Question 14.3: Two-span beam with overhang is subjected to force P and unif......

The given structure has two redundant constraints. There is possible the different schemes of structural failure. Let us consider one of them. The progressive increase of the loads leads to the appearance of the plastic hinge at one of the supports, so the structure becomes statically indeterminate of the first degree. Further increase of the loads leads to the appearance of the plastic hinge at another support. Finally, plastic hinge happens at the last support. It means that the entire continuous beam is being transformed into two simply supported beams subjected to given loads and plastic moments M_y  at the supports A, B and C as shown in Fig. 14.8b. Direction of moments M_y  is shown according to location of extended fibers in elastic state. The order of appearance of the plastic hinges on supports depends on the relationships between force P and load q as well as geometrical parameters of the beam.

This failure mechanism allows developing the theory of plastic analysis of continuous beams subjected to several loads. However, it does not mean that exactly the above sequence of formation of plastic hinges will be realized. For example, if load q is small, then plastic hinges can appear first of all at the supports A, B and at point K, and after that at support C and in the second span. Real emergence sequence of plastic hinges may be defined only after determination of limit load as shown below.

Reaction of support A and bending moment at the point K of the first span caused by force P as well as plastic moments at supports A and B equals

\begin{aligned}& R_A=R_A(P)+R_A\left(M_y^A\right)+R_A\left(M_y^B\right)=\frac{P b}{l_1}+\frac{M_y^A}{l_1}-\frac{M_y^B}{l_1}=\frac{P b}{l_1} \\& M_K=M_K\left(R_A\right)-M_K\left(M_y^A\right)=\frac{P a b}{l_1}-M_y\end{aligned}        (a)

Bending moment at the middle point of the second span caused by distributed load  q  and moments at supports B and C equals

M=\frac{q l_2^2}{8}-M_y      (b)

The following increase of the load leads to the appearance of the plastic moments within the first and second spans. In the limit state these bending moments must be equal to plastic moments; therefore expressions (a) and (b) should be rewritten for both spans as follows:

For first span \quad \frac{P a b}{l_1}-M_y=M_y \quad or \quad \frac{P a b}{l_1}=2 M_y      (c)

For second span \frac{q l_2^2}{8}-M_y=M_y \quad or \quad \frac{q l_2^2}{8}=2 M_y      (d)

Equation (c) and (d) show that in limit state, the limit bending moment caused by external load equals twice plastic moment. This can be presented geometrically as shown in Fig. 14.8c. Limit plastic moments  (L P M) are shown by two horizontal dotted lines. These lines show that limit moments for supports and for any section of the beam are equal, may be negative or positive, however, they cannot be more than M_y .

Now we can fit a space between two (L P M) lines by bending moment diagrams caused by external load for each simply supported beam. This procedure is called equalizing of bending moments and can be effectively applied for any continuous beam.

For the first span, (c) allows calculating the limit force P

\frac{P_{\mathrm{lim}} a b}{l_1}=2 M_y \rightarrow P_{\lim }=\frac{2 M_y l_1}{a b}=\frac{2 \times 60(\mathrm{kNm}) \times 7(\mathrm{~m})}{3(\mathrm{~m}) \times 4(\mathrm{~m})}=70 \mathrm{kN}     (e)

For the second span, (d) allows calculating the limit distributed load q

\frac{q_{\mathrm{lim}} l_2^2}{8}=2 M_y \rightarrow q_{\mathrm{lim}}=\frac{16 M_y}{l_2^2}=\frac{16 \times 60(\mathrm{kNm})}{36\left(\mathrm{~m}^2\right)}=26.67 \mathrm{kN} / \mathrm{m}

Now we need to take into account the condition of simple loading as well as limiting force P=70 \mathrm{kN}  and load q=26.67 \mathrm{kN} / \mathrm{m} . Knowing the distributed load we can calculate, according condition P=2 q l_2 , corresponding limit force P

P_{\lim }=2.0 \times 26.67(\mathrm{kN} / \mathrm{m}) \times 6(\mathrm{~m})=320 \mathrm{kN}>70 \mathrm{kN}        (f)

For part CD

\frac{q_{\lim } l_3^2}{2}=M_y \rightarrow q_{\mathrm{lim}}=\frac{2 M_y}{l_3^2}=\frac{2 \times 60(\mathrm{kNm})}{2^2\left(\mathrm{~m}^2\right)}=30 \mathrm{kN} / \mathrm{m},

which leads to

P_{\lim }=2.0 \times 30 \times 6=360 \mathrm{kN}>70 \mathrm{kN}           (g)

Thus, limit distributed load q in both cases leads to the limit force P which can not be accepted. The final limit load is governed by minimum value given by formulae (e), (f) and (g), so the limit load P_{\lim }=70 \mathrm{kN} , and corresponding

q_{\lim }=\frac{P_{\lim }}{2 l_2}=\frac{70}{2 \times 6}=5.83 \mathrm{kN} / \mathrm{m}

Discussion

On the basis of obtained numerical results we can explain the order of appearance of the plastic hinges. The limit load P=70 \mathrm{kN}  and corresponding load q are determined from the conditions of appearance of plastic hinges at the supports A, B  and at point K of application force P. Thus, the failure of the structure in whole is defined by a failure of the span AB  because this simply supported beam is being transformed in mechanism. In this case, with further increase of the load q , the relationship P=2 q l_2  is not held anymore, since load P cannot reach the value greater than P_{\text {lim }} . It is obvious that span BC  still can resist to increased load q , however, the structure in whole is differ from the original one.

The problem of determination of limit load for continuous beam with given bearing capacity has unique solution.