# CIVE 270 - Lab Manual FALL 2019 ```CIV E 270
Mechanics of Deformable Bodies
LABORATORY MANUAL
Fall 2019
Version 2
CIV E 270
MECHANICS OF DEFORMABLE BODIES
LABORATORY MANUAL VERSION 6.4
(Revised: January 2011)
THE DEPARTMENT OF CIVIL &amp; ENVIRONMENTAL
ENGINEERING
UNIVERSITY OF ALBERTA
EDMONTON, ALBERTA
CIV E 270
MECHANICS OF DEFORMABLE BODIES
LABORATORY MANUAL VERSION 4.2
(Revised: December 2011)
Text:
Hibeller, R.C. 2011, Mechanics of Materials, SI Edition, Pearson.
INSTRUCTOR:
SECTION:
NAME:
STUDENT ID:
LAB SECTION:
THE DEPARTMENT OF CIVIL &amp; ENVIRONMENTAL ENGINEERING
UNIVERSITY OF ALBERTA
MARKING SCHEME
1.
The labs are marked generally out of 10. The basis for marking is:
(i)
general presentation
(ii)
engineering content
(iii)
discussion, where applicable
2.
The labs are due at the end of each session period unless informed otherwise by the instructor.
3.
You may wish to keep a record of your marks below.
LAB NO.
TITLE
MARK
1
Seminar - Axial Force, Shear Force, and Bending Moment Diagrams
2
Tension Test
3
Compression Test
4
5
Torsion Test
6
Seminar - Bending Stresses
7
Bending Test
8
Seminar - Transformation of Stress - Mohr's Circle
9
Combined Stresses
10
Beam Deflections
11
Buckling of Columns
TOTAL
STANDARD LAYOUT FOR LAB
REPORTS IN CIV E 270
General Presentation
1.
Include the lab write-up as part of the lab.
2.
Use engineering paper.
3.
Use pencil.
4.
Present the content in a logical manner (provide sufficient space).
5.
Label diagrams and figures properly.
6.
Graphs should have both axes labeled with units and should have a title block.
7.
Use straight edges and french curves where required.
8.
Answers should be in engineering units.
Length - mm
Stress - MPa
Area - mm2
S, Z - mm3
Moment - kN&middot;m
I, J - mm4
9.
Report answers with the correct number of significant digits.
10.
BE NEAT!
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 1
Name:
I.D.:
Date:
Mark:
Lab Section:
AXIAL FORCE, SHEAR FORCE, AND BENDING MOMENT DIAGRAMS
1.1
Step by Step Procedure
1.
Select a Free Body - i.e. isolate beam or portion of it.
2.
Draw the Free Body Diagram
(i) draw the free body
(ii) identify all known forces
(iii) identify all possible unknown reactions
(iv) label and dimension diagram
3.
Determine Reactions
(i)
from the free body diagram, set up the equations of statics,
ΣFx = 0, ΣFy = 0, ΣM = 0.
(ii)
solve for the unknown reactions
(iii) check solution
4.
Using the method of sections and/or load, shear force and bending moment
relationships, draw the shear force an bending moment diagrams.
The method of sections uses statics to examine a free body to the left or right of any
section to determine the axial force, shear force and bending moment at the section.
1.2
Relationships Among Load, Shear and Bending Moment
1.2.1
(i) The slope of the shear force diagram is equal to the load intensity:
Laboratory 1 Civ E 270
2
dV
= −w
dx
(ii) The change in shear between two locations x1 and x2 along the beam is equal
to the area under the load diagram between x1 and x2
x2
V( x 2 ) − V( x1 ) = ∫ − w dx
x1
dV
dx
Shear diagram
0
0
constant
constant
constant
sloping straight line
f(x)
f(x)
slope changes with x
+ve
-ve
-ve
+ve
w has 2 intensities
2 values
cusp
left and right of point
∞
jump
1.2.2 Shear Force, Bending Moment Relationships
(i) The slope of the bending moment diagram is equal to the shear force
dM
= V
dx
(ii) The change in moment between two locations x1 and x2 along the beam is
equal to the area under the shear force between x1 and x2
M ( x 2 ) − M ( x1 ) =
x2
∫ V dx
x1
3
Laboratory 1 Civ E 270
Shear force, V
dM
dx
Moment diagram
0
0
max. or min.
constant
constant
sloping straight line
f(x)
f(x)
slope changes with x
+ve
+ve
-ve
-ve
V has 2 intensities
left and right
2 values
concentrated moment
∞
cusp
jump
1.3 Sketch Axial Force, Shear Force and Bending Moment Diagrams
The members shown on the following pages are idealizations of various structural
components or machine elements. As a first step in the design of these members calculate the
reactions and sketch;
(i)
free body diagram (F.B.D.)
(ii)
axial force diagrams (if applicable) (A.F.D.)
(iii) shear force diagrams (S.F.D.)
(iv) bending moment diagrams. (B.M.D.)
Label maximum and minimum values and their locations, points of tangency, and points
of zero slope and locations where the curve crosses the axis. Do any needed calculations on
engineering paper and hand in with your lab.
Note: Draw diagrams perpendicular to the member.
Laboratory 1 Civ E 270
4
Laboratory 1 Civ E 270
5
Laboratory 1 Civ E 270
6
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 2
Name:
I.D.:
Date:
Mark:
Lab Section:
TENSION TEST
2.0 OBJECTIVES
•
To illustrate different types of tensile behaviour by performing tension tests on aluminum
and steel bars.
•
To determine significant mechanical properties for aluminum and steel.
•
To demonstrate three different devices for measuring strain.
2.1 GENERAL BACKGROUND
The tension test is a fundamental material test to determine the properties of materials
needed for the design of structures. Typical stress vs. strain diagrams for aluminum and steel are
shown in Figures 2.1 and 2.2, respectively. In addition to showing typical shapes for the
diagrams, the figures define specific mechanical properties such as the modulus of elasticity, the
proportional limit, the ultimate strength and the stress and strain at fracture. Figure 2.2 also
shows the upper and lower yield points for low carbon steel. The most significant yield strength
parameter for structural steel is the static yield stress, defined as the stress just subsequent to
yielding that is maintained when the strain rate is zero. Of significance, as well, is the strainhardening strain, which marks the end of the yield plateau.
Because aluminum does not have a well-defined yield point, a yield strength is obtained by
the offset method as shown in Figure 2.3. A line is drawn parallel to the initial portion of the
stress vs. strain diagram and offset by a strain of 0.2% (0.002). The yield strength is defined as
the stress corresponding to the point of intersection of the offset line and the stress vs. strain
curve.
Typical tensile fractures are shown in Figure 2.4.
Laboratory 2 Civ E 270
2
2.2 REFERENCES
•
Beer, Johnston, DeWolf and Mazurek, 2009, Mechanics of Materials, 4th Edition in SI
Units, McGraw Hill, Toronto, Chap. 2, p. 47-55.
•
ASTM A370-09, Standard Methods and Definitions for Mechanical Testing of Steel
Products, American Society for Testing and Materials, Philadelphia, 2009.
2.3 PREPARATION OF TEST SPECIMENS
1.
The aluminum specimen is 600 mm in length and the steel specimen is prepared based
on the ASTM A370 standard.
2.
A 200 mm gauge length is lightly marked with a center punch on each specimen.
3.
A pair of strain gauges are mounted on opposite faces of each coupon.
2.4 TEST PROCEDURE
2.4.1
Aluminum
1.
Measure and record the diameter of the specimen and the gauge length.
2.
Place and align the specimen in the testing machine.
3.
Connect the strain gauges, extensometer and MTS load and stroke outputs to the data
acquisition system.
4.
grips are properly biting into the specimen and that slip will not occur during the test.
5.
6.
7.
At approximately 5000 and 7000 microstrain, measure the static stress level.
8.
9.
After fracture, record the final length for the 200 mm gauge length and the diameter
at the narrowest point. Examine and sketch the fracture surface, paying particular
attention to the texture and shape.
2.4.2
Flat Steel Bar
1.
Measure and record the width and thickness of the flat bar and the gauge length.
2.
Place and align the specimen in the testing machine.
Laboratory 2 Civ E 270
3
3.
Connect the strain gauges, extensometer and MTS load and stroke outputs to data
acquisition system.
4.
5.
6.
7.
Take several static yield stress readings along the yield plateau.
formation of L&uuml;ders' lines on the specimen.
8.
9.
After fracture, record the final length for the 200 mm gauge length and the width and
thickness at the narrowest point. Examine and sketch the fracture, paying particular
attention to the texture and shape of the fracture surface.
2.5 RECORD OF TEST DATA AND ANALYSIS OF RESULTS
2.5.1
(a)
Aluminum
Initial diameter, Di =
mm
Initial area, Ai
mm2
(b)
Gauge length, L =
(c)
Fracture
(d)
=
mm
=
kN
=
kN
Final gauge length =
mm
Final diameter
mm
=
Description and sketch of fracture surface:
Observe the
Laboratory 2 Civ E 270
(e)
4
Stress-strain data for the aluminum specimen is presented in two graphs to be handed
out in the lab period. The first shows the elastic and initial plastic behaviour at an
expanded scale and the second shows the complete stress-strain behaviour.
Using the curves based on extensometer data, report the following material properties
and show them on the graphs.
Modulus of elasticity,
Proportional limit,
0.2% offset static yield stress
Ultimate strength,
Ultimate strain,
Percent elongation at fracture
(f)
E
σp
σys
σu
εu
=
=
=
=
=
=
MPa
MPa
MPa
MPa
mm/mm
%
Using the modulus of elasticity, E, as obtained by students in your class and recorded
in the table shown, compute the mean, standard deviation (maintain at least five
significant figures) and coefficient of variation. (The differences in the values of E
that are reported arise from differences in plotting and slope calculations by different
persons using identical test data).
E (MPa)
1 n
Mean, x =
∑x
n i =1 i
Standard deviation, s =
1
(
n( n − 1)
Coefficient of variation, V=
where
( ) (
n ∑ x2i − ∑ x i
)
(s / x)
)
2
x 100%
xi = observation
n = total number of observations
2.5.2 Flat Steel Bar
(a)
Initial width, w =
Initial area, Ai =
(b)
Gauge length, L =
(c)
Fracture
mm, initial thickness, t =
mm2
mm
kN
mm
5
Laboratory 2 Civ E 270
=
kN
Final gauge length =
Final width
mm
=
mm, final thickness =
mm
(d) Description and sketch of fracture surface:
(e)
Stress-strain data for the steel specimen is presented in two graphs to be handed out
in the lab period. The first shows the elastic and initial plastic behaviour at an
expanded scale and the second shows the complete stress-strain behaviour.
Using the curves based on extensometer data, report the following material properties
and show them on the graphs.
Modulus of elasticity,
E
=
MPa
Upper yield point,
σyu
=
MPa
Lower yield point,
σyl
=
MPa
Static yield stress,
σys
=
MPa
Yield strain,
εy
=
mm/mm
Strain hardening strain,
εst
=
mm/mm
Ratio of strain hardening
strain/yield strain,
εst/εy
=
Ultimate strength,
σu
=
MPa
Ultimate strain,
εu
=
mm/mm
Percent elongation at fracture
=
%
True stress at fracture
=
MPa
2.6 Comment on the differences, if any, between strain readings obtained from the
extensometer and from the strain gauges. What advantages and disadvantages can you see
to these two methods of strain measurement?
Laboratory 2 Civ E 270
Figure 2.1
Stress vs. Strain Diagram for Aluminum
Figure 2.2
Stress vs. Strain Diagram for Steel
6
Laboratory 2 Civ E 270
Figure 2.3
Offset Yield Strength for Aluminum
Figure 2.4
Typical Tensile Fractures for Metals (Ref. Davis, Troxell and Hauck,
&quot;The Testing of Engineering Materials&quot;, Figure 8.20, p. 140.)
7
Laboratory 2 Civ E 270
8
3.0 Design Problem
Member AB in the structure shown consists of a rod made of the same grade of aluminum as the
tested. Using a factor of safety of 2.5 on the ultimate stress, determine:
(a) the required diameter of rod AB,
(b) the required cross-sectional area of link EF.
Assume that the tensile and compressive strengths are the same.
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 3
Name:
I.D.:
Date:
Mark:
Lab Section:
COMPRESSION TEST
3.0
3.1
3.2
OBJECTIVES
•
To obtain stress vs. strain diagrams for concrete and wood.
•
To plot a stress vs. strain diagram for a clay sample.
•
To determine significant mechanical properties from these diagrams.
•
To illustrate the calculation of the secant and tangent moduli of elasticity.
•
To apply the test results to a practical problem.
REFERENCES
•
Davis, Troxell and Hauck, 1982. The Testing of Engineering Materials. 4th Edition,
McGraw Hill, Toronto, Chap. 9, pp. 151-163.
•
Canadian Standards Association, 2004. Methods of Test and Standard Practices for
Concrete. National Standard of Canada, CAN/CSA-A23.2-04.
•
Canadian Standards Association, 2004. Design of Concrete Structures. National
•
Canadian Wood Council, 2005. Wood Design Manual. Ottawa.
TEST PROCEDURE
3.2.1
Concrete Cylinder and Wood Specimen
1.
Measure and record the initial gauge length and cross-sectional dimensions.
2.
Mount the LVDT with bracket on the specimen.
3.
Align the specimen in the testing machine.
4.
Apply an initial load of approximately 1/10th of the calculated failure load.
5.
Unload and set instrumentation to zero.
Laboratory 3 Civ E 270
2
6.
Load continuously at approximately 2 kN/s and record voltage reading from the
LVDT associated with predetermined load levels.
7.
8.
Examine and sketch the failure. Typical failure modes are shown in Figures 4.1 for
concrete and 4.2 for wood.
3.3 RECORD OF TEST DATA AND ANALYSIS OF RESULTS
3.3.1 Concrete cylinder
Initial diameter,
Di
=
mm,
Initial area,
Ai
=
mm2
(b)
Gauge length,
L
=
mm
(c)
The data acquisition system is set up to calculate and record the strains, ε =
(a)
ΔL
L
, and
stresses, f'c = P/A. At the end of test record the maximum applied load;
(d)
kN.
Describe and sketch the failure:
(e) A graph of the stress vs. strain curve for the concrete will be handed out. Note the
compressive strength, f' c , corresponding to the ultimate load, i.e.,
f' c = Pu/Ai =
and the corresponding strain εo =
MPa
mm/mm.
Compute the test value of the secant modulus of elasticity at 0.4 f' c :
Laboratory 3 Civ E 270
3
MPa
Ecs =
(f)
CSA Standard, Design of Concrete Structures CAN/CSA-A23.3-04 gives two
empirical relationships, based on a large number of tests, for the modulus of elasticity
of concrete to be used in design:
⎛ γ ⎞
Ec = (3300 f' c + 6900) ⎜ c ⎟
⎝ 2300 ⎠
where
1.5
γc = mass density kg/ m3
f ' c = compressive strength, MPa
and E c = 4500 f' c for the normal weight concrete with γc = 2400 kg/m3
From the latter compute Ec =
MPa
How well does this approximation agree with the experimental result from (e)?
(Calculate the test/predicted ratio for the modulus of elasticity.)
3.3.2 Wood Specimen
Initial width, bi
=
mm
Initial depth, di
=
mm
Initial area, Ai
=
mm2
(b)
Gauge length, L
=
mm
(c)
In general, the compressive strength of wood is a function of:
(a)
(i) the grade and species of wood
(ii) how the specimen is loaded (parallel or perpendicular to the grain)
(iii) service condition (wet, dry, type of exposure)
(iv) treatment of wood (preservatives, fire-retardant)
(vi) slenderness.
4
Laboratory 3 Civ E 270
(d)
Pu =
kN.
(e)
Describe and sketch the failure:
(f)
A graph of the stress vs. strain curve for the wood will be handed out. Note the
compressive strength, Fc, corresponding to the ultimate load
Fc = Pu/Ai =
MPa
and the corresponding strain εu =
mm/mm
Compute the test value of the tangent modulus of elasticity at 0.5 Fc. This is found
from the stress vs. strain curve by drawing a tangent to the curve at a stress of 0.5 Fc.
Et =
MPa
Laboratory 3 Civ E 270
3.4
5
APPLICATION OF RESULTS
The short timber post resting on the concrete footing
shown in the figure must carry a load of 220 kN. Both the
post and the footing are to be square in plan. Assume that
the footing is thick enough so as to deliver a uniformly
distributed stress to the soil under the full area of the
concrete. Determine the cross-sectional dimensions of the
post and the footing. Use the following three stress
limitations:
1. The maximum stress in the wood should be based on
the ultimate compressive strength that you measured in
the test conducted in this laboratory and a factor of safety of 2.25.
2. The allowable bearing stress in the concrete under the post must not exceed 37.5% of the
ultimate compressive strength that you measured in this laboratory.
3. A reasonable first estimate for the allowable soil bearing capacity for clay is its ultimate
compressive strength obtained from a compression test on a small clay sample (also known as
the “unconfined compression strength”). A reasonable value for the unconfined compression
strength of Athabasca clay (commonly found in around Alberta) is 105 kPa.
6
Laboratory 3 Civ E 270
Shear Cone
Shear Plane
Shear Cone with
Splitting above
Crushing
Wedge Split
Shearing
Splitting
Shearing and
Splitting
Figure 3.2 Types of failure of wood under compression parallel to the grain (ASTM D143).
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 4
Name:
I.D.:
Date:
Mark:
Lab Section:
INDETERMINATE PRISMATIC ELEMENTS SUBJECTED
TO AXIAL LOADS AND TEMPERATURE CHANGES
4.1 GENERAL BACKGROUND
There are many practical applications in which the use of free-body diagrams and
application of the equations of equilibrium are not sufficient to determine the internal forces in a
member or component. Such problems are termed statically indeterminate. In order to solve
these problems, additional information is required, and this comes from consideration of the
geometry of the problem and from the relationships between forces and deformations. This
laboratory session will address these issues. It is limited to cases involving axial forces,
including the effects of temperature. Later in the course, we will deal with other members (e.g.,
beams) that are also statically indeterminate.
In general, we must meet the following conditions:
1.
Equilibrium conditions for the system must be met, both locally and globally.
2.
Geometrical compatibility among the various components of the structural system
must be satisfied.
3.
The stress vs. strain relationships for the materials involved must be complied with.
Laboratory 4 Civ E 270
2
4.2 EXAMPLES
Example 1
Given: A steel rod fits loosely into a
copper tube. Initially, the copper is
400.00 mm long and the steel is 400.06
mm long. Determine the stresses in the
steel and in the copper when an axial
force P = 125 kN is applied.
Solution: If the force P = 125 kN does
not close the gap, there will be no load in
the copper and the stress in the steel can
be calculated directly. However, if the
force P = 125 kN does close the gap, the
load will be shared between the steel and
the copper.
The only equation of
equilibrium available (ΣFy = 0) is not sufficient to identify how much of the force goes into the
steel and how much goes into the copper.
Force required to close gap:
PL
we can solve for the value of P (=P1) that will just close the gap (Δ) of
AE
ΔAE
0.06 mm as P1 =
L
Using Δ =
P1 =
0.06 mm x 1000 mm 2 x 200x10 3 N / mm 2
= 30x10 3 N = 30 kN
400.06 mm
Since the force required to close the gap (30 kN) is less than the total force that will be applied
(125 kN), both the steel and the copper will be loaded (stressed) when the total force is applied.
After the gap is closed, the additional load beyond the 30 kN, i.e., 125-30 = 95 kN is shared
between the steel and the copper. Compatibility of the system requires that the steel and the
copper components of the system deflect the same amount –
Δc = Δs
or,
Pc L c
Ac Ec
=
Ps L s
As Es
and, since L c ≅ L s we can write
Pc = Ps
Ac Ec
As Es
=
1500 mm 2 x 120x10 3 MPa
1000 mm 2 x 200x10 3 MPa
Ps
Laboratory 4 Civ E 270
3
or, Pc = 0.90 Ps
We also require (equilibrium) that Ps + Pc = 95 kN
Thus, Ps + 0.90 Ps = 95 kN
Solving, Ps = 95 kN / 1.9 = 50.0 kN
and, from Eq. 1, we obtain Pc = 0.90 Ps = 45.0 kN
Finally, the total load in the steel is 30 kN + 50 kN = 80 kN and the stress in the steel can be
calculated as
σs =
Ps
As
=
80x10 3 N
1000 mm 2
= 80.0 MPa
The load in the copper is 45 kN, and the stress in the copper can be calculated as
σc =
Pc
Ac
=
45x10 3 N
1500 mm 2
= 30.0 MPa
Example 2:
The rods shown are mounted between two rigid
boundaries. The system undergoes a rise in
temperature of 100oC. Determine the force in the
rods.
Thermal expansion will close the gap and the
boundary constraint will result in internal
compressive forces.
The equilibrium condition ΣFH = 0 identifies for
us that the force in the first rod must be equal to the
force in the second rod.
The compatibility condition is:
gap = elastic deformation + thermal expansion
or, 0.5 mm = Δforce + Δtemp
The force-deformation relations give Δforce as:
Δforce =
− P x 300 mm
2
1000 mm x 200000 MPa
+
− P x 600 mm
2000 mm 2 x 100000 MPa
Laboratory 4 Civ E 270
4
The thermal deformation Δtemp is obtained as:
Δtemp = α ΔT L = 6.5 x 10-6 /&deg;C x 100&deg;C x 300 mm + 13.0 x 10-6 x 100 x 600 = 0.975 mm
Substituting into the compatibility condition for Δtemp and Δforce, the force P is obtained as
105.6 kN in compression.
4.3 EXERCISE
1.
Links AB and CD have cross-sectional areas of
80 mm2 and 30 mm2, and modulii of elasticity
70 GPa and 210 GPa, respectively. Determine
the force and stress in each link. Assume
Hooke’s law applies and neglect the weight of
the rigid bar EF.
2.
A column for a steel building is made by encasing a
W150 x 22 wide-flange steel shape in concrete. The
concrete helps to protect the steel from the heat of a
fire and also shares in carrying the load of 250 kN.
During a fire, the loaded column lengthens to 1 mm
longer than its original unloaded length. What is the
temperature increase of the column during the fire?
Assume that the temperature in the concrete and the
steel are the same and that the temperature increase
will not be sufficient to affect the material properties
significantly.
Steel
Concrete
Area
2840 mm2
70 685 mm2
E
205x103 MPa
20x103 MPa
α
11.7x10-6 /&deg;C
10.8x10-6 /&deg;C
Laboratory 4 Civ E 270
3.
5
An assembly consists of a steel bolt and an aluminum collar. The pitch of the
single-threaded bolt is 3 mm and its cross-sectional area is 600 mm2. The cross-sectional
area of the collar is 900 mm2. Calculate the stresses in the aluminum collar and in the steel
bolt for:
(a) the nut is brought to a snug position and then given an additional 1/8 turn
(b) the nut is brought to a snug position and subject to a temperature increase of 50 oC
(c) 1/8 turn beyond snug and 50 oC temperature increase simultaneously
Laboratory 4 Civ E 270
6
Blank Page
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 5
Name:
I.D.:
Date:
Mark:
Lab Section:
TORSION
5.1
GENERAL BACKGROUND
When a straight prismatic bar is subjected solely to moment about its longitudinal axis, it
twists about that axis, shearing stresses are developed, and a single stress resultant, the twisting
moment T, is developed on each transverse cross section. The straight prismatic bar of circular
cross section is the most common structural element subjected primarily to torsion. (Consider,
for example, the transmission shaft used in conjunction with most engines.) Because of
symmetry, planes normal to the axis of the bar remain plane during deformation, and the
shearing stress τ varies linearly with ρ, the radial distance from the axis of the shaft.
Using equilibrium and compatibility relationships, the elementary theory of torsion can
be described as:
τ=
and
where
φ=
Tρ
J
or
τ max =
Tc
J
TL
GJ
J
=
the polar moment of inertia
=
π d4
1 4
for a solid circular shaft
πc =
2
32
=
1
π 4
d 2 − d 14
π c 42 − c14 =
2
32
c
=
the radius of a solid circular shaft
c2
=
the outside radius of a hollow circular shaft
c1
=
the inside radius of a hollow circular shaft
(
)
(
)
for a hollow circular shaft
and d, d1, and d2 are the corresponding diameters
Laboratory 5 Civ E 270
5.2
2
DESIGN OF TRANSMISSION SHAFTS
The most common use of circular shafts is to transmit mechanical power from one device
or machine to another, for example, the transmission shaft of an automobile. The power is
transmitted through the rotary motion of the shaft, and the amount of power transmitted depends
on the magnitude of the torque and the speed of rotation.
Power (P) = Torque (T) x Angular velocity (ω)
or P (W or N&middot;m/sec) = T (N&middot;m) x 2πf (rad/sec)
where f = frequency, Hz (cycles/sec)
The transmission shafts can be designed using the elastic torsion formulas in Sec. 5.1
after the torque, T, applied to the shafts has been determined according to the above
relationships.
5.3
THIN-WALLED HOLLOW SHAFTS
Based on the equilibrium conditions of an element of the thin-walled tube cut out
between two sections a distance dx apart, the shear flow, q, is found to be constant along the
circumference of thin-walled hollow shafts. By integrating the internal torque, produced by the
shear stress, along the centerline of the cross section, the total torque, T, can be obtained as
shown in the figure. It can be expressed as:
T=qx2A
where A = area enclosed by the centerline
or τ =
q
T
=
t
2A t
where τ is the average shear stress through the thickness.
5.4
EXAMPLE PROBLEMS
Example 1: Statically Determinate Shaft
A solid shaft, as shown, is 80 mm diameter and is
made of brass with G = 40 x 103 MPa. Determine the
maximum shearing stress in the shaft and the total
rotation of end C.
A
Laboratory 5 Civ E 270
3
a)
Draw the free body diagram
b)
Calculate reaction, TA
ΣT = -TA + 25 k&middot;m -10kN&middot;m = 0
TA = 15 kN&middot;m
c)
Draw the torque diagram using the section method
d)
Calculate the maximum shearing stress
τ max
c=
=
Tmax c
J
, Tmax = 15 kN&middot;m (from the torque diagram)
80
1
1
= 40 mm , J = π c 4 = π ( 40 )4 = 4.021 &times; 106 mm 4
2
2
2
τ max
=
15 x 10 6 N mm ( 40 mm)
4.021 x 10 6 mm 4
= 149 MPa
Laboratory 5 Civ E 270
e) Determine the total rotation at C
φc =
φC/B + φB/A =
TAB L AB
GJ
4
+
TBC L BC
GJ
=
15 x 106 N • mm x 1 x 103 mm
−10 x 106 N • mm x 2 x 103 mm
+
40 x 103 MPa x 4.021 x 106 mm4 40 x 103 MPa x 4.021 x 106 mm4
=
0.093 - 0.124 = -0.031 rad. = -1.8&deg;
Example 2: Statically Indeterminate Shaft
The pulley at B is attached to two rigid supports at A and C by solid shafts AB and BC.
The shafts are made of steel with G = 75 x 103 MPa. Determine the maximum shear stress in the
shafts and the angle of twist of the pulley under a torque of 6 kN&middot;m applied to the pulley.
a)
Draw the free body diagram
Laboratory 5 Civ E 270
b)
5
Write the equilibrium equation
ΣT = -TA + 6 kN&middot;m - TC = 0
c)
Write the condition for compatibility of deformations
φB/A = φB/C = φ
TA x 250
TC x 500
=
1
1
G x π(25) 4
G x π(35) 4
2
2
TA = 0.52 TC
d)
Solve the compatibility condition together with the equilibrium equation for the
unknown reactions.
Thus, TA = 2.05 kN&middot;m, TC = 3.95 kN&middot;m
e)
Calculate the maximum shearing stress
τ max = TA x 25 or TC x 35 ,
1
π(25) 4
2
1
π(35) 4
2
whichever is larger
Shaft AB governs the design τmax = 83.5 MPa
f)
Determine the angle of twist of the pulley
φ = φB/A = φB/C = 0.64&deg;
5.5
EXERCISE
1.
The stepped shaft shown is made of two materials: steel and aluminum. The allowable
shear stresses are 170 MPa and 100 MPa for the steel and aluminum, respectively, and
o
the maximum allowable angle of twist at the free end is 8 . Check if the shaft can
safely carry the given loads and satisfies the deformation criterion. Use
G = 77 000 MPa for steel and G = 26 000 MPa for aluminum.
Laboratory 5 Civ E 270
2.
6
A steel rod to which three disks are attached is shown below. The rod is fixed against
rotation at its left end, but free to rotate in a bearing at its right end. Each disk is 300 mm in
diameter. Calculate the maximum shear stress in each segment of the rod. What is the
angle of twist of the section A relative to the fixed section E? Also draw a diagram for
angle of twist vs. position, showing the angle of twist for every position in the rod.
(G = 80 GPa)
F
E
300 m
m
20 mm
D
300 m
m
15 mm
C
200 m
m
B
10 mm
200 m
m
A
10 mm
Bearing
50 N
250 N
100 N
3.
Determine the maximum shear stress in the shafts in the motor-pump combination. The
motor delivers 75π kW at 330 rpm. Assume no power loss due to friction in the bearings or
in the gears. Shafts AB and BC are made of steel (G = 77 GPa), and each has a diameter of
75 mm.
4.
A tube has the semicircular shape shown. If the stress
concentration at the corners is neglected, what torque
will cause a shear stress of 40 MPa?
Laboratory 5 Civ E 270
5.
7
A circular and a square tube are made of the same material. The tubes have the same length,
thickness and cross-sectional area and are subjected to the same torque. What is the ratio of
the maximum shear stress in the circular tube to the maximum shear stress in the square
tube?
Laboratory 5 Civ E 270
8
Blank Page
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 6
Name:
I.D.:
Date:
Mark:
Lab Section:
BENDING STRESSES
6.1 GENERAL BACKGROUND
Beam bending theory is based on the hypothesis that bending a straight prismatic element
introduces curvature. This leads naturally to the recognition that certain fibres along the cross
section of the element will shorten while other fibres will elongate. This establishes the neutral
plane as the plane of fibres that neither shorten nor elongate. Next, the assumption that a cross
section normal to the neutral plane will remain plane and normal during bending deformation
leads naturally to the observation that strain distribution on the cross section is linear and
establishes the relation between the radius of curvature, ρ, and the strain, ε:
ε
1
=−
ρ
y
Laboratory 6 Civ E 270
2
Considerations of equilibrium between the stress distribution and the axial force, P, and
the bending moment, M, lead to the following relations:
∫ σ dA = P
A
∫ σ y dA = M
A
The first equation establishes the position of the neutral axis. When there is no axial force in the
element, the neutral axis passes through the centroid of the cross-sectional area. The second
equation leads to the stress distribution on the section. If the material is linear elastic, the stress
distribution is linear and the stresses are identified as:
σ=−
My
I
If the stress vs. strain response is not linear, the stress distribution is not linear, but the
equations of equilibrium are still valid. Thus, for a given strain distribution we can construct the
corresponding stress distribution and integrate the stress distribution using the moment
equilibrium equation to obtain the corresponding moment M. Alternatively, the stress
distribution can be broken into blocks. Each block represents a force located at the centroid of
the block. The summation of the moment of the forces about the neutral axis gives the total
moment.
6.2 EXERCISE
and bending moment diagrams. Determine the required
depth of the beam if the maximum allowable stress is
10 MPa. The cross-section is rectangular with a width of
150 mm.
15 kN
6 kN/m
2m
1m
3
Laboratory 6 Civ E 270
2.
The roof of an industrial building is to be supported by wide-flange beams spaced at 1.2 m
on center with a 6 m span (L=6 m). The design load on the roof is 12 kPa. For the typical
beam shown below, calculate the maximum bending stress.
102 mm
w
6
x
4
Ix = 65x10 mm
3
3
Sx = 415x10 mm
d=313 mm
L
W 310 x 33
3.
The overhanging beam shown below is made of cast iron for which the allowable stresses
are 40 MPa in tension and 100 MPa in compression. Draw the shear force and bending
moment diagrams in terms of wo. What is the maximum uniformly distributed load that can
be applied on the beam?
w0
80mm
N.A.
180mm
1.2 m
3m
1.2 m
6
IN.A. = 50x10 mm
4
Cross-section
(a) depth d of the stem for a value of
y = 50 mm as indicated in the
figure;
(b) the applied moment;
20
600
60 mm
200 mm
An elastic-perfectly plastic (elasto-plastic)
material has a yield strength σy = 30 MPa.
For the T-section and the indicated depth
of yielding, determine:
100 mm
100 mm
y = 50 mm
neutral
axis
yielded zones
d60
4.
(c) the fully plastic moment of this
T-section.
20
200 mm
4
Laboratory 6 Civ E 270
Blank Page
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 7
Name:
I.D.:
Date:
Mark:
Lab Section:
BENDING TEST
7.0 OBJECTIVES
•
To verify the assumption that flexural strains vary linearly with the distance from the
neutral axis.
•
To illustrate elastic and inelastic bending behaviour.
•
To calculate the modulus of elasticity and the yield strength using bending tests.
•
To predict member strength based on measured material properties and simple beam
theory.
•
To apply the test results to a practical problem.
7.1 REFERENCES
•
Beer, F.P., Johnston, Jr., E.R, DeWolf, J.T. and Mazurek, D.F. 2009. Mechanics of
Materials. 5th Edition in SI Units. McGraw-Hill Inc., &sect; 4.1-4.4 and 4.8- 4.10.
7.2 TEST PROCEDURE
7.2.1 Aluminum I-Beam, Elastic Bending Test
1.
Place and align the I-beam in the testing machine, as shown schematically in
Figure 7.1.
2.
Connect the strain gauges to the strain indicator.
3.
Apply an initial load of about 10 kN to seat the beam and the load and reaction
devices.
Laboratory 7 Civ E 270
2
4.
5.
Record strain readings for gauges at locations 1 through 5 for loads of 10 kN, 20 kN,
30 kN and 40 kN.
6.
7.2.2 Rectangular Steel Beam, Inelastic Bending Test
1.
Measure and record the cross-sectional dimensions of the test specimen.
2.
Place and align the specimen in the testing machine, as shown schematically in
Figure 7.2.
3.
Connect the strain gauges and LVDT’s to the data acquisition system.
4.
specimen.
5.
6.
7.
Continue loading until strain values at the extreme tension or compression fibres are
at least five times the yield strain.
7.3 RECORD OF TEST DATA AND ANALYSIS OF TEST RESULTS
7.3.1 Aluminum I-Beam, Elastic Bending Test
(a)
The locations of strain gauges 1 through 5, cross-sectional dimensions, geometric
properties and material properties are given in Figure 7.1.
(b)
Record the strain gauge readings on Data Sheet 7.1.
(c)
Compute theoretical strains at gauge locations 1 through 5 for loads of 20 kN and
40 kN based on material and geometric properties and the elastic flexural formula
developed in Chapter 4 of Beer, Johnston and DeWolf.
(d)
Based on the strains measured by gauges 1 and 5 at a load level of 40 kN, calculate
the modulus of elasticity E and compare it with the value of E given in Fig. 7.1.
(e)
Plot both the measured (experimental) and computed (theoretical) strain distributions
for loads of both 20 kN and 40 kN. Use a scale of 25 mm = 250 microstrain (με) on
the x-axis and use full scale for the cross-sectional depth on the y-axis. Label the
axes and curves. All these plots are to be done on the same graph.
Laboratory 7 Civ E 270
3
7.3.2 Rectangular Steel Beam, Inelastic Bending Test
(a)
(b)
The measured (average) dimensions of the rectangular bar are as follows.
beam depth,
d=
mm
beam width,
b=
mm
The material properties of this beam have been determined from a uniaxial tension
test on a coupon made from a bar of the same heat.
Modulus of elasticity, E =
MPa
Static yield strength, σy =
MPa
σu =
MPa
Ultimate strength,
(if available)
(c)
Using the measured cross-sectional dimensions, calculate the values of A, Ix, Sx, and
Zx, where x is the axis of bending.
(d)
Draw a free body diagram of the beam and use this to derive an equation for the
maximum moment, M, as a function of the machine load P.
(e)
Two strain gauges have been attached to the top and bottom of the rectangular section
to measure extreme fibre strain, as shown in Fig. 7.2. Write the equation that relates
extreme fibre strain to curvature of the beam.
(f)
Predict values of My and Mp based on measured material properties and calculated
values Sx and Zx. Calculate the corresponding loads, Py and Pp, that cause yielding
of the extreme fibres and yielding of the whole cross-section.
(g)
A graph of the moment vs. curvature behaviour for the beam will be handed out. The
graph contains two curves. In the first, moment is calculated as in part (d). In the
second, a correction for the second order effects is included in the moment
calculation. At large curvatures, the load points move inwards and the support points
move outwards, resulting in an increase in the shear span. Calculate the magnitude of
the second order effect for this test.
(h)
Label the locations of Mytest, Mypredicted, Mptest and MPpredicted on the corrected
moment vs. curvature plot.
(i)
Determine the slope of the initial linearly elastic portion of the moment vs. curvature
plot. Compare this to the product EI.
7.4 DISCUSSION
7.4.1
Comment on the measured and computed strain distributions obtained in the elastic
bending test.
Laboratory 7 Civ E 270
7.4.2
4
Compare the yield and maximum moments obtained in the Inelastic Bending Test
with your predicted values. Comment on the differences.
7.5 APPLICATION OF RESULTS
A manufacturer produces 12 m long mild steel bars having the same cross-section and
material properties as the section used for the inelastic bending test. It is proposed to load the
bars onto trucks using a crane that will pick up the bars only at their centres. The stresses during
this operation are not to exceed 50% of the yield strength. Determine the maximum moment,
calculate the maximum stress in the cross-section (neglect shear stresses), and compare this with
the yield strength. Use the yield strength given in Section 7.3.2(b) and the density of rolled steel,
γ = 7850 kg/m3. Is the proposed loading procedure acceptable? If not, suggest a solution to the
manufacturer's problem. Suggest improvements even if the procedure is acceptable.
a = 25.4 mm; b = 76.3 mm
d = 101.7 mm; t = 6.5 mm
w = 4.7 mm
Ix = 2.542 x 106 mm4
E = 72 000 MPa
Figure 7.1 Schematic of set-up of elastic bending test and cross-section of aluminum I-beam
Figure 7.2 Schematic of set-up of inelastic bending test and cross-section of steel beam
Laboratory 7 Civ E 270
5
Name:
Date:
Data Sheet 7.1
Strain Indicator Readings, ε (10-6 mm/mm), (με)
Gauge
No.
0
10
20
30
40
0
1
2
3
4
5
Measured Strains
Strain Indicator Readings, ε (10-6 mm/mm),
(με)
Gauge
No.
1
2
3
4
5
10
20
30
40
Average
Zero
Laboratory 7 Civ E 270
6
Blank Page
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 8
Name:
I.D.:
Date:
Mark:
Lab Section:
TRANSFORMATION OF STRESS – MOHR'S CIRCLE
8.0 INTRODUCTION
This laboratory session treats the problem of transformation of stress at a point in a
body. Only the case of plane stress is covered, that is, σz = τxz = τyz = 0. (Refer to Fig. 7.1
in the textbook for nomenclature). This means that two faces of the general infinitesimal
cubic element are free of stress. Section 8.1 of this write-up reviews the transformation
equations and Section 8.2 introduces the Mohr's circle representation of these equations.
Section 8.3 sets out the exercise to be performed by the student. The reference material in
the text is Chapter 7, p. 422-465.
An engineer calculates the stresses at a point in a member subjected to tensile,
compressive, torsional, or bending forces using principles developed in a course like
Civ E 270. This is done in the most convenient way, that is, an element is chosen for
investigation in a way that suits the analysis. For example, the stresses in a bar under axial
tension are calculated on a face normal to the longitudinal axis of the member. However,
such a calculation does not necessarily identify the conditions of maximum stress at the
point. The maximum stress may occur on an element with a different orientation. Having
made the initial calculations for stress, the engineer must have the &quot;tools&quot; that enable an
examination of the stresses for elements at the same point, but which have other orientations.
8.1 TRANSFORMATION EQUATIONS
Given: The state of stress is known at a certain location in a body (Fig. 8.1). These
stresses have been obtained from an analysis of the structural member or element, starting
with the forces acting on the member. The orientation of the element relative to the member
is known and, as described above, is chosen in a way that is convenient for the analysis.
Required: Establish the state of stress at the same point, but with the element rotated
arbitrarily, as shown in Fig. 8.2. (A determination of the &quot;critical&quot; orientation will eventually
be made.) In referring to the stresses on the faces of the original element, the subscripts x
and y are used. For the stresses acting on the faces of the rotated element, the subscripts x'
and y' are used.
Laboratory 8 Civ E 270
2
Procedure: The equations of equilibrium are used to establish the stresses in the
general orientation, Fig. 8.2. Because these equations involve a summation of forces, not
stresses, the resultant forces on each face must be expressed in terms of the stresses. The
resulting equations can then be solved for the unknown stresses. Certain trigonometric
transformations are also used, and the stresses for the general case can be developed as:
σ x' =
σx + σy
2
τ x'y' = −
σ y' =
+
σx − σy
2
σx + σy
2
−
σx − σy
2
cos 2θ + τxy sin 2θ
(Eq. 8.1: text eqn. 7.5)
sin 2θ + τxy cos 2θ
σx − σy
2
Figure 8.1
(Eq. 8.2: text eqn. 7.6)
cos 2θ - τxy sin 2θ
(Eq. 8.3: text eqn. 7.7)
Figure 8.2
8.2 MOHR’S CIRCLE REPRESENTATION OF THE TRANSFORMATION
EQUATIONS
Equations 8.1 and 8.2 are the parametric equations of a circle. As shown by Otto Mohr
(1835-1918), and described in text section 6.3, Equations 8.1 and 8.2 can be manipulated to
obtain
(σ x ' − σ ave ) 2 + τ x2'y' = R 2
(Eq. 8.4: text eqn. 7.11)
where
σ ave =
and
σx +σy
2
(Eq. 8.5: text eqn. 7.10)
3
Laboratory 8 Civ E 270
⎛ σx − σy ⎞
⎟
R = ⎜⎜
⎟
2
⎝
⎠
2
+ τ 2xy
(Eq. 8.6: text eqn. 7.10)
The circle can be obtained by first drawing a set of rectangular coordinates as shown
in Fig. 8.3. Referred to this coordinate system, the circle is the locus of all points having
coordinates ( σ x ' , τ x 'y ' ) satisfying Equation 8.4. Any given value of the angle θ establishes
a unique point (point M in Fig. 8.3) on the circle. The coordinates of point M describe the
stress conditions ( σ x ' , τ x 'y ' ) on the plane in the body that corresponds to that orientation,
that is, the plane perpendicular to the x’ axis.
τ
Clockwise
D
M (σ , τ )
x' x'y'
τmax
R
C
B
A
(σave, 0)
σ
Compression
σ
Tension
σmin
E
τ
Counter clockwise
σmax
Figure 8.3
8.2.1 Special Features Shown by the Mohr’s Circle Representation
It is self-evident that the conditions represented by points A and B (Fig. 8.3) are the
maximum and minimum normal stresses. It is also observed that these values of maximum
and minimum normal stress occur when the shear stress is zero. These particular stresses are
called the principal stresses and the planes on which they act are the principal planes. The
points on the circle corresponding to the principal stresses (A and B) are at the ends of a
diameter of the circle, that is, they are 180&deg; apart. It can be shown from Equations 8.1 – 8.3
that a rotation θ between two planes requires a rotation 2θ between the corresponding points
on the Mohr’s circle. It follows, therefore, that the principal stresses act on planes that are
90&deg; apart in the body.
Points D and E represent conditions of maximum shear stress. On the Mohr’s circle,
they are always 90&deg; away from the principal stress points, so that the planes on which they
4
Laboratory 8 Civ E 270
act in the body are at 45&deg; from the principal planes. Note also that there are normal stresses,
equal to σave, acting on the planes of maximum shear stress.
8.2.2
Construction of the Mohr’s Circle
The mathematical relationship between the transformation Equations (8.1 – 8.3) and
the Mohr’s circle dictates the procedure one must follow in drawing and using the circle.
The procedure is as follows:
a)
Sketch the element for which the normal and shearing stresses are known or have
been calculated (Fig. 8.4). Show the actual directions of the stresses.
b)
Establish a set of rectangular coordinate axes for plotting the Mohr's circle
(Fig. 8.5). The horizontal axis is used to plot normal stresses σ and the vertical
axis is used to plot shear stresses τ. Label the axes and show the sign
conventions.
c)
Locate the centre of the Mohr's circle on the σ (horizontal) axis, as the point with
abscissa σave=
σ x + σy
.
2
d)
Plot the point on the circle corresponding to the stress conditions on the face
perpendicular to the x axis (ie, σx and τxy). The normal stress value is plotted as
positive or negative depending on whether the stress is tensile or compressive.
However, a special rule that is specific to the plotting of the Mohr's circle must be
used to distinguish between shear stresses that tend to rotate the element
clockwise versus those tending to rotate it counter-clockwise. (On any element
there will always be two of each). If the shear stress is clockwise, the point is
plotted above the σ axis and if it is counter-clockwise, the point is plotted below
the σ axis. Thus, in the present example, the point X on the Mohr's circle in
Fig. 8.5, is the point associated with the plane perpendicular to the x axis.
e)
Following the same general procedure outlined in (d), plot the point on the Mohr's
circle associated with the plane perpendicular to the y axis. This gives point Y in
Fig. 8.5. Note that it and point X are at the opposite ends of a diameter.
f)
Having established both the centre and a diameter of the circle, the complete
circle can now be drawn.
g)
With the geometry of the circle established, the values of the stresses acting on
any element can be calculated. For example, consider the rotated element in
Fig. 8.2. The face perpendicular to the x' axis is oriented an angle θ counterclockwise from the face perpendicular to the x axis. The corresponding point X'
on the circle is obtained by starting at X and rotating through an angle 2θ. Note
that the direction of rotation on the circle is the same as the direction of rotation
between the planes. The point Y' at the other end of the diameter from X' is
associated with the plane perpendicular to the y' axis. The coordinates of points
5
Laboratory 8 Civ E 270
X' and Y' can be determined from the geometry of the circle using trigonometry.
To determine the direction of the normal and shear stresses on these planes, the
rules used in step (d) must be followed.
h)
The values the maximum shear stress and the principal stresses can be determined
by considering the vertical and horizontal diameters respectively. The orientation
of the corresponding planes can also be determined.
Figure 8.4
τ
Clockwise
Y
Y'
C
σ
Tension
σ
Compression
X'
2θ
X
σx
(right-hand vertical face)
τ
Counter clockwise
Figure 8.5
τxy (right-hand
vertical face)
6
Laboratory 8 Civ E 270
8.3 EXERCISE
8.3.1 For the following stress state, draw Mohr’s circle, then determine the magnitude
and direction of the principal stresses and the maximum shear stresses. In each
case, draw the element with the stresses shown.
σ y = 20 MPa
σ x = 100 MPa
τ xy = -30 MPa
8.3.2 At a point in a plate, the stresses are σx = 100 MPa, σy = 0, τxy = 60 MPa. A
o
weld runs across the plate at this location at θ = 45 . The allowable tensile stress
in the plate is 130 MPa and the allowable tensile stress in the weld (normal to
the weld axis) is 97.5 MPa.
(a) Is the plate safe?
(b) Is the weld safe?
θ = 45
A
σy
o
σx
A
τ xy
7
Laboratory 8 Civ E 270
8.3.3 A 100 mm x 200 mm wood cantilever beam is used to carry the loads as shown
in the figure.
(a) Calculate the normal and shearing stresses acting at point B of the section
at the support.
(b) Show these stresses on an element at B and draw the corresponding Mohr’s
circle. Determine the coordinate of the centre of the circle and the radius.
(Hint: in order to show the stress state on a single element, it must be oriented
with one side parallel to the shear stress resultant.)
z
z
A
1 kN/m
B
y
50 mm
x
P = 0.5 kN
2m
y
100 mm
Section at the Support
200 mm
25 mm
8
Laboratory 8 Civ E 270
Blank Page
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 9
Name:
I.D.:
Date:
Mark:
Lab Section:
COMBINED STRESSES
9.0 OBJECTIVES
•
To measure strains on a hollow tubular specimen loaded in bending, and to use these in
conjunction with beam flexure theory to determine values for three elastic constants;
Young's modulus, E, Poisson's ratio, ν, and the shear modulus, G. The material used is
the aluminum alloy 6061-T6.
•
To measure strains on the specimen when it is loaded in combined bending and torsion.
To use the measured strains for the two load cases, along with the theory for bending and
for torsion, in order to verify the principle of superposition.
•
To use Mohr's circle to transform stresses in two dimensions.
9.1 REFERENCE
•
Beer, F.P., Johnston, E.R. Jr., DeWolf, J.T. and Mazurek, 2009. Mechanics of Materials,
5th Edition in SI Units, McGraw-Hill Inc., Chapter 8.
9.2 TEST PROCEDURE
1.
Record the diameter and wall thickness of the tube.
2.
Identify and record the load to be applied to the combined stress specimen.
3.
Connect the strain gauges to the strain indicator.
4.
Balance the gauges and take the initial zero readings.
5.
Apply the load at location A, as shown in Figure 9.1. Record the strain gauge readings.
6.
Apply the load at location B, as shown in Figure 9.1.
7.
Record the strain gauge
Laboratory 9 Civ E 270
2
9.3 ANALYSIS OF RESULTS
9.3.1
Calculation of Strains
When the specimen is under load, the strain indicator readings for each gauge give the
strains relative to the readings obtained when the specimen is not loaded. Often there is some
drift in the zero readings during the course of the experiment, and for this reason zero readings
each gauge, the average of the two zero readings is subtracted from the strain indicator reading.
The results of this calculation are to be presented in a tabular form on Data Sheet 9.1, where the
original readings were recorded. In addition, use the principle of superposition to calculate the
strains due to torsion alone.
9.3.2
Flexural Stresses
When the load is at location A, the specimen is loaded in bending and shear only.
Gauges 1 and 2 are located on the bottom of the tube and measure, respectively, the longitudinal
and transverse strains due to bending.
The analysis for the flexural loading consists of the following steps, which are to be
(a)
Draw a free body diagram for the cantilever beam AC, shown in Figure 9.1, with the
load applied at location A. Draw the shear force and bending moment diagrams and
determine the shear force and bending moment at section C.
(b)
Compute the theoretical stress at the location of gauge 1.
(c)
On a sketch of a two-dimensional square element located at gauge 1, indicate the
longitudinal and transverse directions of the specimen and show the stresses that act
on the element. Draw the Mohr's circle for this state of stress.
(d)
The strain readings from gauges 1 and 2 can be used to determine an experimental
value for Poisson's ratio, ν. Calculate this value of ν.
(e)
The calculated stress from part (b) and the associated measured strain can be used to
determine an experimental value for Young's modulus, E. Calculate this value of E.
(f)
The third elastic constant, the shear modulus G, is dependent on ν and E. Using the
experimental values obtained for both ν and E, calculate the value of G.
9.3.3
Torsional Stresses
When the load is applied at location B, the specimen is in a state of combined flexure and
torsion. When the load is applied at location A, the specimen is in a state of flexure only. In
each case, the deformation of the specimen is in the linear elastic range, meaning that the
deflections and rotations are proportional to the applied load. Therefore, the principle of
superposition may be used, that is, the following relationship applies:
Laboratory 9 Civ E 270
3
strains due to torsion only = strains due to combined flexure − strains due to flexure
direct shear and torsion
and direct shear
For a two-dimensional element with its edges oriented parallel and perpendicular to the
axis of the tube, the torsional loading results in shearing stresses as shown in the figure below.
The corresponding shearing strains are shown by the dashed lines.
Electrical resistance strain gauges can only detect extension or contraction. For this
reason, gauges 3, 4, 5 and 6 are mounted in the direction of the principal stresses, i.e., at 45&deg; to
the longitudinal axis of the tube.
The analysis for the torsional stresses consists of the following steps:
(a)
Compute the torque at C when the load is at B.
(b)
Compute the theoretical maximum shear stress due to torsion on a cross-section at C.
(c)
Draw an element at section C, which is at mid-height of the tube on the side away
from the loading arm. Orient the element so that its top and bottom edges are parallel
to the axis of the tube. Show the theoretical stresses acting on this element due to the
torsional loading only. Use Mohr's circle to calculate the stresses acting on an
element that is oriented at 45&deg; relative to the original element and show these on a
sketch of the second element. Based on these results you can deduce the expected
signs of the strains measured by gauges 3 and 4. Do these agree with the signs of the
strains determined from the measurements? Note that a strain gauge measures
changes in length parallel to the axis of the gauge.
(d)
Repeat all the steps in item (c) for an element located at section C, which is at
mid-height and on the same side of the specimen as the loading arm. In this case, the
gauges considered are 5 and 6.
(e)
The theoretical normal stresses in the directions of gauges 5 and 6 have been
determined in step (d). These are to be used, along with Hooke's law, to calculate
theoretical values of the normal strains ε5 and ε6 in the directions of gauges 5 and 6,
respectively. Be sure to use the correct form of Hooke's law, remembering that the
element is in a bi-axial state of stress. The values of E and ν to be used are those
calculated in Sections 9.3.2(d) and (e).
Recall that the generalized expression for Hooke’s law and the relationship between
E, G, and ν are
Laboratory 9 Civ E 270
σy
σ
σ
− ν z;
εx = x − ν
E
E
E
(f)
4
G =
E
2 (1 + ν)
Calculate the ratios of measured strain value to predicted strain value for gauges 3 to
6. Record these in the table on page 7.
9.3.4 Combined Stresses
Under the combined bending and torsional loads, the directions of the principal stresses
will be different from those found in the previous section for torsion alone. Strain gauges 7 and
8 are mounted at the top of the tube at section C in the theoretically predicted directions for the
The analysis for the combined stresses consists of the following:
(a)
Draw an element located at the position of gauges 7 and 8. Take the top and bottom
edges to be parallel to the axis of the tube. On the element, show all the stresses that
act due to both bending and torsion.
(b)
For the state of stress from part (a), draw Mohr's circle. Use Mohr's circle to compute
the principal stresses and the direction in which each principal stress acts. Draw the
element on which the principal stresses act and indicate the angle θ p between the
bottom edge of the element and the axis of the tube.
9.4
(c)
Having determined the theoretical principal stresses, Hooke's law may again be used
to determine the associated normal strains. These strains correspond to those
measured by gauges 7 and 8.
(d)
Calculate the ratio of the measured strain to the predicted strain for each of gauges 7
and 8; see attached tables on page 7.
DISCUSSION
1.
Where was superposition used in this laboratory exercise? Based on the results
obtained, discuss the validity of the principle of superposition. As part of this
discussion, indicate the limitations of this principle.
2.
Discuss the differences between measured and computed values of strain. For
example, consider the significance of the observed differences and suggest possible
reasons for them.
5
Laboratory 9 Civ E 270
Dimensions in mm
Figure 9.1 Combined Stress Specimen
Figure 9.2 Location of Strain Gauges
6
Laboratory 9 Civ E 270
Name:
Date:
Data Sheet 9.1
Combined Stresses
Dimensions of Specimen:
Outside diameter:
67.90 mm
Wall thickness:
2.04 mm
P = 362.78 N
Gauge Number
1
Initial Zero
A
B
Final Zero
Average Zero
2
3
4
5
6
7
8
7
Laboratory 9 Civ E 270
Measured Strains
Position
Strain 10-6 mm/mm
Gauge Number
1
2
3
4
5
6
7
8
A
B
Torsion only
= B-A
Computed Strains
Strain 10-6 mm/mm
Gauge Number
1
2
3
4
5
6
7
8
Bending Only
—
—
—
—
—
—
—
—
Bending plus
Torsion
—
—
—
—
—
—
Torsion Only
—
—
—
—
Ratio of Measured to Computed Strains
Gauge Number
1
2
3
4
5
6
7
8
Bending Only
—
—
—
—
—
—
—
—
Bending plus
Torsion
—
—
—
—
—
—
Torsion Only
—
—
—
—
8
Laboratory 9 Civ E 270
Blank Page
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 10
Name:
I.D.:
Date:
Mark:
Lab Section:
BEAM DEFLECTIONS
10.0 OBJECTIVES
•
To measure the deflections of a member due to bending.
•
To review the calculation of beam deflections.
•
To verify experimentally the validity of beam deflection theory.
10.1 REFERENCE
Beer, F.P., Johnston, E.R. Jr., J.T. DeWolf and D.F. Mazurek, 2006,
Materials, 5th edition in SI units, McGraw-Hill Inc., Chapter 9.
10.2 EQUIPMENT
Each group will receive the following material:
•
1500 mm aluminium straightedge
•
1 wooden metre stick
•
Steel mass of approximately 1 kg
•
1 wood support with flat top
•
1 wood support with notch in top
•
2 dowels (roller and pin)
•
1 spirit level
•
1 steel ruler
•
Digital caliper (for Part 1 and Part 3))
Mechanics of
Laboratory 10 Civ E 270
2
The properties of the aluminium straight edge are:
E = 71 000 MPa
Measure the width and thickness at three different locations along the straight edge and use
the average for your calculations. Use the table below to record individual measurements.
Width =
mm
mm
Thickness =
Measurement 1 Measurement 2 Measurement 3
Average
Width (mm)
Thickness (mm)
10.3 TEST PROCEDURE
This lab consists of three parts. Each group is expected to complete all parts and they can
be completed in any order. Each group member should participate.
In this section you will investigate the deflection of simply supported beams of varying lengths
1.
Set up your apparatus with the aluminium straightedge (but excluding the mass) to the
required central span length (Lspan in Figure 10.1), keeping an equal overhang on each
end. See the first column of Table 10.1 for the required central span length. Scribed
lines on one side of the aluminium straightedge indicate the required position of the
supports.
2.
Using your steel ruler, determine the distance from the surface of the table (to be used
as your reference plane in this exercise) to the straightedge at mid-span (note that a
line has been scribed at midspan for your convenience). Record the values in Table
10.1.
3.
Place the mass at mid-span with the short dimension of the mass parallel to the span
length.
4.
Once again, measure the distance from the surface of the table to the straightedge at
mid-span and record in Table 10.1.
5.
Repeat for 4 span lengths approximately evenly spaced in a range between 900 and
1200 mm.
Laboratory 10 Civ E 270
3
6.
For each span length, calculate the measured and expected mid-span deflection from
the mass only and record in Table 10.1. Use the modulus of elasticity given above and
the section dimensions that you have measured and recorded above. Show your
calculations on engineering paper to be attached to your lab report.
7.
In the space provided, plot the mid-span deflection under the mass only vs. span
length. Plot both the measured and calculated deflections. Comment about the
discrepancy between the two sets of deflections and identify possible sources of error.
Roller
Pin
L span
Figure 10.1 Test 1 Configuration
Part 2 – Deflection of a Simply Supported Beam Under Non Symmetrical Loading
In this section you will investigate the deflection of simply supported beams of varying lengths
under a non symmetrical point load. Your objective is to find the approximate location of the
maximum deflection and to find its magnitude. The procedure is as follows:
1.
Set up your apparatus with the aluminium straightedge (but excluding the mass) to a
central span length (Lspan in Figure 10.1) of 1000 mm, keeping the overhang on each
end about the same length. Place the mass at 200 mm from the pin support as shown in
Figure 10.2.
2.
Measure the distance from the top of the table to the beam at mid-span and record this
value in Table 10.2.
3.
Using the spirit level, determine the location where the slope of the beam is
approximately zero, i.e. point of maximum deflection, and determine the distance from
this location to the roller. Record this measurement and measure the distance to the
table surface at this location.
4.
Measure and record the height of the beam above the table at locations 10 mm on each
side of the location determined in step 3 (refer to Figure 10.2) to determine whether
the location found in step 3 is the point of maximum deflection.
5.
Using either one of the methods presented in class, determine the location of
maximum deflection. Calculate the expected deflections using the same cross-section
and material properties used in the previous exercise. Compare the results of your
calculations with the measured values and comment.
Laboratory 10 Civ E 270
4
Weight of mass =
Pin
10
10
N
Measure distance
to top of table
Roller
x
200
1000
Figure 10.2 Test 2 Configuration
Part 3 – Experimental determination of the modulus of elasticity
In this part of the lab, your group will determine the modulus of elasticity, E, of a wooden metre
stick from deflection measurement. Follow the following procedure.
1. Determine the average cross-section dimensions of the meter stick using the digital
calipers at four (4) locations along the length. Using the average dimensions, calculate
the moment of inertia, I, of the cross-section about the axis of bending. Use Table 10.3
to record this information.
2. Setup the apparatus to a central span of 900 mm, keeping an equal overhang at each
end (see Figure 10.3).
3. Measure the height above the table at mid-span under self weight and record this value
in Table 10.4.
4. Place the steel mass at mid-span and measure the additional deflection due to live load.
5. Plot a bending moment diagram for the steel mass at mid-span only (ignore the self
weight).
6. Plot the corresponding curvature diagram. Hint: recall that curvature = M/EI.
7. Using the moment area theorem and the curvature diagram, find an expression that
relates the mid-span deflection to the modulus of elasticity, E.
8. Calculate the modulus of elasticity.
9. Select another span length and repeat steps 3 to 8. Comment on any variation in the
result.
Laboratory 10 Civ E 270
5
Weight of mass =
50
450
450
N
50
Figure 10.3 Test 3 Configuration
Table 10.1 – Deflection of Beams
Central span
length
(mm)
900
1000
1100
1200
Mid-span height
Self-weight
(mm)
Self-weight plus
mass
(mm)
Measured Δmid
Predicted Δmid
Mass only
(mm)
Mass only
(mm
Measured
Predicted
6
D mid (mm)
Laboratory 10 Civ E 270
900
1000
1200
1100
L central span (mm)
Height after
mass
removed
(mm)
Measured
deflection
(mm)
Predicted
deflection
(mm)
Measured
Predicted
x – 10
——
——
——
——
x + 10
——
——
——
——
Location
Height to
straightedge with
mass
(mm)
Under mass
x
Mid span
Distance x from roller to point of maximum deflection =
mm
Laboratory 10 Civ E 270
7
Table 10.3 – Dimensions of wooden beam cross-section
Width (mm)
Height (mm)
1
2
3
4
Average
Average Moment of Inertia =
mm4
Table 10.4 – Deflection of wooden beam
Central span
length
(mm)
900
Height above table
(self-weight only)
(mm)
Height above table
(with mass)
(mm)
Measured
deflection
(mm)
Calculated
E
(MPa)
Laboratory 10 Civ E 270
8
Blank Page
University of Alberta
Department of Civil and Environmental Engineering
Civ E 270
Laboratory 11
Name:
I.D.:
Date:
Mark:
Lab Section:
BUCKLING OF COLUMNS
11.1 OBJECTIVES
•
To demonstrate the effect of support conditions on the load-carrying capacity of a
column.
•
To study the effect of slenderness ratio on the load-carrying capacity of a column.
11.2 TEST PROCEDURE
11.2.1 Model Column Test— Elastic Buckling
An Aluminium column is tested with various support conditions and the measured buckling
capacity will be compared with the predicted capacity calculated using measured
dimensions. Use EAl = 68 000 MPa for calculations related to part 11.2.1 of this lab.
A schematic of the test setup for this part of the lab is shown in Figure 11.1. The aluminium
column is pinned at the top and bottom (rounded ends supported in circular grooves) and
up to three intermediate lateral supports are provided by knife edges. When a lateral
support is not needed, the cap screws used to hold the knife edges in place are removed
(use the allen key provided) and the knife edges are removed. For the lateral support at
mid-height, you should simply slide out the support plate with the knife edges.
Use the following procedure for this part of the lab:
1.
Remove the column from the test apparatus and measure and record its cross-section
dimensions in Table 1 of section 11.3. You should measure the dimensions at least at
three locations along the length and record the average values. Use an electronic
caliper for these measurements. Your measurements should be recorded in section
11.3.
Laboratory 11 Civ E 270
2
Figure 11.1 Model Column Test Set-Up
2.
Measure and record (also in Table 1) the full length of the column using the 1500 mm
aluminium straightedge provided. This measurement should be recorded to within
0.5 mm.
3.
Remove all the intermediate lateral supports (the knife edges at the top and bottom
lalteral supports and the entire support assembly at mid-height). NOTE: the spacings
between the lateral supports shown in the figure are only approximate. You should
measure the actual spacings when you conduct your experiments.
4.
Power the scale and zero with the bottom end bearing block in place. Note that the
scale reads only in pounds (lb.) or in kilograms (kg). Select pounds since it will give
you a slightly better resolution and convert later into newtons (1 lb. = 4.448 222 N).
5.
screw so that the column just fits between the top and bottom bearing blocks). Once
the column is in place, turn the loading screw (you should hold the top bearing block
since it has the tendency to spin when you turn the loading screw) and observe the
change in load. Allow the column to buckle and record the maximum load in Table 2.
6.
Replace the lateral support at mid-height and adjust the knife edges so that they touch
the column lightly on each side. Measure the distance from the knife edges to the top
of the column and to the bottom of the column. Repeat step 5 and record the load.
Note the axis about which buckling takes place and observe where the points of
inflection are located.
Laboratory 11 Civ E 270
3
7.
Place all three lateral supports and adjust all the knife edges so that they touch the
column lightly on each side. Measure the distances between the knife edges and
record these dimensions. Repeat step 5 and record the load. Note the axis about which
buckling takes place and observe where the points of inflection are located.
8.
Remove the middle lateral support assembly. Once again, measure accurately the
distance between the lateral supports and the end supports. Load the column as
described in step 5 and record the load. Note the axis about which buckling takes
place and observe where the points of inflection are located.
9.
11.2.2 Full-Size Column Test — Concentric Load
NOTE: Since this part of the laboratory requires the use of a high capacity universal testing
machine, the tests will be conducted by a qualified technician. You should focus on
the experiment and ask questions to make sure you understand the procedure. The
test results that you need for your calculations will be provided by your instructor
and the technician.
1.
Record the actual cross-sectional properties of the HSS (Hollow Structural Section).
2.
Measure and record the column lengths between pins (points of rotation).
3.
The column specimen will be placed and aligned in the test apparatus and the testing
machine.
4.
5.
6.
Repeat the same procedure for all five column specimens.
11.3 RECORD OF TEST DATA
11.3.1 Model Column
Table 1 – Measured (average) dimensions of the model column
Width, b
(mm)
Depth, d
(mm)
Full length
(mm)
Moment of Inertia (mm4)
Strong axis
Weak axis
Laboratory 11 Civ E 270
4
Table 2 – Test effective length and buckling load
Step*
Distance between
supports (mm)
Effective length,
KL† (mm)
(lb.)
(N) ‡
5
6
7
8
*
5 – no intermediate supports; 6 – column with lateral support at mid-heigth only; 7 – column
with 3 intermediate lateral supports; 8 – column with 2 intermediate lateral supports.
†
The effective length can be taken as the longest unsupported segment.
‡
1 lb. = 4.448 222 N
11.3.2 Full-Size Columns
(a)
The measured (average) dimensions of the test specimens (HSS 51 x 25 x 3.2) are as
follows:
width,
b
=
mm
depth,
d
=
mm
thickness,
t
=
mm
Laboratory 11 Civ E 270
5
Because it is difficult within the time available to calculate the cross-sectional properties exactly
(the rounded corners must be accounted for), we will simply use the values tabulated for this
section. These are Ix = 0.122x106 mm4, Iy = 0.040x106 mm4, and A = 418 mm2
(b)
The results of a stub column test on a 200 mm long specimen tested previously are
given in the form of the stress vs. strain curve. The specimen was tested in
compression using flat ends. Because of residual stresses, test results from stub
column tests (a short column test) represent the characteristics of the cross section
more accurately than do the results of tension coupon tests.
(c)
Record the test buckling loads in table 3.
Table 3 – HSS pinned end column test
Nominal Length
(mm)
Effective Length
KL (mm)
KL/ry
Measured Pcr
(kN)
cr
(MPa)
1500
1200
900
600
300
200 (stub column test)
Notes:
1.
The effective length of the stub column test specimen is one-half the nominal length
because the specimen is tested between two flat surfaces and acts as a fixed-fixed
column with an effective (buckling) length factor of 0.5.
2.
The effective length factor for all other columns is 1.0 (pinned-pinned)
3.
The effective column length of the pinned ended columns is 64 mm greater than the
actual column length because of the additional length added to the column by the knife
edge fixtures.
Laboratory 11 Civ E 270
6
11.4 ANALYSIS OF TEST RESULTS
11.4.1 Model Column
(a)
Using the measured cross-sectional dimensions, calculate the moment of inertia about
the weak axis and the strong axis and record the results in Table 1.
(b)
Using the applicable moment of inertia and effective length, calculate the elastic
buckling load for each column, Pcr, and record in Table 4 as the predicted Pcr.
(c)
Compare the calculated buckling loads with the test buckling loads. Show the ratio of
measured to predicted load for each case.
Table 4 – Comparison of test results with predicted capacity – model column
Step
Effective length,
KL† (mm)
Measured Pcr†
Predicted Pcr
(N)
(N)
Measured/Predicted
5
6
7
8
† Obtained from Table 2.
(d)
In the space below, comment about the reasons for the differences between the
measured and the predicted capacities:
Laboratory 11 Civ E 270
7
11.4.2 Full-Size Column
(a)
Using the stub column results, draw the static stress versus strain curve. Use the static
curve to calculate E and to estimate the proportional limit stress (p), the static yield
stress (ys), and the tangent modulus (ET) values at several stress levels between p
and ys. Present the calculations for ET in Table 5 and Table 6. Note that Table 6 is
used to record the calculation of the Euler buckling stress for various values of
slenderness, Le/r. A sufficiently wide range of Le/r should be selected to get critical
stresses from about 10% of the yield stress to about 10% above the yield stress.
(b)
Plot (use a full page) the theoretical Euler (Table 6) and tangent modulus (Table 5)
buckling curves on a stress () versus slenderness ratio (KL/r = Le/r) graph using the
results of your calculations reported in Tables 5 and 6. Note that the tangent modulus
curve intersects the Euler curve at a stress equal to the proportional limit (p). Both
curves should be plotted as lines.
(c)
On the same graph plotted in (b), plot the experimental results as individual points.
Table 5 – Tangent modulus calculations

(MPa)
Rise
(MPa)
Run
()
Et 
Rise
Run x 10 6
(MPa)
Le

r
2 Et

Laboratory 11 Civ E 270
8
Table 6 – Euler Buckling Calculations
Le
r
L
 cr   2 E  e 
 r 
2
```