When A Beam Or Bar Is Fixed At Both Ends, And Loaded In The Middle
Description
This section is from the book "The English And American Mechanic", by B. Frank Van Cleve. Also available from Amazon: The English And American Mechanic.
When A Beam Or Bar Is Fixed At Both Ends, And Loaded In The Middle
Rule
Multiply the value of the material by 6 times the breadth and the square of the depth in inches, and divide the product by the length in feet.
Note
When the beam is loaded uniformly throughout its length, the result must be doubled.
Example - What weight will a bar of cast iron, 2 inches square and 5 feet in length, support in the middle, without permanent injury?
225X2X6X22=10800, whlch, / 5=2160 lbs.
Or, if the Dimensions of a Beam or Bar are Required to Support a Given Weight in the Middle, Between the Fixed Ends.
Rule
Divide the product of the weight and the length in feet by 6 times the value of the material, and the quotient will give the product of the breadth and the square of the depth.
Example
What dimensions will a cast iron square bar 5 feet In length require to support without permanent injury a stress of 2160 lbs? 2160X5 10800
--------=------=8, whlch, / 2 ins. for the assumed breadth,=4, and√4- 225X6 1350 2 ins the depth.
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