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Introduction to Wood, Soils, and Steel: Strength of Materials, Timber Design, Soil Mechanics, Steel Design

This lecture provides an overview of the basic knowledge required in strength of materials, timber design, soil mechanics, and steel design. It covers topics such as bending members, deflection members, shear members, column members, and bearing problems. The lecture also reviews timber design, with a focus on bearing perpendicular to the grain and deflection limits. It also includes a review of soil mechanics, covering vertical and lateral stresses.

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Introduction to Wood, Soils, and Steel: Strength of Materials, Timber Design, Soil Mechanics, Steel Design

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  1. WOOD, SOILS, AND STEEL INTRO KNOWLEDGE BASE REQUIRED: STRENGTH OF MATERIALS TIMBER DESIGN SOIL MECHANICS STEEL DESIGN REVIEW OF TIMBER DESIGN • BENDING MEMBERS • DEFLECTION MEMBERS • SHEAR MEMBERS • COLUMN MEMBER • BEARING PROBLEM

  2. REVIEW OF TIMBER: BEARING PERPENDICULAR TO THE GRAIN- fc(perp) P lb lb+3/8 where lb= bearing length Note: When the bearing length is less than 6 in. and when the distance from the end of the beam to the contact area is more than 3 in., the allowable bearing stress may be increased by Cb.

  3. The deformation limit of .04 inch. is provided by ASTM D143 provides adequate service in typical wood-frame construction. Special Cases In some designs where the deformation is critical, a reduced value can be applied. ( WWPA P.9 Table F) Deflection can be designed for a reduce limit of .02 in. (also refer to P.251 in text) Fc (perp .02) = 0.73 Fc (perp .04) + 5.60 Sample Problem: Given a Hem-Fir Select Structural with 11,000#s on supports: a) check for the bearing of a cantilever support. b) Assume critical deflection for heavy impact loads at end of cantilever.

  4. 4x8 3.5” > 3” 2 - 2x12 1.5” 1.5” Fc(perp) = 405 psi lb= 3” therefore we can increase bearing stress, but lets be conservative and use lb as recommended N.G.

  5. We have to increase bearing V 11,000 Req’d Area= ----- = ---------- = 27 sq in. Fc(perp) 405psi add 2-2X12 X 12 A= 6 X 3.5 = 21 sq in < 27 sq in NG 2-3X12 X 12 A=[(2X1.5)+(2X2.5)](3.5)= 28 sq in > 27 sq in OK b) 4x8 bearing problem is O.K., now solve for critical deflection with limit of .02 inch F’c(perp .02) = 0.73 (405) + 5.60 F’c(perp .02)= 301.25 psi Req’d Area = 11000/301.25 =36.5 sq in add 2- 4x12x12 A=[(2X3.5)+(2X1.5)](3.5)= 35 sq in N.G. use 2- 6x12 - 49 sq in. or a steel plate 3.5X10.5

  6. LECTURE #4 REVIEW OF SOIL MECHANICS • VERTICAL STRESSES • LATERAL STRESSES BASIC SOIL MECHANICS REVIEW: = UNIT WEIGHT OF SOIL (PCF, KN/m3) = SATURATED UNIT WEIGHT OF SOIL = BOUYANT UNIT WEIGHT OF SOIL = UNIT WEIGHT OF WATER(62.4PCF,9.81 KN/m3)

  7. LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: - = VERTICAL STRESSES: = VERTICAL STRESS (PSF, TSF,KN/m2) (TOTAL STRESS) (EFFECTIVE STRESS) CALCULATE TOTAL AND EFFECTIVE Ka=0.5

  8. LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: VERTICAL STRESSES: EFFECTIVE VERTICAL STRESSES:

  9. LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL STRESSES: LATERAL FORCE:

  10. LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL FORCES:

  11. LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL FORCES: TO FIND RESULTANT SUM FORCES: R TO FIND RESULTANT LOCATION TAKE MOMENT:y

  12. LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL FORCES: 236.25# 1575# 1140# TO FIND RESULTANT SUM FORCES: R TO FIND RESULTANT LOCATION TAKE MOMENT:y

  13. LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL FORCES: R=236.25+1575+1140=2951.25# LATERAL ARM:

  14. LECTURE #4(CONT) BASIC STEEL DESIGN REVIEW: For temporary structures we use primarily involves: Rolled Sections - W: sections Angle Sections Channel Sections Very Rarely - Steel Joist Decking(thin gages) Trusses Elements include Girders, Beams, Columns, & Struts (angles &channels)

  15. LECTURE #4(CONT) BASIC STEEL DESIGN REVIEW: Review of AISC Construction Manual: Allowable Stresses Most commonly used is A36 Steel Steel Design Procedures a) Use ASD - Allowable Stress Design Procedure (Basic Allowable Stress For A36 Steel) Tension- Ft=.6 Fy=14,400psi Shear - Fv= .4 Fy=32,400psi Bearing - Fp=.9Fy=32,400 psi Bending - Fb = .66Fy=23,760psi Based on Compact Section Compression=Fa (variable depending on unbraced length)

  16. LECTURE #4(CONT) BASIC STEEL DESIGN REVIEW: Review of AISC Construction Manual (cont): Compact vs. Non- Compact Compact Beam are rolled beam that can achieve the plastic moment. Stress Distribution of I Beam

  17. LECTURE #4(CONT) BASIC STEEL DESIGN REVIEW: • Review of AISC Construction Manual (cont): • Compact Sections- • are symmetrical about the y-y axis • webs/flanges must have certain web thickness ratios • compression flange must be adequately braced • against lateral buckling Design Procedure for Bending 1. Determine the Maximum Bending Moment 2. Compute the Required Section Modulus based on allowable stresses Fb=.66Fy (compact) or Fb=.60 Fy (non compact) 3.Lightest weight section is the most economical

  18. LECTURE #4(CONT) BASIC STEEL DESIGN REVIEW: • Review of AISC Construction Manual (cont): • Compression Members: • lc- distance between spacing of lateral braces • When brace spacing is less than or equal to lc • then Fb=.66Fy • Lu - maximum unsupported length • When brace spacing is greater than Lc but less than • Lu then Fb=.60Fy • When brace spacing is greater than lu, Fb is not • determined and Total allowable moment has to be • taken from a chart.

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