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Dial Indicator #
E, ksi
I, in^4
Beam height, in
L, in
10000
2,5
3
90
1
2
3
Measured Deflections (inches)
Dial Indicator Position, x
(inch)
22,5
45
67,5
Applied Loads
Loading Case / Test 1
P1 (lbs)
P2 (lbs)
100
100
Measured Deflections (inches)
Loading Case / Test 1 Loading Case / Test 2 Loading Case / Test 3
-0,0765
-0,1075
-0,0765
-0,0420
-0,0535
-0,0345
Applied Loads
Loading Case / Test 2 Loading Case / Test 3
100
0
0
100
-0,0360
-0,0555
-0,0435
CIVE 302
Spring 2013
Dr. Dowell
T. Johnson
Lab 8. Beam Displacements
When forces or moments act on an object, they may either act normal to a plane (axial),
transverse to a plane (shear), or impart rotation to a plane (bending). Each of these three types
of forces causes a respective type of deformation, shown below in Figure 8.1, and can be
understood by looking at the change in shape of a block. Axial forces shrink or expand the face
on which they act, shear forces translate the face on which they act, and moments rotate the
face on which they act.
Figure 8-1: Different Types of Deformation
Beams are some of the most well-known structures in engineering practice, resisting applied
loads by bending between reaction points. This behavior is often caused by means of point or
distributed loads acting on the member which cause shear, so beams experience both shear
forces and bending moments. Beams by themselves generally do not carry axial forces, though
they may in certain frame systems. Knowing this, we can thus reason that the deflected shape
of a beam incorporates both shear and flexural deformations.
For most beams, flexural deformations dominate the deflected shape of the member and
accurately capture force-displacement behavior; thus, shear deformations are treated as
negligible . To understand how flexural deformations work, let us look at two common types:
cantilever beams and simply supported beams. Figure 8-2 shows a cantilever beam with point
load P applied at the right end of the member and fixed support at the left end of the member.
The right end of the beam has no reaction and is free to both rotate and translate; the left side,
however, is fixed from all rotation and translation. In comparison, the simply supported beam
shown in Figure 8-3 is pinned at its left end and has a roller at its right end, with vertical load P
applied at one-third of the span length. The left support is a pin that cannot translate in any
direction, while the right support is a roller that prevents vertical displacement but allows
longitudinal displacement.
y
MA A
P
x
B
RA
L
Deflected
shape
Figure 8-2: Deflected Shape of a Cantilever Beam in Bending
y
Pin
A
RA
P
x
B
C
L/3
Deflected
shape
L
Figure 8-3: Deflected Shape of a Simply Supported Beam in Bending
2
Roller
RB
Pin and roller details permit member rotations at both ends of the simply supported member.
The roller support is required to ensure that axial forces are not developed from vertical loading
of the beam, allowing for very small longitudinal movements to relieve any membrane forces
(tension or compression) that would otherwise develop.
Bending moments can be related to the physical curvature in the beam by dividing by the
flexural rigidity of the beam, where rigidity is the product of a beam’s material resistance and
geometric resistance to deformation. For flexure, the modulus of elasticity E defines the
material resistance and section moment of inertia I defines the geometric resistance. This is
given as
M

EI
(8-1)
From calculus, we know that the curvature of a function is the change in slope with respect to x

d
dx
(8-2)
Furthermore, by definition slope is equal to the change in y with respect to x, so

dy
dx
(8-3)
3
Applying these relationships to one another, we can substitute Eq. (8-3) into Eq. (8-2) which
results in
 dy 
d 
d  dy  d 2 y
dx
       2
dx
dx  dx  dx
(8-4)
Now placing this definition of curvature into Eq. (8-1) yields
M d2y

EI dx 2
(8-5)
If EI is taken to the other side of the equation the expression is written
d2y
EI 2  M
dx
(8-6)
The expression given above in Eq. (8-6) is the basic equation to determine vertical
displacements of beams with applied loading. Once the bending moment M is given as a
function of x, a differential equation develops that is readily solved by direct integration. The
first integral provides the slope at any location x and the second integral results in
displacements at any position x along the member length.
4
For the cantilever beam given in Figure 8-2, reactions at the left end consist of the vertical shear
RA and bending moment MA of P and PL, respectively, which may be derived from statics. At
any location x along the member length the bending moment is given as
M  RA x  M A  Px  PL
(8-7)
When this moment is substituted into Eq. (8-6) the following differential equation results
EI
d2y
 Px  PL 
dx 2
(8-8)
To understand the next step this can be written as
EI
d  dy 
   Px  PL
dx  dx 
(8-9)
And since dy over dx is the slope of the beam, this is written
EI
d
 Px  PL 
dx
(8-10)
5
Finally dx is taken to the right of the expression, giving
EId  Px  PLdx
(8-11)
Now both sides of Eq. (8-11) are integrated, resulting in the slope expression
 x2

EI   P  PLx   C1
 2

(8-12)
Because there are no limits on the integrals a constant of integration C1 is required. This
constant is found based on known boundary conditions, and represents EI times the slope at
the x = 0 position. In this case, the slope at the beginning of the beam is zero because of the
fixed wall of the cantilever. At the start of the beam x = 0, and this is substituted into Eq. (8-12)
to determine the constant as
 02

EI 0   P
 PL0  C1
 2

(8-13)
Therefore constant C1 = 0. Thus the final slope expression is
 x2

EI   P  PLx 
 2

(8-14)
6
Note that this is actually the slope of the beam multiplied by EI. If slope is desired then E and I
are moved to the other side of the expression giving


1  x2
 P  PLx 
EI  2

(8-15)
The force P can also be brought out of the parentheses resulting in


P  x2
  Lx 
EI  2

(8-16)
This is the final slope (or rotation) expression that is valid for the entire length of the member.
For the cantilever example of Figure 8-2 the slope will be negative at all points, which is
consistent with the deflected shape shown in the figure. This can be reasoned mathematically
by observing that for each positive step of dx along the length of the beam dy is always
negative. Alternatively stated, as position x increases to the right, in the positive direction, the
vertical displacement y increases in the downward direction – negative.
The expression given above provides the slope of the beam for any given location x along the
member length, but vertical displacements are also desired. Integration of this slope
expression provides the displacement equation. In Eq. (8-17) the slope is written again, but in
differential form and E and I are brought back to the left of the equation, giving
7
EI
 x2

dy
 P  Lx 
dx
 2

(8-17)
Now dx is moved to the right of the expression, giving
 x2

EI dy   P  Lx dx
 2

(8-18)
Integration of both sides of this expression allows vertical displacements y to be found
 x3
x2 
EIy  P  L   C2
2
6
(8-19)
Here a second constant of integration C2 developed and is determined from known boundary
conditions. Similar to C1, C2 represents EI times the initial displacement at the x = 0 position.
Because the cantilever is fixed at its left end, there is no vertical displacement y at x is zero,
therefore
2
 03

0 
  C2
EI 0  P
L
2 
 6
(8-20)
From this expression the second constant is found to be zero, and the displacement equation
can be written
8
 x3
x2 
EIy  P  L 
2
6
(8-21)
Modulus of elasticity E and moment of inertia I can be moved to the right of the expression,
resulting in
y
P  x3
x2 
  L 
EI  6
2
(8-22)
This is the final equation for the displaced shape of the cantilever beam and is valid for the full
member length, as is the slope expression given in Eq. (8-16). For a cantilever beam both
constants of integration are always zero. A beam with different boundary conditions, however,
causes one or more of the two integration constants to be non-zero – do not naturally assume
both to be zero.
This approach to solving displacements is called double integration and works well if one
moment expression can be used to define the beam moments for the full member length. If
different moment expressions are required for different regions of the beam, though, the
method becomes cumbersome and difficult to manage. Each discontinuity in the slope of the
moment diagram requires the consideration of a new moment expression, meaning that two
additional boundary conditions and two additional constants of integration must be considered.
For example, if two point loads are applied to a simply supported beam then there are three
different moment regions, resulting in six constants of integration. This also results in three
separate displacement equations and three separate slope equations, one for each region.
9
The constants can only be found by setting up and solving six simultaneous equations, which is
a significant amount of work and highly unfavorable to hand calculations.
A mathematical variation of the double integration approach, however, can be used to address
the above concerns. Using expressions called singularity functions, we can solve the simply
supported beam shown in Figure 8-3 with a single applied point load P. Singularity functions
allow one moment expression to be written that is valid for the entire member length. Thus,
what would be two regions for double integration is condensed into a single region and, thus,
only two constants of integration need to be considered, eliminating the need to solve several
simultaneous equations for numerous C values. As shown below, both constants of integration
can be determined from the boundary conditions of the beam.
Based on statics, the reaction at the left support RA is equal to two-thirds of the applied load P,
and the reaction at the right support RB is one-third of P. When x ≤ L/3 the bending moment is
M  RA x 
2
Px
3
(8-23)
And when L/3 ≤ x ≤ L the bending moment can be derived as
L 2
1
1

M  RA x  P x    Px  Px  PL  PL  x 
3 3
3
3

10
(8-24)
Using singularity functions the two moment expressions given above can be written as a single
expression as
M  RA x  P x 
L
2
L
 Px  P x 
3
3
3
(8-25)
The angular brackets are singularity functions, and they are different from regular functions in
that they have rules that follow them and must be obeyed. The rules can be stated as
Furthermore, singularity terms may be integrated according to the following rule
xa
 x  a dx  n  1
n
n1
C
Examining this expression logically, for any distance along the beam x that is less than the
distance to a, the function does not exist. When this distance x becomes greater than the
distance a, however, the function assumes the form of a normal polynomial expression.
Similarly, integration of singularity functions follows the same rules as polynomial integration.
Note that terms should not be brought into or removed from singularity functions – the
singularity brackets do not behave as parenthesis operators.
11
Understanding these rules, it is clear from viewing Eq. (8-25) that this one expression satisfies
both moment equations given for the two separate regions in Eqs. (8-23 and 8-24). In this case
the value a is replaced by L/3, the distance along x at which the point load P is applied.
Using the moment expression derived via singularity functions, the double integration approach
may be applied in the same manner as the cantilever example in order to determine
displacements. As with the moment expression, both a single slope and displacement
expression are valid for the entire member length.
d2y 2
L
EI 2  Px  P x 
dx
3
3
(8-26)
The first integral gives the slope expression of
dy 2 2 P
L
EI
 Px 
x
dx 6
2
3
2
 C1
(8-27)
Since the slope is not known at any location along the length, the constant of integration is not
yet determined and another integration will be required.
EIy 
1 3 P
L
Px 
x
9
6
3
3
 C1 x  C2
(8-28)
12
Now both constants of integration are found from known displacement boundary conditions.
At the left end of the member there is no vertical displacement (@ x = 0, y = 0), and the
singularity term is zero (x < a = L/3). Therefore EI 0  1 3 P0  0  C1 0  C2 9 (8-29) From this C2 must be zero, which is what we expect given that a reaction at the x = 0 location prevents the beam from deflecting at that point. Considering our second boundary condition, the right end of the beam when x = L the vertical displacement is also zero, giving 3 1 P2  3 EI 0  PL    L   C1 L  9 6 3  (8-30) Note that since the x value of L is larger than a of L/3, the singularity term turns on and acts like a regular function. Now the constant C1 is solved for giving C1   5 2 PL 81 (8-31) Thus the final slope and displacement equations are 13 EI dy 2 2 P L  Px  x dx 6 2 3 1 P L EIy  Px 3  x 9 6 3 2 5 PL2 81 (8-32) 5 PL2 x 81 (8-33)  3  Note that both slope and displacement equations contain a singularity function, and calculations of slopes or displacements at a point must properly follow the rules for these terms. For example, the singularity term is not included for slope and displacement values at x = L/4, but it is included if x = L/2. Any number of singularity terms can be included to define the moment expression. For the beam given in Figure 8-4, the moment expression is defined using two singularity terms as M  RA x  P x  L 2L L 2L P x  Px  P x  P x 3 3 3 3 (8-34) Integration two times of this expression provides the slope and displacement equations, both of which will included two singularity terms. 14 y Pin A P P x D C B Deflected shape P L/3 Roller P L/3 L Figure 8-4: Simply Supported Beam with Multiple Point Loads Experiments Aluminum beams with 90 inches between simple supports (length L = 90 inches) are to be loaded with different weights and displacements measured at ¼ points. The beam has moment of inertia I = 2.5 in4 and modulus of elasticity E of 10,000 ksi. Loads are applied to the beams at ⅓ locations (30 inches and 60 inches) using load hangers and pre-measured weights. Displacements are measured with dial indicators. After the beam is set on the supports and the load hangers are placed, the dial indicators are positioned at the ¼ locations (22.5 inch spacing). The displacement gages are then zeroed before the weights are placed on the hangers, but note that the sensitivity of the equipment may make it difficult to completely zero each indicator. If this occurs, record the initial reading of the gage and use it as your zero reference point. Using this approach the displacement indicators will measure vertical displacements associated with only the applied loads. Test 1. Apply 100 pounds at 30 inches and at 60 inches in 20 pound increments (40 pound total increments and total of 200 pounds applied). Measure the displacement after each incremental loading. Also measure the displacements upon removal of each load increment. Test 2. Apply 100 pounds at 30 inches (and no load at 60 inches) in 20 pound increments. Measure the displacement after each incremental loading. Also measure the displacements upon removal of each load increment. 15 Test 3. Apply 100 pounds at 60 inches (and no load at 30 inches) in 20 pound increments. Measure the displacement after each incremental loading. Also measure the displacements upon removal of each load increment. REQUIRED CALCULATIONS: - Predicted versus measured displacements at 22.5, 45, and 67.5 inches for each load increment in Tests 1, 2, and 3. - Bending moment diagram for the 100 lb point load case for Tests 1, 2, and 3. - Maximum tensile and compressive bending stresses the beam experiences for Tests 1, 2, and 3. Compare these against an assumed yield stress of 40 ksi. REQUIRED PLOTS: Using the vertical axis as displacement and the horizontal axis as position x along member length), plot the following for the 20, 60, and 100 pound load cases (3 plots total): - For Test 1, predicted displacements at 2.5 inch increments along the length of the beam. Plot individual measured displacements on the same graph at the 22.5, 45, and 67.5 inch marks as points. - For Test 2, predicted displacements at 2.5 inch increments along the length of the beam. Plot individual measured displacements on the same graph at the 22.5, 45, and 67.5 inch marks as points. - For Test 3, predicted displacements at 2.5 inch increments along the length of the beam. Plot individual measured displacements on the same graph at the 22.5, 45, and 67.5 inch marks as points. 16 REQUIRED DISCUSSION: Discuss how the theoretical and measured displacements compare. If they are different explain why. Do they trend in the same direction? Upon unloading, do the measured results follow the results from loading at the same applied force level? Are there any differences between predicted displacements upon loading and unloading, or are they theoretically identical? Compare the results from Test 1 at the dial indicator locations to the sum of the results of Tests 2 and 3 at those locations. Note anything of interest from your calculations, and state any mathematical relationships you may observe. Definitions: Modulus of elasticity E (ksi). This is the initial linear slope of the stress-strain diagram. Moment of Inertia (in4). A property of the beam cross-section geometry that affects curvature, stress and strain. Dial indicator. A displacement-measuring device that behaves like a spring-operated potentiometer. The plunger is spring loaded and the entire gage is typically connected to a stand with a magnet that allows it to be easily positioned along the beam. Prior to applying loads to the beam, the plunger positioned such that it will not run out of stroke during the experiment and the digital dial indicator is set to zero displacement. If they indicator cannot be fully zeroed then a reading is made prior to load application and again after the load is applied. The displacement of interest is found from the difference in the two displacement readings, treating the initial reading as the effective zero mark. 17 Purchase answer to see full attachment