Appendix – Derivation of the section modulus Z
In this appendix, we derive the section modulus Z of an elastic object. For simplicity, we
initially consider a rectangular solid, as shown in Fig. (1A), which has no external
forces acting on it. In Fig. (1A), the two lines AB and CD are parallel.
The application of a transverse stress, as shown by the solid arrows in Fig. (1B), causes
the object to flex. In our discussion, we assume that the left end of the object remains in
its original orientation and that the right end is flexed downwards. Note that the amount
of flexure shown is exaggerated (to make the induced changes easier to view).
The downward flex of the object will cause its top half to be under tension and its bottom
half to be under compression. The surface exactly at the midpoint remains unchanged in
length by the flexure. The line labeled “NA” in Fig. (1B) is the same length as in the
unflexed object and is known as the neutral axis.
When the object is stressed in this manner, the direction of the line AB is not changed but
the direction of the line CD is rotated. In Fig. (1B), the original direction of CD is shown
by the line C’D’. The applied stress has caused the top half of the object to stretch. The
maximum stretch (and maximum strain) is at the top of the object and is denoted as δ, the
distance between C and C’. The applied stress has caused the bottom half of the object to
compress. The maximum compression (and maximum strain) is at the bottom of the
object (the distance between D and D’) and is also equal to δ.
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By definition, the stress at the neutral axis is zero. The strain in the object is a linear
function of the distance from the neutral axis. Because the stress and strain are connected
by the elastic modulus, the stress is also a linear function of the distance from the neutral
axis.
The tension and compression in the two halves of the object due to the externally applied
forces cause the object to generate reactive internal stresses, as shown in Fig. (2A).
The tensile stress applied to the top half of the object induces an internal stress to the left
in that part of the object. Likewise, the compressive stress applied to the bottom half of
the object induces an internal stress to the right in that part of the object.
The external forces acting on the object generate a “clockwise” torque on the object. The
induced internal stresses in the object generate a torque acting on the object in the
“counterclockwise” direction. This internal torque stops the clockwise rotation of the
object. The magnitude of the internal torque which can be generated by the object
determines the ultimate strength of the object.
The internal stress in the object is largest at the outer edges of the object, as shown in
Fig. (2B). This maximum stress is denoted as σM.
The stress σ at a point which is a distance y from the neutral axis is given by
 y
R
σ =  σ M .
(1)
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Figure (3A) shows the object with a circular cross section as viewed from a perspective
rotated by 90o from the earlier figures. The line NA is the neutral axis, a line that is
under neither tension nor compression. The stress at a given point on the cross section of
the object is given by Eq. (1). Note that it is determined by the perpendicular
distance from the neutral axis NA to the given point (i.e., by the y-coordinate).
Consider the infinitesimal area dA shown in Fig. (3A). The magnitude of the force dF on
this area dA is given by
 y
dF =  σ M dA
R
(2)
and the corresponding magnitude of the internal torque dτ on this area dA is
 y2
 y
dτ =  σ M (dA) y = 
R
 R

σ M dA .

(3)
The magnitude of the net internal torque is then given by
 y2 
σ
τ = ∫  σ M dA = M
R
 R
∫y
2
dA .
(4)
This integral is known as the second moment of the area:
I = ∫ y 2 dA .
(5)
Equation (4) becomes
τ=
σM
R
I=
I
σ M ≡ Zσ M
R
,
(6)
where the section modulus Z is defined to be the second moment of the area I divided by
R, the distance from the neutral axis to the outer edge of the object. (Note that the section
modulus is sometimes denoted as S.)
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The section modulus is the quantity which determines a object’s ability to resist external
forces which flex the object. For a fixed value of the maximum stress σM applied to the
object, an object with a larger Z will generate a larger internal torque. The larger Z is, the
greater the ability of the object to resist the external torque.
For the sake of simplicity, let us initially consider the object to have a circular crosssectional area and no marrow cavity, as shown in Fig. (3B). The symmetry of this
example facilitates the evaluation of the integral for the second moment of the area:
R
I = ∫ y 2 dA = 2 ∫ y 2 R 2 − y 2 dy =
−R
πR 4
4
.
(7)
The corresponding section modulus is:
πR 4
Z=
I
πR 3
= 4 =
.
R
R
4
(8)
Theropod femora are hollow because of the marrow cavity. A model for the crosssectional area of a theropod femur is shown in Fig. (4). The second moment of the
area for a hollow object is
I=
π
(R
4
4
)
− Ri4 ,
(9)
where R is the outer radius of the object and Ri is the radius of the marrow cavity. The
corresponding section modulus is:
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Z=
π
(R
4R
4
)
− Ri4 .
(10)
However, a bone such as a femur does not have perfectly symmetrical cross sections.
The appropriate section modulus Z can then be calculated by a numerical integration if
the cross-sectional data are available.
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Appendix Figure captions:
Figure 1: (A) A rectangular section of a object subject to no external forces. (B) The
flexed object is shown by the solid lines. The solid arrows show the external forces
applied to the object. The line AB has remained parallel to its direction in the unflexed
object. The line CD has rotated compared to its direction in the unflexed object (shown
by line C’D”). The distances of stretching at the very top and of compression at the very
bottom are equal and shown by δ. NA shows the neutral axis which is neither stretched
or compressed.
Figure 2: (A) Internal reactive stresses of the object are to the left above the neutral axis
(NA) and to right below the neutral axis. (B) The axis CD has been rotated
counterclockwise so that it is vertical and defines the y-axis for the coordinate system.
The radius of the object is R. The internal reactive stresses are indicated by the arrows.
Figure 3: (A) The cross section of the object of radius R. Note that this view is rotated by
90o from the previous figures. The neutral axis (NA) is used to define the x-axis. In this
view, the force due to the internal stress is out of the page for the positive values of y
(shown by “+”) and into the page for negative values of y (shown by “-“). (B) The cross
section of the object of radius R. Because of the symmetry of the stress field, the
infinitesimal area dA is expressed as 2 R 2 − y 2 dy.
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Figure 4: Cross section of a circular, hollow object with outer radius R and inner radius
R i.
21
Figure 1
22
Figure 2
23
Figure 3
24
Figure 4
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