4
4.1
Rigid body statics
Moments
z
y
M
O
x
dx Fy
d
r
F
rB rAA
O
y
M
F
dy Fx
x
F
B
force line of action
Figure 12: Moment definitions.
This section serves to review the definition of moments caused by forces and moment couples.
More details can be found in (Hibbeler and Yap 2013) 4.1-4.6.
A moment acts about a so-called moment axis which passes through a point, e.g. O as shown
in Fig. 12. A moment is a vector which is indicated either by a double-arrow or a bent vector. It
is defined as the vector cross product of the distance vector r from point O with respect to the
force line of action and the force vector F
M = rA × F = rB × F
(25)
where the principle of transmissibility applies, i.e. distance vector r can point at any point lying
on the force line of action. The distance vector are computed as follows:
rA = (Ax − Ox ) ex + (Ay − Oy ) ey + (Az − Oz ) ez
(26)
rB = (Bx − Ox ) ex + (By − Oy ) ey + (Bz − Oz ) ek
(27)
As a vector, each moment has a magnitude and a direction. Its magnitude is given by
M = dF
(28)
where F is the force with its force line of action passing O with a perpendicular distance d, i.e.
the shortest distance. d is the so-called moment arm or lever arm. Its direction, the moment
axis, is the normal vector on the plane containing both, vector F and d. The direction of moment
vector can be found from the right-hand-rule:
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� The forefinger points from O to the force line of action, the middle finger points in direction
of the force and the thumb indicates the moment axis about which the moment occurs.
� With the thumb pointing in direction of the moment, the finger tips indicate the sense of
orientation of the moment about its axis, represented by the thumb.
Moments are often expressed in terms of their Cartesian components - so-called scalar analysis:

 

Mx
dy Fz − dz Fy
M = M x e x + My e y + Mz e z =  M y  =  d z F x − d x F z 
Mz
dx Fy − dy Fx
(29)
The sign of the moment components is determined either from permutation circle or using the
right-hand-rule.
In case of slender members such as beams, their axis is usually taken as the x-axis such that
the Cartesian moment component Mx refers to a torsional moment (e.g. caused by a force couple
twisting the beam), whereas components My and Mz refer to flexural moments causing the beam
to bend.
z
-F
O
B
rB
r
rA
A
F
-F
d
F
y
x
Figure 13: Moment couple definition.
Moment couples are caused by a pair of equal parallel forces acting in opposite directions and
separate by a perpendicular distance as illustrated in Fig. 13:
M = rA × F − rB × F = (rA − rB ) F = r × F
or alternatively taking the perpendicular distance d of both force line actions
M = d F.
A moment couple is a free vector, because it only depends on the positioning of the forces relative
to each other but not where they are located in space.
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4.2
Equilibrium equations
Usually, structural analysis is performed on the undeformed structure assuming the deformation
to be very small, i.e. the structure stays rigid and the direction of forces and moment arms are
unchanged. This simplification is meaningful for most engineering materials like steel or concrete.
An exception are slender structures subject to axial compression forces where the analysis on the
deformed configuration may be necessary to include secondary moments due to force eccentricities.
The equilibrium of the rigid body or parts of a corresponding free body system considers forces,
moments caused by forces as well as moment couples (Hibbeler and Yap 2013 5.1):
�
�
F = 0
M(O) = 0
(force resultant is zero)
(moment resultant is zero)
or using Cartesian components we find 6 scalar equilibrium equations (Hibbeler and Yap 2013
5.6):
�
Fx = 0
�
Fz = 0
�
Fy = 0
�
Mx(O) = 0
�
Mz(O) = 0
�
My(O) = 0
In two dimensions, the statics of a rigid body is governed by only three equilibrium equations
(Hibbeler and Yap 2013 5.3):
�
�
�
Fx = 0
Fy = 0
Mz(O) = 0
or alternatively,
�
�
�
Fx = 0
Mz(A) = 0
Mz(B) = 0
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y
Mz
FBD
x
M=Fd
A
MAz
P
F
A
L/2
P
Ay
L/2
F
M=Fd
Ax
B
L/2
(a)
P
MAz
d
Ax
B
L/2
P
F
Ay
F
(b)
Figure 14: Equivalence of force couple and moment couple.
Since a moment couple is a free vector, it must be equally incorporated in the moment equilibrium equation of a beam regardless what point is considered. In order to illustrate this, we want
to evaluate besides the force equilibrium firstly the moment equilibrium about point A:
�
�
Fx = 0 = Ax → Ax = 0
Fy = 0 = Ay − P → Ay = P
�
Mz(A) = 0 = MAz + F · d − P · L → MAz = −F · d + P · L
�
Mz(B) = 0 = MAz + F · d − Ay · L → MAz = −F · d + P · L .
then secondly, the moment equilibrium about point B:
For both points the two moment couples, that is the externally applied moment M = F · d and the
support reaction moment MAz , equally enter the equilibrium equations. The reason for this can
be further elucidated considering the action of an equivalent force couple on the beam as shown
in Fig. 14(b):
�
�
�
Fx = 0 = Ax − F + F → Ax = 0
Fy = 0 = Ay − P → Ay = P
Mz(A) = 0 = MAz + F ·
d
d
+ F · − P · L → MAz = −F · d + P · L
2
2
then secondly again, the moment equilibrium about point B:
�
Mz(B) = 0 = MAz + F ·
d
d
+ F · − Ay · L → MAz = −F · d + P · L
2
2
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Clearly, the action of the force couple equally enters the moment equilibrium about point A and
B, respectively.
Two-force members are structural members such as the rods in a truss structure which are
only subjected to forces at both ends. Due to force and moment equilibrium these end forces need
to be opposite equal in magnitude and direction. They are acting along the same force line of
action which is the line connecting both end points of the member.
Three-force members are structural members are subjected to only three forces (including
externally applied and support forces alike). The state of moment equilibrium for these members
can only be fulfilled if these three forces form a concurrent force system where all forces intersect
at one point or a parallel force system where all forces are parallel to each other.
For more details the reader is referred to Hibbeler and Yap 2013 5.4.
4.3
Support reactions
Generally, structures and structural members are classified according to their geometry and the
kind of loads applied to them. In Sec. 3 we have encounter structural members such as cables and
rods which are subject to forces acting along the member’s axis. Structural members exhibiting
two small dimensions (thickness direction) and are subject to forces which act transverse (perpendicular) to the members axis are called beams. Curved beams are referred to as archs and two
rigidly connected and to each other inclined beams are called frames. If only one dimension is
small, then the member is referred to as plate, shell or membrane. Plates are thicker than a shell
and membrane, because they are required to sustain transverse shear and flexure whereas shells
and membrane are usually exposed to internal forces acting in the planes tangential (parallel) to
the surface.
Structures are connected to their surroundings via supports such that they are fixed at a certain
location. In this sense, supports prevent the structure at selected points/regions from translating
and rotating by means of counter-acting reaction forces and moments:
� Translation into a certain direction is prevented by a support reaction force in opposite
direction.
� Rotation about a certain point is prevented by a support reaction moment in opposite
orientation.
The magnitude of the support reactions depends on the magnitude of the applied loadings.
The three most common support types schematically depicted in Fig. 15 are (1) fixed support
preventing translation in any direction and rotation about any axis; (2) pin support preventing
only translation in any direction but not the rotation; (3) roller support preventing only the
vertical translation. For the purpose of analysis the resultant support reaction forces are usually
considered in Cartesian component representation.
There is a specific number of Cartesian support reaction force/moment components which is
minimum necessary to keep a structure in equilibrium and prevent it from moving, i.e. evading
deformation under applied loads. The number of support reactions equals to the number of
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y
Mz
x
MAz
Ax
fixed
Ay
Ax
pinned
Ay
roller
Ay
Figure 15: 2D support reactions.
Figure 16: Kinematic structural system due to insufficient support reaction forces in the two-dimensional
case.
equilibrium equations, i.e. three in 2D and six in 3D, referring to a externally statically determinate
system. However, if the support reactions are placed such that the whole structure can translate
in a certain direction or rotate about certain point without undergoing deformation - so-called
rigid body motion, then the system is called kinematic or unstable which means it is improperly
constrained. In absence of support reaction moments, this is the case if
i) in 2D, all support reaction forces of the structure are concurrent at one of the points of
support, e.g. at point A as shown in Fig. 16, or in 3D, all support reactions intersect a
common line as shown in Fig. 17.
ii) all support reaction forces of the structure are parallel.
iii) some of the structure’s components form a collapsible mechanism.
If the magnitude and the direction of the support reaction force resultant is unknown, then two
unknown Cartesian force components arise in 2D. However, if the direction of the support reaction
force resultant is known, e.g. for a smooth slope or collar-rail support, then both Cartesian force
Page 21
Figure 17: Kinematic structural system due to insufficient support reaction forces in the threedimensional case.
Figure 18: Pole plan.
components can be expressed in terms of the resultant force, effectively reducing the number of
unknows by one.
The graphical kinematical analysis of structures involves a so-called pole plan. It requires to
identify major and relative poles of rotation:
� A major pole of rotation defines a fixed pole around which a rigid body rotates. Pin supports
are usually major poles. In case of roller supports the major pole lies on the line normal to
the direction of the roller.
� A relative pole of rotation defines a point around which two rigid bodies relatively rotate.
Common hinges are usually relative poles of rotation.
A structure has a collapsible mechanism if
a) it consists of two connected structural elements where both have a major pole and a common
relative pole, and all three lie on a straight line as shown in Fig. 18(a).
b) it consists of three connected structural elements and all relative poles lie on a straight line as
depicted in Fig. 18(b).
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On the other hand, if a structure has got more support reactions than equilibrium equations, i.e.
more than three in 2D and more than six in 3D, then the system is called statically indeterminate.
The term indeterminate refers to the fact that the unknown support reactions cannot be solved
anymore by solely making use of the equilibrium equation but must supplemented by deformation
compatibility conditions.
For more details the reader is referred to (Hibbeler and Yap 2013) 5.2, 5.5, 5.7 and 6.6.
4.4
Distributed loading
4.4.1
General distributed loading
Figure 19: General distributed loading.
In many cases, the external loading on a structure is distributed over a larger area and cannot
be treated as a concentrated load anymore. These kinds of distributed loadings can act (1)
perpendicular to the surface as shown in Fig. 19a), e.g. water pressure on a tank wall or the
weight of sand lying on a horizontal structure; (2) parallel to surface, e.g. friction of a block
sliding over the surface; (3) with an angle to the surface, e.g. the lift and drag on an aircraft wing.
In all three cases the loading is measured in Pa or Nm−2 , i.e. force per area.
The resultant differential force dFR caused by a distributed load p is obtained by multiplying
it by a differential area element dA, i.e. dFR = p(x, z) · dA. Integrating dFR over the entire area
where p is applied, the total resultant force exerted by a distributed load p is found:
FR =
�
dFR =
A
�
p(x, z) dA
(30)
A
which corresponds to the ”volume” under the distributed-loading diagram p vs. x, z.
Since FR and p(x, z) as acting across the plate are required to be equivalent in their effect
on the statics of the member, the moment caused by both about any point needs to be equal.
Consequently, the location (x̄, z̄) of the resultant force of a general distributed loading is found
by equating the moment components Mx and Mz , e.g. about the origin O, caused by the resultant force FR and by the distributed force acting on a differential area element, p(x, z) dA, and
Page 23
integrated it over the entire area A:
�
�
x · p(x, z) dA
x dV
A
V
x̄ = �
≡ �
,
p(x, z) dA
dV
A
V
z̄ =
�
A
�
z · p(x, z) dA
p(x, z) dA
A
≡
�
z dV
�V
.
(31)
dV
V
In this sense, it is clear that the force line of action of the resultant FR passes through the centroid
of the ”volume” under the distributed-loading diagram as shown in Fig. 19b).
(x,z)
R
z
Figure 20: Distributed loading acting a thin strip with constant width.
If p(x, z) only varies in one dimension and is constant in the other one, e.g. constant in width
direction of a beam, i.e. the z-direction, we can multiply p with the beam’s thickness b and we
obtain a distributed load per line w(x) = p · b measured in Nm−1 . The magnitude of its resultant
as illustrated in Fig. 20b) is consequently obtained by integrating along this line:
�
�
FR =
dFR =
w(x) dx
(32)
L
L
which is the ”area” under distributed-loading diagram w vs. x depicted in Fig. 20c).
As before, the moments caused by the resultant force FR and by the distributed force acting
on a differential line element, w(x) dx, and integrated it over L must be equivalent providing an
expression to obtain x̄, force point of action of FR :
�
−x̄ · FR = −
x · w(x) dx
(33)
L
and substituting for FR Eq. (32)
�
x · w(x) dx
x̄ = L�
.
w(x) dx
L
(34)
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Note: x̄ represents the x-coordinate of the centroid of the area under loading diagram w vs. x as
depicted in Fig. 20c).
For more details the reader is referred to (Hibbeler and Yap 2013) 4.9. and 9.4.
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