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Breaking Through-Welding
Lesson 10
Math Activities: Lessons A, B, and C
Lesson A:
There are two different measurement systems in common use-the US conventional system and the SI
metric system. The US conventional system is used primarily in the United States. The SI metric
system is the standard in much of the rest of the world, and is also used in the United States.
Conversions between the US conventional system and the SI metric system may need to be made
occasionally. Common conversions are listed on the chart below. Use the chart to answer questions on
the following page.
Lesson A Questions:
Notice that to make conversions from US conventional to metric, you must find the
proper unit of measure and multiply the conventional measurement by the
multiplier in column 4 to calculate the metric conversion. If you want to go from
metric to conventional, you would divide the value by the amount in column 4.
1. If you need to convert the volume of a cylinder from in³ to mm³, what is your
multiplier?
2. If you have a pipe that is 46 inches (in) long, what would the length of the pipe be in
meters (m)?
3. A medium pressure generator produces acetylene at a gauge pressure of 15 psi. What
would this reading be in kPa?
4. You are welding a butt joint and have 2 pieces of mild steel that each measure 1/4” x
3” x 6”. Convert each of these measurements to millimeters (mm).
5. The area of a sheet of metal is 8 m² and you need to convert this measurement to ft².
What is the area of this metal in ft²?
Lesson B:
Word Problems:
1. A welder spends 3 ¼ hours cutting and grinding steel to shape and 1 ½ hours
positioning, welding, and finishing parts to complete a job. How many hours did the job
require?
2. Four fillet welds and 32 spot welds are required to complete the welding on a parts bin.
How many spot welds are required to produce 124 parts bins?
3. A welder has a sheet of cold rolled steel that is 96” in length. Four pieces were cut
from the sheet. They were 16”, 21”, 3.75”, and 4.25”. What is the length of the
remaining piece?
4. Volume of a cylinder can be determined by the following equation:
V = .7854 x D² x h (D is diameter, h is height)
What is the volume of a cylinder with a diameter of 16” and a height of 64”?
5. As a welder, you have been hired to build a cylindrical tank that will hold 160 ft³ of
liquid. Your friend suggests that you make the tank with a diameter of 4’ and a height of
12’. Will this tank be sufficient for the amount of liquid specified?
Lesson C: Stress
Stress is the effect of an external force applied upon a solid material. Every machine part
or structural member is designed to safely withstand a certain amount of stress. The type
of stress is determined by position and direction. The five types of stress are tensile,
compressive, shearing, bending, and torsional.
1. Tensile Stress:
St = P
Where: St = tensile stress in pounds per square inch
A
P = force in pounds (lb)
A = area in inches squared ( in²)
What is the tensile stress of a 12,000 lb force applied to a piece of flat stock with a crosssectional area of .75 in²?
2. Compressive Stress:
Sc = P
Where: Sc = compressive stress in pounds per square inch (lb/in²)
A
P = force in pounds (lb)
A = area in inches squared (in²)
What is the compressive stress of a 10,000 lb. force applied to a piece of flat stock with a
cross-sectional area of .75 in²?
3. Shearing Stress:
Ss = P
Where: Ss = shearing stress in pounds per square inch (lb/in²)
A
P = force in pounds (lb)
A = area in inches squared (in²)
What is the shearing stress of a 40,000 lb force applied to a piece of flat stock with a
cross-sectional area of .75 in²?
4. Bending Stress:
Sb = Mc
Where: Sb = bending stress in pounds per square inch (lb/in²)
Z
M = maximum bending movement in inch-pounds (in-lb)
c = distance from neutral axis to farthest point in cross section in
inches (in.)
Z = section modulus in cubic inches (in³)
What is the bending stress of a 1” square solid shaft subjected to a bending moment of
1200 in-lb.? From the neutral axis to the cross-sectional area c is 0.75” and the section
modulus is 0.053.
5. Torsional Stress
T = P x L Where: T = torque in inch-pounds (in-lb)
P = force in pounds (lb)
L = distance in inches (in.)
What is the torque of a 240 lb. force applied over a distance of 16”?
Answer Key: Lesson 10 – Math Activities
Lesson A:
1. 16387
2. Multiply by .0254 to get 1.17 m
3. Multiply by 6.8948 to get 103.42 kPa
4. Multiply each inch measurement by 25.4 to get 6.35mm x 76.2 mm x 152.4 mm
5. Divide the metric measurement by .092903 to get 86.11 ft².
Lesson B:
1. Add to get 4 ¾ hours.
2. Multiply the number of spot welds for one bin by the number of bins needed to
get 3968 spot welds.
3. Add together the lengths of the 4 pieces and subtract the total from the sheet
length to get 51” of remaining sheet metal.
4. V = .7854 x (16 x 16) x 64
V = .7854 x 256 x 64
V = 12.867.99 in³
5. No. A tank with these measurements will hold 150.80 ft³ of liquid. Your tank
needs to hold 160 ft³, so it would be too small.
V= .7854 x (4 x 4) x 12
V = .7854 x 16 x 12
V = 150.80 ft³
Lesson C: Stress
1. St = 12,000 = 16,000 lb/in²
.75
2. Sc = 10,000 = 13,3333.33 lb/in²
.75
3. Ss = 40,000 = 53,333.33 lb/in²
.75
4. Sb = 1200 x 0.75
0.053
5. T = 240 x 16
= 16,981 lb/in²
= 3840 in-lb