FREE BODY DIAGRAM CONSTRUCTION BASICS AND TUTORIALS

TIP ON HOW TO CONSTRUCT FREE BODY DIAGRAM

Construction of Free-Body Diagrams Tutorials

The full procedure for drawing a free-body diagram which isolates a body or system consists of the following steps.

Step 1. Decide which system to isolate. The system chosen should usually involve one or more of the desired unknown quantities.

Step 2. Next isolate the chosen system by drawing a diagram which represents its complete external boundary. This boundary defines the isolation of the system from all other attracting or contacting bodies, which are considered removed. 

This step is often the most crucial of all. Make certain that you have completely isolated the system before proceeding with the next step.

Step 3. Identify all forces which act on the isolated system as applied by the removed contacting and attracting bodies, and represent them in their proper positions on the diagram of the isolated system. Make a systematic traverse of the entire boundary to identify all contact forces. 

Include body forces such as weights, where appreciable. Represent all known forces by vector arrows, each with its proper magnitude, direction, and sense indicated. Each unknown force should be represented by a vector arrow with the unknown magnitude or direction indicated by symbol. 

If the sense of the vector is also unknown, you must arbitrarily assign a sense. The subsequent calculations with the equilibrium equations will yield a positive quantity if the correct sense was assumed and a negative quantity if the incorrect sense was assumed. 

It is necessary to be consistent with the assigned characteristics of unknown forces throughout all of the calculations. If you are consistent, the solution of the equilibrium equations will reveal the correct senses.

Step 4. Show the choice of coordinate axes directly on the diagram. Pertinent dimensions may also be represented for convenience. 

Note, however, that the free-body diagram serves the purpose of focusing attention on the action of the external forces, and therefore the diagram should not be cluttered with excessive extraneous information. 

Clearly distinguish force arrows from arrows representing quantities other than forces. For this purpose a colored pencil may be used.

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF HOMOGENEOUS SOLIDS TUTORIALS

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF HOMOGENEOUS SOLIDS
Tutorials On Center of Gravity and Mass Moment of Inertia of Homogeneous Solids

This is a reference on the center of gravity and mass moment of inertia for typical homogeneous solids. These are helpful reference in solution of civil engineering problems.

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF SPHERE

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF HEMISPHERE

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF CIRCULAR DISK

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF THIN RING

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF CYLINDER

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF CONE

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF THIN PLATE

CENTER OF GRAVITY & MASS MOMENT OF INERTIA OF SLENDER ROD

SHEAR AND BENDING MOMENT DIAGRAMS BASICS AND TUTORIALS

SHEAR AND BENDING MOMENT DIAGRAMS BASIC INFORMATION
A Tutorials on Shear and Bending Moment Diagram? How To Make Shear and Bending Moment Diagram


In order to plot the shear force and bending moment diagrams it is necessary to adopt a sign convention for these responses. A shear force is considered to be positive if it produces a clockwise moment about a point in the free body on which it acts.

A negative shear force produces a counterclockwise moment about the point. The bending moment is taken as positive if it causes compression in the upper fibers of the beam and tension in the lower fiber. In other words, sagging moment is positive and hogging moment is negative.

The construction of these diagrams is explained with an example given in Figure 2.4.

The section at E of the beam is in equilibrium under the action of applied loads and internal forces acting at E as shown in Figure 2.5. There must be an internal vertical force and internal bending moment to maintain equilibrium at Section E.

The vertical force or the moment can be obtained as the algebraic sum of all forces or the algebraic sum of the moment of all forces that lie on either side of Section E.

The shear on a cross-section an infinitesimal distance to the right of pointAisC55 k and, therefore, the shear diagram rises abruptly from 0 to C55 at this point. In the portion AC, since there is no additional load, the shear remainsC55 on any cross-section throughout this interval, and the diagram is a horizontal as shown in Figure 2.4. 

An infinitesimal distance to the left of C the shear is C55, but an infinitesimal distance to the right of this point the 30 k load has caused the shear to be reduced to C25. 

Therefore, at point C there is an abrupt change in the shear force from C55 to C25. In the same manner, the shear force diagram for the portion CD of the beam remains a rectangle. In the portion DE, the shear on any cross-section a distance x from point D is 

               S = 55 − 30 − 4x D 25 − 4x

which indicates that the shear diagram in this portion is a straight line decreasing from an ordinate of C25 at D to C1 at E. 

The remainder of the shear force diagram can easily be verified in the same way. It should be noted that, in effect, a concentrated load is assumed to be applied at a point and, hence, at such a point the ordinate to the shear diagram changes abruptly by an amount equal to the load.

In the portion AC, the bending moment at a cross-section a distance x from point A isM D 55x. Therefore, the bending moment diagram starts at 0 at A and increases along a straight line to an ordinate of C165 k-ft at point C. 

In the portion CD, the bending moment at any point a distance x from C is M D 55.x C 3/ − 30x. Hence, the bending moment diagram in this portion is a straight line increasing from 165 at C to 265 at D. In the portion DE, the bending moment at any point a distance x from D is M D 55.x C 7/ − 30.X C 4/ − 4x2=2. 

Hence, the bending moment diagram in this portion is a curve with an ordinate of 265 at D and 343 at E. In an analogous manner, the remainder of the bending moment diagram can be easily constructed.

Bending moment and shear force diagrams for beamswith simple boundary conditions and subject to some simple loading are given in Figure 2.6.