Iowa Public Television

Challenge Question Number 2 for Draftsperson #1201

This sketch illustrates a typical plate detail used in our towers. This plate would be used as a stiffener inside the fluted column. It is also a platform used to climb the ladders to the top of the tank. The dimensions in the center of the plate show the opening reqired for the ladder to go through the platform. The numerical dimensions are what we are typically given and the dimensions shown as letters are those that the drafter must calculate to complete the detail for the shop.

Clearly describe what you expect from the students:
Given that the piece is cut from 3/16” thick steel plate and the weight of steel is .2836 pounds per cubic inch, calculate the piece weight.

In order to give the teacher some guidance in evaluating the student's project(s); list some tips that may help to assess the student's work:
Weight = Total area x Thickness x Weight/Cu. In.

Calculate area:
Divide overall piece into small, easier to calculate parts.

Total Area = Area 1 + Area 2 + Area 3 + Area 4 – Area 5

Area 1 = H x 4
= 176.1110 x 4
= 704.4440 Sq. In

Area 2 = H x E
= 176.1110 x 19.3952
= 3415.7081 Sq. In.

Area 3 = 2 x ((F x E)/2)
= 2 x ((13.9227 x 19.3952)/2)
= 2 x 135.0168
= 270.0336
Area 4 = Area of circular segment – area between center and chord

Calculate circular segment area
Area =  x R2 x (/360º)
=  x 174.8752 x (A/360)
= 3.14159 x 30581.2656 x .1982
= 19039.8947 Sq. In.

Calculate area from chord to center
Area = (Base x Height)/2
= C x (174.875 – D)/2
= 203.9564 x (174.875 – 32.8127)/2
= 203.9564 x 142.0623 / 2
= 14487.2576 Sq. In.

Area 4 = 19039.8947 – 14487.2576
= 4552.6371 Sq. In.

Area 5 = ( x 152 / 2) + (18 x 30)
= (3.14159 x 225 / 2) + 540
= 353.4292 + 540
= 893.4292 Sq. In.

Total area = 704.4440 + 3415.7081 + 270.0336 + 4552.6371 – 893.4292
= 8049.3936 Sq. In.

Weight = 8049.3936 Sq. In x .1875 In. x .2836 #/Cu. In.
= 8049.3936 x .1875 x .2836
= 428.0265#