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WOOD DESIGN REVIEW

WOOD DESIGN REVIEW. 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. REVIEW OF TIMBER:. BEARING PERPENDICULAR TO THE GRAIN- fc(perp).

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WOOD DESIGN REVIEW

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

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

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

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

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

  12. REVIEW LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL FORCES: 472.5# 1575# 1500# TO FIND RESULTANT SUM FORCES: R TO FIND RESULTANT LOCATION TAKE MOMENT:y

  13. REVIEW LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL FORCES: R=472.5+1575+1500=3547.5# LATERAL ARM:

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