Question 15.7: The calculation of deflections in a truss.

Suppose that we wish to calculate the deflection, \Delta_{2}, in the direction of the load, P_{2}, and at the joint at which P_{2} is applied in a truss comprising n members and carrying a system of loads P_{1}, P_{2}, \ldots, P_{k}, \ldots, P_{r}, as shown in Fig. 15.15. From Eq. (15.39)

\delta\left(\sum\limits_{j=1}^{n}\int_{0}^{F_{j}}\delta_{j}\,\mathrm{d}F_{j}-\sum\limits_{k=1}^{r}\Delta_{k}\,P_{k}\right)=0       (15.39)

the total complementary energy, C, of the truss is given by

C=\sum\limits_{j=1}^{n} \int_{0}^{F_{j}} \delta_{j} \mathrm{~d} F_{j}-\sum\limits_{k=1}^{r} \Delta_{k} P_{k}        (i)

From the principle of the stationary value of the total complementary energy with respect to the load P_{2}, we have

\frac{\partial C}{\partial P_{2}}=\sum\limits_{j=1}^{n} \delta_{j} \frac{\partial F_{j}}{\partial P_{2}}-\Delta_{2}=0      (ii)

from which

\Delta_{2}=\sum\limits_{j=1}^{n} \delta_{j} \frac{\partial F_{j}}{\partial P_{2}}      (iii)

Note that the partial derivatives with respect to P_{2} of the fixed loads, P_{1}, P_{3}, \ldots, P_{k}, \ldots, P_{r}, vanish.

To complete the solution we require the load-displacement characteristics of the structure. For a non-linear system in which, say,

F_{j}=b\left(\delta_{j}\right)^{c}

where b and c are known, Eq. (iii) becomes

\Delta_{2}=\sum\limits_{j=1}^{n}\left(\frac{F_{j}}{b}\right)^{1 / c} \frac{\partial F_{j}}{\partial P_{2}}        (iv)

In Eq. (iv) F_{j} may be obtained from basic equilibrium conditions, e.g. the method of joints, and expressed in terms of P_{2}; hence \partial F_{j} / \partial P_{2} is found. The actual value of P_{2} is then substituted in the expression for F_{j} and the product \left(F_{j} / b\right)^{1 / c} \partial F_{j} / \partial P_{2} calculated for each member. Summation then gives \Delta_{2}.

In the case of a linearly elastic structure \delta_{j} is, from Sections 7.4 and 7.7, given by

\delta_{j}=\frac{F_{j}}{E_{j} A_{j}} L_{j}

in which E_{j}, A_{j} and L_{j} are Young’s modulus, the area of cross section and the length of the j th member. Substituting for \delta_{j} in Eq. (iii) we obtain

\Delta_{2}=\sum\limits_{j=1}^{n} \frac{F_{j} L_{j}}{E_{j} A_{j}} \frac{\partial F_{j}}{\partial P_{2}}       (v)

Equation (v) could have been derived directly from Castigliano’s first theorem (Part II) which is expressed in Eq. (15.35)

{\frac{\partial U}{\partial P_{j}}}=\Delta_{j}        (15.35)

since, for a linearly elastic system, the complementary and strain energies are identical; in this case the strain energy of the j th member is F_{j}^{2} L_{j} / 2 A_{j} E_{j} from Eq. (7.29). Other aspects of the solution merit discussion.

U={\frac{P^{2}L_{0}}{2A E}} (7.29)

We note that the support reactions at A and B do not appear in Eq. (i). This convenient absence derives from the fact that the displacements, \Delta_{1}, \Delta_{2}, \ldots, \Delta_{k}, \ldots, \Delta_{r}, are the actual displacements of the truss and fulfil the conditions of geometrical compatibility and boundary restraint. The complementary energy of the reactions at A and B is therefore zero since both of their corresponding displacements are zero.

In Eq. (v) the term \partial F_{j} / \partial P_{2} represents the rate of change of the actual forces in the members of the truss with P_{2}. This may be found, as described in the non-linear case, by calculating the forces, F_{j}, in the members in terms of P_{2} and then differentiating these expressions with respect to P_{2}. Subsequently the actual value of P_{2} would be substituted in the expressions for F_{j} and thus, using Eq. (v), \Delta_{2} obtained. This approach is rather clumsy. A simpler alternative would be to calculate the forces, F_{j}, in the members produced by the applied loads including P_{2}, then remove all the loads and apply P_{2} only as an unknown force and recalculate the forces F_{j} as functions of P_{2} ; \partial F_{j} / \partial P_{2} is then obtained by differentiating these functions.

This procedure indicates a method for calculating the displacement of a point in the truss in a direction not coincident with the line of action of a load or, in fact, of a point such as \mathrm{C} which carries no load at all. Initially the forces F_{j} in the members due to P_{1}, P_{2}, \ldots, P_{k}, \ldots, P_{r} are calculated. These loads are then removed and a dummy or fictitious load, P_{\mathrm{f}}, applied at the point and in the direction of the required displacement. A new set of forces, F_{j}, are calculated in terms of the dummy load, P_{\mathrm{f}}, and thus \partial F_{j} / \partial P_{\mathrm{f}} is obtained. The required displacement, say \Delta_{\mathrm{C}} of \mathrm{C}, is then given by

\Delta_{\mathrm{C}}=\sum\limits_{j=1}^{n} \frac{F_{j} L_{j}}{E_{j} A_{j}} \frac{\partial F_{j}}{\partial P_{\mathrm{f}}}      (vi)

The simplification may be taken a stage further. The force F_{j} in a member due to the dummy load may be expressed, since the system is linearly elastic, in terms of the dummy load as

F_{j}=\frac{\partial F_{j}}{\partial P_{\mathrm{f}}} P_{\mathrm{f}}     (vii)

Suppose now that P_{\mathrm{f}}=1, i.e. a unit load. Equation (vii) then becomes

F_{j}=\frac{\partial F_{j}}{\partial P_{\mathrm{f}}} 1

so that \partial F_{j} / \partial P_{\mathrm{f}}=F_{1, j}, the load in the j th member due to a unit load applied at the point and in the direction of the required displacement. Thus, Eq. (vi) may be written as

\Delta_{\mathrm{C}}=\sum\limits_{j=1}^{n} \frac{F_{j} F_{1, j} L_{j}}{E_{j} A_{j}}     (viii)

in which a unit load has been applied at \mathrm{C} in the direction of the required displacement. Note that Eq. (viii) is identical in form to Eq. (ii) of Ex. 15.6.

In the above we have concentrated on members subjected to axial loads. The arguments apply in cases where structural members carry bending moments that produce rotations, shear loads that cause shear deflections and torques that produce angles of twist. We shall now demonstrate the application of the method to structures subjected to other than axial loads.