CTC / MTC 222 Strength of Materials

1. CTC / MTC 222Strength 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/As • 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, σb • σb = Applied Load/Bearing Area = F/Ab • Area Ab is the area over which the load is transferred • For flat surfaces in contact, Abis the area of the smaller of the two surfaces • For a pin in a close fitting hole, Abis the projected area, Ab = 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 = FL / 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 = Eα (∆T)

10. Chapter Four – Torsional Shear Stress and Deformation • For a circular member, τmax = Tc / J • T = applied torque, c = radius of cross section, J = polar moment of inertia • Polar moment of Inertia, J • Solid circular section, J = πD4 / 32 • Hollow circular section, J = π(Do4 - Di4 )/ 32 • Expression can be simplified by defining the polar section modulus, Zp= J / c, where c = r = D/2 • Solid circular section, Zp = πD3 / 16 • Hollow circularsection, Zp = π(Do4 - Di4 )/ (16Do) • Then, τmax = T / Zp

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 • Applicationof 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: • AT ̅Y̅ = ∑ (Ai yi ) where: • AT = total area of composite shape • ̅Y̅ = distance to centroid of composite shape from some reference axis • Ai = area of one component part of shape • yi = distance to centroid of the component part from the reference axis • Solve for ̅Y̅ = ∑ (Ai yi ) / AT • Perform calculation in tabular form • See Examples 6-1 & 6-2

15. Moment of Inertia ofComposite Shapes • Perform calculation in tabular form • Divide the shape into component parts which are simple shapes • Locate the centroid of each component part, yi 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, Ii • Compute the distance, di = ̅Y̅ - yi of the centroid of each part from the overall centroid • Compute the transfer term Ai di2 for each part • The overall moment of inertia IT , is then: • IT = ∑ (Ii + Ai di2) • 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) • σmax = 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, ct , is different than distance to bottom fiber cb • σtop = M ct / I • σbot = M cb / I

17. Section Modulus, S • Maximum Stress due to bending • σmax = M c / I • Both I and c are geometric properties of the section • Define section modulus, S = I / c • Then σmax = M c / I = M / S • Units for S – in3 , mm3 • Use consistent units • Example: if stress, σ, is to be in ksi (kips / in2 ), 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 • Stop = I / ctop • Sbot = I / cbot

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 = AP ̅y̅ , where: • AP = 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 AP from the overall centroidal axis • Units of Q are length cubed; in3, mm3, m3,

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 Rm / t ≥ 10, pressure vessel is considered thin-walled • Stress in wall of thin-walled sphere • σ = p Dm / 4 t • Longitudinal stress in wall of thin-walled cylinder • σ = p Dm / 4 t • Longitudinal stress is same as stress in a sphere • Hoop stress in wall of cylinder • σ = p Dm / 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