

y

z ΔA

τmax

τave

h

y

τ = tq q =

IVQ⋅

y t

34




Shear Flow (along the vertical plane z in the direction of x) regarding a particle with an arbitrary curved cutting surface

∑Fx = 0 : q =I

VQ [kN/cm] Shear Flow per unit length

conclusion: shear flow along the vertical plane z will be derived equivalently to the shear flow along the horizontal plane x Shearing Stress (along the vertical plane z in the direction of x)

τ =tq

ItVQ = [kN/cm²] Shear Stress per area

τzy = τxz equal shearing stresses on mutually

perpendicular planes

y

y

z ΔA

¯

τzx τxz

at particle

35




example 4.2 Determine the shear stress along the z-plane of the welding seams in the edges of the rectangular cross section. given: h = 120 mm b = 60 mm t = 5 mm Vd = 10.0 kN

Iz = 4cm4.30912

³11512

³126=

⋅−

⋅

Q = ³cm38.1475.50.55.0yA =⋅⋅=⋅Δ

q = cm/kN465.0cm4.309

³cm38.14kN0.10I

VQ4 =

⋅=

(q is the total shear flow acting on the particle with two cutting planes)

τ ²cm/kN465.0cm5.0

cm/kN465.021

tq

21

===

Shearing Stress in thin walled members variation and orientation of the shear flow q on a member subjected to a vertical shear force V:

t h

b

y

z

y

z y ¯

ΔA

V

q

V

q

V

q

36




Shear Center Thin walled members in unsymmetric loading problem: applying the vertical force V perpendicular to the centroidal axis causes a moment of torsion (twisting) around the x-axis of the member solution: applying the vertical force V at a certain distance to web center axis of

the member, the shear center equilibrium: 0HHFz =−=∑ 0VVFy =−=∑

0eVhHMx =⋅−⋅=∑ V

hHe ⋅=⇒

I

VQq = dsqHb

0∫ ⋅= dsqV

h

0∫ ⋅=

example 4.3: channel member (see figure above) determination of shear center:

IVQq =

2htsyAQ ⋅⋅=⋅=

I2htsVq

⋅⋅⋅⋅

=⇒

I4²bhtV²s

21

I2htVds

I2htsVdsqH

b

0

b

0

b

0 ⋅⋅⋅⋅

=⋅⋅⋅

=⋅⋅⋅⋅⋅

=⋅= ∫∫

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛⋅⋅+

⋅+

⋅=⋅+= ∑

2

iii2htb

12³tb2

12³ht²yAII t³ is very small, will be neglected

( )b6h12

²ht²htb21

12³htI +

⋅=⋅⋅+

⋅=

( ) b6h²b3

b6h²htV412²b²htV

VI4²b²htV

VhHe

+=

+⋅⋅⋅⋅⋅⋅⋅⋅

=⋅⋅⋅⋅⋅

=⋅

=⇒

external load (action) = internal load (reaction)

= q V

H

H

h

e

V

y

z

s b

t

37

Strength of Materials 5 Torsion 1/8



5 Torsion

5.1 Torsion of Circular Shafts Equilibrium equations at particle: ∑Mx = 0 Summation of Moments about the axis of the member (torque) internal moment = external moment

derivation of the torsion formula:

dA²

cdA

cT

A

max

A

max ⋅ρ⋅τ

=⎟⎠⎞

⎜⎝⎛ ρ⋅⋅τρ

= ∫∫

dA²IA

p ⋅ρ= ∫ [cm4]

x

Mx=T

Φ

L

geometric assumptions for the derivation of the shear formula on a circular shaft member with a torque applied:

• a plane section perpendicular to the axis of the member remains plane • within the elastic limit, Hooke’s Law is applied thus shear stress and strain

(corresponding to the angle of twist Φ) are in proportional correlation

• shearing strains (thus shearing stress) vary linearly from the central axis

• parallel planes perpendicular to the axis of the member remain in a constant distance (L)

T = torsional moment, torque external moment = internal moment polar moment of inertia (constant property of cross-sectional area)

dA

τmax

ρ

c

maxcτ

ρ=τ

38




example 5.1 A cantilever element with a hollow cross section with is subjected to 3 torques. Determine the maximum shearing stress. cross section: Ø = 120 mm, t = 6 mm material: steel G = 81 000 N/mm²

( ) ( ) 44444p cm10.7004.56

2bc

2I =−

π=−

π=

²cm/kN85.12cm1.700

cm6kNcm1500I

cT4

pmax =

⋅=

⋅=τ

polar moment of inertia Ip for a solid circular member:

32d

2c

42d³2dA²I

44c

0

c

0

4

A

pπ

=π

=ρ

π=ρ⋅πρ=⋅ρ= ∫∫

for a hollow circular member:

( )44c

b

c

b

4

A

p bc24

2d³2dA²I −π

=ρ

π=ρ⋅πρ=⋅ρ= ∫∫

t³c2Ip π≈ (for b≈ c)

τmax

b

c t

τmax c

elastic torsion formula

pmax

IcT ⋅

=τ maximum shear stress

maxp cI

Tτ

ρ=

ρ⋅=τ general shear stress

M1,d=15 kNm

M2,d=20 kNm

M3,d=10 kNm

1.0 m 2.0 m 2.0 m

M [kNm]

15

-5

5

39




Angle of Twist by Hooke’s Law for shearing strain: G⋅γ=τ

p

maxI

cT ⋅=τ

GIcT

G p

maxmax

⋅

⋅=

τ=γ

by geometry: cx ⋅ΔΦ=Δ⋅γ angles measured in radians [rad]

cxGIcT

p⋅ΔΦ=Δ⋅

⋅⋅

GIxT

p ⋅

Δ⋅=ΔΦ

example 5.2 Determine the rotation at the free end of the cantilever element of example 5.1.

GILT

p ⋅

⋅=Φ

ΔΦ

Δx γ

c

[ ] [ ]°Φ=π

⋅Φ2360rad

²cm/kN8100cm1.700cm100kNcm500cm200kNcm500cm200kNcm1500

GILT

4p ⋅

⋅+⋅−⋅=

⋅=Φ ∑

⋅

°==Φ 53.2rad044.0

40




y

z q ds

ds

0 r

ds

r

5.2 Torsion of Thin-walled Members (closed sections) Hollow section of arbitrary shape with a varying wall thickness ti. assumption: shear stresses are evenly distributed across the wall thickness precondition: shear stresses on mutually perpendicular planes are equal τxy = τyx ; τxz = τzx dxtF 111 ⋅⋅τ= dxtF 222 ⋅⋅τ= 0Fx =∑ : 21 FF = q = shear flow (shear force per meter 2211 tt ⋅τ=⋅τ⇒ = q of the perimeter) q = equal on all cutting planes of the respective element ∫ ∫ ⋅⋅=⋅= dsrqdsrqT T = torque r ds = 2 Atriangle ∫ ⋅=⋅⇒ A2dsr

) =A

) area to center line of perimeter

qA2T ⋅⋅=

A2

Tq )⋅

=

A)

dx

t1

t2

τ1 τ1

τ2

τ2

F1

F2 particle

tA2T⋅⋅

=τ ) min

max tA2T⋅⋅

=τ ) shear stress

pIGLT⋅⋅

=Φ ( )∫⋅

=

tdsA2I

2

p

)

angle of twist

41




example 5.3 The beam of a bridge structure consists of a hollow trapezium section. The structure is subjected to an unsymmetrical load situation causing a torque. Determinate the maximum shear stress due to torsion such as the maximum angle of twist at the free end. T = - b F constant torsional moment along L

( ) ²b23bbb2

21A =+=

)

tmin = t :

tb3F

t²b3Fb

tA2T

max ⋅⋅=

⋅⋅⋅

=⋅⋅

=τ ) additional shear stress due to bending

is neglected in this example

( )52t³b9

tb5

212

tb

t2b2

²b232

tdsA2I

2

2

p+

⋅⋅=

⋅++

⎟⎠⎞

⎜⎝⎛ ⋅

=⋅

=

∫

)

( )

t²bGLF

952

t³b9G52LFb

IGLT

p ⋅⋅⋅+

=⋅⋅⋅+⋅⋅

=⋅⋅

=Φ

F

L t t

2t b

b

2b F

42




Distribution of shear stress in a thin-walled rectangular member subjected to a torque:

5.3 Torsion of Noncircular Members (open sections)

Distribution of shear stress in a rectangular member subjected to a torque:

τ

L

τmax

t

p

maxmax I

tT ⋅=τ shear stress formula

∑⋅≈3iip th

31I polar moment of inertia

L L

Members of same behaviour towards an applied torque:

wall thickness : t

43




example 5.4 Compare the torsional strength and stiffness of a thin-walled tube of circular cross section with and without a longitudinal slot. a) closed section:

t³R2t³c2Ip π=π≈

t²R2T

IcT

pmax

π=

⋅=τ

b) open section (slotted): L = 2πR

³Rt32th

31I 3

iip π=⋅∑≈

²Rt2T3

ItTp

maxmax π

=⋅

=τ

ratio of shear stress a) to b):

1 : tR3

(ratio of maxτ )

ratio of stiffness a) to b)

1 : 3

Rt

31

⎟⎠⎞

⎜⎝⎛

(ratio of pI )

R t

44




r

table 5.1: summery of formulas related to torsion Torsion Shear Stress

Angle of Twist Polar Moment of Inertia

Circular Shafts

Thin Walled Members (closed sections)

Non Circular Members (open sections)

[ ] [ ]°Φ=π

⋅Φ2360rad

pmax

IcT ⋅

=τ

maxp cI

Tτ

ρ=

ρ⋅=τ

pIGLT⋅⋅

=Φ

minmax tA2

T⋅⋅

=τ )

tA2T⋅⋅

=τ )

τmax

b

c t

( )44p bc

2I −

π=

t³c2Ip π≈ (for b≈ c)

τmax c

ρ

pmax

IcT ⋅

=τ

maxp cI

Tτ

ρ=

ρ⋅=τ

pIGLT⋅⋅

=Φ

32d

2cI

44

pπ

=π

=

p

maxmax I

tT ⋅=τ

pItT ⋅

=τ

∑⋅≈3iip th

31I

pIGLT⋅⋅

=Φ

h h

wall thickness : t

A)

( )∫⋅

=

tdsA2I

2

p

)

pIGLT⋅⋅

=Φ

t

45

Strength of Materials 6 Stress Analysis 1/10



6 Stress Analysis

6.1 Compound Stresses, Superposition of Stresses

Summery of formulas for the stress analysis (linear-elastic):

• normal stresses due to axial force:

AFσx =

• normal stresses due to bending:

z

z

z

zx

SM

IyM

−=⋅

−=σ

y

y

y

yx

SM

IzM=

⋅=σ

• shearing stresses due to shear force in a beam:

tIQV

y

yyxy

⋅⋅

=τ

tIQV

z

zzxz

⋅⋅

=τ

• shearing stresses due to torque:

pI

T ρ⋅=τ circular shafts

tA2

T⋅⋅

=τ ) closed thin walled members

σ(-)

My (+)

z

y

x

’tensile fibre for Mz’ Mz (+)

My(+)

convention for algebraic sign ±

z(+)

σ(-) σ(+)

y(+)

σ(+)

Mz(+)

46




Superposition of stresses

• considering single, individual load situations for each internal load reaction • summation of stresses due to the algebraic sign convention (±) • maximum and minimum stresses are found at the respective positions of a

cross section Limitation of superposition

• considering internal reaction force results according to Theory 1st Order neglects the effect on internal reaction forces caused by deflection (e.g. beam subjected to bending plus axial force)

Superposition of normal stresses design analysis

y

y

z

zx

IzM

IyM

AFσ ⋅±

+⋅±

−±= allowmax σ≤σ

Superposition of shearing stresses design analysis TV τ±τ±=τ allowmax τ≤τ Special problems concerning combined loading of bending moment and axial force how to avoid tension (open gap) in a member with an eccentric load (e.g. dam, masonry wall):

+

P e

=

σ(-)

σ(-)

σ(-)

σ(+)

condition: ( ) 0²hb6eP

hbP

SM

AP

=⋅⋅⋅

+⋅

−=+−=σ

6he =⇒

3h 3

b b

h

zone of applicable resultant force to meet the condition

47




example 6.1 A rectangular beam member is subjected to unsymmetrical bending and an eccentric compressive force. Determinate the maximum and minimum normal stresses within the section at the fixed support and indicate the position of the neutral plane. b = 50 mm h = 100 mm L = 1000 mm φ = 30° M = 1 kNm F = 10 kN

443

y cm17.104cm12

510I =⋅

=

443

z cm67.416cm12105I =⋅

=

kNcm50M21MsinM y ==⋅ϕ=

136.6kNcm5cmFM321

2hFMcosM z −=⋅−⋅−=−⋅−= ϕ

y

y

z

zx

IzM

IyM

AFσ ⋅±

+⋅±

−±=

A: ²cm/kN64.2cm17.104

cm5.2kNcm50cm67.416

cm5kNcm6.136²cm50

kN1044x =

⋅+

⋅−−−=σ

B: ²cm/kN24.0cm17.104

cm5.2kNcm50cm67.416

cm5kNcm6.136²cm50

kN1044x =

−⋅+

⋅−−−=σ

C: ²cm/kN64.0cm17.104

cm5.2kNcm50cm67.416

cm5kNcm6.136²cm50

kN1044x −=

⋅+

−⋅−−−=σ

D: ²cm/kN04.3cm17.104

cm5.2kNcm50cm67.416

cm5kNcm6.136²cm50

kN1044x −=

−⋅+

−⋅−−−=σ

position of neutral plane, graphical solution:

x z

y

φ M

F L

b

h

A B

C D

+

+

- -

A B

C D

z

y

48




example 6.2 The beam of a bridge structure consists of a hollow trapezium section. The structure is subjected to an unsymmetrical load by an applied vertical force F. Determinate the maximum shear stress due to the applied load. shear stress due to torque: T = - b F constant torsional moment along L

( ) ²b23bbb2

21A =+=

)

tmin = t :

tb3F

t²b3Fb

tA2T

max,T ⋅⋅−=

⋅⋅⋅

−=⋅⋅

−=τ )

shear stress due to bending: largest shear stress occurs in the web

tb5F

tb252

FA

V

webave,V

⋅⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅⋅

==τ approximation formula

superposition:

78.0tb

Ftb5

Ftb3

FVTmax ⋅

⋅=

⋅⋅+

⋅⋅=τ±τ±=τ

F

L t t

2t b

b

2b F

shear flow due to torsion T(+) (q evenly distributed)

shear flow due to applied shear force V(+) (maximum q in the web)

+ maxτ

49




6.2 Analysis of plane stress Transformation of plane stress plane stress = stress components in all directions (σx, σy,τxy, τxy) of an isolated element (dx, dy, t) Unlike the vector of a force (F [kN]), a stress vector (σ,τ [kN/cm²]) is to be multiplied by the respective area (dA) of a face to be applicable to mathematic operations (addition, subtraction, multiplication). Equations of equilibrium on an isolated wedge of an infinitesimal element (dx, dy, t): definition: area of the inclined plane BC = dA, area AB = sinθ dA, area AC = cosθ dA

∑ = 0F 'x : =⋅σ dA'x ∑ = 0F 'y : =⋅τ dA'y'x applying the same procedure to an inclined plane at an angle of θ+π/2, the normal stress σy' is derived.

Using the correlations xyxy τ=τ , ( )θ+=θ 2cos121²cos , ( )θ−=θ 2cos1

21²sin ,

θ=θθ 2sincossin2 , θ=θ−θ 2cos²sin²cos the equations for the transformation of plane stress are obtained:

σy

σx

y

x

τxy τyx

+

+

σy

σx

y

x

τxy

τyx

θ y

θ

σy

σx

τyx

τx'y' σx' x'

y'

τxy

x A B

C

( ) ( )( ) ( ) θθ⋅τ+θθ⋅σ+

θθ⋅τ+θθ⋅σ

cossindAsinsindA

sincosdAcoscosdA

yxy

xyx

( ) ( )( ) ( ) θθ⋅τ−θθ⋅σ+

θθ⋅τ+θθ⋅σ−

sinsindAcossindA

coscosdAsincosdA

yxy

xyx

50




adding equation (6.1) and (6.2): = constant Principal Stresses The plane of maximum and minimum normal stress is found by differentiating the equations for transformation (6.1) with respect to θ and equalizing the derivative set to zero:

( ) 02cos22sin221

d xyyx'x =θτ+θσ−σ−=θ

σ (6.4) hence

(6.5) Both angels of incline, θ1 and θ1 + π/2, meeting above condition are denoted by the principal directions indicating the principal planes. Applying the angle functions the principal stresses are simplified: (6.6)

σy

σx

y

x

τxy τyx

θ

τy'x' τx'y'

σy'

σx’

x'

y'

equations for the transformation of plane stress

( ) ( ) θτ+θσ−σ+σ+σ=σ 2sin2cos21

21

xyyxyx'x (6.1)

( ) ( ) θτ−θσ−σ−σ+σ=σ 2sin2cos21

21

xyyxyx'y (6.2)

( ) θτ+θσ−σ−=τ 2cos2sin21

xyyx'y'x (6.3)

yx'y'x σ+σ=σ+σ

yx

xy1

22tan

σ−σ

τ=θ

2xy

2yxyx

2,1 22τ+⎟⎟

⎠

⎞⎜⎜⎝

⎛ σ−σ±

σ+σ=σ

51




Equation (6.4) is equal to equation (6.3). Since the principal directions (θ1) are obtained by equalizing equation (6.4) to zero it is concluded that: on planes on which maximum and minimum normal stresses occur (principal stresses), no shearing stresses are existent (σ1 = max, σ2 = min, τx'y' = 0). Maximum Shearing Stresses Differentiating equation 5.3 and equalizing the derivate to zero:

( ) 02sin22cosd xyyx

'y'x =θτ+θσ−σ−=θ

τ (6.7) hence

(6.8) or with (6.6): (6.9) substitution of (6.8) into (6.1) or (6.2):

( )yx21' σ+σ=σ (6.10)

Thus maximum shearing stresses occur on planes that are not necessarily free of normal stress, σ' = σx' = σy'.

1

22tan

12tanθ

−=θ

hence directions of 2θ2 and 2θ1 are perpendicular, or directions of maximum normal (θ1) and maximum shearing stresses (θ2) are 45° apart. example: a state of pure shear can be transformed in a state of equal but opposite

principle normal stresses under an incline of θ = 45°

xy

yx2

22tan

τ

σ−σ−=θ

2xy

2yx

max 2τ+⎟⎟

⎠

⎞⎜⎜⎝

⎛ σ−σ±=τ

( )21max 21

σ−σ±=τ

τxy

τyx σ1 = τxy σ2 = -τxy

52




6.3 Mohr’s Circle Equations (6.1) or (6.2) and (6.3) can be represented graphically. Transforming the equations:

( ) ( ) θτ+θσ−σ+=σ+σ−σ 2sin2cos21

21

xyyxyx'x

( ) θτ+θσ−σ−=τ 2cos2sin21

xyyx'y'x

squaring and adding both equations and simplifying:

( ) 2xy

2yx2

'y'x

2

yx'x 221

τ+⎟⎟⎠

⎞⎜⎜⎝

⎛ σ−σ=τ+⎥⎦

⎤⎢⎣⎡ σ+σ−σ (6.11)

since σx, σy and τxy are given constants in a problem they are summarised as:

2xy

2yx

2²r τ+⎟⎟

⎠

⎞⎜⎜⎝

⎛ σ−σ=

with (5.10), ( )yx21' σ+σ=σ , equation (5.11) is written as:

( ) ²r²' 2'y'x'x =τ+σ−σ (6.12)

Equation (5.12) is representing a circle or radius r in the σ,τ-plane, having its center at (σ',0). The ordinate of a point on the circle is the shearing stress τx'y', the abscissa is the normal stress σx'. The circle is called Mohr’s Circle of stress. The state of stress under any arbitrary plane of incline is represented graphically. Constructing Mohr’s Circle of Stress with σx, σy and τxy as given values:

• set up a coordinate system, horizontal axis = σ, vertical axis = τ • plot both stresses for σx and σy on the σ-axis respecting the algebraic sign (+/-) • plot the shearing stresses τxy using the opposite sign at σx (e.g. (-), below the

σ-axis, for τxy being positive) and the correct sign at σy (e.g. (+) for for τxy (+)) • connect both points by a straight line, the point of intersection with the

abscissa is the center of the circle, now the circle can be drawn The state of plane stress of an element is represented by the drawn circle. Any plane of incline is represented by a point on the circle. The angle of incline of the respected plane towards the initial x,y-system is equal to half of the value of the counter-clockwise rotation (2θ) on the circle.

( )yx21' σ+σ=σ

σ

τ

(+) σx

σy (-)

σ' 2θ

(+) τxy

τxy (+)

53




Special states of stress on Mohr’s Circle: a) state of axial tension σ' = σ0/2 τmax = σ0/2 b) state of pure shear σx = τ0 σy= -τ0 c) hydrostatic state of stress σx = σy = σ' τ = 0

( )yx21' σ+σ=σ

b)

σy= -τ0

45°

σx= τ0 τ0

σ

τ

σx= τ0 σy= -τ0

σ'

τ0

a)

σ

τ

σx = σ0 (=σ1) σy = 0

σ'

τmax

σ0

σ' = σ0/2

τmax = σ0/2

45°

σ'

θ

σ'

σ

τ

σ'

c) σ'

σ'

54




6.4 Failure Theories A member of a certain material subjected to an axial load can be easily tested in a tensile test. Thus the yield load and the ultimate load can be obtained. An allowed tension is defined to ensure a safe design. For any member in an bi- or tri-axial state of stress determining its load capacity and the parameter for a safe design is not that easy. A material specimen in a tensile test might well break along its shearing plane (45° incline, τmax = σ/2) and not along the plane of normal stress (perpendicular). Materials that are weak in shearing strength are expected to fail along the shearing planes (45° in pure tension or compression, 90° in pure torsion). Examples for materials showing such behaviour: mild steel, concrete or loam (in compression). Materials being weak in tensile strength will fail along the planes of normal stress (90° in pure tension or compression, 45° in pure torsion). Examples for materials showing such behaviour: sandstone, chalk. To enable a safe design, a certain stress limit condition has to be defined to be compared to the allowable stress being obtained by the tensile test. Only the maximum distortion energy theory will be mentioned here without indicating its derivation. It is denoted as the von Mises yield condition:

allow2xyyx

2y

2x 3 σ≤τ+σσ−σ+σ

Principle stress trajectories and crack pattern for a rectangular beam subjected to bending:

45°

tension

compression

F F

55

Strength of Materials 7 Deflection of Beams 1/4



7 Deflection of Beams

7.1 Sequence of Equations for the Deflection of elastic Beams Geometric relations, static and equilibrium conditions are taken into account to set up a sequence of equations. Recalling the image and relations for small deflections

geometry: ρc

dxδxε maxx, ==

ρ

=εy

x (7.1)

since ε⋅=⋅

=σ EI

yM,

IEyM⋅⋅

=ε (7.2)

(6.1)=(6.2): κ=⋅

=ρ IE

M1 curvature

due to analytic geometry, curvature is defined as:

( )[ ]232'v1

''v1

+=

ρ v' = θ = slope - since v' is very small:

v = deflection - due to M being

IEM

²dxv²d''v1

⋅===

ρ positive if oriented downward, hence:

IEM''v⋅

−=

considering small deflections only:

θ−='v

A

A1 B1

initial shape

M M

deflected shape

v

B

initial shape

deflected shape

neutral axis

dx+δx dx

c y y

ρ

x

initial shape

deflected shape v

x

v'

-θ

56




Recalling the static differential relations

VdxdM'M == and )x(q

dxdV'V −==

summery:

Sequence of equations for the deflection of beams (linear-elastic): v(x) deflection of elastic curve θ(x) = - v'(x) slope of elastic curve M(x) = - v''(x)EI = κ(x) EI = θ'(x)EI moment – curvature relation V(x) = - v'''(x)EI = M'(x) shear force q(x) = v'v(x)EI = V'(x) lateral load per unit length

57




7.2 Table of Deflections and Slopes for common situations

L

q

L

P

L/2 L/2

L

P

a b

P

L

L

q

system & load max deflection v slope at end θ

EI384qL5)2/L(vv

4

max ⋅⋅

==

EI3³PL)L(vvmax ⋅

==

EI8qL)L(vv

4

max ⋅==

( )[ ]³xx²b²LEIL6

Pbvmax

−−

⋅⋅

=

EI48PL)2/L(vv

3

max ⋅== EI16

²PL)L()0(⋅

=θ−=θ

EI24³qL)L()0(⋅

=θ−=θ

EI2²PL)L(

⋅=θ

EI6³qL)L(

⋅=θ

58




example 7.1 A steel beam, consisting of a standard I-section is subjected to a dead load of g = 8 kN/m² and a traffic load of p = 5 kN/m². The spacing between the beams is 3.0 m. The total allowed deflection equals to L/300.

a) Determine whether the maximum vertical deflection of the beam is within the allowed range

b) Determine the vertical camber to be applied to the beam to achieve a plane system due to pure self weight of the structure. Does the system now meet the required criterion?

c) The camber of the beam is to be replaced by a suspension cable at the centre of the system. Determine the pretension force in the cable.

given: L = 5.0 m, Iz = 5740 cm4, Esteel = 21 000 kN/cm²

a) gbeam = 8 * 3.0 = 24.0 kN/m linear force

kNm63.158

²gLM g == bending moment

cm62.1cm5740²cm/kN21000384

)cm500(cm/kN10245EI384

gL5v 4

424

gmax, =⋅⋅⋅⋅⋅

=⋅

⋅=

−

deflection

pbeam = 5 * 3.0 = 15.0 kN/m

kNm0.258

²pLM p ==

cm01.1cm5740²cm/kN21000384

)cm500(cm/kN10155EI384

pL5v 4

424

pmax, =⋅⋅⋅⋅⋅

=⋅

⋅=

−

cm67.1300/500vcm63.2vvv allowpmax,gmax,max ==>=+= not sufficient!

b) height of camber = =gmax,v 1.62cm cm67.1vcm01.1vvvv allowpmax,gmax,gmax,max =<=++−= OK

c) gmax,

3

Pmax, vcm62.1EI48

PLv ==⋅

= deflection due to force at L/2

cm62.1)cm500(

cm5740²cm/kN2100048vL

EI48P 3

4

3

⋅⋅=

⋅=

kN98.74P = pretension force

L

59

Strength of Materials 8 Buckling 1/5



8 Buckling Buckling is a sudden loss of stability that occurs to a member subjected to a compressive load. The system failure is caused by infinitesimal small deflections due to small imperfections being inherent in each structure. It relates to the geometry of the system (dimensions, boundary conditions, type of cross section) and the material applied (elastic modulus).

8.1 Stability of Equilibrium A vertical rigid bar (no bending) having a torsional spring of stiffness k at its support is subjected to a vertical load P. The system is displaced by a small (infinitesimal) amount.

P L sinθ ≈ P L θ moment caused by vertical force (small displacement)

k θ restoring moment by torsional spring

k θ = P L θ neutral stability - equilibrium condition

k θ > P L θ stable

k θ < P L θ unstable Pcr = k/L critical buckling load

8.2 Euler Formula for the pin-ended column A column with a flexural rigidity of EI with pinned supports, being free to rotate about both ends is subjected to a vertical load P. An imperfection of the system causes bending of the column (M) and horizontal deflection at its centre (v). M = -P v bending moment due to deflection

EI

vPEIM''v ⋅

=−= differential equation for the elastic curve

EIP² =λ applying λ, transforming the equation:

0v²''v =λ+ being equal to an equation for simple

harmonic motion having the solution:

xcosBxsinAv λ+λ=

L

P P

θ

L

P

v

60




v(0) = v(L) = 0 A and B are determined from the boundary condition

v(0) = 0 = A sin 0 + B cos 0 hence B = 0 and v(L) = 0 = A sin λL excluding the solution A = 0, the equation is satisfied by λL = nπ hence π=⋅ nL)EI/(P and

²LEI²Pcr

π= Euler Formula, critical load for the pin-ended column

The Euler Formula for the pin-ended column is the fundamental case for the buckling analysis. Buckling will take place in direction of the least moment of inertia of the respective cross section.

8.3 Euler Formulas for various boundary conditions Due to the boundary conditions set for a member subjected to a compressive load, the differential equation v'' = -M/(EI) has different solutions. The solutions can be generalised and transformed to resemble the fundamental case of the Euler Formula for pin-ended columns by introducing Le as the effective length. In the analysis the effective length Le is used instead of the actual column length L.

Le = KL K = effective length factor ²L

EI²Pe

crπ

=

L = Le

P

Le = 2L

P

L Le = 0.7L

P

L Le = 0.5L

P

L

1 2 3 4

K = 2

²L4EI²Pcr

π=

K = 1

²LEI²Pcr

π=

K = 0.7

²LEI²2Pcr

π=

K = 0.5

²LEI²4Pcr

π=

61




example 8.1 A pin-ended steel column consists of a hollow rectangular cross section. At its top end the column is supported by horizontal bracings in y- and z-directions. At a height of 6.0 m a horizontal bracing is attached in direction of the z-axis.

a) Determine the critical buckling load Pcr of the system b) The support at the bottom of the column is changed to a fixed support.

Determine Pcr of the new system. given: rectangular hollow section 200 x 100 x 5 [mm]

Iz = 1522.42 cm4, Iy = 512.42 cm4

ESteel = 21000 kN/cm²

a) buckling plane x-y: cm1000Le =

kN5.315)²cm1000(

cm42.1522²cm/kN21000²²L

EI²P4

e

zcr =

⋅⋅π=

π=

buckling plane x-z, upper part:

kN8.663)²cm400(

cm42.512²cm/kN21000²²L

EI²P

4

e

ycr =

⋅⋅π=

π=

buckling plane x-z, lower part: cm600Le =

kN0.295)²cm600(

cm42.512²cm/kN21000²²L

EI²P

4

e

ycr =

⋅⋅π=

π=

min(Pcr) = 295.0 kN

b) buckling plane x-y: cm0.700m0.107.0Le =⋅=

kN0.644)²cm700(

cm42.1522²cm/kN21000²²L

EI²P

4

e

zcr =

⋅⋅π=

π=

buckling plane x-z, upper part: cm400Le =

kN8.663)²cm400(

cm42.512²cm/kN21000²²L

EI²P

4

e

ycr =

⋅⋅π=

π=

buckling plane x-z, lower part: cm420m0.67.0Le =⋅=

kN0.602)²cm420(

cm42.512²cm/kN21000²²L

EI²P

4

e

ycr =

⋅⋅π=

π=

min(Pcr) = 602.0 kN

P

z

y

x

4m

6m

z

y

62




8.4 Limitations of the Euler Formulas The derivation of the Euler Formula is based on elastic material behaviour. Thus it is only applicable within the linear-elastic range of the material. A closer look is to be taken at the stress caused by the applied vertical load of the column.

²LEI²Pe

crπ

= introducing a new definition:

2rAI ⋅= hence:

AIr = radius of gyration, transforming Pcr:

²L

²EAr²Pe

crπ

=

2e

crcr )r/L(

E²AP π

==σ critical stress

r/Le slenderness ratio

Euler Hyperbola: critical stress versus slenderness ratio, applicable within the

linear-elastic range The Euler Hyperbola provides a general solution to determine the critical stress for any column according to its slenderness ratio. For long columns (large ratio of slenderness), the Euler Hyperbola can generally be applied. Exceeding the linear elastic range of the material, the proportional limit is reached, hence the Euler Hyperbola can no longer be used. The graph representing the critical stress therefore approaches the limit stress of the material (e.g. the yield point). Thus short column failure is not a failure due to stability but due to the strength of the applied material. Further criteria considering the buckling analysis of a column of a certain material can be found in the respective national codes.

63




limit stress of material

e.g. σy.p., σult

critical stress σcr

slenderness ratio Le/r intermediate column range

short column range

long column range

buckling elastically buckling plastically

2e

cr )r/L(E²π

=σ Euler Hyperbola

no b

uckl

ing,

mat

eria

l fai

lure

stable design

unstable design

proportional limit

σ

ε

yield point

0

figure 8.1: critical stress diagram, Euler Hyperbola

64

AAU Strength of Materials Lecture Notes 09-05-21

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