山東電纜橋架安裝過(guò)程中，用到的彎頭一般有三種：一種是水平彎，一是垂直上下彎，還有一種是“之”字彎。廢話不多說(shuō)，以下舉例說(shuō)明這些彎頭制作方法及公式：

There are generally three types of elbows used in the installation process of cable trays in Shandong: horizontal bending, vertical up and down bending, and "zigzag" bending. Without further ado, the following examples illustrate the methods and formulas for making these elbows:

例:400X150的橋架做45°“之”字彎。

Example: Make a 45 ° zigzag bend on the 400X150 cable tray.

從橋架側(cè)面取15cmX0.8在橋架底部畫(huà)點(diǎn)將這段距離等分,然后三個(gè)點(diǎn)連接就成一個(gè)45°的等腰三角形(另一邊用同一方法)，然后將這個(gè)三角切掉就形成一個(gè)45°彎，然后再用要做的高度X1.41為里邊的頂點(diǎn)再畫(huà)三角形，方向與前面那個(gè)相反，做出來(lái)就是一個(gè)45°之字彎。90°彎是取要做的那一面的2倍畫(huà)三角，例如400X150做水平彎頭,則取400X2畫(huà)三角切掉就行了，如果要做垂直彎頭則取150X2。

Take 15cmX0.8 from the side of the bridge and draw a point at the bottom of the bridge to divide the distance equally, then connect the three points to form a 45 ° Isosceles triangle (the same method is used for the other side), then cut the triangle to form a 45 ° bend, and then draw a triangle with the height to be done X1.41 as the top of the inside, and the direction is opposite to the front one, so that it is a 45 ° zigzag bend. A 90 ° bend involves drawing a triangle twice the size of the side to be made. For example, if 400X150 is used as a horizontal bend, 400X2 is used to draw the triangle and cut it off. If a vertical bend is to be made, 150X2 is used.

橋架總結(jié)出來(lái)的公式：

The formula summarized by the bridge:

比如45度的坡度彎頭，你只要記住這句話就知道怎么做了“爬坡45度，橋架高度乘以0.8中分劃三條線為一個(gè)彎，斜邊是爬坡高度乘以1.4130度是橋架高度乘以0.8除以45乘以30中分劃三條線為一個(gè)彎，斜邊是爬坡高度乘以2水平90度彎是橋架寬度乘以0.8中分劃線，斜邊(內(nèi)邊)600以上200乘以1.41600以下100乘以1.41”。

For example, for a 45 degree slope elbow, you only need to remember this sentence to know how to do it: "Climbing a 45 degree slope, multiplying the height of the bridge by 0.8, dividing the three lines into one bend. The slope is the climbing height multiplied by 1.4130 degrees, dividing the bridge height by 0.8, dividing it by 45, and dividing it by 30. The slope is the climbing height multiplied by 2. The horizontal 90 degree bend is the bridge width multiplied by 0.8, and the slope (inner edge) is 600 above 200 times 1.41600 below 100 times 1.41.

制作方法及步驟

Production method and steps

1、制作層彎頭：

1. Make the first layer of elbow:

層兩個(gè)彎頭的做法同“一”中彎頭的做法，兩個(gè)彎頭間距a值由現(xiàn)場(chǎng)確定，但需滿(mǎn)足電纜彎曲半徑的要求。

The practice of the two elbows on the first layer is the same as that of the elbow in "1". The distance between the two elbows, a, is determined on site, but it must meet the requirements of the cable bending radius.

2、制作第二層彎頭：

2. Make the second layer elbow:

（1）計(jì)算第二層橋架個(gè)彎頭的定位尺寸e：

(1) Calculate the positioning dimension e of the first elbow of the second layer bridge:

先量出層橋架定位尺寸d值，則e=d+btg1/2α

First measure the positioning dimension d of the first layer bridge, then e=d+btg1/2 α

式中：b為層間距，角α值同“一”中角α值

In the formula: b is the interlayer spacing, and the angle α The value is the same as the middle corner of 'one' α value

（2）依照“一”的方法制作第二層彎頭的第1個(gè)切口α；

(2) Make the first cut of the second layer of elbow according to the "one" method α；

（3）計(jì)算兩個(gè)彎頭間距c值：c=2btg1/2α+a式中：b為層間距，a為層橋架兩個(gè)彎頭的間距

(3) Calculate the distance c between two elbows: c=2btg1/2 α+ In formula a, b is the interlayer spacing, and a is the spacing between the two elbows of the first layer of bridge

（4）依照“一”的方法制作第二層彎頭的第2個(gè)切口α；

(4) Make the second cut of the second layer of elbow according to the method of "one" α；

（5）計(jì)算補(bǔ)償尺寸h：h=2btg1/2α

(5) Calculate compensation size h: h=2btg1/2 α

3、制作第三層彎頭：

3. Make the third layer elbow:

制作方法同第二層彎頭

The production method is the same as the second layer elbow

f=e+btg1/2αg=2btg1/2α+c

F=e+btg1/2 α G=2btg1/2 α+ C

注：多列等間距等寬橋架任意角彎頭的制作方法與多層等間距等高橋架任意角彎頭的制作方法相同。

Note: The production method for any angle bend of multi column equally spaced and equally wide bridge is the same as that for any angle bend of multi-layer equally spaced and equally high bridge.

三、多層“之”形彎頭的作法：

3、 The practice of multi-layer "zigzag" bends:

與“二”類(lèi)似，只是a、c、g值相等，所切的兩個(gè)α角在橋架的兩邊

Similar to 'two', except that the values of a, c, and g are equal, and the two tangent α Corner on both sides of the bridge

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