Mechanics of Solids and Statics

Elastic constants and yield criteria

For an isotropic linear elastic material, only two elastic constants are independent, and the others follow from the standard relations among Young's modulus E, shear modulus G, bulk modulus K and Poisson's ratio \nu. The key relation is E = 2G(1+\nu), so for a given G the value of E grows as \nu grows. Thermodynamic stability restricts Poisson's ratio to -1 < \nu \le 0.5, so the largest possible Young's modulus occurs at \nu = 0.5, giving E_{max} = 3G. Poisson's ratio is the negative ratio of lateral strain to axial strain under uniaxial loading, so an axial strain of 0.001 accompanied by a lateral strain of -0.0003 gives \nu = 0.3. The bulk modulus is the ratio of hydrostatic stress to volumetric strain, K = -p/(\Delta V/V), and it should not be confused with compressibility, which is its reciprocal, or with Young's modulus, which applies to uniaxial normal stress.

A yield criterion states the combination of stresses at which a material begins to yield. In the von Mises criterion, yielding starts when the equivalent stress equals the uniaxial yield strength, \sigma_{eq} = \frac{1}{\sqrt{2}}\sqrt{(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2}. For plane stress with only \sigma_1 and \sigma_2 nonzero, this reduces to \sigma_{eq} = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2}, and the cross term -\sigma_1\sigma_2 must not be dropped. For example, with \sigma_1 = 150 MPa and \sigma_2 = 50 MPa, the equivalent stress is \sqrt{17500} \approx 132.3 MPa.

The Tresca criterion instead uses the maximum shear stress, and yielding begins when \tau_{max} = (\sigma_{max} - \sigma_{min})/2 reaches half of the uniaxial yield strength. The intermediate principal stress has no effect on this condition, so \sigma_y = \sigma_{max} - \sigma_{min}. For principal stresses of 120 MPa, 80 MPa and 20 MPa, the Tresca yield strength is therefore 100 MPa.

Rigid body equilibrium

A rigid body in a plane is in equilibrium when the sum of all force components and the sum of all moments about any point are zero, giving three scalar equations: \sum F_x = 0, \sum F_y = 0 and \sum M = 0. Unknown support or contact forces are found by writing these equations in the most convenient order, so that each equation contains as few unknowns as possible.

A smooth contact surface can transmit only a force normal to the surface, because frictionless surfaces cannot provide tangential reaction. Resolving the weight of a body resting on a smooth incline gives the normal reaction N = W\cos\theta, where \theta is the incline angle measured from the horizontal. For a body of weight 50 N on a smooth incline of 40^\circ, the normal reaction is about 38.3 N.

A pure couple produces no net force, so the two supports of a simply supported beam carry equal and opposite vertical reactions. Moment equilibrium about one support then fixes the other reaction, R = M/L for a span L, and the reaction does not depend on where along the span the couple is applied. Truss joints and reactions are solved with the same equations, and the sign convention for internal forces is taken up in the truss concept.

Thin-walled cylinder pressure stresses

A thin-walled closed cylinder under internal pressure p carries two principal membrane stresses. The circumferential, or hoop, stress is \sigma_h = \frac{pd}{2t}, where d is the mean diameter and t is the wall thickness, and the longitudinal stress is \sigma_L = \frac{pd}{4t}. The hoop stress is therefore exactly twice the longitudinal stress, \sigma_h = 2\sigma_L, a relation that holds for a closed cylinder but not for a sphere, where the two stresses are equal.

Because the radial stress is essentially zero at the surface of a thin wall, the maximum shear stress is \tau_{max} = \sigma_h/2 = \frac{pr}{2t}, where r is the mean radius. When designing a vessel from allowable limits, both the normal-stress limit and the shear-stress limit must be checked, and the lower resulting pressure governs. For example, with r = 0.5 m, t = 5 mm and p = 1 MPa, the hoop stress is 100 MPa and the maximum shear stress is 50 MPa.

The axial change in length follows from Hooke's law for biaxial stress, \varepsilon_L = (\sigma_L - \mu\sigma_h)/E. Substituting the stresses gives \delta L = \frac{PdL}{4tE}(1-2\mu), which is equivalent to \frac{PdL}{2tE}\left(\frac{1}{2}-\mu\right).

Cantilever beam deflection

A cantilever is fixed at one end, so the bending moment is largest at the support and the curvature there is \kappa = M/EI, where EI is the flexural rigidity. For a uniformly distributed load w over the full length L, the fixed-end moment is wL^2/2, which gives the curvature at the support directly. A consistent unit check is essential here, since a load given in kN/m must be converted to N/m before combining with EI in N·m^2.

The standard results for a cantilever of length L are: tip deflection \frac{PL^3}{3EI} for a point load P at the free end, and \frac{wL^4}{8EI} for a uniform load. When the load acts at an intermediate point at distance a from the support, the segment beyond the load stays straight, so the free-end deflection is the deflection at the load point plus the slope there times the overhang b. Those two quantities are \frac{Pa^3}{3EI} and \frac{Pa^2}{2EI}, respectively. For a point load at midspan, the free-end deflection is \frac{PL^3}{24EI} + \frac{PL^3}{16EI} = \frac{5PL^3}{48EI}.

Truss member forces

A truss is a set of pin-jointed straight members whose external loads are applied only at the joints, so each member carries only axial force. The method of joints isolates a joint and applies \sum F_x = 0 and \sum F_y = 0 to it. A joint with only two unknown member forces can be solved directly, which is why such joints are the usual starting point.

The sign convention is that tension is positive and compression is negative, and the sign of the computed force tells you which of the two the member carries. A member that pulls the joint toward its other end is in tension, and a member that pushes the joint away is in compression. A roller support transmits only a reaction perpendicular to its surface, so the reactions for a truss with a roller are found from the moment equations before the joints are analysed. For a joint loaded symmetrically by two members each inclined at \theta to a vertical load W, each member force has magnitude W/(2\cos\theta).