B

Figure 3.4

A loaded body.

A HA HA

VP

ly VB VB

P

HP ly

HB HB

B

lx

y

x

Figure 3.5

Free body diagram of the simple loaded body.

’HP ’ HA ’ HB = 0

VP + VB = 0.

The above expressions represent two equations with three unknowns (HA , HB and

VB ), being insuf¬cient to determine these. However, the sum of the moments with

respect to an arbitrary point has be zero as well. Computing the resulting moment

with respect to point A gives

’2ly HB ’ ly HP ’ lx VP = 0.

This yields one additional equation such that the unknown reaction forces may be

determined.

The above procedure is not unique in the sense that different points can be used

with respect to which the sum of moments should be required to be zero. For

instance, rather than using the sum of moments with respect to point A, the sum

of moments with respect to point P could have been used.

43 3.3 Free body diagram

Suppose that

lx = 2 [ m] , ly = 1 [ m] , HP = 20 [ N] , VP = 10 [ N] .

Substitution of these values into the equation above, renders:

’20 ’ HA ’ HB = 0

10 + VB = 0

’2HB ’ 20 ’ 20 = 0.

Clearly, from the second equation it follows immediately that VB = ’10 [N],

while from the last equation it is clear that HB = ’20 [N], which leaves the ¬rst

equation to calculate HA as HA = 0 [N]. Hence we have the solution:

HA = 0 [N], HB = ’20 [N], VB = ’10 [N].

Example 3.4 An example of a statically indeterminate case appears if the rolling support at

point A of Fig. 3.4 is replaced by a hinge as in point B. In that case an additional

reaction force in the vertical direction must be introduced at point A, see Fig. 3.6.

With the same values of the parameters as before, force and moment equilibrium

yields

’20 ’ HA ’ HB = 0

10 + VA + VB = 0

’2HB ’ 20 ’ 20 = 0.

VA VA

A

HA HA

VP

ly VB VB

P

HP ly

HB HB

B

lx

y

x

Figure 3.6

Free body diagram of the simple loaded body, statically indeterminate case.

Static equilibrium

44

The horizontal reaction forces HA and HB can, incidentally, still be calculated,

giving, as before:

HA = 0 [N], HB = ’20 [N].

But, there is insuf¬cient information to compute VA and VB . In fact, there are only

three equations to determine the four unknowns (HA , HB , VA and VB ).

The nature of the support de¬nes the possible set of reaction forces that have to

be introduced. In the above given two-dimensional statically indeterminate exam-

ple the supports are assumed to be hinges or pin-connections. Effectively this

means that no moments can be exerted on the support and only forces in the x-

and y-direction have to be introduced.

Example 3.5 An example of a ¬xed support is given in Fig. 3.7(a), showing a beam that is

clamped at one end and loaded at a distance L from this ¬xation (cantilever beam).

The reaction forces and moment can be computed by enforcing force and moment

equilibrium. The sum of the forces in the x- and y-direction has to be equal to zero:

HA = 0

’F + VA = 0,

while the sum of the moments has to be zero, for instance, with respect to point A:

MA ’ LF = 0.

L F

A

y

x

(a)

F

VA VA

HA

HA

MA MA

(b)

Figure 3.7

Free body diagram of bar ¬xed at one end and loaded at the other.

45 3.3 Free body diagram

ez

Fz Mx

a

A Fy

My ey

Fx

P

P

b

ex Mz

(a) (b)

Figure 3.8

A beam construction loaded by a force P and the free body diagram.

Example 3.6 Consider the beam construction, sketched in Fig. 3.8(a), loaded by a force P. The

beam is clamped at point A and we want to determine the reaction loads at point A.

First of all a coordinate system is introduced and a free body diagram of the loaded

beam construction is drawn, as in Fig. 3.8(b). The applied load is represented by

the vector:

P = ’Pez .

The reaction force vector on the beam construction at point A is denoted by F and

is decomposed according to:

F = Fx ex + Fy ey + Fz ez ,

while the reaction moment vector at point A is written as

M = M x e x + My e y + Mz e z .

The requirement that the sum of all forces is equal to zero implies that

F + P = 0,

and consequently

Fx = 0, Fy = 0, Fz ’ P = 0.

The requirement that the sum of all moments with respect to A equals zero

leads to:

M + d — P = 0,

Static equilibrium

46

where the distance vector d is given by

d = bex + aey ,

hence

d — P = ( bex + aey ) — ( ’Pez )

= bPey ’ aPex .

Consequently

Mx ’ aP = 0, My + bP = 0, Mz = 0.

Example 3.7 Consider the man sketched in Fig. 3.9 who is lifting a weight. We would like to

compute the force F M in the muscle connecting the upper arm to the shoulder. A

basis {ex , ey } is introduced with the origin in the joint, point J. The basis vector ex

has the direction of the arm, while basis vector ey is perpendicular to the arm (see

¬gure). The forces W = ’Wey , due to the weight of the arm, and W 0 = ’W0 ey

due to the lifted weight are both supposed to be known. The reaction force in the

joint F J and the force in the muscle F M are both unknown. However, the direction

of the force in the muscle is known since this force is oriented with respect to the

arm at an angle θ. Consequently

F J = FJx ex + FJy ey ,

FM

ey

A C

B

θ

FJ ex

J

W

W0

a

b

c

Figure 3.9

Lifting a weight.

47 Exercises

while the force in the muscle is given by

F M = ’FM cos( θ) ex + FM sin( θ) ey .

Notice that both the x- and y-component of the joint reaction force, FJx and FJy ,

respectively, are unknown (the joint is modelled by a hinge), while for the muscle

only the magnitude of the muscle force FM is unknown.

Application of the force balance in the x- and y-direction yields

FJx ’ FM cos( θ) = 0

and

FJy + FM sin( θ) ’W ’ W0 = 0.

Force equilibrium supplies two equations for three unknowns, hence moment

equilibrium needs to be enforced as well. With the points A, B and C located

at xA = aex , xB = bex and xC = cex , respectively, moment equilibrium with

respect to point J requires: