Presentation on theme: "WOOD, SOILS, AND STEEL INTRO KNOWLEDGE BASE REQUIRED: STRENGTH OF MATERIALS STEEL DESIGN SOIL MECHANICS REVIEW OF TIMBER DESIGN BENDING MEMBERS DEFLECTION."— Presentation transcript:
WOOD, SOILS, AND STEEL INTRO 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) + 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.
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) + 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
REVIEW OF SOIL MECHANICS VERTICAL STRESSES LATERAL STRESSES 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 )
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)
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)
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
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
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.