【導(dǎo)讀】筋，使其達(dá)到“實(shí)用、安全、經(jīng)濟(jì)、美觀”的要求。1）．平面圖：底層平面，標(biāo)準(zhǔn)層平面，頂層平面；2）．立面圖：主立面，背立面，側(cè)立面；3）．剖面圖：主要剖面，樓梯剖面；4）．屋面、樓地面、墻面等細(xì)部做法。1）．結(jié)構(gòu)方案的布置，主要構(gòu)件尺寸的初定原則及取值；框架結(jié)構(gòu)設(shè)計(jì)已成為必須面對(duì)的現(xiàn)實(shí)之一。框架結(jié)構(gòu)是由梁柱桿系構(gòu)成，能夠承受豎向和水平荷載作用的承重結(jié)構(gòu)體系。水平荷載作用下，框架柱承擔(dān)水平剪力和柱端彎矩，并由此產(chǎn)生水平側(cè)移，使用人數(shù)在四人及四人以上時(shí)，廁所于盥洗應(yīng)分割設(shè)置。時(shí)，應(yīng)采用管道煤氣或液化氣做燃料，并應(yīng)有良好的排煙氣設(shè)施。寢室之間的交往空間設(shè)計(jì)，將“交往”的概念層層融入其中。首先必須要有足夠的空間滿足交往活動(dòng)的需求；再者，還應(yīng)有一些設(shè)施來(lái)激發(fā)、誘導(dǎo)人們的交往行為。

　　

【正文】 idered to be basic random variables. Therefore, it can be shown that Thus, substituting Eqs. (15)and (16)into Eqs. (13) and (14), each term of Eq.(21) can be evaluated. can also be derived by taking the partial derivatives withrespect to Pu and Mu as As discussed earlier, only steel members where RC shear walls are not present are checked for the strength limit states. The steel members are expected to be weaker in strength in this case. Thus, although the parameters in Eqs. (13) and (14) are expected to be influenced by the presence of shear walls, the partial derivatives with respect to the random variables related to shear walls, namely, Ec and ν, need not be evaluated. For the serviceability limit state represented by Eq. (20), for the steel frame elements it can be shown that and where . For the serviceability limit state, the partial derivatives with respect to Ec and ν of the RC shear walls are zero. Therefore, the partial derivatives need not be calculated. Evaluation of Jacobians As discussed previously, the four Jacobians in Eq. (11)need to be puted. Jy,x and its inverse are easy to pute due to the triangular nature of the transformation. Since s is not an explicit function of x, Js,x=0. However, the other two Jacobians of the transformation, ., Js,D and JD,x, are not easy to pute since s, D, and x are implicit functions of each other. The adjoint variable method (Arora and Haug 1979。 Ryu et al. 1985) is used to pute the product of the second term in Eq. (11) directly, instead of evaluating its constituent parts. An adjoint vector λ can be introduced such that where KT=K+Ksh and K and Ksh were defined in Eqs. (2)and (4). It can be shown that (Gao 1994) For the strength performance criteria represented in Eqs. (13) and (14), =0, and is already derived in Eq. (22). Js,D needs to be evaluated at this stage. Normally, when the strength performance functions are considered, the internal force vector σ is the only contribution to the load effect s and can be expressed as s＝ Aσ, where A is the transformation matrix with constant elements. Thus, one can obtain where d is the nodal displacement vector in the global coordinate for the element and can be expressed as From the knowledge of the deterministic finiteelement algorithm, the relationship between s and d can be shown to be Where where (i＝ 1,2)＝ rotation at two nodes。 and ＝ displacements of an element in the two global coordinates。 and l=the original length of a member. Using Eq. (29), the derivative of the internal force with respect to the displacements can be defined as Where (33) (34) where L=deformed length and can be calculated as L= All the other variables were defined , Eq.(27) can be evaluated at this stage. In Eq. (29), is easy to obtain since the explicit dependence of F on the basic random variables is known, assuming the external load is not affected by the structural response, and can be derived by modifying Eq. (3) as Where and can be expressed as Since and Rd0 are not functions of the basic random variables, the derivative of R with respect to x can be expressed as Where can be expressed for a beamcolumn frame element as Where (39) and The λ evaluated from the serviceability and strength performance criteria can then be associated with the FEM algorithm by substituting λ into the last part of Eq. (26). As a result, the second part of Eq. (26) yields known quantities which are required to evaluate the gradient of the performance functions in Eq. (9) . To consider the presence of shear walls, the derivatives of internal forces, in Eq. (26), with respect to Ec and ν need to be evaluated. They can be derived as where Ec and ν=modulus of elasticity and Poisson’s ratio for shear walls, respectively, and each derivative of Eq. (42) can be shown to be and where All the matrices and parameters in Eqs. (43) and (44)were defined in Eqs. (5) through (8). The gradient of a limitstate function is now available in explicit form for a frame and shear wall structural system. Therefore, the reliability index and the corresponding probability of failure can be evaluated using the FORM analysis presented in Eqs. (9) through (12). Numerical Examples To investigate the effect of shear walls on the overall reliability, both a frame without shear walls and a frame with shear walls are studied in this study. All loads are applied statically. The reliability of the frame with and without shear walls is evaluated using the proposed algorithm. The accuracy of the proposed algorithm is established using Monte Carlo simulations. Reliability Analysis of Frame without Shear Walls A twostory twobay frame shown in Fig. 2 (Fig. 1 without the shear walls) is considered first. A36 steel is used. The statistical characteristics of the crosssectional and material properties required for the reliability analysis are given in Table 3. The frame is subjected to dead, live, and horizontal loads. The statistical properties of these loads are given in Table 3. For the strength limit state, the reliability of the most critical beam at node e and the most critical column at node c are evaluated using the proposed algorithm with the performance functions represented by Eqs. (13) and (14). For the serviceability limit state, the horizontal drift of the top floor at node a and the vertical deflection of the beam at the midspan at node d are checked. In Eq. (20), the prescribed horizontal drift at the top floor is considered to not exceed h/400, where h is the height 006Ff the frame. Thus, is equal to in this example. Similarly, the prescribed vertical deflection in the midspan of the beam is considered to be l/360 under the unfactored live load, where l is the span length of the beam. In this case, is c