After reading previous entry and understanding the meaning of distributed forces, how to calculate the equivalent resultant to the distributed load, you should try to solve next exercises.
EXERCISE 1: Calculate the supports reactions at A and B for the beam subjected to the asymmetric load distributions.
EXERCISE 2: The symmetric simple truss is loaded as shown. Determine the reactions at wall support A and the tension in cable CD.
DISTRIBUTED FORCES
In the previous sections we treated all forces as concentrated along
their lines of action and at their points of application. This treatment provided a reasonable model for those forces. Actually, “concentrated” forces do not exist in the exact sense, since every external force applied mechanically to a body is distributed over a finite contact area, however small.
What is more, distributed forces are a particular case of a coplanar parallel system of forces. The body force due to the gravitational attraction of the earth (weight) is by far the most commonly encountered distributed force. Figure shows (Meriam & Kreige 7th edition):
Figure shows a statically determinate beam that is supporting a general distributed load. The intensities qi(x) of a distributed load may be expressed as force per unit of length of beam. The intensity may be constant or variable, continuous or discontinuous. The intensity of the loading in Figure is constant from A to B and variable from O to A and from B to C. Although the intensity itself is not discontinuous at A and B, the rate of change of intensity dq/dx is discontinuous.
Any distributed load can be replaced by a concentrated load with a magnitude equal to the area under the load curve and a line of action passing through the centroid of the area. Therefore, we need to determine the equivalent force of the distributed force, i. e. the resultant, and the centre of gravity where it will be applied. Once the distributed loads have been reduced to their equivalent concentrated loads, the external reactions acting on the beam may be found by a straightforward static analysis as developed in Chapter 6 (equilibrium).
After reading previous entry and understanding the meaning of stability, rollover and stabilising moments, you should try to solve next exercise.
A homogeneous cylinder of radius R has a cylindrical hole of diameter R as shows in Figure. Cylinder can roll down the inclined plane without slipping. Determine the maximum angle θ of the inclined plane because the cylinder be in equilibrium. Obtain also the angle between the horizontal and the diameter AB in this situation.
NOTE: you can send your solution by tutorial tool in UACloud for including it in the continuous evaluation.
Any system of forces may be replaced by a resultant force R and a resultant moment (couple of a force) M at any point of the space. We call this couple vectors (R, M) as system “torsor”. Figure (from Meriam & Kraige 7th edition) shows a system of forces acting on a rigid body.
The point O selected as the point of concurrency for the forces is arbitrary, and the magnitude and direction of M depend on the particular point O selected. The magnitude and direction of R, however, are invariant, i.e. are the same no matter which point is selected.
If a system of forces acting on a body are such that the resultant force and resultant moment at point O are mutually perpendicular, the whole system can be replaced by a single force acting along a new line of action such that it causes the same moment with respect to O. We now examine the resultants for several special system of forces:
STABILITY AND TOPPLING
Let’s consider a 2D situation. A rigid body under the action of forces (which are necessarily coplanar) can only be in equilibrium if the system of forces seen in the free-body diagram can be reduced to three concurrent forces that meet at a point that is inside the body.
Three concurrent forces at point C which is inside the rigid body: EQUILIBRIUM (upper figure).
Three non-concurrent forces inside the rigid body: 
THERE IS NO EQUILIBRIUM (bottom figure). Therefore, if a body is not in equilibrium, we can check if it is prone to overturning or stable against toppling. For this, we calculate the resultant of all external forces and draw it at the point where its line of action cuts the line of action of the weight. Assume the rigid body can rotate about point O, we call rollover moment the expression:
and stabilising moment the expression:
If the rollover moment is larger than the stabilising moment, the rigid body will turn over.
A rigid body is said to in equilibrium if the resultant of all forces and all their moments taken about any and all points are zero.
Free body diagram (FBD): depiction of an object with ALL the external forces acting on it.
Reactions coming from the supports: usually a body is constrained against motions using supports. It is essentially to correctly estimate the number and type of reactions that a support can provide.
http://www.slideshare.net/kevyn52/equilibriumofrigidbody
A slide-share presentation about equilibrium can help you to understand the key concepts of this chapter:
ROTATIONAL DYNAMICS: EXERCISES
A mass M = 120 kg hangs from a beam which is held by a cable.
iling, as shown in Figure. If the angle between the rope and the vertical is θ = 30º and the radius of the circle is r = 35 cm:
Chapter 4: RIGID BODIES III
Massachusetts Institute of Technology have developed Open Course Wares (OCW) where you can follow a “Classical Mechanics course” by Internet.
One of the most important magnitudes in architecture is the moment of a force or torque. And this magnitude will be extremely important in Chapter 5: Equilibrium.
Linear momentum, i.e. p = m·v, has a rotational analogue that it is called angular momentum, L. For a symmetrical object rotating about a fixed axis through the centre of mass (CM), the angular momentum is L = I·ω where I is the moment of inertia and ω is the angular velocity about the axis of rotation. The SI units for L are kg·m^2/s, which has no special name.
We saw in Chapter 2: Dynamics laws and applications that Newton’s second law can be written more generally in terms of momentum ΣF = Δp/Δt. In a similar way, the rotational equivalent of Newton’s second law which is Στ = I·α, can also be written more generally in terms of angular momentum Στ = ΔL/Δt where Στ is the net torque acting to rotate the object and ΔL is the change in angular momentum in a interval time Δt.
Angular momentum is an important concept in physics because, under certain conditions, it is a conserved quantity. If Στ = ΔL/Δt on an object is zero then ΔL = 0, so L does not change. This is the law of conservation of angular momentum for a rotating object:
The total angular momentum of a rotating object remains constant if the net torque acting on it is zero.
Worked example: moment-of-inertia-of-a-disc-from-a-falling-mass
Worked example: massive-pulley-problems
MisConceptual Questions
Video lecture: Rotational dynamic example
Problem set number 5
If you have some doubts, you can watch next video related to rotational dynamic (Professor Michel van Biezen):
If you were satisfied with this example, you can check more video lectures on the webpage:
Chapter 4: RIGID BODIES II
Chapter 4. Rigid bodies (I)