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本文計算的是一个三層三间的屋架?衙繉拥淖∽拥淖芗袅退麄兊母叨鹊某朔e喚做这一層的剪力矩,那么这个剪力矩是和同一層的柱子的两端的弯矩的总和值相等的燃俣恳粚拥募袅氐扔?.为便利超見,譬如我們先假定第n層的剪力矩等于1。把这一層的各柱子的兩端的固定端弯炬求得,再把他們在本層的柱子的兩端的各結点分配一下。因为我們用的集体分配系数和隔点傳遞系数是根据整个屋架計算的緣故,有一部分的結点的弯矩是不完全平衡的,同时第n—1和n+1層的柱子的两端出现了新的弯炬,因此这两層也出現了新的剪力矩了。在另一方面,第n層的剪力矩已經不等于1,却等于一个新值α了。但是本層和上下两層的柱子两端的弯矩和上下两層的剪力矩和这个新剪力矩值α的中間是有一定的比例的。为了要本層的剪力矩值仍旧变成1,我們拿α除所有弯矩值和剪力矩值,因此我求得一系列的新值,这些新值就喚做n層的剪力矩的影响值嬎阄菁苁蔽覀兡檬实钡氖党烁鲗拥募袅氐挠跋熘?叫各厨的剪力矩的总和差不多和他們的实在值相等,然后進行分配弯矩。分配之后,各層的剪力矩和他們的实在值必定还有一定的距离或一定的差数。因此我們再用另一批适当的新数值乘各層的剪力矩的影响值,叫各層的剪力矩的总和差不多和这些差数相等,然后分配弯矩。这样一遍一遍的算下去等到求得剪力矩和他們的实在值中间没有差数或相差很小才歇手。我們計算这个例題时僅僅算了两遍,最大誤差已經小到0.075%,因此就沒有再算下去了。
This paper calculates a three-story three-story roof truss. The rushing moment is the same as the column on the same floor. If the sum of the bending moments at both ends is the same, then the value of the sum of the moments is the same. For the sake of convenience, for example, we first assume that the shear moment on the nth floor is equal to 1. Obtain the fixed end bends of the ends of the columns of this layer and assign them to the junctions at the ends of the columns of this layer. Because we use the collective distribution coefficient and the transfer coefficient of the partition based on the calculation of the entire truss, the bending moments of some of the nodes are not fully balanced, and the ends of the columns of the n-1th and n+1th layers appear at the same time. With the new bending torch, new shearing moments have also appeared on these two floors. On the other hand, the shear moment of the nth layer is not equal to 1 but is equal to a new value α. However, the bending moment at both ends of the column and the upper and lower columns, and the shear moments of the upper and lower layers, and the value of the new shear moment value α have a certain proportion. In order to make the shear moment value of this layer still be 1, we divide all the moment values and shear moment values by α, so I obtain a series of new values that call the influence of the shear torque of the n-layer. The total sum of the shear moments of the chefs is almost equal to their actual value, and then the bending moments are distributed. After distribution, the shear moment of each layer and their actual value must have a certain distance or a certain difference. Therefore, we use another set of appropriate new values to multiply the influence of the shear moments of the layers, so that the sum of the shear moments of the layers is approximately equal to these differences, and then the bending moments are assigned. This way, count and wait till the difference between the shear torque and their actual value is not a small difference or a small difference. We calculated this example only two times, the maximum error has been as small as 0.075%, so it did not count down.