Example 85 Frame

An asymmetric uniform, frame is pinned at supports A and G and is subjected to a system of factored loads as shown in Figure 8.15. Assuming the Xverticai.ioad=1.7 and ^-horizontal loads=1.4 determine the required plastic moment of resistance Mp of the section.

Figure 8.15

Number of diienLes-of-iniliiwrtmnasy It, = [(3w + r) - 3«] = 1(3 * 3) + <l) - (3 x 4)] = 1 Number ofpoi&iMe hinjie positions ft = 5 (B, C. D, E and F)

(i/i. 3 beam ntcchnnisrns an<11 sway mcchanism)

Kinematic Method:

Consider each independent mechanism separately.

Mechanism (i): Beam ABC

JJH Id)

Mechanism (Jt)t Beam GDE

II hN

II hN

Mote

I ntcrnal work is (tone al a 11 lull u>ll [nisi t ji>us. No inltniiil work is done ;il support A. The signs of the rotations indicate tension inside or :i w eoujiic Ac Hw. ~~' <5= 3.5£?

liiCcninl Wsrt [>onc = liMcnal Wort IhiK

IMfS-liSff ,WP- tO.SkNm

Internal Work Done = Lxlcrnal Work lione

Combinations:

Consider the independent mechanisms, their associated work equations and Mp values as shown in Figure 8.16:

Figure 8.16

It is evident from inspection of the collapse mechanisms that the hinges located at C and E can be eliminated since in some cases the rotation is negative whilst in others it is positive. The minimum number of hinges to induce total collapse is one more than the number of redundancies, i.e. (ID+1)=2 and therefore the independent mechanisms should be combined to try and achieve this and at the same time maximize the associated Mp value. It is unlikely that mechanism (i) will be included in the failure mechanism since its associated Mp value is relatively small compared to the others. It is necessary to investigate several possibilities and confirm the resulting solution by checking that the bending moments do not exceed the plastic moment of resistance at any section.

When combining these mechanisms the hinge at C will be eliminated and the resulting Mp value can be determined by adding the work equations. It is necessary to allow for the removal of the hinge at C in the internal work done since in each equation an (Mp9) term has been included, but the hinge no longer exists. A total of 2Mp must therefore be subtracted from the resulting internal work, i.e.

It is possible that this is the true collapse mechanism, however this would have to be confirmed as indicated above by satisfying conditions (ii) and (iii) in Section 8.1.2.

An alternative solution is also possible where the hinges at C and E are eliminated, this can be a achieved if mechanism (v) is combined with mechanism (iii).

In mechanism (v) p=O.50 (see the sway calculation above) and hence the total rotation at joint E=-(0+p)=-1.50. If this hinge is to be eliminated then the combinations of mechanisms (iii) and (v) must be in the proportions of 1.5:1.0. (Note: when developing mechanism (v) the proportions were 1:1).

The total value of the internal work for the eliminated hinge=(2x1.5Mp)=3.0Mp, i.e.

The +ve rotation indicates tension inside the frame at point D and the -ve rotation indicates tension outside the frame at point F.

This is marginally higher than the previous value and since there does not appear to be any other obvious collapse mechanism, this result should be checked as follows:

Figure 8.17

Consider I Ik1 equilibrium ol'llie frame between F and CI:

Consider the equilibrium of (lie frame on the right- hand side,it l>: +veJ)^ - 0 + 5S.7 - (21,0 x 3.0) + (19,57 * 6.0) - {Vq x 4.0) -0

Consider llie complete structure:

+ve —= 0 fix + 21.0 + 21-0 - 19.57 = 0 Ih " - 22.43 kN ■*■

Bending moment al B Mk = + (22.43 * L.5) = + 33.65 kNin i Mf

Bending moment m C .1k = + (22.43 « - (21.0 k 1.5} - +■ 3S.79 kMin <. \fT

Eiending moment at E Mr. = - (19.57 K 6.0) M21.Ox 3.0) = -54.42 kNio £ Mf

Figure 8.18

The three conditions indicated in Section 8.1.2 have been satisfied: i.e.

Mechanism condition: minimum number of hinges required=(ID+1)=2 hinges,

Equilibrium condition: the internal moments are in equilibrium with the collapse loads,

Yield condition: the bending moment does not exceed Mp anywhere in the frame.

It is often convenient to carry-out the calculation of combinations using a table as shown in Table 8.1; eliminated hinges are indicated by EH in the Table.

Independent and Combined Mechanisms for Example 8.5

Hinge

(i)

(ii)

(iii)

(iv)

(v)=(ii)+(iv)

(vi)=(v)+1. 5 (iii)

Position

B(Mp)

+2.09

-

-

-

-

-

C(Mp)

-9

-9

-

+9

EH (2.0Mp9)

EH (2.0Mp9)

D (Mp)

-

+2.09

-

-

+2.09

+2.09

E (Mp)

-

-9

+9

-0.59

-1.59

EH (3.0Mp9)

F (Mp)

-

-

-2.09

-

-

-3.09

External Work

31.59

136.09

63.09

63.09

199.09

293.56»

Internal Work

3.0Mp9

4.0Mp9

3.0Mp9

1.5Mp9

5.5Mp9

10.0Mp9

Eliminated hinges

-

-

-

-

2.0Mp9

Mp0

-

-

-

-

3.5Mp9

5.0Mp6

Mp (kNm)

10.5

34.0

21.0

42.0

56.86

58.70

Table 8.1 Static Method:

This frame can also be analysed readily using the static method since it only has one degree-of-indeterminacy. When using this method the frame can be considered as the superposition of two frames; one statically determinate and one involving only the assumed redundant reaction as shown in Figure 8.19. Applying the three equations of equilibrium to the two force systems results in the support reactions indicated.

Table 8.1 Static Method:

This frame can also be analysed readily using the static method since it only has one degree-of-indeterminacy. When using this method the frame can be considered as the superposition of two frames; one statically determinate and one involving only the assumed redundant reaction as shown in Figure 8.19. Applying the three equations of equilibrium to the two force systems results in the support reactions indicated.

Figure 8.19

The iiftll valtrc of 1 lie rendions and bending moments = [Frame (i) 4 Frame (Li)]: e.g.

Equations can be developed for each of the five possible hinge positions in terms of the two frames as follows:

Equation (1)

Mc - + H2,0 y 3.0) - (21,0 x [ .5) - f 3.QIf{,) = + 94.5 - 3.0//<;

Equation (2)

M, = + (42,0 x 3.0)4- 03.06 ^ 4.0) -<21.0 * 1.5)- (3,0 * Bo)- (4.0 * 0,375//(i)

Equation

Equation (4)

Equation (5)

As indicated previously, only two hinges are required to induce total collapse. A collapse mechanism involving two hinge positions can be assumed and the associated equations will each have two unknown values, i.e. HG and Mp and can be solved simultaneously.

The value of the bending moment at all other hinge positions can then be checked to ensure that they do not exceed the calculated Mp value. If any one does exceed the value then the assumed mechanism was incorrect and others can be checked until the true one is identified.

Assume a mechanism inducing hinges at D and E as in (v) above.

Equation (7)

Check the value of the moments at all other possible hinge positions.

Ma = + 63.0- \.SHc, ~ + 63.0 - (1.5 x 19.98) = + 33.03 kNmi Aip JUfc = + 94.5 - 3.0/Zçj - + 94,5 - (3.0 x 19,98) - + 34,56 kNin i

Since the bending moment at F is greater than Mp this mechanism does not satisfy the 'yield condition' and produces an unsafe solution.

The reader should repeat the above calculation assuming hinges develop at positions D and F and confirm that the true solution is when Mp=58.7 kNm as determined previously using the kinematic method.

+1 0

Post a comment