Stress
Stress is "force per unit area" - the ratio of applied force F to cross section area - defined as "force per area".
- tensile stress - stress that tends to stretch or lengthen the material - acts normal to the stressed area
- compressive stress - stress that tends to compress or shorten the material - acts normal to the stressed area
- shearing stress - stress that tends to shear the material - acts in plane to the stressed area at right-angles to compressive or tensile stress
Tensile or Compressive Stress - Normal Stress
Tensile or compressive stress normal to the plane is usually denoted "normal stress" or "direct stress" and can be expressed asσ = Fn / A (1)
where
σ = normal stress ((Pa) N/m2, psi)
Fn = normal component force (N, lbf (alt. kips))
A = area (m2, in2)
- a kip is a non-SI unit of force - it equals 1,000 pounds-force
- 1 kip = 4448.2216 Newtons (N) = 4.4482216 kilonewtons (kN)
Example - Tensile Force acting on a Rod
A force of 10 kN is acting on a circular rod with diameter 10 mm. The stress in the rod can be calculated as
σ = (10 103 N) / (π ((10 10-3 m) / 2)2)
= 127388535 (N/m2)
= 127 (MPa)
Example - Force acting on a Douglas Fir Square Post
A compressive load of 30000 lb is acting on short square 6 x 6 in post of Douglas fir. The dressed size of the post is 5.5 x 5.5 in and the compressive stress can be calculated as
σ = (30000 lb) / ((5.5 in) (5.5 in))
= 991 (lb/in2, psi)
Shear Stress
Stress parallel to the plane is usually denoted "shear stress" and can be expressed asτ = Fp / A (2)
where
τ = shear stress ((Pa) N/m2, psi)
Fp = parallel component force (N, lbf)
A = area (m2, in2)
Strain
Strain is defined as "deformation of a solid due to stress" and can be expressed asε = dl / lo
= σ / E (3)
where
dl = change of length (m, in)
lo = initial length (m, in)
ε = unit less measure of engineering strain
E = Young's modulus (Modulus of Elasticity) (N/m2 (Pa), lb/in2 (psi))
- Young's modulus can be used to predict the elongation or compression of an object.
Example - Stress and Change of Length
The rod in the example above is 2 m long and made of steel with Modulus of Elasticity 200 GPa. The change of length can be calculated by transforming (3) as
dl = σ lo / E
= (127 106 Pa) (2 m) / (200 109 Pa)
= 0.00127 (m)
= 1.27 (mm)
Young's Modulus - Modulus of Elasticity (or Tensile Modulus) - Hooke's Law
Most metals deforms proportional to imposed load over a range of loads. Stress is proportional to load and strain is proportional to deformation as expressed with Hooke's lawE = stress / strainModulus of Elasticity, or Young's Modulus, is commonly used for metals and metal alloys and expressed in terms 106 lbf/in2, N/m2 or Pa. Tensile modulus is often used for plastics and is expressed in terms 105 lbf/in2 or GPa.
= σ / ε
= (Fn / A) / (dl / lo) (4)
where
E = Young's modulus (N/m2) (lb/in2, psi)
Shear Modulus
S = stress / strain
= τ / γ
= (Fp / A) / (s / d) (5)
where
S = shear modulus (N/m2) (lb/in2, psi)
τ = shear stress ((Pa) N/m2, psi)
γ = unit less measure of shear strain
Fp = force parallel to the faces which they act
A = area (m2, in2)
s = displacement of the faces (m, in)
d = distance between the faces displaced (m, in)
Elastic Moduli
Elastic moduli for some common materials:| Material | Young's Modulus | Shear Modulus | Bulk Modulus | |||
|---|---|---|---|---|---|---|
| 1010 N/m2 | 106 lb/in2 | 1010 N/m2 | 106 lb/in2 | 1010 N/m2 | 106 lb/in2 | |
| Aluminum | 7.0 | 10 | 2.4 | 3.4 | 7.0 | 10 |
| Brass | 9.1 | 13 | 3.6 | 5.1 | 6.1 | 8.5 |
| Copper | 11 | 16 | 4.2 | 6.0 | 14 | 20 |
| Glass | 5.5 | 7.8 | 2.3 | 3.3 | 3.7 | 5.2 |
| Iron | 9.1 | 13 | 7.0 | 10 | 10 | 14 |
| Lead | 1.6 | 2.3 | 0.56 | 0.8 | 0.77 | 1.1 |
| Steel | 20 | 29 | 8.4 | 12 | 16 | 23 |
