AAU Strength of Materials Lecture Notes 09-05-21
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Transcript of 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
Department of Civil Engineering
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
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 1 Stress 2/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 1 Stress 3/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 1 Stress 4/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 1 Stress 5/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 1 Stress 6/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 1 Stress 7/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 1 Stress 8/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 1 Stress 9/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 2/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 3/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 4/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 5/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 6/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 7/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 8/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 9/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 10/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 2 Strain 11/11
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 3 Pure Bending of Beams 2/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 3 Pure Bending of Beams 3/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 3 Pure Bending of Beams 4/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 3 Pure Bending of Beams 5/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 3 Pure Bending of Beams 6/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 3 Pure Bending of Beams 7/9
Department of Civil Engineering
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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
Strength of Materials 3 Pure Bending of Beams 8/9
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 3 Pure Bending of Beams 9/9
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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
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 4 Shear in Beams 2/7
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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
Strength of Materials 4 Shear in Beams 3/7
Department of Civil Engineering
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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
Strength of Materials 4 Shear in Beams 4/7
Department of Civil Engineering
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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
Strength of Materials 4 Shear in Beams 5/7
Department of Civil Engineering
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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
Strength of Materials 4 Shear in Beams 6/7
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 4 Shear in Beams 7/7
Department of Civil Engineering
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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
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 5 Torsion 2/8
Department of Civil Engineering
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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
Strength of Materials 5 Torsion 3/8
Department of Civil Engineering
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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
Strength of Materials 5 Torsion 4/8
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 5 Torsion 5/8
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 5 Torsion 6/8
Department of Civil Engineering
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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
Strength of Materials 5 Torsion 7/8
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 5 Torsion 8/8
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 6 Stress Analysis 2/10
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 6 Stress Analysis 3/10
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 6 Stress Analysis 4/10
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 6 Stress Analysis 5/10
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 6 Stress Analysis 6/10
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 6 Stress Analysis 7/10
Department of Civil Engineering
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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
Strength of Materials 6 Stress Analysis 8/10
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 6 Stress Analysis 9/10
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 6 Stress Analysis 10/10
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 7 Deflection of Beams 2/4
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 7 Deflection of Beams 3/4
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 7 Deflection of Beams 4/4
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 8 Buckling 2/5
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 8 Buckling 3/5
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 8 Buckling 4/5
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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
Strength of Materials 8 Buckling 5/5
Department of Civil Engineering
ADDIS ABABA UNIVERSITY Faculty of Technology material by Karsten Schlesier
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