#### Installation — business terrible - 1 part

September 8th, 2015

For the hybrid element, the total rotation 0 at an end is comprised of two parts (see Fig. 1). The first part, 0e, represents the end rotation of the elastic beam-column, while Bp represents the plastic rotation of the hinge. Thus, at the full-yield 0=0e+0pu. To simulate the spread of plasticity of an actual inelastic beam-column, the 0 of the hybrid element must be identical to that of an actual beam-column.

Consider an actual inelastic beam-column under linear distributed first-order moments (see Fig. 9a). Define moment gradient k=±|M2/M1/, where M2 and M1 are the bending moments at Ends 1 and 2, respectively, and M2 < M1. Moment gradient к is positive for double curvature and negative for single curvature (-1<k<1). The lengths of the plastic zones at Ends 1 and 2 are L1 and L2, respectively. The corresponding curvature distribution along the member is illustrated in Fig. 9b. The plastic curvature begins at the section where moment is equal to Myc. Figure 9c shows the distribution of plastic curvature alone at the onset of full-yield at End 1. Note that the identical relationship between M and фр at End 1 and in Fig. 8b. Amongst the total rotation End 1 undergoes, the portion of plastic rotation comes from the plastic curvatures along the length of the member. Assume that Bpu of the hinge-spring is equal to the plastic rotation 0p at End 1when the end is at the onset of full-yield.

The conjugate beam method is used to evaluate the plastic rotation at End 1 for the beam-column under a curvature loading shown in Fig. 9c. The conjugate beam is under the loading of plastic curvatures alone since only plastic rotation is of concern. The end rotation of the actual beam-column is equal to the shear at the same section of its conjugate beam.

For calculating 0pu, moments Myc and Mpc shall be computed from Eqs. 10 and 11, respectively. This can be seen clearly from the fact that the load path from the section with zero moment along the member length to End 1 follows line D-Op in Fig. 2 (N remains unchanged along the member length).

It is desirable to correlate Bpu to the moment gradient к because plastic zones within a member are dependent on the moment distribution along the length (see Fig. 9a). In the following Bpu values corresponding to four particular moment gradient к are determined first. Then, the interpolation method is used to compute Bpu value for any moment gradient к.

For the convenience of deriving formulations for 0pu, the ratio у is defined as

Then, Bpu values corresponding to four different moment gradient к are found as follows:

(1) k=1, double curvature. M1=M2=Mpc and фp=фpu at both end sections. The lengths of plastic zones are L1=L2=0.5(1-y)L. The centroid of curvature loading at each end is x = ®Lu while the area of plastic curvature loading at each end is equal to (юзxфpuxLl). Therefore, the shear force at End 1 of the conjugate beam under the plastic curvature loading is equal to

0 pu =®зФ puL1 (l – 2 x) L = ®2®39e y (1 – y)[1 -®4 (1 – y)]|1 ~{n Ny)“ J(L d) (k=1) (14)

(2) k=0, single curvature. M1=Mpc and фp = фpu at End 1, while M2=0 and фp=0 at End 2. Plastic zones are L1=(1-y)L at End 1 and L2=0 at End 2. Hence, the shear force at End 1 of the conjugate beam under the plastic curvature loading is equal to

0 pu = ®зФ puL1 (L – x) L = 2®2 ®3?S у (l“УХ1-®4 (l – УЙ1 ~{N Ny У J(L d) (K=°) (15)

(3) к= – X, single curvature. M1=Mpc and фр= фри at End 1, and M2=Myc and фр=0 at End 2. Then Li=L at End 1 and L2=0 at End 2. The shear force at End 1 of the conjugate beam under the plastic curvature loading is equal to

®pu = ®зФ puL (L – X) L = 2®2®3 l1 -®4 fey 1 ~{n Ny У J(L d) (K= -^) (16)

(4) k= -1, single curvature. M1=M2=Mpc and ф^ф^ along the length of the member. The conjugate beam is under uniform plastic curvature loading. The shear force at the end of the conjugate beam is

0 pu =Ф puL 2 = «^6 у [l ~{n Ny У J(L d) (k= -1) (17)

Equations 14 to 17 are for a general beam-column. The factors ю2, ю3, ю4, a, q, and у are dependent on the sectional properties of the beam-column. Fig. 9 Conjugate beam method for computing end rotation: (a) linear moment distribution and plastic zones (b) total curvature distribution and curvature loading for conjugate beam (c) plastic curvature distribution when End 1 is at the incipience of full-yield. |

I-Sections

This section is to illustrate how to use Eqs. 14 through 17 to determine 0pu-к relation for wide flange I-sections bending about strong axis (i. e. W-sections in CISC 2004). For W-sections: it is found that й2=0.9, Юз=0.308, and o>4=0.21 using the moment-curvature-thrust curves given by Chen (1971); the average section shape factor q =1.14; the peak residual stress is commonly taken to be 30% of yielding strength (i. e., Fr=0.3Fy).

It is found that Bpu is not sensitive to N/Ny ratios when k=1 and 0. Thus, Eqs. 16 and 17 are simplified respectively for W-sections as

0.116 £y (L d) (k=1) |
(18) |

0.233£y (L d) (k=0) |
(19) |

Assuming a moderate axial force N=0.2Ny for a general beam-column, then q [1 – (N/Ny)a] = 1.0 in Eqs. 16 and 17, which are respectively simplified as

>pu = 0.438єy(id) (K= -0.5) |
(20) |

0 pu = 0.9e y (id) (K= -1) |
(21) |

Based on Eqs. 18 through 21, a piecewise 0pu-K curve is established for W-sections in Fig. 10. Using the interpolation method, a unique Qpu value is available corresponding to the к value of an element. The moment gradient к of an element at the first-yield of one hinge-spring is used to determine the Qpu value for that spring. After the Qpu value is computed for a particular hinge-spring, its value will remain unchanged during the rest of the loading history.

Fig. 10 Relation between moment gradient к and Bpu value for W-sections |