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1 Level II: Section 03 Design Principles Section Downloads 2 Section Downloads Handouts & Slides can be printed Version 2.0 Course binders are available for purchase Not required Download & Print TTT II Sec 03 Slides TTT II Sec 03 Handouts 3 1

2 Outline Statics & Strength of Materials Review Force Resolution Mechanics of Materials 4 Statics & Strength of Materials Review The Design Process Load vs. Resistance or Actual vs. Allowable Determine all loads & load cases that will be subjected on the structure Design structure to resist all these loads & load cases load 23 x load Factor Factor of of Safety Safety of of Three Two 6 2

3 The Design Process Load vs. Resistance Various loads 7 Three Basic Design Principles 8 Three Basic Design Principles Statics Mechanics of Materials Triangulation 9 3

4 Statics Bodies at Rest Forces Moments Static Equilibrium 10 Forces Direction of Force 11 Axial Forces Act through the length of the truss member. axial Truss Member Compression axial 12 4

5 Axial Forces Act through the length of the truss member. axial Truss Member axial Tension 13 Transverse Forces Truss Member 14 Truss Action 2X X X + X 2X = 0 X 15 5

6 Moments Force that produces a rotation on a member and subsequent bending stresses. a L P 16 Moments M = P x L Moment (ft-lbs) Force (lbs) Length (ft) P L Torque 17 Moment Equilibrium M board = M wall 18 6

7 Truss Action Bending Moment 19 Static Equilibrium Used to calculate unknown forces acting on a structure lbs R 1 R lbs 1500 lbs 20 Static Equilibrium 3000 lbs R 1 Σ Forces = 0 R

8 Static Equilibrium 3000 lbs R 1 = 1500 lbs R 2 = 1500 lbs Σ Vertical Forces = R 1 R 2 = = R 1 + R 2 R 1 = R 2 R 1 = 3000/2 = 1500 lbs R 2 = 3000/2 = 1500 lbs 22 Equations of Equilibrium Σ Vertical Forces = 0 Σ Horizontal Forces = 0 Σ V = 0 (for Σ Moments vertical forces) = 0 Σ H = 0 (for horizontal forces) Σ M = 0 (for moments) Three Fundamental Requirements of Static Equilibrium 23 Vertical Forces + Σ V = 0 R 1 = R R 1 - R 2 = R 1 = 0 R 1 = 4025 # R 2 = 4025 # 8000 # w50 # 30 ft R 1 1/2 1/2 R

9 Vertical Forces + Σ M A = (15) = 120,000 ft-lbs 50(15) = 750 ft-lbs - 30(R 2 ) ft-lbs 8000 # Σ M A = Σ 0 M= B = 120, (R 2 ) R 2 = 4025 ft-lbs R 1 = 4025 ft-lbs 50 # A 15 ft 30 ft Σ M B = 0 B R 1 R 2 25 Horizontal Forces + Σ H = 0 R 1 + R 2 W = 0 W R 1 R 2 26 Moments + Σ M c = 0 : 500(17) ft-lbs + 8 (R A ) = 0 R A = ft-lbs 500 # R? A R A A 8 ft C 25 ft 17 ft B 27 9

10 Free Body Diagrams Statically Indeterminate 28 Quiz 1 Force Resolution 10

11 Force Resolution ΣV & ΣH are only valid for vertically and horizontally acting forces. vertical? horizontal 31 Force Resolution F magnitude F V F H 32 Force Resolution F hypotenuse opposite θ adjacent 33 11

12 Force Resolution Example 1 34 Force Resolution Example 1 35 Force Resolution Example 1 sin θ = opposite/hypotenuse opposite = sin θ x hypotenuse F v = sin 30 x F F v = 0.5 x 500 lb F v = 250 lb? 36 12

13 Force Resolution Example 1 cos θ = adjacent/hypotenuse adjacent = cos θ x hypotenuse F H = cos 30 x F F H = x 500 lb F H = 433 lb? 37 Force Resolution Example 1 38 Force Resolution Example 2 ΣV = 0 & ΣH =

14 Force Resolution Example 2 40 Force Resolution Example θ Section 02: Truss Math 41 Force Resolution Example 2 sin θ = opposite/hypotenuse opposite = sin θ x hypotenuse C v = sin x C C v = x C C v = (C) 42 14

15 Force Resolution Example 2 cos θ = adjacent/hypotenuse adjacent = cos θ x hypotenuse C H = cos x C C H = x C C H = (C) 43 Force Resolution Example 2 + ΣV = 0 -R + C v = 0 - R (C) = 0 + ΣH = 0 T - C H = 0 T (C) = (C) = 0 C = # T ( ) = 0 T = # 1000 # 44 Force Resolution Example # 2000 # 1000 # 45 15

16 Quiz 2 Mechanics of Materials Mechanics of Lumber 48 16

17 Mechanics of Lumber Lumber has been used for centuries Easily worked Plentiful Relatively strong Difficult material to engineer Anisotropic Different properties in different directions 49 Wood Cells II 50 Lumber Design Values 51 17

18 Lumber Design Values Six Strength Properties Compression Parallel to Grain Compression Perpendicular to Grain Tension Parallel to Grain Bending Shear Modulus of Elasticity 52 Evaluation of Wood Strength Properties 53 National Design Specification (NDS ) 54 18

19 Southern Pine Use Guide 55 Western Lumber Product Use Manual 56 US Span Book for Major Lumber Species

20 ANSI/TPI Lumber Stresses Lumber Design Values F b - Bending Stress F t - Tension Stress F v - Shear Stress F cii - Compression Stress Parallel to Grain F c - Compression Stress Perpendicular to Grain E Modulus of Elasticity 59 Bending Stress F b Load 60 20

21 Tension Stress F t 61 Shear Stress F v Involves the application of stress from 2 opposite directions causing portions of an object to move in parallel but opposite directions. 62 Shear Stress F v 63 21

22 Compression Stress Parallel to Grain F cii 64 Compression Stress Perpendicular to Grain F c 65 Modulus of Elasticity E = 3,200,000 psi 3.2E stress strain 66 Courtesy of Arizona State University 22

23 Quiz 3 Feedback 23