1. CTC / MTC 222 Strength of Materials Final Review

2. Final Exam Tuesday, December 13, 3:00 – 5:00 30% of grade Graded on the basis of 30 points in increments of ½ point Open book May use notes from first two tests plus two additional sheets of notes Equations, definitions, procedures, no worked examples Also may use any photocopied material handed out in class Work problems on separate sheets of engineering paper Hand in test paper, answer sheets and notes stapled to back of answer sheets

3. Course Objectives To provide students with the necessary tools and knowledge to analyze forces, stresses, strains, and deformations in mechanical and structural components. To help students understand how the properties of materials relate the applied loads to the corresponding strains and deformations.

4. Chapter One – Basic Concepts SI metric unit system and U.S. Customary unit system Unit conversions Basic definitions Mass and weight Stress, direct normal stress, direct shear stress and bearing stress Single shear and double shear Strain, normal strain and shearing strain Poisson’s ratio, modulus of elasticity in tension and modulus of elasticity in shear

5. Direct Stresses Direct Normal Stress,  σ = Applied Force/Cross-sectional Area = F/A Direct Shear Stress,  Shear force is resisted uniformly by the area of the part in shear  = Applied Force/Shear Area = F/A s Single shear – applied shear force is resisted by a single cross- section of the member Double shear – applied shear force is resisted by two cross- sections of the member

6. Direct Stresses Bearing Stress, σ b σ b = Applied Load/Bearing Area = F/A b Area A b is the area over which the load is transferred For flat surfaces in contact, A b is the area of the smaller of the two surfaces For a pin in a close fitting hole, A b is the projected area, A b = Diameter of pin x material thickness

7. Chapter Two – Design Properties Basic Definitions Yield point, ultimate strength, proportional limit, and elastic limit Modulus of elasticity and how it relates strain to stress Hooke’s Law Ductility - ductile material, brittle material

8. Chapter Three – Direct Stress Basic Definitions Design stress and design factor Understand the relationship between design stress, allowable stress and working stress Understand the relationship between design factor, factor of safety and margin of safety Design / analyze members subject to direct stress Normal stress – tension or compression Shear stress – shear stress on a surface, single shear and double shear on fasteners Bearing stress – bearing stress between two surfaces, bearing stress on a fastener

9. Chapter Three – Axial Deformation and Thermal Stress Axial strain ε, ε = δ / L, where δ = total deformation, and L = original length Axial deformation, δ δ = F L / A E If unrestrained, thermal expansion will occur due to temperature change δ = α x L x ∆T If restrained, deformation due to temperature change will be prevented, and stress will be developed σ = E α (∆T)

10. Chapter Four – Torsional Shear Stress and Deformation For a circular member, τ max = Tc / J T = applied torque, c = radius of cross section, J = polar moment of inertia Polar moment of Inertia, J Solid circular section, J = π D 4 / 32 Hollow circular section, J = π (D o 4 - D i 4 ) / 32 Expression can be simplified by defining the polar section modulus, Z p = J / c, where c = r = D/2 Solid circular section, Z p = π D 3 / 16 Hollow circularsection, Z p = π (D o 4 - D i 4 ) / (16D o ) Then, τ max = T / Z p

11. Chapter Five – Shear Forces and Bending Moments in Beams Sign Convention Positive Moment M Bends segment concave upward  compression on top

12. Relationships Between Load, Shear and Moment Shear Diagram Application of a downward concentrated load causes a downward jump in the shear diagram. An upward load causes an upward jump. The slope of the shear diagram at a point (dV/dx) is equal to the (negative) intensity of the distributed load w(x) at the point. The change in shear between any two points on a beam equals the (negative) area under the distributed loading diagram between the points.

13. Relationships Between Load, Shear and Moment Moment Diagram Application of a clockwise concentrated moment causes an upward jump in the moment diagram. A counter-clockwise moment causes a downward jump. The slope of the moment diagram at a point (dM/dx) is equal to the intensity of the shear at the point. The change in moment between any two points on a beam equals the area under the shear diagram between the points.

14. Chapter Six – Centroids and Moments of Inertia of Areas Centroid of complex shapes can be calculated using: A T ̅Y̅ = ∑ (A i y i ) where: A T = total area of composite shape ̅Y̅ = distance to centroid of composite shape from some reference axis A i = area of one component part of shape y i = distance to centroid of the component part from the reference axis Solve for ̅Y̅ = ∑ (A i y i ) / A T Perform calculation in tabular form See Examples 6-1 & 6-2

15. Moment of Inertia of Composite Shapes Perform calculation in tabular form Divide the shape into component parts which are simple shapes Locate the centroid of each component part, y i from some reference axis Calculate the centroid of the composite section, ̅Y̅ from some reference axis Compute the moment of inertia of each part with respect to its own centroidal axis, I i Compute the distance, d i = ̅Y̅ - y i of the centroid of each part from the overall centroid Compute the transfer term A i d i 2 for each part The overall moment of inertia I T, is then: I T = ∑ ( I i + A i d i 2 ) See Examples 6-5 through 6-7

16. Chapter Seven – Stress Due to Bending Positive moment – compression on top, bent concave upward Negative moment – compression on bottom, bent concave downward Maximum Stress due to bending (Flexure Formula) σ max = M c / I Where M = bending moment, I = moment of inertia, and c = distance from centroidal axis of beam to outermost fiber For a non-symmetric section distance to the top fiber, c t, is different than distance to bottom fiber c b σ top = M c t / I σ bot = M c b / I

17. Section Modulus, S Maximum Stress due to bending σ max = M c / I Both I and c are geometric properties of the section Define section modulus, S = I / c Then σ max = M c / I = M / S Units for S – in 3, mm 3 Use consistent units Example: if stress, σ, is to be in ksi (kips / in 2 ), moment, M, must be in units of kip – inches For a non-symmetric section S is different for the top and the bottom of the section S top = I / c top S bot = I / c bot

18. Chapter Eight – Shear Stress in Beams The shear stress, , at any point within a beams cross-section can be calculated from the General Shear Formula:  = VQ / I t, where V = Vertical shear force I = Moment of inertia of the entire cross-section about the centroidal axis t = thickness of the cross-section at the axis where shear stress is to be calculated Q = Statical moment about the neutral axis of the area of the cross- section between the axis where the shear stress is calculated and the top (or bottom) of the beam Q is also called the first moment of the area Mathematically, Q = A P ̅y̅, where: A P = area of theat part of the cross-section between the axis where the shear stress is calculated and the top (or bottom) of the beam ̅y̅ = distance to the centroid of A P from the overall centroidal axis Units of Q are length cubed; in 3, mm 3, m 3,

19. Shear Stress in Common Shapes The General Shear Formula can be used to develop formulas for the maximum shear stress in common shapes. Rectangular Cross-section  max = 3V / 2A Solid Circular Cross-section  max = 4V / 3A Approximate Value for Thin-Walled Tubular Section  max ≈ 2V / A Approximate Value for Thin-Webbed Shape  max ≈ V / t h t = thickness of web, h = depth of beam

20. Chapter Twelve – Pressure Vessels If R m / t ≥ 10, pressure vessel is considered thin- walled Stress in wall of thin-walled sphere σ = p D m / 4 t Longitudinal stress in wall of thin-walled cylinder σ = p D m / 4 t Longitudinal stress is same as stress in a sphere Hoop stress in wall of cylinder σ = p D m / 2 t Hoop stress is twice the magnitude of longitudinal stress Hoop stress in the cylinder is also twice the stress in a sphere of the same diameter carrying the same pressure