Centroids moments of inertia

University of Manchester
School of Mechanical, Aerospace and Civil Engineering
Mechanics of Solids and Structures
Dr D.A. Bond
Pariser Bldg. C/21
e-mail: d.bond@umist.ac.uk
Tel: 0161 200 8733

UNIVERSITY OF
MANCHESTER
1st YEAR LECTURE NOTES
MECHANICS OF SOLIDS AND STRUCTURES
SEMESTER 2

§ 11: CENTROIDS AND MOMENTS OF AREA
§ 12: BEAM SUPPORTS AND EQUILIBRIUM
§ 13: BEAM SHEAR FORCES & BENDING MOMENTS
§ 14: BENDING THEORY

1

Centroids and Moments of Area

11. CENTROIDS AND MOMENTS OF AREAS
11.1 Centroid and First Moment of Area
11.1.1 Definitions
The Centroid is the geometric centre of an area. Here the area can be said to be concentrated, analogous to
the centre of gravity of a body and its mass. In engineering use the areas that tend to be of interest are cross
sectional areas. As the z axis shall be considered as being along the length of a structure the cross-sectional
area will be defined by the x and y axes.
An axis through the centroid is called the centroidal axis. The centroidal axes define axes along which the
first moment of area is zero.
The First Moment of Area is analogous to a moment created by a Force multiplied by a distance except this
is a moment created by an area multiplied by a distance. The formal definition for the first moment of area
with respect to the x axis (QX):

Q X = ∫ ydA (11.1)

Similarly for the first moment of area with respect to the y axis (QY) is:

QY = ∫ xdA (11.2)

where x, y and dA are as defined as shown in Figure 11.1.
Y x dA

C
y
y

x X

Total Area = A

Figure 11.1
The X and Y subscripts are added to indicate the axes about which the moments of area are considered.
11.1.2 Co-ordinates of the Centroid
The centroid of the area A is defined as the point C of co-ordinates x and y which are related to the first
moments of area by:

Q X = ∫ ydA = yA

QY = ∫ xdA = x A (11.3)

An area with an axis of symmetry will find its first moment of area with respect to that axis is equal to zero
i.e. the centroid is located somewhere along that axis. Where an area has two axes of symmetry the centroid
is located at the intersection of these two axes

2


11.1.3 Example: Centroid of a Triangle
Determine the location of the centroid of a triangle of base b and height h.
Y dA
x

dx
^
Y
x C h
^
X

y

b X

Figure 11.2
Ans: x = 2b/3 and y = h/3
11.1.4 Centroid of a Composite Area
Where an area is of more complex shape a simple method of determining the location of the centroid may be
used which divides the complex shape into smaller simple geometric shapes for which the centroidal
locations may easily be determined. Consider Figure 11.3 which shows a complex shape of Area A made
from three more simple rectangular shapes of Areas A1, A2 and A3. As the centroids of the rectangular shapes
are easily determined from symmetry the locations of their respective sub-area centroids are used to calculate
the location of the centroid of the composite shape.
A1
Y
C1
x

C

A2 C2
y

C3

A3 X

Figure 11.3
Q Y = ∫ xdA

= ∫ xdA + ∫ xdA + ∫ xdA
A1 A2 A3

= x 1 A1 + x 2 A 2 + x 3 A 3
= xA
The same method can be used to calculate the y-wise location of the centroid of the composite area.

3


11.1.5 Example: Centroid of a L section

A1 x=
(b + ht
2
)
2(b + h )
Y

h 2 + t (b + 2h )
y=
x 2(b + h )
h C

C1 y

t
C2

A2 X
t
b

Figure 11.4
11.2 Second Moment of Area
11.2.1 Definitions
The second moment of the area about the x axis (IX) is defined as:

I X = ∫ y 2 dA (11.4)

and the second moment of the area about the y axis (IY) similarly as:

I Y = ∫ x 2 dA (11.5)

Some texts refer to second moment of Area as Moment of Inertia. This is not technically correct and Second
Moment of Area should be preferred.
11.2.2 Example: Rectangle (of dimensions b × h)
Derive an expression for the second moments of Area for a rectangle with respect to its centroidal axes (Use
the symbol ^ to indicate centroidal axes and properties with respect to these axes).
The centroid is easily located by using intersecting axes of symmetry.
ˆ dy
Y
dA

h/2
y
C
ˆ
X

h/2

bh 3
Ans: I X =
b
ˆ
12
Figure 11.5

4


The solution for second moment of area for a rectangle is frequently used as many composite shapes are
broken into rectangular sections to determine their composite second moments of area. The rule is often
recalled as:
The second moment of area of a rectangle about its horizontal centroidal axis is equal
to one-twelfth its base (b) multiplied by its height (h) cubed.
Similarly I Y may be determined to be equal to b3h/12.
ˆ

11.2.3 Relationship to Polar Second Moment of Area
The Rectangular Second Moments of Area IX and IY are able to be related to the Polar Second Moment of
Area about the z axis (J) which was introduced in the section on Torsion.
I X + I Y = ∫ y 2 dA + ∫ x 2 dA
( )
= ∫ y 2 + x 2 dA

= ∫ r 2dA

I X + IY = J Z (11.6)
11.2.4 Radii of Gyration
The radius of gyration of an Area A with respect to an axis is defined as the length (or radius r) for which:

I X = ∫ y 2 dA = rX A
2

I Y = ∫ x 2 dA = rY2 A
(11.7)
J Z = ∫ r dA = r A
2
Z
2

As for the First Moment of Area, the X, Y and Z subscripts are added to indicate the axes about which the
second moments of area or radii of gyration are considered.
11.2.5 Parallel Axis Theorem
If the axes system chosen are the centroidal axes the Second Moments of Area calculated are known as the
Second Moments of Area about the centroidal axes. Such axes are often annotated differently to other
axes indicating that they are centroidal axes. In this course the symbol ^ shall be used. If the second moment
of area about another set of axes is required then the Parallel Axis Theorem may be used rather than having
to recalculate the Second Moments of Area.
Y dA

ˆ
Y
C ˆ
y
ˆ
X
Total Area = A
y
d

X

Figure 11.6

5


ˆ
Figure 11.6 shows an area with a centroid at C (with a centroidal x axis shown as X ) and for which the
Second Moment of Area with respect to the X axis is required. The X axis is a distance d away from the
centroidal axis.
I X = ∫ y 2dA

= ∫ (y + d ) dA { as y = y + d}
2
ˆ ˆ

= ∫ y 2 dA + 2d ∫ y dA + ∫ d 2dA
ˆ ˆ

= ∫ y 2 dA + ∫ d 2dA
ˆ {as ∫ y dA = 0 about centroidal axes}
ˆ

I X = I X + Ad 2
ˆ (11.8)
This demonstrates that if the Second Moment of Area is known around an area’s centroidal axis the Second
Moment of Area of that area about another axis distance d from the centroidal axis is simply the sum of the
Centroidal Second Moment of Area and the product Area × d2.
This theorem applies to the Second Moments of Area IX and IY as well as to the Polar Second Moment of
Area provided the appropriate centroidal values are used.
11.2.6 Example: Second moment of Area of a Rectangle about its base axis
For the rectangle shown; determine its second moment of area about its base and left edge axes (X and Y).

ˆ
Y
Y

h/2

C
ˆ
X

h/2

X
b
Figure 11.7

bh3 hb3
Ans: I X = , IY =
3 3

6


11.2.7 Second Moment of Area for a Composite Section
Consider the composite area shown in Figure 11.8. To determine the Second Moment of Area of such a
complex structure a similar approach to that used for calculating the centroids of complex areas is used.
Follow these steps:
i. Determine the Centroids of the sub-Areas
ii. Calculate the Second Moments of Area of the sub-Areas about their centroidal axes
iii. Use the Parallel axis theorem to move sub-Area Second Moments of Area to axis of interest
iv. Sum the contributions of each sub-Area to the overall Second Moment of Area.
A1
Y
d1 C1

A2
d2
C2
d3
A3
C3

X
Figure 11.8
The validity of the above approach can be seen below for determining IY of the area in Figure 11.8:
IY = ∫ x 2 dA

= ∫ x 2 dA + ∫ x 2 dA + ∫ x 2 dA
A1 A2 A3

= ∫ (ˆ + d ) dA + ∫ (ˆ + d 2 ) dA + ∫ (ˆ 3 + d 3 ) dA
2 2 2
x 1 1 x 2 x
A1 A2 A3

= ∫ (ˆ
x 2
1 x )x2 ˆ ( 2
ˆ2 )
2
+ 2 ˆ 1d1 + d12 dA + ∫ ˆ 2 + 2 x2 d 2 + d 2 dA + ∫ x3 + 2 x3d 3 + d 32 dA
ˆ ( )
2

A1 A2 A3

(
ˆ
1
ˆ
1 1 )(
= I Y A + A d + I Y A + A2 d + I Y A + A d
2
ˆ
2
2
2 )( 3
2
3 3 ) as ∫ (2 xd )dA = 2d ∫ (ˆ )dA = 0
ˆ x
A3 A3

This method is often well suited to a tabular layout or a spreadsheet.
11.3 Tabulated Centroids and Second Moments of Area
Many text books list the locations of standard area centroids and provide the Second Moment of Area around
these centroids. The departmental databook has such a table and will be allowed for use in exams
therefore students should become familiar with the use of this table.
11.4 Units
First Moment of Area has units of Length3.
Second Moment of Area has units of Length4.

7


11.4.1 Example: A Regular I section
Derive an expression for the second moment of Area for a regular I beam with respect to its centroidal x axis.

Y ˆ
Y

t

ˆ
X
a
t

t

X
b
Figure 11.9
The I section may be represented as being comprised of a rectangle of dimensions b×(2t+a) from which two
smaller rectangles of dimensions ½(b-t)×a have been taken out all of which have the same x-wise centroidal
axis. The total second moment of area is then simply the sum of all the contributions (with the missing areas
being subtracted).
b(a + 2t )
3 1
(b − t )a 3
IX =
ˆ − 2. 2
12 12
b(a + 2t ) − (b − t )a 3
3
=
12
This solution could also have been derived by considering the three rectangles separately and using parallel
axis theorem although there would have required significantly more work.

8


11.4.2 An Unsymmetrical I section
Consider an unsymmetrical section shown below. The section is symmetrical about the vertical centroidal
Ŷ

( ) axis only. The y-wise position of centroid is to be found so that the second moment of area about its x-
wise centroidal axis can be determined. In examples such as this where the component is constructed from
"regular sub-areas" it is best to follow a tabular method as shown below.
5cm
Y

y3 3 1.5cm

y2 2 6cm

y
^
1cm X

y1 1 1cm

X
10cm
Figure 11.10
The section above is divided into three rectangular areas, (1), (2), (3). The bottom x-axis is used as datum.
The tabular method of finding the centroid and the second moment of area are demonstrated in the following
Table.
Section Area (Ai) yi (Ay )i d = y - yi Ad2 IX i
ˆ I X i + A i d i2
ˆ

i (cm2) (cm) (cm3) (cm) (cm4) (cm4) (cm4)
1 10 0.5 5 3.208 102.913 0.833 103.746
2 6 4 24 -0.292 0.512 18 18.512
3 7.5 7.75 58.125 -4.042 122.533 1.406 123.939
Totals 23.5 87.125 246.197

y=
∑ (Ay) i

∑A i

= 3.708cm
I X = ∑ I X i + ∑ A i d i2
ˆ ˆ

= 246.197 cm 4
2
Use first three columns to find y before proceeding to calculate d, Ad etc.

9

Beam Supports and Equilibrium

12. BEAM SUPPORTS AND EQUILIBRIUM IN BENDING
12.1 Introduction
12.1.1 What is beam bending?
Tension, compression and shear are caused directly by forces. Twisting and bending are due to moments
(couples) caused by the forces.

Tension F F
⇒ F F

Compression F F
⇒ F F

⇒
V
Shear V V
V

Torsion T T ⇒ T T

Bending M M
⇒ M M

Figure 12.1 Structural Deformations
Bending loads cause a straight bar (beam) to become bent (or curved). Any slender structural member on
which the loading is not axial gets bent. Any structure or component that supports the applied forces
(externally applied or those due to self weight) by resisting to bending is called a beam.
12.1.2 Eraser Experiment
What is the basic effect of bending? Mark an eraser on the thickness face with a longitudinal line along the
centre and several equi-spaced transverse lines. Bend it. The centre line has become a curve.
Question:
• What happens to the spacing of the transverse lines?
Bending causes compression on one side and extension on the other. By inference there is a section which
does not extend or shrink. This is called the Neutral Plane. On the eraser this will be the central longitudinal
line. Consistent with extension and compression, bending must cause tensile (pulling) stresses on one side
and compressive (pushing) stresses on the other side of the neutral plane.
Bending is predominantly caused by forces (or components of forces) that are act perpendicular to the axis of
the beam or by moments acting around an axis perpendicular to the beam axis.

10


12.2 Representation of a beam and its loading
Uniformly
Concentrated or Distributed Loads Beam
Point Load (UDL) Non-uniform
Distributed Load
Moment
y WC
x wDE -wFG
MH
B J

A C D E F G H z

RBy RJy

Reaction Loads
Figure 12.2: Typical representation of a beam
For a schematic diagram (suitable for a FBD), normally only a longitudinal view along the centre line (the
locus of the centroids of all the transverse sections, called the Centroidal Axis) is used to represent the beam
(see Figure 12.2). Vertical (y-direction) forces acting on the beam will be assumed to act at the centre line,
but normal to it. Concentrated forces that act at specific points, such as W at C, are shown as arrows.
Distributed loads are shown as an area (or sometimes as a squiggly line) to represent a load distributed over a
given length of the beam. Distributed loads have dimensions of force per unit length. Moments are
represented by a curved arrow.
Question:
• What is the most common form of distributed load?

For reference, a Cartesian co-ordinate system (xyz) consistent with the right hand screw rule is always used.
The origin can be located at any convenient point (usually an end or the centre of the beam). The z-axis is
aligned along the axis of the beam, the y-axis in the direction of the depth of the beam and the x-axis in the
direction of the beam width (into the page). When representing a beam on paper the y and z planes are
normally drawn in the plane of the page and the x axis is perpendicular to the page. The bending forces and
moments considered in this 1st year course will only act in the yz-plane (i.e. the plane of the page).
Become accustomed to this axis system as it is common to most analyses in future years.
12.3 Supports for a beam and their schematic representation
12.3.1 Introduction
A beam must be supported and the reactions provided by the supports must balance the applied forces to
maintain equilibrium. Types of support and their symbolic representations are given in the following
sections.
12.3.2 Simple support
A simple support will only produce a reaction force perpendicular to the plane on which it is mounted (see
Figure 12.3). Simple supports may move in the plane on which they are mounted but prevent any motion
perpendicular to this plane. Simple supports do not produce forces in the plane on which they are mounted
and moments are not restrained by a simple support. So for a simply supported beam the axial displacements
and rotations (which cause slope changes) at the supports are unrestrained (i.e. in Figure 12.3 the beam is

11


free to move along the z-axis and rotate about the x axis). Imagine these supports as being similar to the
supporting wheel of a wheelbarrow.
y y Beam
A
z z
RAy
Support y-direction
Reaction Load

Figure 12.3: Simple support reaction loads
12.3.3 Hinged or Pinned end support
Hinged or pinned supports provide similar support to simple supports with the addition of support in the
plane on which they are mounted i.e displacements in the axial direction are prevented (in Figure 12.4 the
beam is only free to rotate about the x axis). Imagine these supports as being the same as the connection at
the top of a grandfather clock pendulum.

y y Beam
A
RAz
z Support z-direction z
Reaction Load
RAy
Support y-direction
Reaction Load

Figure 12.4: Hinged/Pinned support reaction loads
12.3.4 Fixed or built-in end support
Fixed end supports (also called encastre) support moments in addition to lateral and axial forces. No axial,
lateral or rotational movements are possible at a built in end (i.e in Figure 12.5 the end is not able to move in
either the y or z direction nor can it rotate about the x axis). Imagine these supports as being like the
connection of a balcony onto a building.

y MA
Support Reaction y
Moment Beam
A
RAz
z Support z-direction z
Reaction Load
RAy
Support y-direction
Reaction Load

Figure 12.5: Fixed support reaction loads and moments
12.4 Distributed loads
To simplify the analysis of a distributed load it is usually easier to replace the distributed load with a point
load acting at an appropriate location. As the units of distributed loads are load per unit length the equivalent
point load may be determined by statics.

12


Area under
w(z) = Aw
w(z)
We
y y e

z z

L L

dAw = w(z).dz
w(z)
y

dz z

z
Figure 12.6: Replacing a distributed load with an equivalent point load
For the two cases to be equivalent the sum of the forces in the y and z directions have to be the same and the
sum of the moments about any point have to be the same. Considering forces in the y direction first:
L

∑ Fy = ∫ w( z ) dz = Aw = We
0

That is, the equivalent point force of a distributed load (We) is equal to the area under the w(z)
function (Aw).
Now consider moments about the origin:
L L

∑ M o = ∫ w( z ).z dz = ∫ z dAw = We .e
0 0

Note the similarity between this equation and Equation (11.3) for the first moment of area which allows the
previous expressions to be re-written as:
L

∫ z dA
0
w = z Aw

We .e = Aw .e
∴e = z
That is, the point along the beam at which the effective force (We) must act is at the centroid of the area
(Aw) under the distributed load curve w(z).

13


12.5 Equilibrium considerations for a beam
Consider a beam carrying loads as shown in the figure below. The right hand support at B is a simple support
and can only carry vertical forces. All the horizontal force components have to be supported by the left hand
(hinged) support, at A.
y
c

W M
w
A B

C z

l1

l2
l3

l4
L

Figure 12.7 A beam hinged at the left and simply supported on the right, loads as shown
Consider a point C where the left and right hand parts of the beam are to be separated into two free body
diagrams. To maintain equilibrium in the separated sections additional forces and moments must be applied
at the new ends to keep both sections of the beam in the same geometry as when the beam was intact. These
forces and moments are known as the axial and shear forces and bending moments at position C. These
forces and moments determine how a beam deforms under loading. To determine these forces and moments
the support reactions must first be obtained from the conditions of equilibrium of forces and moment for the
whole beam. Then the forces at the point C (shear force and bending moment) may be obtained from force
and moment equilibrium of the part of the beam to the left or to the right of the section. The left and right
hand parts and all the possible forces acting the two new ends are shown in the free body diagrams below.
12.5.1 The support reactions
The support forces are obtained from the conditions of equilibrium of forces and moments on the whole
beam so DRAW a FBD of the beam and apply equilibrium conditions.
W M
y w
B

RAz A z
We is equivalent
RAy point load to RBy
distributed load w
We

Figure 12.8: FBD of entire beam used to calculate support loads and moments
Equilibrium conditions require:
∑F z = 0, ∑F y = 0, ∑M = 0
First with the condition MA = 0, we get the vertical support reaction at B.
∑

14


∑M A =0

(
= −W .l1 − w l3 − l 2 ) (l 3 + l2
2
)
− M + RBy .L

W .l1 +
2
(
w 2 2
l3 − l2 + M )
⇒ RBy =
L
Equilibrium of forces in the vertical direction, Fy = 0, gives the vertical support force at A:
∑

∑F y =0
= −W − w(l3 − l 2 ) + RBy + RAy
⇒ RAy = W + w(l3 − l 2 ) − RBy
∑ Fz = 0 provides the axial support force at A.
∑F z =0
= R Az
Note: If any of the forces calculated are negative then they act in the opposite sense to that assumed in the
FBD.
12.6 Sign Conventions
The forces and moments that act on a beam at point C (MC, FC and VC) are assigned positive or negative
signs depending on the face that they act on. If the face they act on has a normal in the positive z direction
then positive forces and moments are in the positive y or z directions or as defined by the right hand rule. If
the face has a normal in the negative z direction then a positive force or moment is in the opposite direction.
This sign convention is shown below.

AXIAL FORCES F +ve F F -ve F

SHEAR FORCES V +ve V V -ve V

BENDING MOMENTS M M M M
+ve -ve

w w

DISTRIBUTED LOADS +ve -ve

Figure 12.9 Sign convention for Axial Forces, Shear Forces, Bending Moments and distributed loads
12.6.1 The forces at point C
If the beam is cut at point C (at a distance c from A) then for equilibrium we require equal and opposite
forces FC and VC as well as equal and opposite moments MC, acting at the severed sections of the two parts
of the beam. MC, FC and VC are provided in the complete beam by the internal stresses in the material
of the beam.
• MC = Sum of moments due to all forces to one side of the point C (including support forces) is called
the BENDING MOMENT acting on the vertical face of the beam at position C.

15


• FC = Sum of all x-direction (axial) forces to one side of the point C is called the AXIAL FORCE
acting on the vertical face of the beam at position C.
• VC = Sum of all the y-direction (transverse/shear) forces to one side of the point C is called the
SHEAR FORCE acting on the vertical face of the beam at position C.

y c
y
W
w M
VC
A VC B
MC MC
RAz FC PC

z z
l2-c RBy
RAy l3-c
c-l1
C l4-c
L-c
C

Figure 12.10: Left Hand and Right Hand FBDs for beam sectioned at C
Thus, in the above problem, MC, FC and VC can be found by considering the LH end FBD to be:

M C = − R Ay .c + W ( c − l1 ) = − w(l3 − l2 )c +
Wl1c wc 2 2 Mc
L
+
L
l3 − l 2 +
L
− Wl1 ( )
VC = − R Ay + W = − w(l3 − l2 ) + 1 +
Wl
L 2L
w 2 2 M
l3 − l2 +
L
( )
FC = − R Az = 0
or by using the RH end FBD to be:
 (l + l ) 
M C = w(l3 − l2 ) 3 2 − c  + M − RBy (L − c ) = − w(l3 − l2 )c + 1 +
Wl c wc 2 2 Mc
l3 − l 2 + − Wl1 ( )
 2  L L L

VC = − w(l 3 − l2 ) + RBy = − w(l3 − l2 ) + 1 +
Wl
L 2L
w 2 2 M
l3 − l 2 +
L
( )
FC = 0
Note that the numerical quantities of bending moment, axial force and shear forces must be the same in
magnitude and sense (sign) at a section irrespective of whether they are calculated considering the free body
diagrams of the beam to the left or to the right of the section.
The importance of MC, FC and VC are that the beam’s performance at position C is directly related to these
forces i.e., the stress, the deflection and the local rotation (angle) of the beam are all determined by these
moments and forces.

16


12.6.2 Example (easy): Beam forces at mid span for a cantilever beam
Determine the beam bending moment, shear and axial forces at mid span (C) for the beam and loading
shown in Figure 12.11.
2kN/m 3kN
y 1m
A
B

C

2m

2kN/m
y
VC
MC
RAz A
FC

MA RAy C z

Figure 12.11: A cantilever beam with concentrated and distributed loads
Answers: RAy = 5 kN, MA = 7 kN.m, FC = 0kN, MC = 3 kN.m, VC = -3 kN

12.6.3 Example (more difficult): Beam forces at mid span for a simply supported beam
Determine the beam bending moment, shear and axial forces at mid span (C) for the beam and loading
shown in Figure 12.12.
2kN 2kN/m
y 1/3m 1/3m

A B

C

2m

2kN
y
1kN/m VC
MC
RAz A
FC

RAy C z

Figure 12.12: A simply supported beam with concentrated and distributed loads
Answers: RAy = 59/27 kN, RBy = 31/27 kN, FC = 0kN, MC = 7/9 kN.m, VC = -2/27 kN

17


12.7 Relationships between M, V and Distributed loads
12.7.1 Relationship between M and V
The bending moment and the shear force at a given section are not independent of each other. The mutual
relationship between these quantities is derived by considering the equilibrium of a small length of the beam
between z and z+ z. Assume that the Bending Moment M and Shear Force V vary along the length of the
δ

beam such that at z+ z the Bending Moment is M+ M and the Shear Force is V+ V. This element is shown
δ δ δ

in Figure 12.13. Note that the shear forces and bending moments as shown are all positive.

M M+ M
y
z
V V+ V

z z
Figure 12.13 FBD of beam element to related M and V
Consider the moment equilibrium about the left hand end of the element
∑ M = (V + δV ).δz + M − (M + δM ) = 0
⇒ δM = Vδz + δVδz ≈ Vδz
In the limit when z approaches zero this reduces to:
δ

dM
=V
dz
or
(12.1)
M = ∫ Vdz
Note that the moments are taken about the left hand end of the element and that the second order quantity,
the product of V and z, is neglected because it will be negligibly small.
δ δ

Equation (12.1) states that the variation of bending moment with z will have a slope/gradient equal to the
value of the shear force.

18


12.7.2 Relationship between V and a Distributed Load
Consider the equilibrium of a small length of the beam between z and z+ z upon which a Distributed Load
δ

(w) is applied. Assume that the Shear Force V varies along the length of the beam such that at z+ z the Shear
δ

Force is V+ V. This element is shown in Figure 12.14.
δ

w w+ w
y

z
V V+ V

z z
Figure 12.14: FBD of a beam element to relate Q and a UDL
Equilibrium of forces in the y direction gives:

= −V + (V + δV ) + w.δz + δwδz = 0
1
∑F y
2
1
⇒ δV = − wδz − δwδz ≈ − wδz
2
In the limit when z approaches zero this reduces to:
δ

dV
= −w
dz
or
(12.2)
V = ∫ − wdz
Note that the second order quantity, the product of w and z, is neglected because it will be negligibly
δ δ

small.
Therefore if a small element of a beam has no distributed load on it then w = 0 and the shear force in the
section must be constant. If w is non-uniform (i.e. w = w(z)), this analysis assumes that dz is so small that the
distributed load across dz may be considered uniform at a level defined by w(z). Thus this equations is valid
whether the distributed load is uniform or a function of z (i.e. w may equal w(z)).
Equations (12.1) and (12.2) can be used to check the consistency of predicted shear force and bending
moment variation along a beam.
Combining the previous two equations gives the key relationship:

dV d 2 M
−w= = (12.3)
dz dz 2
19


12.7.3 Example: Simply supported beam with a Uniformly Distributed Load
Determine the shear force and bending moment at mid-span of the beam shown in Figure 12.15 using:
a. The Free Body Diagram approach of section 12.6.1, and
b. Equation (12.3)
y
5kN/m
A B

2m

Figure 12.15 Simply supported beam with a Uniformly Distributed Load

Answers: Vmid-span = 0kN, Mmid-span = -2.5kN/m

20

Shear Forces and Bending Moments

13. SHEAR FORCE & BENDING MOMENT DIAGRAMS
13.1 Introduction
Bending moments cause normal tensile and compressive stresses simultaneously in different parts of a beam
section. Shear forces cause shear stresses that try to cut the beam. The magnitudes of bending moment and
shear forces generally vary from one section to another in a beam. As shown in the previous section, the two
quantities are dependent on each other. Graphs showing the variation of M and V along the length of the
beam are called Bending Moment (BM) and Shear Force (SF) diagrams. The BM and SF diagrams help to
identify the critical sections in beams where bending moments and shear forces are highest.
13.2 Examples of SF and BM diagrams
13.2.1 Example: Cantilever beam with a concentrated load
Consider beam AB fixed at A, carrying a concentrated load (-W) at any position C (distance z = c from A).
W W
y c y c

MA
A B
z z
C RAz
L
RAy

Figure 13.1 Cantilever beam with a concentrated load and its FBD
• To draw the SF and BM diagrams first calculate the support reactions by considering force and
moment equilibrium conditions. DRAW the FBD of the beam (see Figure 13.1).
∑F ⇒z RAz = 0
∑F ⇒ y RAy = W
∑M ⇒ A M A = Wc
• Determine the shear force and bending moment relationships for each section of the beam - where
sub-length boundaries are defined by point loads, moments and the start/finish of Distributed
Loads. Again use a FBD for each section (see Figure 13.2).
W
y z y z-c
c
M M
MA MA
z z

V z V
RAy RAy
Figure 13.2: Sub-lengths (A ≤ z ≤ C) and (C ≤ z ≤ B) FBDs

In length (A ≤ z ≤ C): In section (C ≤ z ≤ B):
V (z ) = − R Ay = −W V (z ) = R Ay -W = 0
M (z ) = M A − R Ay z = Wc − Wz = W (c − z ) M (z ) = M A − R Ay z + W (z-c ) = Wc − Wz + W (x-c ) = 0

21

Beam Bending Theory

• Plot the shear force (V) and Bending Moment (M) diagrams along the length of the beam.
M

Wc
V
A C B
A C B
z z

-W

Figure 13.3: Shear Force and Bending Moment Diagrams for cantilever beam with concentrated load
• Check consistency of shear force and bending moment expressions between sections and at locations
where concentrated loads/moments are applied or where UDLs start or finish.

Hints & Checks for BM and SF diagrams
• Notice that the magnitudes of the shear force and bending moments at A (i.e. V(0) & M(0)) are equal
to the magnitudes of the end reaction load and moment
• Note that the Shear Force (V) is equal to the negative value of the end reaction load (V(0) = -RAy) as
by our definition for shear forces, RAy is acting in the –ve direction. That is, in the +ve y direction but
on a face with a normal in the –ve z direction.
• Whenever there is a concentrated load or bending moment applied to a beam the corresponding Shear
Force and Bending Moment diagrams should show a step of the same magnitude. In this example the
steps at A due to the point load and moment RA and MA are from zero to -W and Wc respectively.
• Notice that the shear force and bending moment at a free end are zero.
• Note that the bending moment varies linearly from Wc to 0 over the distance of z = c in the region 0 ≤
z ≤ c; i.e., it has a constant gradient of -W as expected due to V being equal to -W throughout that
section of the beam (see equation (12.1)).

22

Beam Bending Theory

13.2.2 Example: Cantilever beam with a Uniformly Distributed Load.
Consider beam AB fixed at A, carrying a uniformly distributed load w positioned between C and D as shown
in Figure 13.4.
a c
y y

w MA w

z z
A C D B RAz
RAy
L

Figure 13.4 Cantilever beam with a UDL and its FBD
• To draw the SF and BM diagrams first calculate the support reactions by considering force and
moment equilibrium conditions. DRAW the FBD of the beam (see Figure 13.4).
∑F z ⇒ R Az = 0
∑F y ⇒ RAy = − wc { = Area under Distributed Load curve}
 c
∑M A ⇒ M A = − wc  a+  { = Moment created by effective force acting at centroid}
 2
• Determine shear force and bending moment relationships for each sub-length of the beam. In this
example sub-lengths are defined by the reaction loads/beam ends and the start and finish of the UDL.
Again use a FBD for each sub-length (see Figure 13.5).
z
y y z y z-a
z
z-a z-a-c
a a
M w M w M
MA MA MA
z x z

V V V
RAy RAy RAy

Figure 13.5: FBDs for the sections (A ≤ z ≤ C), (C ≤ z ≤ D) and (D ≤ z ≤ B) respectively.
In section (A ≤ z ≤ C):
V (z ) = − RAy = wc
 c  c
M (z ) = M A − RAy z = -wc a +  + wcz = wc  z-a − 
 2  2
Again M(0) = MA as MA is a concentrated moment input to the beam at the left hand end. This is one
good check to see that derived expression for M is correct.

23

Beam Bending Theory

In section (C ≤ z ≤ D):
V (z ) = − RAy − w( z − a ) = w(− z + a + c )
w(z − a )
( ) ( )
2
w w w
M (z ) = M A + RAy z − = 2 zc-2ac − c 2 − z 2 − 2az + a 2 = − (− z + a + c )
2

2 2 2 2
Note: that M(z) could have been more easily derived using RHS FBD as it would not have included MA

In section (D ≤ z ≤ C):
V (z ) = − RAy − wc = 0
 c
M (z ) = M A − RAy z − wc z − a −  = 0
 2
V M

wc
A C D B
z z
A C D B
-(wc2)/2

-wc(a+c/2)

Figure 13.6: Shear Force and Bending Moment diagrams for a cantilever beam with a UDL
• To check consistency of results note that in section C to D, V varies linearly with slope –w as expected
from equation (12.2) and in all sections dM/dx = V.
• By varying a and c any particular case of a cantilever beam with a uniformly distributed load can be
solved.

24

Beam Bending Theory

13.2.3 Example: Non-uniform distributed load on a cantilever
wo
y

z

L

Figure 13.7: Non-uniform distributed load on a cantilever
Derive an expression for the variation of shear force and bending moment in a cantilever beam loaded by a
non-uniform distributed load as shown in Figure 13.7. Try this using both the FBD approach and by using
equations (11.1) and (11.2).

Answers:

V (z ) =
2L
( ) (
z − L2 and M ( z ) = o z 3 − 3 L2 z + 2 L3
wo 2 w
6L
)

25

Beam Bending Theory

13.2.4 Example: Simply supported beam AB of length L.
Consider a beam, pin supported at one end and simply supported at the other (this combination is often
referred to as simply supported). A concentrated load (-W) acts at distance c from A.
W
y c

A B

C

L

Figure 13.8: A simply supported beam with concentrated load
• Determine support reactions using the FBD of the entire beam:
∑F x ⇒ RAz=0
Wc
∑M A ⇒ RBy=
L
Wc W (L-c )
∑F y ⇒ RAy= W-RBy = W −
L
=
L
• Derive expressions for Shear Force and Bending Moment in each section:
W
y z y z-c
c
M M

z z

V V
RAy RAy

Figure 13.9 Simply supported beam with a concentrated load FBDs

In section (A ≤ z ≤ C): In section (C ≤ z ≤ B):
− W (L − c ) Wc
V (z ) = − RAy = V (z ) = − RAy + W = − RBy =
L L
− W (L-c )z − W (L-c )z − Wc(L-z )
M (z ) = − RAy z = M (z ) = − RAy z + W ( z-c ) = + W (z-c ) =
L L L
V M

Wc
L A C B

A B z z
-W
(L − c ) - Wc
(L − c )
L C L

Figure 13.10: Shear force and Bending Moment diagrams for a simply supported beam with a concentrated load

26

Beam Bending Theory

13.2.5 Example: A simply supported beam with a UDL.
a c
y
w

A B

C D

L

Figure 13.11 Simply supported beam with a UDL and the beam FBD
• Determine support reactions using the FBD of the entire beam:
∑F z ⇒ RAz = 0
wc  c
∑M A ⇒ RBy =
L 
a + 
2
wc   c 
∑M B ⇒ RAy =
L 
 L −  a + 

 2 

• Derive expressions for Shear Force and Bending Moment in each section:
z
z
z-a
y y y z
a
M w M

z z z
M
V V V
RAy RAy RBy

Figure 13.12: FBDs for beam sections (A ≤ z ≤ C), (C ≤ z ≤ D) and (D ≤ z ≤ B) respectively.

In section (A ≤ z ≤ C): In section (C ≤ z ≤ D): In section (D ≤ z ≤ C):
− wc   c  − wc   c  wc  c
V (z ) =  L − a +  V (z ) =  L −  a +   + w( z − a )
 V (z ) = a + 
L 
 2 
 L   2 
 L  2
− wc  c  − wc  c  w( z − a ) M (z ) =
− wc  c
 a + ( L − z )
2
 
M (z ) =  L − a + z
 M (z ) =  L −  a +  z +
L   2 
 L 
 2 
 2 L  2

V M

(
wc a + c
2
)
L C A C E D B

A D B z z
- wc
L
( (
L− a+c
2
)) Mmax

Figure 13.13: Shear force and Bending Moment diagrams for a simply supported beam with a UDL

• How might Mmax be determined?

27

Beam Bending Theory

13.3 Principle of Superposition
13.3.1 Theory
The support reactions and fixing moments as well as shear forces and bending moments (and all other
mechanical entities such as stresses and displacements) at a given section (or point) due to the individual
loads can be calculated separately and summed up algebraically to obtain the total effect of all the loads
acting simultaneously. This is applicable to conservative (i.e. linearly elastic) systems only.
13.3.2 Example: Cantilever beam with a concentrated load and UDL
Consider a cantilever beam with a concentrated load (N) and a UDL (w) as shown in Figure 13.14. This can
be separated into two more simple to analyse cantilever loading cases: a single concentrated load and a single
UDL. The separate results for these two loading cases may be added together to obtain the results for the
complete beam (provided the beam remains in the linear elastic region of its stress-strain curve). This can
greatly simplify analysis as separate, simple expressions for SF and BM may be obtained for each of the
loading cases and then be added together to obtain the SF and BM expressions for the overall beam.
d e
y
N
c
w

z
A C D E B

L

d e
y y
N
c
w

z z
A C D E B A C D E B

Concentrated Load Case (CL) UDL Case (UDL)

Figure 13.14: Example of Principle of Superposition
The results of sections 13.2.1 and 13.2.2 may be used to quickly write the SF and BM expressions for the
combined load case:
In section (A ≤ z ≤ C):
V (z ) = V( CL ) + V( UDL ) = N + we
 e
M (z ) = M ( CL ) + M ( UDL ) = N ( z − c ) + we z-d − 
 2
In section (C ≤ z ≤ D):
V (z ) = V( CL ) + V( UDL ) = we
 e
M (z ) = M ( CL ) + M ( UDL ) = we z-d − 
 2

28

Beam Bending Theory

In section (D ≤ z ≤ E):
V (z ) = V( CL ) + V( UDL ) = w(− z + d + e )

M (z ) = M ( CL ) + M ( UDL ) = (− z + d + e )2
w
2
In section (E ≤ z ≤ B):
V (z ) = V( CL ) + V( UDL ) = 0
M (z ) = M ( CL ) + M ( UDL ) = 0
Note that there are more sections in the overall beam than in either of the individual beams. Simply add the
appropriate expressions from each separated load case to determine the overall expression for each section.
This method may also be used to determine the support loads and moments for the beam. In the example
above:
R Ay = R Ay( CL ) + R Ay( UDL ) = − N − we
 e
M A = M A( CL ) + M A( UDL ) = − Nc − we d + 
 2
Check these results using FBDs for the complete beam.
13.4 Summary of Procedure for drawing SF and BM diagrams:
i. Find support reactions by considering force and moment equilibrium conditions.
ii. Determine shear force and bending moment expressions.
iii. Plot shear force (V) vs z and bending moment (M) vs z to appropriate scales.
iv. Check for consistency of sign convention and agreement of values at the supports and at the ends of
the beam and that equations (11.1) and (11.2) are valid along the length of the beam.
13.5 Macauley1 Notation
13.5.1 Introduction
Frequently it is beneficial to use a single expression for the shear force and bending moment distributions
along a beam rather than the collection of sub-length expressions.
Consider a cantilever beam on which two concentrated forces and a UDL are acting.
Y W1 W2 W1 W2
w w
a MA

z RAz
A B
b
RAy
L

Figure 13.15: Beam for which Macauley expression is to be derived

1
Macauley W.H. Note on the deflection of beams, Messenger of Mathematics, Vol 48 pp 129-130. 1919.

29

Beam Bending Theory

There are three lengths of the beam in which the bending moment is different:
0≤ z≤a M = M A − R Ay z
a≤ z≤b M = M A − R Ay z − W1 ( z − a )
w(z − b )
2
b ≤ z ≤ L M = M A − R Ay z − W1 (z − a ) − W2 (z − b ) −
2
Clearly the expression for the length b ≤ z ≤ L contains the terms in the other two lengths. To reduce the
tedium of working with three separate equations and sub-lengths, the following notation (due to Macauley)
may be used:
w{z − b}
2
M = M A − R Ay z − W1 {z − a} − W2 {z − b} −
2
In this version of the bending moment equation the terms within { } should be set to zero if the value
within these brackets become -ve.
In the general case:
 0 for z < a
{z − a}n = 
(z − a ) for z ≥ a
n

Thus the last two terms become zero if z < b and the last three terms are zero if z < a.
13.5.2 Derivation of Macauley Shear Force and Bending Moment expressions
Using Macauley parentheses the distributed loading on the beam may be written as:
w = w{z − b}
0

The term in the parentheses is equal to zero if z < b and if z ≥ b the term in the parentheses raised to the
power of zero equals 1. This is essentially an on switch for the distributed load that says that when z < b
there is no distributed load and when z ≥ b the distributed load equals w.
Integrating2 the Macauley expression for distributed load gives an expression for the shear force along the
beam (from Equation 11.3):

∫ w = ∫ w{z − b} = w{z − b} + C1 = −V
0 1
where C1 is a constant of integration
The constant however, has to be evaluated for each sub-length of the beam. This is made fairly straightforward
by recalling that the shear force only has a step change when a point load is applied and that the step change is
equal to the value of the point load. So for the example of Figure 13.15 the constant C1 represents a Macauley
expression for all point loads along the beam:
C1 = RAy + W1 {z − a} + W2 {z − b}
0 0

hence the shear force may be written as:
V = − w{z − b}− C1 = − w{z − b}− R Ay − W1 {z − a} − W2 {z − b}
0 0

2
When integrating a Macauley expression, the whole term within the brackets should be treated as a variable. This is
justified by recognising that only the integration constant varies if the term is expanded out, for example:
{x − a} 2
x2 a2
Treating { x − a } as a variable gives ∫ {x − a}dx = 2
+ c1 =
2
− ax + + c1
2
2

∫ {x − a}dx = ∫ xdx − ∫ adx = 2 − ax + c2
x
Expanding { x − a } gives

a2
∴ c2 = c1 + but both expressions are constant
2

30

Beam Bending Theory

Integrating this term again to obtain an expression for bending moments (from Equation 12.3) gives:

M = ∫ Vdz = −
w
{z − b}2 − R Ay z − W1 {z − a} 1 − W2 {z − b}1 + C2
2
Similar to the relationship between the shear force constant and point loads, the constant C2 represents a
Macauley expression for all point moments along the length of the beam. Note: that anti-clockwise moments
applied to the beam are considered positive as they introduce a +ve step in the variation of M along the beam.
C2 = M A
Combining these last two expressions gives the full Macauley expression for bending moment along the
beam as:
w{z − b}
2
M = M A − R Ay z − W1 {z − a}− W2 {z − b}−
2
13.5.3 Example: Macauley Expression for a beam
Derive the Macauley expressions for shear force and bending moment for the beam and loading shown in
Figure 13.16.
Then use the expressions derived to determine the value and location of the maximum bending moment in
the beam.
a a
y
w
B
A

4a

Figure 13.16 Simply supported beam with a UDL

Answers: V = w{z − a} + {z − 2a}0 {z − a}2 + 3wa z − 15 wa {z − 2a}1
3 wa 15 wa w
− M =
4 4 2 4 4

31

Beam Bending Theory

13.5.4 Use of superposition to simplify Macauley expressions
In the previous examples the bending moment equations have been easy to develop in Macauley notation
because the distributed loads have been open ended i.e., they cease at the end of the bar. There are however
some loading situations that are less easily expressed in this way, particularly closed end distributed loads. A
closed end distributed load is one where the DL ceases or step-changes at some location along the length of
the beam. In this situation superposition may be used to develop a bending moment expression that has
multiple open-ended (i.e. ceasing at the end of the beam) DLs.

Y Y Y
w w w

z
z z

A L/2 B L/2 C L/2 L/2 L/2 L/2

Figure 13.17: A cantilever beam with a UDL over part of the beam
For the cantilever beam shown in Figure 13.17 the closed end distributed load may be replaced by two open
ended distributed loads. The net combination of these two new load cases is equivalent to the original load
case.
To derive the Macauley expression simply add the expressions for both cases together.
w( z ) = wUDL1 + wUDL 2
= − w{z − 0} + w{z − L }
0 0
2

∫ w(z )dz = − w{z − 0} + w{z − L } + C1
1 1
2

( )
C1 = R Ay
UDL1
( )
+ R Ay
UDL 2
=
wL
2
V ( z ) = − ∫ w( z )dz = w{z − 0} − w{z − L } −
1 1 wL
2
2

∫V (z ) dz = 2 {z − 0}
w w
{z − L }2 − wL z + C2
2
−
2
2 2
wL2
C 2 = (M A )UDL1 + (M A )UDL 2 =
8
wL2
M = ∫ V ( z ) dz = {z − 0} − {z − L } −
w w wL
z+
2 2
2
2 2 2 8

32

Beam Bending Theory

14. BENDING THEORY
14.1 Introduction
Bending causes tensile and compressive stresses in different parts of the same cross-section of a beam. These
stresses vary from a maximum tension on one surface to a maximum compression on the other passing
through a point where the stress is zero (known as the neutral point). The maximum stresses are
proportional to the bending moment at the cross-section. As the magnitude of the maximum stress dictates
the load bearing capacity of the beam (i.e. for most engineering applications the stresses should be kept
below yield), it is important to find out how the stresses the bending moments are related. The relationship
between stresses and bending moments will be developed in this section. The analysis is restricted by the
assumptions stated in section 14.2. The assumptions can be relaxed and improved analysis can be made but
this is beyond the scope of the first year course and will be covered in future years.
14.2 Assumptions
• The beam is made of linear-elastic material.
• The cross section of the beam is symmetrical about the plane in which the forces and moments act (i.e.
the YZ plane).
• A transverse section of the beam which is plane before bending remains plane after bending.
• Young's Modulus is same in tension and compression.
• The lateral surface stresses (in the y-direction) are negligible. The lateral stress within the beam and
the shear stresses between adjacent "layers" throughout the depth of the beam are ignored (until next
year).
It is possible to do the analysis without these assumptions. But the algebra becomes very complicated.
14.3 The beam bending equation
14.3.1 Location of the neutral axis
Bending moment generally varies along the length of the beam. However it is reasonable to assume that
bending moment is constant over a very small (infinitesimally small) length of the beam. So the case of such
a small length of a beam subjected to a constant bending moment along its length (known as pure bending)
is analysed below.
A small element of a beam is schematically shown in Figure 14.1. An initially straight beam element a'b'c'd'
is bent to a radius R at point z by bending moments M, to abcd. The layers above line e'f' (i.e. on the convex
side) lengthen and those below e'f' (i.e. on the concave side) shorten. The line e'f' (and ef) is therefore the
layer within the beam that neither lengthens nor shortens i.e. the NEUTRAL AXIS.3 The initial length of
e'f' is also equal to the arc length of ef, i.e:
e'f' = dz = ef = Rθ
Consider the line gh at a distance y from the neutral axis. The original length of gh was the same as for all
other layers within the element i.e dz. The new length of gh may is related to its bend radius (R+y) and bend
angle ( ) so that the new length may be written as (R+y)θ. Therefore as strain is the ratio of change in length
θ

to original length and stress ( ) = strain ( ) × modulus of elasticity (E):
σ ε

Strain in layer gh = z direction strain at distance y from neutral axis (ε z ( y ))
gh − g' h' (R + y ) θ − dz (R + y )θ − Rθ yθ y
⇒ ε z( y ) = = = = =
g' h' dz Rθ Rθ R
and

3
Note: A prime (') is used to indicate points in the undeformed condition.

33

Beam Bending Theory

Stress in layer gh = z direction stress at distance y from neutral axis (σ z ( y ))
Ey
⇒ σ z ( y ) = Eε z =
R
y
z
dz

a' d'
y1
a d
y
g h z
e' f'
y2 e f
b' c'
M b c M

R

θ

Figure 14.1 Element of a beam subject to pure bending
Therefore as E and R are constants for a given position z and bending moment M the variation of stress through
a beam is linear as shown in Figure 14.2. Note in Figure 14.2 positive stress is defined using the convention for
the right hand end of beam.

y z
σz = Ey1/R

y1

σz z

y2
M σz = Ey2/R M
Figure 14.2: Axial stress distribution at z

34

Beam Bending Theory

The effect of the axial stress at any point y on a small cross-sectional area dA is to create a small elemental load
(dF) on the elemental cross-section area dA (of thickness dy and width b(y) - where b(y) is used to indicate that
b may vary with position y).
The stress and load are related by dF = σ dA = σ.b.dy.

y
dA
dy

y x

b(y)

Figure 14.3: Cross-section of beam at z
The total axial load (F) on the cross-sectional face may then be related to the beam cross sectional dimensions
and radius of curvature by integrating across the face:
y2 y2 y y2
bEy E 2 E
F = ∫ dF = ∫ σ .bdy = ∫ dy = ∫ bydy = ∫ ydA
y1 y1
R R y1 R y1

As Figure 14.2 shows, the beam is not actually subjected to any axial load so the total axial load on the beam
must equal zero i.e., F = 0 and with E and R both non-zero the only way this relationship can equal zero is for ∫ y
dA to equal zero.
y2
E
F=
R ∫ ydA = 0
y1

E ≠0&R≠0
y2

∴ ∫ ydA = ∫ ydA = Q X = 0
y1

∫ y dA is by definition, the first moment of area of the cross section (QX) and only equals zero if the axis from
which y is measured (i.e. the X axis) passes through the centroid of the cross-section.
Therefore the neutral axis of a simple beam must pass through the beam cross-section centroid. When
analysing beams the z axis is therefore located along the neutral axis.
14.3.2 The bending equation
Consider now the elemental moments (dM) caused by the elemental loads (dF) about the neutral axis; dM =
dF.y = σ.b.dy.y. The total applied moment (M) may then be found by integrating across the surface:
y2 y2 2 y y2
Ey E 2 2 E
M = ∫ dM = ∫ σ .b.dy.y = ∫ b dy = ∫ by dy = ∫y
2
dA
y1 y1
R R y1 R y1

as IX (the second moment of area about the neutral axis for the cross section) = ∫ y2 dA, the bending moment at
point z may be related to the radius of curvature (R), the Young’s Modulus of the beam (E) and the second
moment of Area of the beam cross-section (IX):

35

Centroids moments of inertia

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