Strength of Materials

Normal Stress and Strain

Normal stress and strain occurs to memebrs that are axially loaded, such as in compression or tension.

Stress: - Stress is the force applied to a material per unit area. - Measured in N/M². - Often denoted with σ - Calculated as the force applied over the cross-sectional area in the direciton it’s applied. We use the original area of the material, and don’t update it as the material deforms.

# Where sigma is stress, F is force, and A is the cross-sectional area
var("sigma", "F", "A")
stress_definition = (sigma == F/A)

Strain: - Strain is the deformation or change in the shape of the material as a result from the applied force. - Strain is a ratio, and so doesn’t really have a unit. - Calculated as the difference in the dimension measured (often length) over it’s original length.

# Where L0 is the original length in the measured direction
# and L is the current length in the measured direction
var("L", "L0", "epsilon")
strain_definition =  (epsilon == (L - L0)/L0)

We determine the stress and strain values of a material experimentally. This is done by applying a known-force to the material and measuring its deformation, then using the equation above to rectify for it’s cross-sectional area. Often this is done with a tensile-strength test.

Once the experiment is completed for a certain material, we plot strain as a function of stress:

• P is the proprtionality limit, where the function stops being linear.
• E is the elastic limit, after we reach this the material is permanently deformed.
• Y is the yield point.
• U is the maximum value of stress on the stress-strain diagram.
• F is the fracture point, where the material is separated into two pieces.

Because we use the original crossectional area and don’t update it, our graphs aren’t 100% realistic, and after the elastic point will make the material seem weaker than it really is. The adjusted version is called the True Stress and True Strain.

To compute this updated version we would have to take into account how stress increases strain and strain decreases stress, with a little bit of calculus we get:

true_stress(stress, strain) = stress*(1+strain)
true_strain(A, A0) = ln(A/A0)

Examples

Example 1

A hollow steel tube with an inside diameter of 100mm must carry a tensole load of 400 kN. Determine the outside diameter of the tube if the stress is limited to 120MN/m²

eqn1 = stress_definition.solve(A)[0]
eqn1 = eqn1(F=400000, sigma=120)
eqn1

A == (10000/3)

inner_area = math.pi*(100/2)**2
inner_area

7853.981633974483

var("outer_radius")
eqn2 = (A == (math.pi*outer_radius**2) - inner_area)

solutions = (eqn1 - eqn2).solve(outer_radius)

for i in solutions:
    print(i.lhs(), float(i.rhs()))

outer_radius -59.67439110662103
outer_radius 59.67439110662103

Naturally we reject the negative solution

outer_radius = 59.67
outer_diameter = 2 * outer_radius
print(round(outer_diameter, 2), "mm")

119.34 mm

Example 2

A homogeneous 800 kg bar AB is supported at either end by a cable as shown in Fig. P-105. Calculate the smallest area of each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel.

# Force from the weight of the beam
F = 800*9.81
# We'll assume its split evenly accross the cables (it should be)
F /= 2

# We'll invert the equation for this problem too
eqn1 = stress_definition.solve(A)[0]
eqn1

A == F/sigma

# Then for bronze
eqn1(F=F, sigma=90)

A == 43.6000000000000

# And for steal
eqn1(F=F, sigma=120)

A == 32.7000000000000

Poisson’s Ratio

As as material elongates due to load, it’s cross-sectional area is reduced. This reduction is called lateral strain and it’s related to the axial strain by Poisson’s Ratio (v). For a circle (or circular thing) Poisson’s Ratio is given by:

$$ v=\frac{\epsilon_{lateral}}{\epsilon_{axial}} = \frac{\Delta D / D_0}{\delta/L_0} $$

The Poisson’s Ratio essentially measures the ratio of downwards deformation to horizontal deformation, something like this:

For a list of the Poisson Ratios of common materials, see the reference section at the end of the notebook.

Hooks Law

In the above graph, both the real and “engineering” functions have a linear portion. The slope of this linear portion is called the elastic modulus denoted by E (also called Young’s modulus). Naturally we can write this section of the stress-strain function as a linear equation:

# Where E is the Elastic Modulus of the material
stress(E, strain) = E*strain

Hooke’s law also has a form relating shear stresses and strains:

# Where v is the poisson ratio of the material
# G is the shear modulus of elasticity
# y is the shear modulus
# and E is the elastic modulus
G(E, v) = E/(2*(1+v))
shear_stress(G, y) = G*y 

Reference For Materials

Poisson Ratios of Common Materials