2. 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

3. Class Sample Problem - Design of Wood member in Bending and Shear A 3x10, #1 Grade, Hem-Fir wood joists having a moisture content of 11% with a span of 22 ft. are to be used to support a roof. The expected life of this structure is 10 years and the joist has no splits. Assuming that the uniform load (DL +LL) on this wood joist is 64 #/ft, will this section satisfy this loading? If not what could we use? From the Structural Analysis in Bending DL + LL= 64 #/ft S(3x10) = 35.65 in 3 (P. 16 WWPA) 22’

4. 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.

5. 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.

6. 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.

7. 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