WOOD DESIGN REVIEW KNOWLEDGE BASE REQUIRED: STRENGTH OF MATERIALS STEEL DESIGN SOIL MECHANICS REVIEW OF TIMBER DESIGN BENDING MEMBERS DEFLECTION MEMBERS.

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WOOD DESIGN REVIEW KNOWLEDGE BASE REQUIRED: STRENGTH OF MATERIALS STEEL DESIGN SOIL MECHANICS REVIEW OF TIMBER DESIGN BENDING MEMBERS DEFLECTION MEMBERS SHEAR MEMBERS COLUMN MEMBER BEARING PROBLEM TIMBER DESIGN

BEARING PERPENDICULAR TO THE GRAIN- fc(perp) REVIEW OF TIMBER: l b +3/8 lblb P where l b = 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.

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) 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.

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

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) F’c(perp.02)= psi Req’d Area = 11000/ =36.5 sq in add 2- 4x12x12 A=[(2X3.5)+(2X1.5)](3.5)= 35 sq in N.G. use 2- 6x sq in. or a steel plate 3.5X10.5

REVIEW OF SOIL MECHANICS VERTICAL STRESSES LATERAL STRESSES REVIEW LECTURE #4 BASIC SOIL MECHANICS REVIEW: = UNIT WEIGHT OF SOIL (PCF, KN/m 3 ) = SATURATED UNIT WEIGHT OF SOIL = BOUYANT UNIT WEIGHT OF SOIL = UNIT WEIGHT OF WATER(62.4PCF,9.81 KN/m 3 )

REVIEW LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: = - VERTICAL STRESSES: = VERTICAL STRESS (PSF, TSF,KN/m 2 ) CALCULATE TOTAL (TOTAL STRESS) Ka=0.5

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

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

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

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

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

REVIEW LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL FORCES: R= =3547.5# LATERAL ARM: