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## Q. 14.8.1

The frame in Fig. 14.8-1(a) is of uniform plastic moment of resistance 33.3 kNm and supports loads of working values shown. Using Baker, Home and Heyman’s equivalent load method, estimate the collapse load factor λ.

## Verified Solution

Referring to Fig. 14.8-1(b), for a simply supported beam the uniformly distributed load (w) and the ‘equivalent’ point loads (wL/4, wL/2, wL/4) produce the same maximum bending moment, namely, wL2/8. According to the method, the distributed load on the beam BD is replaced by the equivalent point loads:

(6.67λ)(6)/2 = 20λ kN at midspan
(6.67λ)(6)/4 = 10λ kN at each end.

The frame under the equivalent loading is shown in Fig. 14.8-1(c), in which the dotted arrows at B and D represent the 10 kN loads which do not affect the analysis and which are therefore neglected in subsequent calculations.
A comparison of Fig. 14.8-1(c) and 14.6-10(a) shows that the collapse mechanism for the frame under the equivalent load must be that of Fig. 14.6-10(d). From Example 14.6-4, the collapse load for the mechanism in Fig. 14.6-10(d) is $P = 3M_p/5a$; therefore, for the frame in Fig. 14.8-1(d), the collapse load factor is given by

10 λ = (3) (33.3)/5 = 20
or              λ = 2

Therefore, as a first approximation, the frame collapses in the mechanism in Fig. 14.8-1(d) at a collapse load factor of 2. Since this mechanism is, in fact, for the equivalent load and is not necessarily the correct mechanism for the actual loads, it can at once be concluded from the unsafe theorem that λ = 2 is an upper bound value on the true load factor. To obtain a lower bound value, the bending moment diagram is drawn for the collapse state, using the free-body diagrams in Fig. 14.8-1(e). It can be seen from the bending moment diagram in Fig. 14.8-1(f) that the yield condition is violated in the beam BD; the location x at which the sagging moment is a maximum is readily found from the condition of zero shear:

13.34.x = 42.2                     or            x = 3.16 m

which is not greatly different from the value of x  = 3 m in the mechanism for the equivalent load. The maximum sagging moment is therefore

(42.2)(3.16) – 13.34(3.16)2/2 – 33.3 = 33.5 kNm

which is quite close to the plastic moment Mp of 33.3 kNm. By the safe theorem, it can at once be concluded that a lower bound value on the true collapse load factor is λ = 2 x 33.3/33.5 = 1.988. That is

1.988        ≤          λ          ≤         2

(lower bound)                       (upper bound)

Suppose the calculations are carried one step further and the collapse mechanism is revised so that the sagging hinge in the beam occurs at x = 3.16 m. Using the rotations and displacements in Fig.14.8-2(a)—see Comment (1) at end—the work equation is:

(6.67λ)(6)(6 – 3.16)ɸ/2 + (10λ)(4 ɸ)
= 33.3[
ɸ + 6ɸ/3.16 + 6ɸ/3.16 + ɸ] giving

λ = 1.994

By the unsafe theorem this is an upper bound value, so that

1.988         ≤            λ          ≤      1.994

(lower bound)                     (upper bound)

Of course, the process of successive approximation can be repeated: a bending moment diagram is drawn for the collapse state in Fig. 14.8-2(a), and a new lower bound value of λ determined; then from a revised value of x, a revised mechanism is investigated leading to a new upper bound value of λ, and so on. However, it is seen from this worked example that the process converges so rapidly that the answer 1.988   ≤   λ ≤  2 given by the equivalent-load mechanism is already very good; the second cycle of calculations, leading to  1.988   ≤   λ ≤  1.994 is hardly necessary.

COMMENT
(1) The displacements and rotations in Fig. 14.8-2(a) can be determined by the method of instantaneous centres explained earlier in Section 6.8. Referring to Fig. 14.8-2(b)

$\qquad \qquad FF^\prime =(IF^\prime )\theta =(AF^\prime)\phi$

Hence $\qquad \qquad \theta =\frac{(AF^\prime)}{(IF^\prime)}\phi = \frac{(B^\prime F^\prime)}{F^\prime D^\prime}\phi =\frac{6-x }{x } \phi$

$\qquad \qquad DD^\prime =(ID^\prime )\theta =(ED^\prime)\psi$

Hence $\qquad \psi = \frac{(ID^\prime)}{(ED^\prime)} \theta =\frac{(ID^\prime)}{(AB^\prime)} \theta =\frac{(D^\prime F^\prime )}{(B^\prime F^\prime)} \theta \\ \qquad =\frac{x}{6-x}\theta = \left(\frac{x}{6-x} \right)\left(\frac{6-x}{x} \right)\phi \\ \qquad =\phi$ as expected.

Hinge rotation at F

$\qquad =\theta +\phi =\frac{6-x}{x} \phi +\phi =\frac{6}{x}\phi$

Hinge rotation at D

$\qquad =\theta +\psi =\theta +\phi =\frac{6}{x} \phi$

Horizontal displacement of joint B = 4ɸ
Vertical component of displacement of joint F, δF

$\qquad =x \theta =(x )\left(\frac{6-x}{x} \right)\phi =(6-x)\phi$

(2) In the above Example, having established that the mechanism for the equivalent load is that of Fig. 14.8-1(d),itmaybe concluded1 that for the actual loading the mechanism is of the general form in Fig. 14.8-2(a) where the sagging hinge position F is defined by the variable x. The work equation is then

$\qquad (10\lambda ) (4\phi )+(6.67\lambda )(6)\frac{(6-x)}{2}\phi$ (The displacement of the centroid of the distributed load is (6 — x)ɸ/2.)

$\qquad \qquad =33.3\left[\phi +\frac{6}{x}\phi +\frac{6}{x}\phi +\phi \right]$

or $\qquad \quad \lambda =\frac{3.33(6+x)}{x(8-x)}$

which is an upper bound value for an arbitrary value of x.
The lowest upper bound is given by

$\qquad \qquad \frac{d\lambda }{dx} =0$ which occurs at x = 3.17 m

Whence $\lambda =\frac{3.33(6+3.17)}{3.17(8-3.17)} =1.995$

When there are many members carrying distributed loads, it will be necessary to minimize A with respect to all the x’s defining the hinge positions. The values of the x ‘s are then obtained from

$\qquad \frac{\partial \lambda }{\partial x_1} =0;\frac{\partial \lambda }{\partial x_2} =0;\frac{\partial \lambda }{\partial x_3} =0$ etc.

For a complex structure, the method is therefore rather tedious, but for simple structures it is very effective —see Example 14.8-2.