Stress / Strain / Stretch !!!!!!!!!!!!Mind your units!!!!!!!!!!!!!!!! Stress = Force per unit of area = Force/Area Units of force are Newtons, (metric) Pounds (Imperial) Stress units -­‐> Ksi or Psi, (Pounds/in^2 – Imperial) GPa or MPa. (Pascals –metric) 1 pascal = 1 n/m^2 Strain = Change in length per unit length. Deformation of a material. (unitless) Strain = Stress/Youngs modulus (Modulus of elasticity or “E”) Stretch (deformation) = Length x Strain Stretch units –> same as length units. Youngs modulus of Elasticity (E) represents how well a material stands up to tension. We will use the values shown in this chart: http://en.wikipedia.org/wiki/Young%27s_modulus#Approximate_values Youngs units are MPa/GPa or Psi/Ksi . Metric or Imperial. PRESSURE conversion factors: (used in STRESS calculation) 1 n/m^2 = 1 pascal MPa = one million pascals, GPa = one billion pascals. 1 MPa = 145.0377 Psi 1 Ksi = One thousand Psi. 1 Psi = .00689475 MPa If you see mm, convert to meters first because metric pressure (pascals) is in units of Newtons per square meter. If you see feet, convert to inches first because Imperial pressure (Psi) is in units of pounds-­‐force per square inch. WEIGHT / FORCE conversion factors: If you see a MASS, you need to convert it to a FORCE… So, If you see KG, convert to Newtons. (N) KG x 9.8 = N (Downward force = mass x gravity) -­‐-­‐-­‐-­‐-­‐:Problems:-­‐-­‐-­‐-­‐-­‐ 1-­‐ A 1.9” diameter steel bar is under 32,000 lbs. of tension. What is the stress? (Simple Stress calculation) 2-­‐ .025m diameter cable is under a load of 3,600,000 N. What is the stress in this cable? (Simple stress calc – Metric.) 3-­‐ You are is required to support a 25,000 lb load with three steel rods. The steel dealer has a deal on 20Ksi rods of all diameters. What diameter rods would be required? (Find required area to support the load using 20ksi material, divide the required area by 3, turn this area into diameter.) 4-­‐ Copper wire is all you have. You know it has a yield strength of 70 MPa, and an ultimate strength of 220Mpa. You have to support a 15,000 N load. What gague wire (AWG) will be needed to support this load infinitely? (Find required area, turn area into diameter, find diameter on AWG chart at the Wiki link) http://en.wikipedia.org/wiki/American_wire_gauge 5-­‐ A steel bar, 2”x1.5”, 10 feet long is put under a load of 20 tons. How much does it stretch? (Turn tons into lbs, turn feet into inches, proceed with the simple Stress,Strain,Stretch calcs.) 6-­‐ A 4.5 meter long steel tube with an OD of 110mm and an ID of 75mm is put under a load of 400Kn. Determine how much it deforms in length. ( turn mm into m. Find total area of the pipe. (Outside Diameter area – Inside Diameter area) then proceed with the simple s,s,s calcs.) 7-­‐ An aluminum rod, 7/8” diameter, 36” long is under an 800kg load. How much will it stretch? (Imperial or Metric? Since you want to find stretch, and the dimensions of the bar are given in imperial, you should go with imperial. So convert KG to lbs (div by 2.2) Then proceed with the simple s,s,s calcs.) 8-­‐ You have a 1.5”x3.5” piece of pine wood (a 2x4), 8 feet long and are about to put 2200 lbs of force on it. How much will it stretch? (simple s,s,s) 9-­‐ A 12mm dia. steel rod, 430mm long is under a load of 900kg, what are the stress, strain and stretch? (Convert to dimensions to Meters, convert mass to Newtons.. Then do simple s,s,s) 10-­‐ A 1 meter long, 1x6cm bar is put under a load of 8008 kg. It is steel with an elastic modulus of 220GPa. What is the stress, strain, and stretch? (Turn cm into M, turn kg into N, then proceed with simple SSS calcs) 11-­‐ A .375” square titanium bar, 5 feet long has deformed .0068 inches under an unkonwn load. What is the load? (Work it backwards... You have stretch, plug it into the stretch calc to find strain. Then plug the strain into the strain calc. to find stress, etc etc..) 12-­‐ What would be the minimum wall thickness allowable for a 30mm outside diameter tube made from 180 GPa steel, to safely support 129,000 Kg? (Find the required area needed to support the load based on the materials “E” (180 GPa). Kg must be turned into N first! Then find the area of a solid 30mm rod.. Then figure out how much material must you remove from the center of this rod to produce a tube with the required total area. Turn area into diameter. Find the difference btwn the 30mm OD and the hole that must be in the middle of the tube. This difference is spread between both sides of the tube, so divide it by two to give you the wall thickness. Got it ? Do a drawing or two to help you keep track of things.)