Analysis of Skeletal Structures.

The evaluation of the stress resultants in members of skeletal frames involves the solution of a number of simultaneous equations.When a structure is in equilibrium, every element or constituent part of it is also in equilibrium. This property is made use of in developing the concept of the free body diagram for elements of a structure.

The portal frame sketched in Fig. 9.16 will now be considered for illustrating the concept. Assuming that there is an imaginary cut at E on the beam BC, the part ABE continues to be in equilibrium if the two forces and moment which existed at section E of the uncut frame are applied externally.The internal forces which existed at E are given by (1) an axial force F, (2) a shear force V and (3) a bending moment M. These are known as stress resultants. The external forces on ABE, together with the forces F, V and M, keep the part ABE in equilibrium; Fig. 9.16(b) is called the free body diagram. On a rigid jointed plane frame there are three stress resultants at each imaginary cut. The part ECD must also remain in equilibrium. This  consideration leads to a similar set of forces F, V and M shown in Fig. 9.16(c). It  will be noted that the forces acting on the cut face E are equal and opposite. If the two free body diagrams are moved towards each other, it is obvious the internal forces F, V and M cancel out and the structure is restored to its original state of equilibrium.As previously stated, equilibrium implies ΣPx = 0; ΣPy = 0; ΣM = 0 for a planar structure. These equations can be validly applied by considering the structure as a whole, or by considering the free body diagram of a part of a structure.

In a similar manner, it can be seen that a three-dimensional rigid-jointed  frame has six stress resultants across each section. These are the axial force, two shears in two mutually perpendicular directions and three moments, as shown in Fig. 9.17.

Free body diagram
Fig. 9.16 Free body diagram

Force and moments in x, y and z directions
Fig. 9.17 Force and moments in x, y and z directions

With pin-jointed frames, be they two- or three-dimensional, there is only one stress resultant per member, viz. its axial load.When forces act on an elastic structure, it undergoes deformations, causing displacements at every point within the structure.

The solution of forces in the frames is accomplished by relating the stress result- ants to the displacements. The number of equations needed is governed by the degrees of freedom, i.e. the number of possible component displacements. At one end of the member of a pin-jointed plane frame, the member displacement has translational components in the x and y directions only, and no rotational displacement. The number of degrees of freedom is two. By similar reasoning it will be apparent that the number of degrees of freedom for a rigid-jointed plane frame member is three. For a member of a three-dimensional pin-jointed frame it is also three, and for a similar rigid-jointed frame it is six.

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