x2 = 4ex + 5ey

F = ex + 2ey .

H

F

x2

x1

s

E

ey

ez ex

The concepts of force and moment

36

2.7 A very simple segmented model of a sitting person is shown in the

schematic visualization in the ¬gure. At the right hand H a force F H and a

moment M H are given by:

F H = 5ex + ez

M H = ey + 3ez ,

related to the Cartesian vector basis {ex , ey , ez }. In addition the positions of

the hand H and shoulder S with respect to the origin at the lower end of the

spine are given:

xH = 4ex ’ 2ey + 5ez

xS = ex ’ 2ey + 4ez .

MH

FH

H

S

O

ey ez

ex

(a) Calculate the moment at the shoulder (point S) as a result of the load

at the hand H.

(b) Calculate the moment at the lower end of the spine O as a result of

the load at the hand H.

3 Static equilibrium

3.1 Introduction

According to Newton™s law the acceleration of the centroid of a body multiplied by

its mass equals the total force applied to the body and there will be a spin around

the centroid, when there is a resulting moment with respect to the centroid. But

in many cases bodies do not move at all when forces are applied to them. In that

case the bodies are in static equilibrium. A simple example is given in Fig. 3.1.

In Fig. 3.1(a) a body is loaded by two forces of equal size but with an opposite

direction. The lines-of-action of the two forces coincide and clearly the body is

in equilibrium. If the lines-of-action do not coincide, as in Fig. 3.1(b), the forces

have a resulting moment and the body will rotate. To enforce static equilibrium a

counteracting moment should be applied to prevent the body rotating, as indicated

in Fig. 3.1(c).

3.2 Static equilibrium conditions

If a body moves monotonously (no acceleration of the centroid, no rate of rotation

around the centroid), the body is in static equilibrium. If the velocities are zero

as well, the body is at rest. In both cases the sum of all forces and the sum of

all moments (with respect to any point) acting on the body are zero. Suppose

that n forces F i (i = 1, 2, ..., n) are applied to the body. Each of these forces will

have a moment Mi with respect to an arbitrary point P. There may be a number

of additional moments M j (j = 1, 2, ..., m) applied to the body. Static equilibrium

then requires that

n

Fi = 0

i=1

n m

Mi + M j = 0. (3.1)

i=1 j=1

Static equilibrium

38

F

F

F F M

F

F

(c) Static equilibrium

(a) Static equilibrium (b) No static equilibrium

Figure 3.1

Examples of satisfaction and violation of static equilibrium.

F3

F1

d3

d1

P

d2 d4

F2

F4

(a) (b)

Figure 3.2

An image and a model of a cell.

Simply demanding that the sum of all the forces is equal to zero, to assure

equilibrium, is insuf¬cient, since a resulting moment may induce a (rate of) rota-

tion of the body. Therefore the sum of the moments must vanish as well. The

moment vectors associated with each force vector are computed with respect to

some point P. However, if the sum of all forces is zero, the sum of the moment

vectors should be zero with respect to any point P.

Example 3.1 Fig. 3.2(a) shows an image of a single cell, that was captured by means of an

atomic force microscope. Cells attach themselves to the supporting surface in a

discrete number of points. The forces acting on these points of the cell are shown

as arrows. In Fig. 3.2(b) a model of the cell is given that can be used to examine the

equilibrium of forces and moments. In this case, the sum of all the forces should

be equal to zero:

4

F i = 0,

i=1

while the sum of the moments with respect to the point P (with an arbitrary

position xP ) due to these forces must be zero as well:

4

( di — F i ) = 0.

i=1

39 3.2 Static equilibrium conditions

Notice that in this particular case there are no externally applied additional

moments.

In the example of Fig. 3.2(b) the moments were determined with respect to

point P. If the moment is computed with respect to another point in space, say R,

having coordinates xR = xP + a, then the moment vector with respect to this point

R is de¬ned by

4

M= ( di ’ a) — F i

i=1

4 4

= ( di — Fi ) ’ ( a — Fi )

i=1 i=1

4 4

= ( di — F i ) ’a — Fi ,

i=1 i=1

which vanishes if the forces are in equilibrium and the sum of the moments with

respect to point P equals 0. This implies that any point can be taken to enforce

equilibrium of moments. This argument can be generalized in a straightforward

manner to any number of forces.

Example 3.2 An example of pure force equilibrium is given in Fig. 3.3. This ¬gure shows an

electron micrograph of an actin network supporting the cell membrane. At the

Figure 3.3

TEM image of the actin network supporting the cell membrane, with forces acting on an

interconnection point.

Static equilibrium

40

intersection point of the network the molecules are (weakly) cross-linked. Within

each of the molecules a (tensile) force is present and at the interconnection point

force equilibrium must apply. In Fig. 3.3 the forces acting on one of the intercon-

nection points have been sketched, the sum of these force vectors has to be equal

to zero.

With respect to a Cartesian coordinate system equilibrium requires that the sum

of all forces in the x-, y- and z-direction is zero. With the decomposition of a force

F, according to F = Fx ex + Fy ey + Fz ez , that is the case if:

n

Fx,i = 0

i=1

n

Fy,i = 0

i=1

n

Fz,i = 0,

i=1

and that the sum of all moments in the x-, y- and z-direction with respect to an

arbitrarily selected point P is zero. Choosing point P to coincide with the point of

application of the forces, immediately reveals that the equilibrium of moments is

trivially satis¬ed.

Equilibrium of forces may also be expressed in column notation, according to:

n

F i = 0.

∼ ∼

i=1

3.3 Free body diagram

A free body diagram serves to specify and visualize the complete loading of a

body, including the reaction forces and moments acting on the body that is sup-

ported in one way or the other. The body may be part of a system of bodies and,

using the free body diagram, the reaction forces on the body under considera-

tion imposed by the other bodies may be identi¬ed. For this purpose the body

is isolated from its surroundings and the proper reaction forces and moments are

introduced such as to ensure equilibrium of the body. Clearly, these reaction forces

and moments are not known a priori, but the equilibrium conditions may be used to

try to compute these unknowns. A distinction must be made between the statically

determinate and the statically indeterminate case.

41 3.3 Free body diagram

Requiring force and moment equilibrium provides for a limited number of equa-

tions only, and therefore only a limited number of unknowns can be determined.

For two-dimensional problems force equilibrium results in two equations, while

the requirement of moment equilibrium supplies only one equation, hence three

independent equations can be formulated. Only if the number of unknown reaction

loads equals three is the solution of the unknowns possible. Likewise, in the three-

dimensional case, imposing force and moment equilibrium generates six indepen-

dent equations, such that six unknown reactions can be computed. If a free body

diagram is drawn and all the reactions can be directly identi¬ed from enforcing

the equilibrium conditions, this is referred to as the statically determinate case.

If the reactions de¬ned on a free body diagram cannot be calculated by

imposing the equilibrium conditions, then this is referred to as the statically

indeterminate case. This is dealt with, if more than three forces or moments

for two-dimensional problems or more than six forces or moments for three-

dimensional problems need to be identi¬ed. It should be noted that the equilibrium

equations do not suf¬ce if in the two-dimensional case more than one moment is

unknown and in the three-dimensional case more than three moments are

unknown.

Example 3.3 As a two-dimensional example, consider the body of a single cell as sketched in

Fig. 3.4, that is loaded by a known force FP , while the body is supported at two

points, say A and B. The support is such that at point A only a force in the horizon-

tal direction can be transmitted. This is represented by the rollers, that allow point

A to freely move in the vertical direction. At point B, however, forces in both

the vertical and horizontal direction can be transmitted from the surrounding to

the body, in the ¬gure indicated by the hinge. A free body diagram is sketched in

Fig. 3.5. The supports are separated from the body. It is assumed that the supports

cannot exert a moment on the body, therefore only reaction forces in the horizon-

tal and vertical direction have been introduced. As a naming convention all forces

in the horizontal direction have been labelled H± (the subscript ± referring to the

point of application), while all vertical forces have been labelled V± . At each of

the attachment points, A and B, reaction forces have been introduced on both the

body and the support. According to the third law of Newton (see Section 2.3):

action = ’ reaction, forces are de¬ned in the opposite direction with respect to

each other, but have equal magnitude. The (three) reaction forces at point A and

point B are, for the time being, unknown. They can be calculated by enforcing

force and moment equilibrium of the body. Hence, both the sum of all forces in

the horizontal direction as well as the sum of all the forces in the vertical direction

acting on the body have to be equal to zero. For this purpose the load FP has been

decomposed into a horizontal force HP and a vertical force VP :

Static equilibrium

42

A

FP

P