STEEL DESIGN KNOWLEDGE BASE REQUIRED STRENGTH OF MATERIALS

STEEL DESIGN KNOWLEDGE BASE REQUIRED: STRENGTH OF MATERIALS TIMBER DESIGN SOIL MECHANICS STEEL DESIGN REVIEW OF TIMBER DESIGN • BEARING PROBLEM REVIEW OF SOIL MECHANICS • vertical stresses • lateral stresses

LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: = - VERTICAL STRESSES: = VERTICAL STRESS (PSF, TSF, KN/m 2) (TOTAL STRESS) (EFFECTIVE STRESS) CLASS PROBLEM: CALCULATE TOTAL LATERAL STRESS, Draw stress diagram, calculate resultant & location q=100 k. Pa or KN/m 2 Ka=0. 307

LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL STRESSES: 30. 7 KN/m 2 17. 7 KN/m 2

LECTURE #4(CONT) BASIC SOIL MECHANICS REVIEW: LATERAL FORCE: R=107. 45+30. 98=138. 43 KN 30. 7 KN/m 2 107. 45 KN 138. 43 KN 1. 62 m 30. 98 KN 17. 7 KN/m 2

LECTURE #4(CONT) BASIC STEEL DESIGN REVIEW: Review of AISC Construction Manual (cont): Compact vs. Non- 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 width-thickness ratios (Spec Section - Table B 5. 1, P. 5 -36) • compression flange must be adequately braced against lateral buckling • When theoretical yield stress F’y is equal to or exceeded by the yielding stress than the section is no longer compact.

LECTURE #4(CONT) BASIC STEEL DESIGN REVIEW: Design Procedure for Bending(using Sx) 1. Determine the Maximum Bending Moment 2. Compute the Required Section Modulus based on allowable stresses Fb=. 66 Fy (compact) [equat. F 1 -1, Chapt. F 1. 1] or Fb=. 60 Fy (for non compact)[equat. F 1 -5, Chapt. F 1. 2] 3. Enter the ASD selection table and find the nearest tabulated value of Sx. The lightest weight and the most economical section that will serve for the yield strength stated is in bold face, and should be selected adjacent to or above the required Sx. 4. Check F’y column to be sure of compact section. 5. Check the lc & lu to make sure of adequate bracing.

LECTURE #4(CONT) BASIC STEEL DESIGN REVIEW: Note: For Members with Compact or Non-compact sections. • Recall lc is the max. unbraced length in feet of the compression flange at which the allowable bending is taken at 0. 66 Fy • Recall lu is the max. unbraced length in feet of the compression flange for which allowable bending stress may be taken at 0. 60 Fy when Cb=1

LECTURE #4(CONT) BASIC STEEL DESIGN REVIEW: Recall l, lc, and lu if l is the given length between lateral supports then, • lc>l use Fb=. 66 Fy (assuming compact section) • lu>l>lc use Fb=. 60 Fy • l>lu use the charts of allowable moments in beams (p. 2 -146 to 2 -212) where Cb=1 Note: Cb is the bending coefficient dependent upon moment gradient (F 1. 3) Cb=1 • when the bending moment at any point within an unbraced length is larger than that at both ends of this length • when frames are braced against joint translation. • when considering a conservative value for cantilever

LECTURE #5 BASIC STEEL DESIGN REVIEW: Given: Select a beam of Fy=36 ksi steel subjected to a bending moment of 125 kip-ft. , having its compression flange braced at 6 ft. intervals. Solution: Assume compact section, Fb=0. 66 Fy=23. 8 ksi 1. Maximum Bending Moment Given in Problem 2. Sx(reqd)=M/Fb=(125 x 12)/23. 8=63 cu. in. 3. Nearest Tabulated Value W 16 x 40 64. 7 W 12 x 50 64. 7 Since W 16 x 40 is bold face, it is the lightest suitable section that will serve for that yield strength.

LECTURE #5 BASIC STEEL DESIGN REVIEW: Given: Select a beam of Fy=36 ksi steel subjected to a bending moment of 125 kip-ft. , having its compression flange braced at 6 ft. intervals. Solution contd: 4. A check of the F’y column shows a dash, indicating that F’y is greater than 65 ksi. Therefore, the shape is compact. 5. From the table, Lc=7. 4 ft. >6. 0 ft. so the bracing is adequate and the assumed allowable stress of. 66 Fy is correct. Use W 16 x 40

LECTURE #5(CONT) BASIC STEEL DESIGN REVIEW: Recall: Mr is the beam resisting moment if shape has compact sections if shape has non compact flanges Design Procedure for Bending(using Mr) 1. Enter the ASD selection table and find the nearest tabulated value of Mr. The lightest weight and the most economical section that will serve for the yield strength stated is in bold face, and should be selected adjacent to or above the required Mr.

LECTURE #5 BASIC STEEL DESIGN REVIEW: 2. Check F’y column to be sure of compact section. 3. Check the lc & lu to make sure of adequate bracing. Given: Select a beam of Fy=36 ksi steel subjected to a bending moment of 125 kip-ft. , having its compression flange braced at 6 ft. intervals. Solution: Assume compact section, Fb=0. 66 Fy=23. 8 ksi 1. Enter Bending Moment Given in Problem Nearest Tabulated Value W 16 x 40 128 k-ft W 12 x 50 128 k-ft Since W 16 x 40 is bold face, it is the lightest suitable section that will serve for that yield strength.

LECTURE #5 BASIC STEEL DESIGN REVIEW: Given: Select a beam of Fy=36 ksi steel subjected to a bending moment of 125 kip-ft. , having its compression flange braced at 6 ft. intervals. 3. Check F’y column to be sure of compact section. OK the same as before 4. Check the lc & lu to make sure of adequate bracing. Lc=7. 4 ft > 6. 0 ft therefore Mr is valid Use W 16 x 40