Beam Deflection
Ryan Fuld
ME3900-001
Villanova University
Performed February 25, 2008
I.
Objective
The objective of this experiment was to experimentally find the reaction forces and deflection of
beams under various loading conditions, and to determine beam stiffness. In addition, results these
values were to be verified by comparing against theoretical values defined by principles of statics and
mechanics of materials. Throughout this process, the factors that affect the deflection and stiffness
were to be identified.
II.
Theory
Deflection, in of itself, is essentially a displacement caused by a loading condition. For the case of a
beam, it is the displacement of the beam in a direction not along its axis. In fact, a beam is defined
in this way; that is, applied loads must be acting perpendicular to the axis of the specimen. When
designing a beam, deflection is generally undesired. As a result, what is known as beam stiffness is a
large design factor in beams. Beam stiffness is defined as the ability of the beam to resist deflection.
All elastic materials have stiffness, and thus, in general terms, it is simply the ability of a material to
resist deformation. An empirical definition for beam stiffness can be constructed once the
deflection of the beam is known. In order to find this, consider a simply supported beam of length
‘L’ with a load ‘P’ acting at a distance ‘x’ from the left end, demonstrated in Fig. 1 below. This load
will cause two occurrences; first, reaction forces will develop at the constraints, and second, the
beam will deflect at all unconstrained points along its axis, resulting in a problem having infinite
degrees of freedom. These reaction forces must exist in order to ensure that the sum of the forces
x
P
L
Figure 1. Simply Supported Beam with Load ‘P’ and dimensions ‘x’ and ‘L’
and moments on the entire beam equate to zero. Utilizing this method, these reactions can be
found theoretically. First, examine the free body diagram, where R1 is the reaction at the left
support and R2 is the reaction at the right support.
Figure 2. Free-Body Diagram of Beam, with Reaction forces ‘R1’ and ‘R2’
Summing moments about the left support, the following is obtained:
L ⋅ R2 − x ⋅ P = 0
(1)
Solving for R2 yields
R2 = P ⋅
x
L
(2)
Now that one reaction is known, the other can be found by setting the sum of forces to zero.
R1 + R2 − P = 0
(3)
x⎞
⎛
R1 = P ⋅ ⎜1 − ⎟
⎝ L⎠
(4)
Rearranging terms and solving for R1 yields:
As can be seen, Eqs. (2) and (4) will allow theoretical values for the reaction forces to be found.
However, perhaps of more interest is finding the maximum beam deflection. As previously stated,
deflection is caused by a load, but is, more specifically, determined by numerous other factors. For
this case, the load will be applied at the center of the beam. When this is done, the beam will bend,
and thus, stresses will occur throughout the beam. Using stress and strain relationships, a
differential equation can be determined. First, Hooke’s Law shows:
σ = E ⋅ε
(5)
Here, E is Young’s modulus for the specific material of interest. For this context, σ is treated as the
maximum stress and ε is treated as the maximum strain. Of course, for this beam, this maximum
occurs at the bottom of the beam, under the load, where the beam fibers are experiencing a tensile
stress. Working off this concept, the maximum stress for a beam can be utilized, wherein
σ=
M ⋅c
I
(6)
where
b ⋅ h3
I=
12
(7)
Here in Eq. (6), c is the distance from the center of the beam to the bottom of the beam, and I is the
moment of inertia of the beam. Additionally, M is the bending moment, which is a function of
beam position. In Eq. (7), b is the base length and h is the depth of the beam. Now, equating Eq.
(5) and (6), one obtains:
E ⋅ε =
M ⋅c
I
(8)
A relationship between c and ε can be established by examining the geometry of the deflected beam.
This relationship is defined as follows:
ε=
c
(9)
ρ
Here, ρ is the radius of curvature of the beam. Substituting Eq. (9) into Eq. (8), and solving for the
radius of curvature yields:
ρ=
E⋅I
M
(10)
However, the radius of curvature can be written in terms of the beam deflection, y, by utilizing the
elastic curve equation. The exact relationship is as follows:
ρ=
1
d2y
dx 2
(11)
Lastly, setting Eqs. (10) and (11) equal yields the differential equation of interest:
d2y
M
=
2
E⋅I
dx
(12)
This equation can be integrated in order to solve for deflection, but first an expression for bending
moment, M, in terms of the position from the left support, x, must be obtained.
A general knowledge of static concepts allows the following diagrams to be constructed without
specific formulations:
Shear Diagram
15
P/2
10
Shear, V(x)
5
0
00
1
2
3
4
L/2 5
6
7
8
9
-5
-P/2
-10
-15
Beam Position, x
Figure 3. Shear Diagram for a simply supported beam of length ‘L’ with a load ‘P’ at the center
Fig. 3 demonstrates the shear diagram, where maximum shear is shown to be at the supports. The
reason for the magnitude of P/2 at each end is essentially due to Eqs. (2) and (4), as the value of x in
those equations is simply L/2. This means that the load is distributed evenly among the reactions,
causing a symmetry. These reaction values will be utilized momentarily. Now examine Fig. 4:
10
L
Moment Diagram
10
PL/4
8
Moment, M(x)
6
4
2
0
0 0
1
2
3
4
5
L/2
6
7
8
9
-2
Beam Position, x
Figure 4. Moment Diagram for a simply supported beam of length ‘L’ with a load ‘P’ at the center
Fig. 4 demonstrates the moment diagram, where maximum moment occurs in the middle, as was
already assumed when the equations for stress and strain were examined. The value of maximum
moment is determined by taking the moment arm of L/2, multiplying by the reaction force. Note
that moment is simply the integral of shear. Of greater interest is the general equation for moment.,
which is a function of beam position, x. This can be done by examining the following cut of the
beam:
Figure 5. Beam Cut
Figure 5 shows the free body diagram of the beam cut, where V is the shear force, which is
inconsequential to the moment equation. Summing moments about the cut gives:
M = R1 ⋅ x
(13)
10
L
Here we specify R1 as P/2, which was explainer previously. Thus, for this case:
P⋅x
2
(14)
d2y
P⋅x
=
2
2⋅E ⋅ I
dx
(15)
M =
When Eq. (14) is utilized in Eq. (12), one obtains:
Using separation of variables, this equation can be integrated a first time. The resulting equation is
as follows:
dy
P ⋅ x2
=
+ C1
dx 4 ⋅ E ⋅ I
(16)
This equation represents the slope of the deflection curve, where C1 is a constant resulting from
integration. However, implementing proper boundary conditions allows this constant to be
determined. Recall that maximum deflection occurs at the center of the beam, or L/2; this
essentially means that the slope at this point is equal to 0. Using this condition gives the following:
C1 =
− P ⋅ L2
16 ⋅ E ⋅ I
(17)
Subbing Eq. (17) back into Eq. (16) yields:
dy
P ⋅ x2
P ⋅ L2
=
−
dx 4 ⋅ E ⋅ I 16 ⋅ E ⋅ I
(18)
This is a particular equation for beam slope when the load acts in the center of the beam. This can
be integrated again in order to obtain the equation for deflection. It is as follows:
y ( x) =
P ⋅ x3
P ⋅ x ⋅ L2
−
+ C2
12 ⋅ E ⋅ I 16 ⋅ E ⋅ I
(19)
Another constant results from integration, specified by C2. However, identifying a boundary
condition for deflection will allow this to be determined. Because there is a support at the left end,
or at x=0, the deflection here must be 0. Using this condition in Eq. (19); that is, setting both y(x)
and x to 0, gives:
C2 = 0
Thus, the particular equation for deflection for these conditions is as follows:
(20)
y ( x) =
P
(
4 ⋅ x 3 − 3 ⋅ x ⋅ L2 )
48 ⋅ E ⋅ I
(21)
This is extremely useful for this particular experiment. That is, once the beam deflection is
determined experimentally, this equation can be used to check the validity of the findings. However,
this equation must first be written for the particular case of deflection at the center. This means that
a value of x=L/2 will be substituted into Eq. (12), giving:
y ( L / 2) =
− P ⋅ L3
48 ⋅ E ⋅ I
(22)
This equation allows beam stiffness to be revisited. Writing Hooke’s Law in an alternate form states
that:
k=
P
δ
(23)
This equation defines stiffness, k, as the magnitude of the ratio of load, P, to deflection, δ.
Recognize this equation as the common expression for spring stiffness, in which stiffness is simply
the applied load divided by the distance stretched. Moreover, all elastic materials are essentially
springs; a beam can be mathematically modeled as a spring if its stiffness is found. Yet, to find the
stiffness of a beam, deflection must be known. However, deflection in this equation is simply the
magnitude of the beam deflection in Eq. (22); we set δ equal to absolute value of y, and substitute
Eq. (22) into Eq. (23), yielding:
k=
48 ⋅ E ⋅ I
L3
(24)
This equation, along with Eqs. (22), (2), and (4), are all theoretical values of variables of interest in
this experiment. Experimental values for the reactions and deflection can be measured using the
SM104 Beam-Apparatus, which is shown in Fig. 6 below. Once deflection is found, the absolute
value can be taken and used as δ in Eq. (23). This allows an experimental value of beam stiffness to
be found.
The procedure for the first part of the experiment, determining the experimental reactions, involves
placing a beam on the apparatus, unlocking the load cells, and applying a load. A load cell, along
with a dial gauge, which will be explained shortly, are shown in Fig. 7 below. The load cell contains
a spring that deflects, and this deflection is displayed on a gauge, which shows the reaction forces.
When locked, this spring is unable to deflect. The load cell can read up to 45 N. The type of beam
used in finding the reactions is inconsequential, as only concepts of statics are involved. Once the
beam was placed on the apparatus, the metal hanger was set on the beam and the two load cells were
zeroed out. After this, the load could be applied to the hanger. Two loads were applied, as 5 N and
as 20 N. The location of the hanger was varied along the beam, first at ¼ span from the left
support, then at ½ span from the left support, and lastly at ¾ span from the left support.
Figure 6. SM104 Beam-Apparatus
Figure 7. Dial Gauge and Load Cell
The process for determining the deflection was done by using reading dial gauge, after locking the
load cells. Here, beam type is of consequence, because, as can be seen in Eq. (22), both modulus of
elasticity and moment of inertia affect the deflection, and thus, stiffness. Five different beams were
used here. The first three beams were steel, while the others were brass and aluminum. However,
the steel beams varied in their depth; the first had a depth, h, of 6.35 mm, the second had h=4.72
mm, and the last had h=3.12 mm. Both the brass and aluminum had a depth, h, of 6.35 mm. Here,
as mentioned, the load was placed in the center of each beam. Six different loads were examined;
they were 0, 10, 15, 20, 25, and 30 N. Weight was added to add up to each load in that order, and
then removed all the way back down to 0, allowing two readings to be obtained for loads except 30
N. Note that this was different for the aluminum beam; here the order of weight was 0, 2, 4, 6, 8,
and 10 N. Here, due to available weights, the 10 N was done separately from the others, meaning
that only two readings were obtained for 0 to 6 N. The dial gauge was utilized in order to read the
deflection of the center of each beam. That is, the very end of the gauge was placed on the top of
the metal hanger in order to read the distance it displaced. It contains smallest increments of 0.1
mm, and can read up to 25 mm.
All beams used had a base length, b, of 19 mm, and an effective length, L, of 700 mm.
III.
Tables and Graphs
Table 1: Analytical and Experimental Data Obtained For Part I of Experiment
Reaction Forces (N)
Total Exper.
Weight W (N) at Location 'a'
Percent
Reaction
along beam
Analytical
Experimental
Forces R1 + R2 Difference
R1
R2
R1
R2
a = 1/4
a = 1/2
a = 3/4
(N)
3.75
1.25
3.75
1.25
5
5
0.00%
15
5
15.5
5.25
20
20.75
3.75%
2.5
2.5
2.75
2.5
5
5.25
5.00%
10
10
10
10
20
20
0.00%
1.25
3.75
1.25
3.75
5
5
0.00%
5
15
5.25
15.25
20
20.5
2.50%
Load (N)
0
10
15
20
25
30
Table 2: Deflection of Steel and Brass Beams (WIDTH b=19 mm)
Mid-span Dial (mm)
Steel
Steel
Steel
Brass
Depth h = 6.35mm
Depth h = 4.72 mm
Depth h = 3.12 mm
Depth h = 6.35 mm
+
+
+
+
0
0
0
0
0
.1
0
0
.9
.95
2.1
2.2
7.2
7.4
1.6
1.65
1.4
1.4
3.2
3.2
10.8
11.1
2.4
2.45
1.8
1.85
4.2
4.3
14.45
14.65
3.2
3.25
2.3
2.3
5.3
5.3
18.0
18.2
4.0
4.0
2.7
N.A.
6.3
N.A.
21.6
N.A.
4.8
N.A.
Table 3: Deflection of An Aluminum Beam (WIDTH b=19 mm)
Dial Gauge Reading (mm)
Aluminum: Depth h = 6.35 mm
Load (N)
+
0
0
0.05
2
0.5
0.6
4
1.05
1.1
6
1.6
1.6
8
2.1
N.A.
10
2.6
N.A.
Material
Steel
Steel
Steel
Brass
Aluminum
Table 4: Comparison of Theoretical and Experimental Stiffness
Modulus, E
Beam Depth, h
I
Theoretical Experimental
(N/mm2)
(mm)
(mm4)
200000
6.35
405.41
11.35
10.99
200000
4.72
166.49
4.66
4.74
200000
3.12
48.09
1.35
1.39
110000
6.35
405.41
6.24
6.25
69000
6.35
405.41
3.91
3.86
Table 5: Deflection and Stiffness for Steel Specimens loaded to 30 N
Beam Depth, h
Magnitude of Deflection
Stiffness
(mm)
(mm)
(N/mm)
6.35
2.7
11.11
4.72
6.3
4.76
3.12
21.6
1.39
1.53
189.00
0.159
Percent
Difference
3.11%
1.75%
3.09%
0.17%
1.36%
IV.
Discussion of Results
The experiment provided fairly accurate results, corresponding with theory to a large degree. In the
first part of the experiment, the reaction forces were calculated. Table 1 demonstrates the results
from this portion of the experiment. As can be seen, the largest discrepancy from theory was 5%
error, occurring during the trial where the 5 N load was placed at mid-span. This corresponded to a
total reaction force of 5.25 N. Even here, the difference is not too substantial. Moreover, there are
possible sources of error that could account for this discrepancy. The most likely source of error
would be an improper location of the hanger; that is, if it was not directly where it was intended to
be, there could have been some changes with the load distribution to the supports. However, this
would only account for individual reaction error, and would not explain why the total was 5.25 N in
this case. This may be explained by an error in the load cells, such as if the spring were stretched
beyond its elastic limit, or something else of that nature. Another possible source of error could be
improper zeroing of the load cells, which would also account for this. In actuality, error likely
resulted from a combination of these factors. Nonetheless, the error encountered was small enough
overall; that is, three of the trials had an error of 0.
The second part of the experiment was also highly valid when compared to theory. Here, the
deflection and stiffness were calculated using multiple beams and loads. Table 2 shows the
experimental deflection data when the first four beams (steel and brass) were loaded (+) and
unloaded (-), while Table 3 shows the deflection of the aluminum beam under different loads. The
maximum deflection for all specimens occurred under the 30 N load, and were 2.7 mm for the
thickest steel, 6.3 mm for the second thickest steel, 21.6 mm for the thinnest steal, 4.8 mm for
bronze, and 2.6 mm for aluminum. Examining the steels verifies that deflection has an inverse
linear relationship with moment of inertia. Note that Table 3 uses lesser loads than Table 2 in
determining deflection. The likely reason for this is the fact that aluminum’s modulus of elasticity is
quite lesser (69,000 N/mm2) than that of steel and brass (200,000 N/mm2 and 110,000 N/mm2,
respectively). Thus, aluminum will yield and leave the elastic range prior to steel or brass. However,
at loads only up to 30 N, yielding is sill not likely. As a matter of fact, in order to verify this, a 30 N
load was tested on the aluminum beam. Here, the dial gauge showed a deflection of 7.9 mm, and
the beam successfully returned to its original state once the load was removed; this number verifies
that deflection has an inverse linear relationship with modulus of elasticity. This means that, in fact,
the same original loading conditions in Table 2 could have been applied to the aluminum beam.
Nonetheless, deflection results would have been nigh the same either way, as deflection has nothing
to do with strength if the specimen is in the elastic range.
Returning to Table 2, it can be seen that upon loading and unloading, there are inconsistencies in
deflection values for identical loads. Most obvious is the case of the steel bar with a 3.12 mm depth.
Here, the difference is as much as 0.3 mm, which occurs at the 15 N load. Notice that this beam
had the smallest moment of inertia out of all the beams examined. By Eq. (22), as the moment of
inertia gets very small, the deflection at the mid-span will become very large. This implies a
correspondence between the inconsistencies and the amount deflected. This can be explained by
the fact that when there is large deflection, and curvature is high, it is more difficult for the specimen
to return to its original state. This can thus be applied to all specimens and loads, providing a
general reason for such inconsistencies.
Examining Fig. 8 through 12 allows an analysis of load vs. deflection to be performed for each
beam. Each graph shows a linear relationship between load and deflection. Here, a trendline was
constructed in Excel, which uses the least squares method to generate an equation describing the
relationship. Here, the slope of each line indicates the stiffness of that specimen, since stiffness is
equivalent to load over deflection, as can be seen in Eq. (23). The slopes each these graphs were
thus taken as the approximate experimental stiffness for each specimen, and were recorded in Table
4. This table also displays theoretical values of beam stiffness, along with percent error between the
two. Theoretical beam stiffness, of course, calculated according to Eq. (24). As can be seen, the
highest percent difference was 3.11%, and the lowest was 0.17%, for the thickest steel specimen and
the bronze specimen, respectively. Reasons for any error likely lie in zeroing methods or recording.
Yet, these numerical values show high accuracy between theory and experimentation, further
validating the methods implemented.
Now examine Figs. 13 and 14. These display the relationship between beam stiffness and moment
of inertia and modulus of elasticity, respectively. Both are seen to be linearly related, as expected
according to Eq. (24). Note that the slope for Fig. 14 is extremely small, which is due to the fact
that the modulus of elasticity is extremely high, numerically, compared to the numerical value of
beam stiffness. Moment of inertia has a greater slope, but is it is still not substantial.
The last relationship examined concerns the beam depth, h. Figs. 15 and 16 plot beam deflection vs.
depth and beam stiffness vs. depth, respectively. The data used for these plots is that of Table 5.
Here, the three steel specimens are presented, this time for the single trial with the 30 N load.
Stiffness is calculated simply by dividing this load by the deflection. Notice here that the final data
set is not one obtained from experimentation. The reason for this is that, by only using the three
points available, appropriate relationships may not be apparent. In order to ensure these clarity of
these relationships, a data point is examined far out of the range of the others. This point is
calculated by considering a specimen under a 30 N load that carries a maximum stress of 700
N/mm2. Using this stress, and the appropriate equations, the beam depth, h, can be found. This
can then be used to find the moment of inertia, and thus the beam deflection and beam stiffness.
The exact procedure for determining this is presented in the Calculations section. Furthermore, as
can be seen in Figs. 15, beam deflection varies with beam depth by approximately a third order
polynomial. A power fit shows an exponent of -2.99, which is essentially -3. This agrees with
theory, as moment of inertia contains the cube of beam depth. And since the moment of inertia is
in the denominator, as can be seen in Eq. (22), this results in a relationship wherein y is proportional
to h-3. In Fig. 16, stiffness is examined, with an opposite relationship. Here, the exponent is 2.99,
which is very close to 3. This agrees with theory as well, as in Eq. (24) it was found that stiffness is
proportional to moment of inertia. Thus, because moment of inertia contains a third power term of
beam depth, stiffness, k, is proportional to h-3. Notice that both depth and length of the beam affect
the deflection and stiffness to the same degree, only in opposite manners; they are on opposite sides
of the operator. This means that increasing or decreasing either of these alone can significantly
affect the amount a beam will deflect.
V.
Experimental Data
The experimental data from the experiment follow.
VI.
Calculations
The following are sample calculations for variables calculated in the experiment.
Theoretical Reaction Forces:
The following are calculations for reaction forces with a 5 N load applied at ¼ span from the left
support. The value of x used in these equations will thus be ¼*L.
Restating Eq. (4), gives:
x⎞
⎛3⎞
⎛ 1⎞
⎛ ¼⋅L⎞
⎛
R1 = P ⋅ ⎜1 − ⎟ = 5 N ⋅ ⎜1 −
⎟ = 5 N ⋅ ⎜1 − ⎟ = 5 N ⋅ ⎜ ⎟ = 3.75 N
L ⎠
⎝4⎠
⎝ 4⎠
⎝
⎝ L⎠
Restating Eq. (2) gives:
⎛¼⋅ L ⎞
⎛1⎞
⎛x⎞
R2 = P ⋅ ⎜ ⎟ = 5 N ⋅ ⎜
⎟ = 5 N ⋅ ⎜ ⎟ = 1.25 N
⎝L⎠
⎝ L ⎠
⎝4⎠
Adding these two reactions together returns the original load, 5 N.
Total Reaction Force Percent Difference
The following is a calculation for the percent difference between the total experimental reaction
forces and the applied load for a load of 20 N applied at ¼ span from the left support.
The values for the experimental reaction forces are taken from Table 1.
The equation for percent difference is as follows:
W − ( R1 + R 2) 20 N − (15.5 N + 5.25 N ) 20 N − (20.75 N )
100%
=
=
= .375 *
= 3.75%
W
20 N
20 N
1
Moment of Inertia
The following is a calculation for the moment of inertia of a steel beam with depth, h, of 6.35mm.
Restating Eq. (7) gives:
b ⋅ h 3 (19mm) ⋅ (6.35mm) 3
I=
=
= 405.41mm 4
12
12
Theoretical Beam Stiffness:
The following is a sample calculation for beam stiffness of a steel beam with depth, h, of 6.35mm.
The modulus of elasticity, E, is 200,000 N/mm2. The moment of inertia, I, used in this calculation
can be taken from the result of the previous calculation. The length of the beam, L, is 700 mm.
Restating Eq. (24) gives:
48 ⋅ E ⋅ I 48 ⋅ (200,000 N / mm 2 ) ⋅ (405.1mm 4 )
k=
=
= 11.35 N / mm
(700mm) 3
L3
Experimental Beam Stiffness Percent Difference:
The following is a calculation for the percent difference between the experimental beam stiffness
and the theoretical beam stiffness of a steel beam with depth, h, of 6.35mm.
The value for experimental beam stiffness is taken from the slope of Graph []. The value for
theoretical beam stiffness used in this calculation can be taken from the result of the previous
calculation.
The equation for percent difference is as follows:
k exp − k
k
=
(10.99 N / mm − 11.35 N / mm)
− 0.36 N / mm
100%
=
= 0.311 *
= 3.11%
11.35 N / mm
11.35 N / mm
1
Deflection of Steel Beam Stressed to a Prescribed Amount:
The following are calculations yielding the deflection of a steel beam under a load of 30 N, stressed
to 700 N/mm2. The first equation used is Eq. (6), as stress is known. However, the unknown
desired is beam depth, h, which is not explicit in Eq. (6). Thus, Eq. (7) must be plugged into Eq. (6),
along with the symbolic value of c, which is simply h/2. This gives:
h
M⋅
M ⋅c
2
σ=
=
I
b ⋅ h3
12
Solving for h in this equation yields:
6⋅M
h=
σ ⋅b
The value of M used in this equation is the maximum moment, occurring at x=L/2. Thus, the
moment is:
PL
M =
, as can be read from Fig. 4.
4
Substituting this into the equation for h gives:
6⋅ P⋅ L
6 ⋅ 30 N ⋅ 700mm
= 1.53mm
h=
=
4 ⋅σ ⋅ b
4 ⋅ 700 N / mm 2 ⋅ 19mm
This can be plugged back into Eq. (7) to find I, giving:
b ⋅ h 3 (19mm) ⋅ (1.53mm) 3
I=
=
= 5.67mm 4
12
12
When plugged into Eq. (22), the deflection can be found:
− P ⋅ L3
− (30 N ) ⋅ (700mm) 3
y ( L / 2) =
=
= −189mm
48 ⋅ E ⋅ I 48 ⋅ (200,000 N / mm 2 ) ⋅ (5.67mm 4 )
VII. Conclusion
The experiment wholly displayed expected relationships, corresponding well with theory. Statics
held true through part one of the experiment, in finding reaction forces. For the second part, from
Table 1, it was gathered early on that deflection has an inverse linear relationship with both modulus
of elasticity and moment of inertia. From Figs. 13 and 14 it was concluded that stiffness does
indeed have an opposite relationship with these factors; that is, it has a direct linear relationship with
modulus of elasticity and moment of inertia. Lastly, relationships between both deflection and
stiffness with beam depth were discovered and displayed in Figs. 15 and 16, showing that deflection
varies inversely to the third power with beam depth, while stiffness varies directly to the third power
with beam depth.