【正文】 ucture as analysis object ,take hinged arch model shown in figure 2. 圖2 a圖3 a 圖2 b 圖3 b 圖2 c圖3 c In Figure 2 the right supports vertical linkage representatives roof beams supporting role, ramps connecting rod on behalf of the board itself thin beam reaction effect which is virtual and approximate equivalent. We would like to calculate two the total pressure of physical project through two plate roof beams and transfer to the ends column, So Anti two numerical difference can be seen as two plates bear along with the plane load and roof beams bear the vertical load pressure. Two Anti power link expressions in Various conditions were given as follows, because the model take units width,so the results is line averageload distribution except it has Focus quality in are bouth represent by N , English leftover subscript s, b, represent the plane along the roof panels and vertical role in the roof beam, g, w, e,represent gravity, air pressure and the level of earthquake separately. d, c, represent distribution of concentrated load or effect separately, In the formula h is thiess of every plate,g is gravitation acceleration, a is roof for the horizontal seismic acceleration value formula, Wk represent the standard value with number footnotes represent every numbered ramp the quality distribution per unit area ,m with english footnotes represent quality of per to two symmetrical slopes, the formula can be more 2a represent situation of vertical gravity load ,these formulas as follows: ? ?? ?39。 39。 39。1 1 1 10 0 1 1 0c o s c o s 38 c o s c o s c o s c o sL AL L m L ALNl h l h l m? ? ? ?? ? ? ????? ? ?? ? ? ?? ?39。10 0 0 0 0 0 0 1 01 39。 10 0 0 0 0c os c os 2 c os c os8 si n c os8 si n c os c os 8 si n c os c os c osl l l l l h m m shNl l h h l h l? ? ? ? ? ? ? ? ? ????? ? ? ? ? ? ??? ?? ? ??? ? ? ? ?? ?? ? 101 1 0 1 11 0 1 0 000 1012111c o s2 c o s c o s 2LLL L L L Lm LL L L m LL L LLLLNhB h L h L LI??????? ???? ?? ? ? ?? ? ? ? ? ?? ???????? ??? ? ? ??? ?? ? ? ?? ?? ?? ?0 01 0 0 1 1 1 0 011 20 0 10 1 0210 0 0 1 101 1 1 1 21c o ssin 2 sin2 sin c o s c o sALh Lm LL L L m L L m a L L L Lhh Lm l mN L L L A hL L kB h L h L?? ??? ? ? ?? ? ? ? ?? ??? ? ? ? ? ? ?? ? ? ? ????? ??? ? ? ? ? ?? ? ? ? ?? ?Figure 2b represent situation of bear wind load, these formulas as follows: ? ?? ?2 2 2 21 1 1221 1 1c os c osc os8 c os c os c os c os wkL h L L S liNaL h h b? ? ? ??? ? ????? ? ? ? ? ? ?? ?2 2 2 22 0 0 11 1122 2212 1 1 0c os c osc os 11c os c os c os c os si n 5 c os si nc os c os si n c oskKL h l w Lwwhw h m LNll A L h L a h L?? ? ??? ?? ? ? ? ? ? ? ?? ? ??? ??? ? ??? ? ? ? ? ???? ??? ??Figure 2c represent situation of role of level earthquake, these formulas as follows: ? ?? ?2 2 2 2 21 0 0 1 1 0220 0 1si n c os si n c os 3 si nc os c os c os c os c os aaL h l L LNL h l hl? ? ? ? ? ? ? ??? ? ? ? ? ? ??? ? ?? ? ?? ? ? ?? ?2 2 2 2 21 0 1 1 1203 2 22 2 2 102 1 0 1si n c os si n c os si n si n si n3c os 2 l n c os5 l n c os c os c os c osal h m l m L mmmNn s l ll g h l h l? ? ? ? ? ? ? ?? ? ?? ? ? ? ?? ?? ? ??? ? ? ? ?? ?001 0 0 1 1 0 1 211 0121 0 0 0 11 1 si n c os2 c os 2 c os c os c osaaLLm L L L n L L Lnh LNL l h l h l? ??? ????? ? ? ?? ? ? ???? ? ? ?? ? ? ?????? ? ? 00 0 0 0201 sin2 c osaaLm L L L hLl???????? ? ?????????? ? ?? ?? ? 200 0 010121 0 01 sinsin c os sin c ossin c os c os 2 sin c osaeLm L L L hLmmNl l h?? ? ? ? ?? ? ? ? ?????? ? ?????? ??????? ? ? ? ?? ?001 0 0 1 0 0 1 2 211 11221 0 0 1si n 1 si n c os2 c os 2 c os c os c os si na aLLL L L L m L L LhLhl L h l h? ? ? ??? ? ? ???? ? ? ?? ? ? ???? ? ? ?? ? ? ?????? When vertical seismic calculation required by Seismic Design Parameters It’s calculate formula generally similar as formula 1 to 4 which only need take gravity g as vertical seismic acceleration a. Above formulas apply to right bearings in figure 2 and also to left when exchange data of two plate. As end triangle of Multislope roof ,for simplify and approximate calculation need, we assume two lines distribution load only produced by roof board of several load, IIII crosssection from figure is took to analysis Long trapezoidal plate two’s end triangle, assuming the structure symmetry approximately, take half of structure to establish model (figure 3). Because linked with the end triangular plate3 plane has great lateral stiffness ,therefore assume the model leftist stronghold along the central ponent around which can not be shifted direction. Central Plate vertical stiffness small, in general gravity load of roughly symmetric midpoint only next movement happened possible, Therefore, the model used parallel twolink connection. Wind loading, and the general role of the earthquake in two slope was roughly antisymmetric,so plate model in the central use fixed hinge bearings which allow rotation and transtlateral force to plate 3near the plate beam. Under plate two triangular area is eaves of vertical beams and plates itself along with plane load distribution is function shown in Figure 1 take the variable x as an argument,assume the distance from position of section II to end part is x0s so the slope level length is y0=x0L2/L3,formula 11 to 14 is the value of Vertical triangle of gravity along the x direction arbitrary location of the two load distribution ,where h3 is Slitting vertical thickness of plate 3. ? ?22020c o s 2 1 2 c o sea am k x L h xN L s h v l x?????? ?????? ? ?21 1 121 0 01s in c o s 2 1 2 c o sm k v L h xN l x h x L V? ? ????? ?????? ? ?220 0 0 0 02 2 21 1 0 0m a x