AAU_strength_of_materials_lecture_notes_09-04-21.pdf

Department of
Civil Engineering
ADDIS ABABA UNIVERSITY
Faculty of Technology
Lecture Notes
Strength of Materials
ceng1002
beta version, 21st April 2009
Karsten Schlesier (Dipl.-Ing.)
for use at Addis Abeba University only!

Strength of Materials 2009
Department of
Civil Engineering
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
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
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
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).
ΔA
ΔF
A
F
=
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

2/9
Department of
Civil Engineering
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).
ΔA
ΔF
lim
σ
0
ΔA→
=
Considering a uniform distribution the normal stress is defined as:
A
F
σ = and dA
σ
F
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?
1
1
A
F
σ = ;
2
2
Α
F
σ = 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

3/9
Department of
Civil Engineering
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).
Deviding 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:
A
F
τ =
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

4/9
Department of
Civil Engineering
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: γ
F
Fd ⋅
= design load = existing load · factor of safety
Α
F
σ d
d = design normal stress, axial loaded
allowed
d σ
σ ≤ design stress ≤ allowable stress
Α
F
τ d
d = design average shear stress
allowed
d τ
τ ≤ design stress ≤ allowable stress
ultimate load
allowable load
5

5/9
Department of
Civil Engineering
example 1.2 - design of axially loaded members
The lattice girder 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:
∑ = 0
MA 50kN
kN
100
12.0m
6.0m
FB =
=
⇒
∑ = 0
V 50kN
F
F B
A =
=
⇒
cutting plane - internal equilibrium at left part:
∑ = 0
M3 100kN
3.0m
6.0m
50kN
S1 −
=
⋅
−
=
⇒
∑ = 0
M2 50kN
3.0m
3.0m
50kN
S3 =
⋅
=
⇒
∑ = 0
V 50kN
sin45
S2 =
⋅
⇒ o
70.71kN
S2 =
⇒
b) 106.06kN
1.5
70.71kN
γ
S
S 2
d
2, =
⋅
=
⋅
=
2
2
2
d
2,
d 18.41kN/cm
cm
(2.4)
106.06kN
Α
S
σ =
=
= 2
allow kN/cm
22
σ =
≤ OK
c) 75kN
1.5
50kN
γ
S
S 3
d
3, =
⋅
=
⋅
=
2
allow
d
3,
d 22kN/cm
σ
Α
S
σ =
≤
= 2
2
allow
d
3,
required 3.41cm
22kN/cm
75kN
σ
S
A =
=
≥
⇒
2.08cm
π
3.41cm
4
d
4
d
π
A
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
1
FA
2
3
4
6

6/9
Department of
Civil Engineering
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:
A
2
F
τ d
d
average,
⋅
=
condition from design analysis:
allow
d
d
average, τ
A
2
F
τ ≤
⋅
=
2
2
allow
d
/
49
.
1
/
6
.
33
2
100
τ
2
F
A cm
kN
cm
kN
kN
=
⋅
=
⋅
≥
⇒
4
d
π
A
2
⋅
= cm
38
.
1
d ≥
⇒
mm
chosen 14
d
: =
⇒
F
fig 1.32: bolt connection
F
A
F/2
F/2
F
7

7/9
Department of
Civil Engineering
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

8/9
Department of
Civil Engineering
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.
A
F
σx = initial situation, normal stress
N = cos φ · F resolution of force F normal force N
V = sin φ · F shear force V
ϕ
ϕ
cos
A
A = area of oblique plane
φ
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
x
A
9

9/9
Department of
Civil Engineering
ϕ
ϕ
2
N cos
A
F
A
N
σ =
=
ϕ
2
x
N cos
σ
σ = normal stress acting on oblique plane
ϕ
ϕ
ϕ
ϕ
ϕ
cos
sin
σ
cos
sin
A
F
A
V
τ x ⋅
=
⋅
=
=
using angle function (2 sinφ cosφ = 2 sinφ):
ϕ
sin2
2
σ
τ x
= shear stress acting on oblique plane
conclusion:
x
N σ
σ
max = o
0
=
ϕ
2
σ
max τ x
= o
45
=
ϕ [ ]
4
π
10

Strength of Materials 2 Strain
1/11
Department of
Civil Engineering
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·ΔL
ΔL/2
F
2A
F
11

2/11
Department of
Civil Engineering
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:
L
ΔL
ε = = 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:
ε
σ
E = ⎥
⎦
⎤
⎢
⎣
⎡
2
cm
kN
A high Modulus of Elasticity therefore represents a hard, rigid material like steel, a
low Modulus of Elasticity a soft, deformable material like rubber.
*
A
F
σa) =
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

3/11
Department of
Civil Engineering
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:
ε
E
σ ⋅
= or
AE
F
E
σ
ε =
=
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):
dx
δx
ε = strain, factor of elongation of finite element
dx
E
A
F
dx
E
σ
dx
ε
ΔL
x
0 x
x
x
x
0 x
x
x
0
x ⋅
=
⋅
=
⋅
= ∫
∫
∫
AE
FL
ΔL =
∑ ⋅
⋅
=
i
i
i
i
E
A
L
F
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

4/11
Department of
Civil Engineering
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
∑ ⋅
⋅
=
i
i
i
i
total
E
A
L
F
ΔL
cm
286
.
0
cm
/
kN
000
7
cm
0
1
cm
00
2
kN
00
1
ΔL 2
2
III =
⋅
⋅
=
cm
057
.
0
cm
/
kN
000
7
cm
0
2
cm
00
1
kN
80
ΔL 2
2
II =
⋅
⋅
=
cm
014
.
0
cm
/
kN
000
7
cm
0
2
cm
00
1
kN
0
2
ΔL 2
2
III =
⋅
⋅
=
cm
357
.
0
014
.
0
057
.
0
286
.
0
ΔLtotal =
+
+
=
100 cm
200 cm 200 cm
I II III
F1=60kN F2=20kN F3=100kN
fig. 2.14
14

5/11
Department of
Civil Engineering
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. 2.21shows 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
developes 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
Alumium: 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

6/11
Department of
Civil Engineering
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
AE
PL
ΔLP −
= compression of rod caused by a negative support force
ΔLT = ΔLP
αT · ΔT · L =
AE
PL
P = - αT · ΔT·AE
σ = - αT · ΔT·E
ΔL
PB (-)
PA (-)
L
fig. 2.31
16

7/11
Department of
Civil Engineering
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
AE
PL
ΔL i
=
m
0
.
10
AE
P
1
Li =
⎟
⎠
⎞
⎜
⎝
⎛
+
⇒ mm
10000
mm
/
N
170000
45mm
10000N
1
L 2
2
i =
⎟
⎠
⎞
⎜
⎝
⎛
⋅
+
⇒
mm
95
.
9986
Li =
⇒ 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

8/11
Department of
Civil Engineering
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
E
σ
E
σ
E
σ
ε z
y
x
x ν
ν −
−
=
E
σ
E
σ
E
σ
ε z
y
x
y ν
ν −
+
−
=
E
σ
E
σ
E
σ
ε z
y
x
z +
−
−
= ν
ν
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

9/11
Department of
Civil Engineering
example 2.4
A solid rectangular steel block is subjected to uniform pressure acting along its
surface. If edge AB is expanded 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 = ( )
ν
ν
ν 2
1
E
p
E
p
E
p
E
p
−
−
=
+
+
− uniform strain
εx = 4
-2
10
3
40mm
mm
10
1.2
AB
x −
⋅
−
=
⋅
−
=
Δ
− = εy = εz strain in x-, y- and z-direction
mm
mm
BC
y y
3
4
10
6
20
10
3 −
−
⋅
−
=
⋅
⋅
−
=
⋅
=
Δ ε total compression in y-direction
mm
mm
BD
z z
3
4
10
9
30
10
3 −
−
⋅
−
=
⋅
⋅
−
=
⋅
=
Δ ε total compression in z-direction
b) εx = ( )
ν
2
1
E
p
−
−
29
.
0
2
1
/
210000
)
10
3
(
2
1
2
4
x
⋅
−
⋅
⋅
−
−
=
−
⋅
−
=
⇒
−
mm
N
E
p
ν
ε
2
/
150 mm
N
p −
= (= -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

10/11
Department of
Civil Engineering
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: ∑ = 0
F : xy
xy τ
τ =
∑ = 0
M0 : yx
xy τ
τ =
( ) ( ) dy
dz
dx
τ
dx
dz
dy
τ yx
xy ⋅
⋅
=
⋅
⋅
area · lever
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 5
fig. 2.51: shearing stress acting
on a finite element
dx
y
x
z
dz
dy
τxy
τyx
τxy
τyx
20

11/11
Department of
Civil Engineering
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
Elastic Modulus:
γ
τ
=
G ⎥
⎦
⎤
⎢
⎣
⎡
2
cm
kN
The material properties E, G and ν are not independent. At this stage the correlation
shall be given without derivation as:
( )
ν
+
=
1
2
E
G
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
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

2/9
Department of
Civil Engineering
Equations of equilibrium concerning σx:
∑ = 0
Fx : 0
dA
σx =
⋅
∫
∑ = 0
My : 0
dA
σ
z x =
⋅
⋅
∫
∑ = 0
Mz : z
M
dA
σ
y
- 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:
ρ
c
dx
δx
ε max
x, =
= (similar triangles)
max
x,
x ε
c
y
ρ
y
ε =
=
(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

3/9
Department of
Civil Engineering
3.3 Normal Stress
Using the proportional correlation of stress and strain:
max
x,
x σ
c
y
σ =
(3.13) in (3.11):
∑ = 0
Fx : 0
dA
y
c
σ
dA
σ
c
y
dA
σ max
max
x ∫
∫
∫ =
⋅
=
⋅
=
⋅
∫ dA
y : first moment of cross section (statical moment)
→ about the neutral axis =0
→ neutral axis = centroidal axis
∑ = 0
Mz : z
2
max
x M
dA
y
c
σ
dA
σ
y =
⋅
=
⋅
⋅ ∫
∫
z
2
max
M
dA
y
c
σ
=
⋅
∫
dA
y
I 2
⋅
= ∫ : second moment of cross section (moment of inertia)
Transformation of (3.14):
z
z
max
I
c
M
σ =
(3.13) in (3.15):
z
z
x
I
y
M
σ =
Introducing:
c
I
S = elastic section modulus
(3.15) becomes:
z
z
max
S
M
σ =
since
I
y
M
ε
E
σ
⋅
=
⋅
= →
I
E
y
M
⋅
⋅
=
ε
recalling (3.12):
ρ
y
ε =
in (3.18): κ
=
⋅
=
ρ I
E
M
1
(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

4/9
Department of
Civil Engineering
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

5/9
Department of
Civil Engineering
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.
( ) h
3
2
h
2
1
3
2
2
a =
⋅
⋅
⋅
=
max
max σ
4
bh
σ
b
2
h
2
1
R =
⎟
⎠
⎞
⎜
⎝
⎛
=
∑ = 0
M : M
σ
6
bh
a
R max
2
=
=
⋅
S
M
6
bh
M
σ 2
max =
= flexure formula, linear elastic
allow
safety
max σ
γ
σ ≤
⋅ e.g. design of beam
→ excursion on inelastic bending, rect. cross section ( el
pl S
5
.
1
S ⋅
= )
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
z
x
I
z
M
I
y
M
A
F
σ
⋅
±
+
⋅
±
−
±
= allow
max σ
≤
σ
For algebraic sign convention, see chapter 6.
a
-σmax
+σmax
M
R
R
y
y
z z
b
h
26

6/9
Department of
Civil Engineering
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
4
3
z cm
6666.7
12
bh
I =
=
3
z
2
z cm
666.7
c
I
6
bh
S =
=
=
2
N
x, cm
kN
0.05
A
N
σ =
=
2
3
z
z
M
x, cm
kN
0.75
cm
666.7
kNcm
500
S
M
σ ±
=
±
=
±
=
Superposition:
allow
2
2
max
σ
kN/cm
0.85
kN/cm
0.8
0.75
0.05
σ =
≤
=
+
= 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

7/9
Department of
Civil Engineering
b) steel cross section T100
σ║,allow = 21.8 kN/cm2
A = 20.9 cm2
4
z cm
79
1
I =
2
N
x, cm
kN
0.48
A
N
σ =
=
2
3
z
z
top
M,
x, cm
kN
20.27
cm
7.26
cm
179
kNcm
500
-
I
y
M
σ =
−
=
= (see sign convention)
2
3
z
z
bottom
M,
x, cm
kN
65
.
7
-
cm
2.74
-
cm
179
kNcm
500
-
I
y
M
σ =
−
=
−
=
Superposition:
allow
2
2
max
σ
kN/cm
.8
1
2
kN/cm
75
.
0
2
27
.
0
2
48
.
0
σ =
≤
=
+
= OK
3.3.2 Position of neutral surface
Position of neutral axis demands: 0
σx =
→ 0
I
z
M
I
y
M
y
y
z
z
=
+
−
→
y
z
y
z
z
y
I
I
z
tanθ
I
M
I
z
M
y ⋅
⋅
=
⋅
⋅
=
→ tanθ
I
I
y
y
z
=
z
→ tanθ
I
I
tan
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

8/9
Department of
Civil Engineering
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 consist 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

9/9
Department of
Civil Engineering
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 0
2
E
E
n
T
s
=
=
transformed dimension: cm
300
15
20
b
n
b I
II =
=
=
new centroidal axis (from top):
∑
∑ ⋅
=
i
i
i
A
y
A
y : cm
18.3
300
1
25
15
300)25.5
(1
25)12.5
(15
y =
⋅
+
⋅
⋅
+
⋅
=
moment of inertia:
)
y
A
(I
I 2
i
i
i
∑ ⋅
+
= :
2
3
2
3
z 2
.
7
)
300
1
(
12
1
300
8
.
5
)
25
15
(
12
25
15
I ⋅
+
⋅
+
⋅
+
⋅
=
4
z cm
47723
I =
stress timber:
2
4
max
t, kN/cm
1.15
cm
47723
cm
18.3
kNcm
30000
I
c
M
σ =
⋅
=
=
stress steel:
2
4
I
max
s, kN/cm
9.68
cm
47723
cm
7.7
kNcm
30000
20
σ
n
σ =
⋅
=
⋅
=
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
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

2/7
Department of
Civil Engineering
Derivation of Shearing Stresses in a Beam
Shear Flow
(along the horizontal plane y in the direction of x)
∑Fx = 0 : ΔH + 0
dA
)
( B
A
A =
⋅
σ
−
σ
∫
Δ I
My
=
σ
ΔH = dA
y
I
M
M C
B
∫ ⋅
−
Q = y
A
dA
y ⋅
Δ
=
⋅
∫
Statical Moment
ΔH = Q
I
M
Δ
[kN]
q =
dx
H
Δ
=
I
Q
x
M
Δ
Δ
[kN/cm] V
dx
dM =
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

3/7
Department of
Civil Engineering
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 = ( ) ³
cm
120
6
10
2
y
A =
⋅
⋅
=
⋅
Δ
q = cm
/
N
04
.
37
cm
1620
³
cm
120
N
500
I
VQ
4
=
⋅
=
F = Nail
/
N
6
.
92
cm
5
.
2
cm
/
N
04
.
37 =
⋅
Shearing Stress Formula
τ =
t
q
It
VQ
= [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

4/7
Department of
Civil Engineering
approximation: τave =
web
A
V
Distribution of Shear Stress
example: Rectangular Cross Section
τ (y1) =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
⎟
⎠
⎞
⎜
⎝
⎛
=
=
⋅
=
⋅
= ∫
∫
Δ
²
y
2
h
I
2
V
2
²
y
I
V
dy
yb
Ib
V
dA
y
Ib
V
It
VQ
1
2
2
/
h
1
y
2
/
h
1
y
A
distribution parabolically
V
dA
A
=
⋅
τ
∫ → τave =
A
V
min τ at y1 = h/2 → τmin = 0
max τ at y1 = 0 → τmax =
A
2
V
3
I
8
²
Vh =
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 =
t
q
q =
I
V
Q⋅
y
t
34

5/7
Department of
Civil Engineering
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 =
t
q
It
VQ
= [kN/cm²] Shear Stress per area
τzy = τxz equal shearing stresses on mutually
perpendicular planes
y
y
z
ΔA
¯
τzx
τxz
at particle
35

6/7
Department of
Civil Engineering
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 = 4
cm
4
.
309
12
³
11
5
12
³
12
6
=
⋅
−
⋅
Q = ³
cm
38
.
14
75
.
5
0
.
5
5
.
0
y
A =
⋅
⋅
=
⋅
Δ
q = cm
/
kN
465
.
0
cm
4
.
309
³
cm
38
.
14
kN
0
.
10
I
VQ
4
=
⋅
=
(q is the total shear flow acting on the particle with two cutting planes)
τ ²
cm
/
kN
465
.
0
cm
5
.
0
cm
/
kN
465
.
0
2
1
t
q
2
1
=
=
=
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

7/7
Department of
Civil Engineering
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: 0
H
H
Fz =
−
=
∑
0
V
V
Fy =
−
=
∑
0
e
V
h
H
Mx =
⋅
−
⋅
=
∑
V
h
H
e
⋅
=
⇒
I
VQ
q = ds
q
H
b
0
∫ ⋅
= ds
q
V
h
0
∫ ⋅
=
example 4.3: channel member (see figure above)
determination of shear center:
I
VQ
q =
2
h
t
s
y
A
Q ⋅
⋅
=
⋅
=
I
2
h
t
s
V
q
⋅
⋅
⋅
⋅
=
⇒
I
4
²
b
h
t
V
²
s
2
1
I
2
h
t
V
ds
I
2
h
t
s
V
ds
q
H
b
0
b
0
b
0
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
=
⋅
⋅
⋅
⋅
⋅
=
⋅
= ∫
∫
( ) ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
+
⋅
+
⋅
=
⋅
+
= ∑
2
i
i
i
2
h
t
b
12
³
t
b
2
12
³
h
t
²
y
A
I
I t³ is very small, will be neglected
( )
b
6
h
12
²
h
t
²
h
t
b
2
1
12
³
h
t
I +
⋅
=
⋅
⋅
+
⋅
=
( ) b
6
h
²
b
3
b
6
h
²
h
t
V
4
12
²
b
²
h
t
V
V
I
4
²
b
²
h
t
V
V
h
H
e
+
=
+
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
⋅
⋅
=
⋅
=
⇒
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
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:
dA
²
c
dA
c
T
A
max
A
max ⋅
ρ
⋅
τ
=
⎟
⎠
⎞
⎜
⎝
⎛
ρ
⋅
⋅
τ
ρ
= ∫
∫
dA
²
I
A
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
max
c
τ
ρ
=
τ
38

2/8
Department of
Civil Engineering
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
( ) ( ) 4
4
4
4
4
p cm
10
.
700
4
.
5
6
2
b
c
2
I =
−
π
=
−
π
=
²
cm
/
kN
85
.
12
cm
1
.
700
cm
6
kNcm
1500
I
c
T
4
p
max =
⋅
=
⋅
=
τ
polar moment of inertia Ip
for a solid circular member:
32
d
2
c
4
2
d
³
2
dA
²
I
4
4
c
0
c
0
4
A
p
π
=
π
=
ρ
π
=
ρ
⋅
πρ
=
⋅
ρ
= ∫
∫
for a hollow circular member:
( )
4
4
c
b
c
b
4
A
p b
c
2
4
2
d
³
2
dA
²
I −
π
=
ρ
π
=
ρ
⋅
πρ
=
⋅
ρ
= ∫
∫
t
³
c
2
Ip π
≈ (for b ≈c)
τmax
b
c
t
τmax
c
elastic torsion formula
p
max
I
c
T ⋅
=
τ maximum shear stress
max
p c
I
T
τ
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

3/8
Department of
Civil Engineering
Angle of Twist
by Hooke’s Law for shearing strain:
G
⋅
γ
=
τ
p
max
I
c
T ⋅
=
τ
G
I
c
T
G p
max
max
⋅
⋅
=
τ
=
γ
by geometry:
c
x ⋅
ΔΦ
=
Δ
⋅
γ angles measured in radians [rad]
c
x
G
I
c
T
p
⋅
ΔΦ
=
Δ
⋅
⋅
⋅
G
I
x
T
p ⋅
Δ
⋅
=
ΔΦ
example 5.2
Determine the rotation at the free end of the cantilever element of example 5.1.
G
I
L
T
p ⋅
⋅
=
Φ
ΔΦ
Δx γ
c
[ ] [ ]
°
Φ
=
π
⋅
Φ
2
360
rad
²
cm
/
kN
8100
cm
1
.
700
cm
100
kNcm
500
cm
200
kNcm
500
cm
200
kNcm
1500
G
I
L
T
4
p ⋅
⋅
+
⋅
−
⋅
=
⋅
=
Φ ∑ ⋅
°
=
=
Φ 53
.
2
rad
044
.
0
40

4/8
Department of
Civil Engineering
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
dx
t
F 1
1
1 ⋅
⋅
τ
= dx
t
F 2
2
2 ⋅
⋅
τ
=
0
Fx =
∑ : 2
1 F
F =
q = shear flow (shear force per meter
2
2
1
1 t
t ⋅
τ
=
⋅
τ
⇒ = q of the perimeter)
q = equal on all cutting planes of the
respective element
∫ ∫ ⋅
⋅
=
⋅
= ds
r
q
ds
rq
T T = torque
r ds = 2 Atriangle ∫ ⋅
=
⋅
⇒ A
2
ds
r
)
=
A
)
area to center line of perimeter
q
A
2
T ⋅
⋅
=
A
2
T
q )
⋅
= A
)
dx
t1
t2
τ1
τ1
τ2
τ2
F1
F2
particle
t
A
2
T
⋅
⋅
=
τ )
min
max
t
A
2
T
⋅
⋅
=
τ ) shear stress
p
I
G
L
T
⋅
⋅
=
Φ
( )
∫
⋅
=
t
ds
A
2
I
2
p
)
angle of twist
41

5/8
Department of
Civil Engineering
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
( ) ²
b
2
3
b
b
b
2
2
1
A =
+
=
)
tmin = t :
t
b
3
F
t
²
b
3
F
b
t
A
2
T
max
⋅
⋅
=
⋅
⋅
⋅
=
⋅
⋅
=
τ ) additional shear stress due to bending
is neglected in this example
( )
5
2
t
³
b
9
t
b
5
2
1
2
t
b
t
2
b
2
²
b
2
3
2
t
ds
A
2
I
2
2
p
+
⋅
⋅
=
⋅
+
+
⎟
⎠
⎞
⎜
⎝
⎛
⋅
=
⋅
=
∫
)
( )
t
²
b
G
L
F
9
5
2
t
³
b
9
G
5
2
L
F
b
I
G
L
T
p ⋅
⋅
⋅
+
=
⋅
⋅
⋅
+
⋅
⋅
=
⋅
⋅
=
Φ
F
L t
t
2t
b
b
2b
F
42

6/8
Department of
Civil Engineering
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
max
max
I
t
T ⋅
=
τ shear stress formula
∑
⋅
≈
3
i
i
p t
h
3
1
I polar moment of inertia
L
L
Members of same behaviour towards an applied torque:
wall thickness : t
43

7/8
Department of
Civil Engineering
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:
t
³
R
2
t
³
c
2
Ip π
=
π
≈
t
²
R
2
T
I
c
T
p
max
π
=
⋅
=
τ
b) open section (slotted):
L = 2πR
³
Rt
3
2
t
h
3
1
I 3
i
i
p π
=
⋅∑
≈
²
Rt
2
T
3
I
t
T
p
max
max
π
=
⋅
=
τ
ratio of shear stress a) to b):
1 :
t
R
3
(ratio of max
τ )
ratio of stiffness a) to b)
1 :
3
R
t
3
1
⎟
⎠
⎞
⎜
⎝
⎛
(ratio of p
I )
R
t
44

8/8
Department of
Civil Engineering
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
⋅
Φ
2
360
rad
p
max
I
c
T ⋅
=
τ
max
p c
I
T
τ
ρ
=
ρ
⋅
=
τ
p
I
G
L
T
⋅
⋅
=
Φ
min
max
t
A
2
T
⋅
⋅
=
τ )
t
A
2
T
⋅
⋅
=
τ )
τmax
b
c
t
( )
4
4
p b
c
2
I −
π
=
t
³
c
2
Ip π
≈ (for b ≈c)
τmax
c
ρ
p
max
I
c
T ⋅
=
τ
max
p c
I
T
τ
ρ
=
ρ
⋅
=
τ
p
I
G
L
T
⋅
⋅
=
Φ
32
d
2
c
I
4
4
p
π
=
π
=
p
max
max
I
t
T ⋅
=
τ
p
I
t
T ⋅
=
τ
∑
⋅
≈
3
i
i
p t
h
3
1
I
p
I
G
L
T
⋅
⋅
=
Φ
h
h
wall thickness : t
A
)
( )
∫
⋅
=
t
ds
A
2
I
2
p
)
p
I
G
L
T
⋅
⋅
=
Φ
t
45

Strength of Materials 6 Stress Analysis
1/10
Department of
Civil Engineering
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:
A
F
σx =
• normal stresses due to bending:
z
z
z
z
x
S
M
I
y
M
−
=
⋅
−
=
σ
y
y
y
y
x
S
M
I
z
M
=
⋅
=
σ
• shearing stresses due to shear force in a beam:
t
I
Q
V
y
y
y
xy
⋅
⋅
=
τ
t
I
Q
V
z
z
z
xz
⋅
⋅
=
τ
• shearing stresses due to torque:
p
I
T ρ
⋅
=
τ circular shafts
t
A
2
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

2/10
Department of
Civil Engineering
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
z
x
I
z
M
I
y
M
A
F
σ
⋅
±
+
⋅
±
−
±
= allow
max σ
≤
σ
Superposition of shearing stresses design analysis
T
V τ
±
τ
±
=
τ allow
max 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:
( ) 0
²
h
b
6
e
P
h
b
P
S
M
A
P
=
⋅
⋅
⋅
+
⋅
−
=
+
−
=
σ
6
h
e =
⇒
3
h 3
b
b
h
zone of applicable resultant
force to meet the condition
47

3/10
Department of
Civil Engineering
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
4
4
3
y cm
17
.
104
cm
12
5
10
I =
⋅
=
4
4
3
z cm
67
.
416
cm
12
10
5
I =
⋅
=
kNcm
50
M
2
1
M
sin
My =
=
⋅
ϕ
=
136.6kNcm
5cm
F
M
3
2
1
2
h
F
M
cos
Mz −
=
⋅
−
⋅
−
=
−
⋅
−
= ϕ
y
y
z
z
x
I
z
M
I
y
M
A
F
σ
⋅
±
+
⋅
±
−
±
=
A: ²
cm
/
kN
64
.
2
cm
17
.
104
cm
5
.
2
kNcm
50
cm
67
.
416
cm
5
kNcm
6
.
136
²
cm
50
kN
10
4
4
x =
⋅
+
⋅
−
−
−
=
σ
B: ²
cm
/
kN
24
.
0
cm
17
.
104
cm
5
.
2
kNcm
50
cm
67
.
416
cm
5
kNcm
6
.
136
²
cm
50
kN
10
4
4
x =
−
⋅
+
⋅
−
−
−
=
σ
C: ²
cm
/
kN
64
.
0
cm
17
.
104
cm
5
.
2
kNcm
50
cm
67
.
416
cm
5
kNcm
6
.
136
²
cm
50
kN
10
4
4
x −
=
⋅
+
−
⋅
−
−
−
=
σ
D: ²
cm
/
kN
04
.
3
cm
17
.
104
cm
5
.
2
kNcm
50
cm
67
.
416
cm
5
kNcm
6
.
136
²
cm
50
kN
10
4
4
x −
=
−
⋅
+
−
⋅
−
−
−
=
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

4/10
Department of
Civil Engineering
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
( ) ²
b
2
3
b
b
b
2
2
1
A =
+
=
)
tmin = t :
t
b
3
F
t
²
b
3
F
b
t
A
2
T
max
,
T
⋅
⋅
−
=
⋅
⋅
⋅
−
=
⋅
⋅
−
=
τ )
shear stress due to bending:
largest shear stress occurs in the web
t
b
5
F
t
b
2
5
2
F
A
V
web
ave
,
V
⋅
⋅
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
⋅
⋅
=
=
τ approximation formula
superposition:
78
.
0
t
b
F
t
b
5
F
t
b
3
F
V
T
max ⋅
⋅
=
⋅
⋅
+
⋅
⋅
=
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)
+
max
τ
49

5/10
Department of
Civil Engineering
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
∑ = 0
F '
x : =
⋅
σ dA
'
x
∑ = 0
F '
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 xy
xy τ
=
τ , ( )
θ
+
=
θ 2
cos
1
2
1
²
cos , ( )
θ
−
=
θ 2
cos
1
2
1
²
sin ,
θ
=
θ
θ 2
sin
cos
sin
2 , θ
=
θ
−
θ 2
cos
²
sin
²
cos
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
( ) ( )
( ) ( ) θ
θ
⋅
τ
+
θ
θ
⋅
σ
+
θ
θ
⋅
τ
+
θ
θ
⋅
σ
cos
sin
dA
sin
sin
dA
sin
cos
dA
cos
cos
dA
yx
y
xy
x
( ) ( )
( ) ( ) θ
θ
⋅
τ
−
θ
θ
⋅
σ
+
θ
θ
⋅
τ
+
θ
θ
⋅
σ
−
sin
sin
dA
cos
sin
dA
cos
cos
dA
sin
cos
dA
yx
y
xy
x
50

6/10
Department of
Civil Engineering
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:
( ) 0
2
cos
2
2
sin
2
2
1
d
xy
y
x
'
x
=
θ
τ
+
θ
σ
−
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
( ) ( ) θ
τ
+
θ
σ
−
σ
+
σ
+
σ
=
σ 2
sin
2
cos
2
1
2
1
xy
y
x
y
x
'
x (6.1)
( ) ( ) θ
τ
−
θ
σ
−
σ
−
σ
+
σ
=
σ 2
sin
2
cos
2
1
2
1
xy
y
x
y
x
'
y (6.2)
( ) θ
τ
+
θ
σ
−
σ
−
=
τ 2
cos
2
sin
2
1
xy
y
x
'
y
'
x (6.3)
y
x
'
y
'
x σ
+
σ
=
σ
+
σ
y
x
xy
1
2
2
tan
σ
−
σ
τ
=
θ
2
xy
2
y
x
y
x
2
,
1
2
2
τ
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ σ
−
σ
±
σ
+
σ
=
σ
51

7/10
Department of
Civil Engineering
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:
( ) 0
2
sin
2
2
cos
d
xy
y
x
'
y
'
x
=
θ
τ
+
θ
σ
−
σ
−
=
θ
τ
(6.7) hence
(6.8)
or with (6.6):
(6.9)
substitution of (6.8) into (6.1) or (6.2):
( )
y
x
2
1
' σ
+
σ
=
σ (6.10)
Thus maximum shearing stresses occur on planes that are not necessarily free of
normal stress, σ' = σx' = σy'.
1
2
2
tan
1
2
tan
θ
−
=
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
y
x
2
2
2
tan
τ
σ
−
σ
−
=
θ
2
xy
2
y
x
max
2
τ
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ σ
−
σ
±
=
τ
( )
2
1
max
2
1
σ
−
σ
±
=
τ
τxy
τyx
σ1 = τxy
σ2 = -τxy
52

8/10
Department of
Civil Engineering
6.3 Mohr’s Circle
Equations (6.1) or (6.2) and (6.3) can be represented graphically. Transforming the
equations:
( ) ( ) θ
τ
+
θ
σ
−
σ
+
=
σ
+
σ
−
σ 2
sin
2
cos
2
1
2
1
xy
y
x
y
x
'
x
( ) θ
τ
+
θ
σ
−
σ
−
=
τ 2
cos
2
sin
2
1
xy
y
x
'
y
'
x
squaring and adding both equations and simplifying:
( ) 2
xy
2
y
x
2
'
y
'
x
2
y
x
'
x
2
2
1
τ
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ σ
−
σ
=
τ
+
⎥
⎦
⎤
⎢
⎣
⎡
σ
+
σ
−
σ (6.11)
since σx, σy and τxy are given constants in a problem they are summarised as:
2
xy
2
y
x
2
²
r τ
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ σ
−
σ
=
with (5.10), ( )
y
x
2
1
' σ
+
σ
=
σ , equation (5.11) is written as:
( ) ²
r
²
' 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.
( )
y
x
2
1
' σ
+
σ
=
σ
σ
τ
(+)
σx
σy
(-)
σ'
2θ
(+)
τxy
τxy
(+)
53

9/10
Department of
Civil Engineering
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
( )
y
x
2
1
' 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

10/10
Department of
Civil Engineering
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:
allow
2
xy
y
x
2
y
2
x 3 σ
≤
τ
+
σ
σ
−
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
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:
ρ
c
dx
δx
ε max
x, =
=
ρ
=
ε
y
x (7.1)
since ε
⋅
=
⋅
=
σ E
I
y
M
,
I
E
y
M
⋅
⋅
=
ε (7.2)
(6.1)=(6.2): κ
=
⋅
=
ρ I
E
M
1
curvature
due to analytic geometry, curvature is defined as:
( )
[ ]2
3
2
'
v
1
'
'
v
1
+
=
ρ
v' = θ = slope - since v' is very small:
v = deflection - due to M being
I
E
M
²
dx
v
²
d
'
'
v
1
⋅
=
=
=
ρ
positive if oriented downward, hence:
I
E
M
'
'
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

2/4
Department of
Civil Engineering
Recalling the static differential relations
V
dx
dM
'
M =
= and )
x
(
q
dx
dV
'
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

3/4
Department of
Civil Engineering
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
EI
384
qL
5
)
2
/
L
(
v
v
4
max
⋅
⋅
=
=
EI
3
³
PL
)
L
(
v
vmax
⋅
=
=
EI
8
qL
)
L
(
v
v
4
max
⋅
=
=
( )
[ ]
³
x
x
²
b
²
L
EIL
6
Pb
vmax
−
−
⋅
⋅
=
EI
48
PL
)
2
/
L
(
v
v
3
max
⋅
=
=
EI
16
²
PL
)
L
(
)
0
(
⋅
=
θ
−
=
θ
EI
24
³
qL
)
L
(
)
0
(
⋅
=
θ
−
=
θ
EI
2
²
PL
)
L
(
⋅
=
θ
EI
6
³
qL
)
L
(
⋅
=
θ
58

4/4
Department of
Civil Engineering
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
kNm
63
.
15
8
²
gL
Mg =
= bending moment
cm
62
.
1
cm
5740
²
cm
/
kN
21000
384
)
cm
500
(
cm
/
kN
10
24
5
EI
384
gL
5
v 4
4
2
4
g
max, =
⋅
⋅
⋅
⋅
⋅
=
⋅
⋅
=
−
deflection
pbeam = 5 * 3.0 = 15.0 kN/m
kNm
0
.
25
8
²
pL
Mp =
=
cm
01
.
1
cm
5740
²
cm
/
kN
21000
384
)
cm
500
(
cm
/
kN
10
15
5
EI
384
pL
5
v 4
4
2
4
p
max, =
⋅
⋅
⋅
⋅
⋅
=
⋅
⋅
=
−
cm
67
.
1
300
/
500
v
cm
63
.
2
v
v
v allow
p
max,
g
max,
max =
=
>
=
+
= not sufficient!
b) height of camber = =
g
max,
v 1.62cm
cm
67
.
1
v
cm
01
.
1
v
v
v
v allow
p
max,
g
max,
g
max,
max =
<
=
+
+
−
= OK
c) g
max,
3
P
max, v
cm
62
.
1
EI
48
PL
v =
=
⋅
= deflection due to force at L/2
cm
62
.
1
)
cm
500
(
cm
5740
²
cm
/
kN
21000
48
v
L
EI
48
P 3
4
3
⋅
⋅
=
⋅
=
kN
98
.
74
P = pretension force
L
59

Strength of Materials 8 Buckling
1/5
Department of
Civil Engineering
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
around 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
v
P
EI
M
'
'
v
⋅
=
−
= differential equation for the elastic curve
EI
P
² =
λ applying λ, transforming the equation:
0
v
²
'
'
v =
λ
+ being equal to an equation for simple
harmonic motion having the solution:
x
cos
B
x
sin
A
v λ
+
λ
=
L
P P
θ
L
P
v
60

2/5
Department of
Civil Engineering
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
=
⋅ n
L
)
EI
/(
P and
²
L
EI
²
Pcr
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
²
L
EI
²
P
e
cr
π
=
L = Le
P
Le = 2L
P
L
Le = 0.7L
P
L
Le = 0.5L
P
L
1 2 3 4
K = 2
²
L
4
EI
²
Pcr
π
=
K = 1
²
L
EI
²
Pcr
π
=
K = 0.7
²
L
EI
²
2
Pcr
π
=
K = 0.5
²
L
EI
²
4
Pcr
π
=
61

3/5
Department of
Civil Engineering
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: cm
1000
Le =
kN
5
.
315
)²
cm
1000
(
cm
42
.
1522
²
cm
/
kN
21000
²
²
L
EI
²
P
4
e
z
cr =
⋅
⋅
π
=
π
=
buckling plane x-z, upper part:
kN
8
.
663
)²
cm
400
(
cm
42
.
512
²
cm
/
kN
21000
²
²
L
EI
²
P
4
e
y
cr =
⋅
⋅
π
=
π
=
buckling plane x-z, lower part: cm
600
Le =
kN
0
.
295
)²
cm
600
(
cm
42
.
512
²
cm
/
kN
21000
²
²
L
EI
²
P
4
e
y
cr =
⋅
⋅
π
=
π
=
min(Pcr) = 295.0 kN
b) buckling plane x-y: cm
0
.
700
m
0
.
10
7
.
0
Le =
⋅
=
kN
0
.
644
)²
cm
700
(
cm
42
.
1522
²
cm
/
kN
21000
²
²
L
EI
²
P
4
e
z
cr =
⋅
⋅
π
=
π
=
buckling plane x-z, upper part: cm
400
Le =
kN
8
.
663
)²
cm
400
(
cm
42
.
512
²
cm
/
kN
21000
²
²
L
EI
²
P
4
e
y
cr =
⋅
⋅
π
=
π
=
buckling plane x-z, lower part: cm
420
m
0
.
6
7
.
0
Le =
⋅
=
kN
0
.
602
)²
cm
420
(
cm
42
.
512
²
cm
/
kN
21000
²
²
L
EI
²
P
4
e
y
cr =
⋅
⋅
π
=
π
=
min(Pcr) = 602.0 kN
P
z
y
x
4m
6m
z
y
62

4/5
Department of
Civil Engineering
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.
²
L
EI
²
P
e
cr
π
= introducing a new definition:
2
r
A
I ⋅
= hence:
A
I
r = radius of gyration, transforming Pcr:
²
L
²
EAr
²
P
e
cr
π
=
2
e
cr
cr
)
r
/
L
(
E
²
A
P 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

5/5
Department of
Civil Engineering
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
2
e
cr
)
r
/
L
(
E
²
π
=
σ
Euler Hyperbola
no
buckling,
material
failure
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_09-04-21.pdf

Recommended

Recommended

More Related Content

Similar to AAU_strength_of_materials_lecture_notes_09-04-21.pdf

Similar to AAU_strength_of_materials_lecture_notes_09-04-21.pdf (20)

Recently uploaded

Recently uploaded (20)

AAU_strength_of_materials_lecture_notes_09-04-21.pdf