Steps for Steel Design

DESIGN OF MEMBERS FOR FLEXURE(NSCP 6TH Edition 2010, Section 506 and AISC 2005 Specifications Chapter F)

Step 1:Determine the following: Mu for LRFD Load Combinations (SEI/ASCE 7, Section 2.3) Ma for ASD Load Combinations (SEI/ASCE 7, Section 2.4)

Step 2:Assume steel section from AISC Manual. Determine its design load due to self-weight and add the corresponding values to the dead load of ASD and LRFD load combinations.

Step 3:(NSCP Table 502.4.1 & AISC Table B4.1)From the assumed section, classify cross-sectional shapes as compact, non-compact or slender, depending on the values of the width-thickness ratios of the flange and web. For I-shaped sections:

If and the flange is continuously connected to the web, the shape is compact. If the shape is non-compact. If the shape is slender.

For flanges,For webs,

Determine the lateral-torsional buckling modification factor for non-uniform moment diagrams when both ends of the unsupported segment are braced, Cb.(NSCP Eq. 506.1-1 & ---)For doubly symmetric members, Step 4:See NSCP Table User Note 506.1.1 and select case that is suitable for the flange and web slenderness. Determine the value of the nominal flexural strength, Mn from the selected case.

Step 4.1: (NSCP Section 506.2)If for both flange and web (compact section) bent about their major axis.

1. Solve for the limiting lengths Lp and Lr:(NSCP Eq. 506.2-5 & AISC Eq. F2-5)(NSCP Eq. 506.2-6 & AISC Eq. F2-6)For doubly symmetric I-shape: Note: if the square root term in Eq. 506.2-4 is conservatively taken equal to 1.0, Eq. 506.2-6 becomes(NSCP Eq. 506.2-7 & AISC Eq. F2-7)

2. Mn shall be the lower value obtained according to the following limit states:

Yielding (NSCP Section 506.2.1)(NSCP Eq. 506.2-1 & AISC Eq. F2-1) Lateral-Torsional Buckling (NSCP Section 506.2.2)a. For , the limit state of LTB does not apply.b. For , (NSCP Eq. 506.2-2 & AISC Eq. F2-2)c. For , (NSCP Eq. 506.2-3 & AISC Eq. F2-3)(NSCP Eq. 506.2-4 & AISC Eq. F2-4)Note: The square root term in Eq. 506.2-4 may be conservatively taken equal to 1.0.

The available flexural strength shall be greater than or equal to the maximum required moment causing compression within the flange under consideration Cb is permitted to be conservatively taken as 1.0 for all cases.

Step 4.2: (NSCP Section 506.3)If (compact web);If (non-compact flange) or(slender flange) bent about their major axis.

1. Mn shall be the lower value obtained according to the following limit states:

Lateral-Torsional Buckling (NSCP Section 506.3.1)(Apply provisions in Section 506.2.2) Compression Flange Local Buckling (NSCP Section 506.3.2)a. For ,(NSCP Eq. 506.3-1 & AISC Eq. F3-1)b. For,(NSCP Eq. 506.3-2 & AISC Eq. F3-2)kc shall not be taken less than 0.35 nor greater than 0.76

Step 4.3: (NSCP Section 506.4)If (compact web) or (non-compact web) bent about their major axis.

1. Solve for the limiting lengths Lp and Lr:(NSCP Eq. 506.4-7 & ---)(NSCP Eq. 506.4-8 & ---)The stress FL, is determined as follows:a. For ,(NSCP Eq. 506.4-6a & ---)b. For ,(NSCP Eq. 506.4-6b & ---)

2. Determine the web plastification factor for compression Rpc:a. For ,(NSCP Eq. 506.4-9a & ---)b. For ,(NSCP Eq. 506.4-9b & ---)

3. Determine the web plastification factor for tension Rtc:a. For ,(NSCP Eq. 506.4-9a & ---)b. For ,(NSCP Eq. 506.4-9b & ---)

4. Mn shall be the lower value obtained according to the following limit states: Compression Flange Yielding (NSCP Section 506.4.1)(NSCP Eq. 506.4-1 & ---) Lateral-Torsional Buckling (NSCP Section 506.4.2)a. For , the limit state of LTB does not apply.b. For , (NSCP Eq. 506.4-2 & ---)c. For , (NSCP Eq. 506.4-3 & ---)(NSCP Eq. 506.4-4 & ---)(NSCP Eq. 506.4-5 & ---)For , J shall be taken as zero. Compression Flange Local Buckling (NSCP Section 506.4.3)a. For , the limit state of LB does not apply.b. For , (NSCP Eq. 506.4-12 & ---)c. For,(NSCP Eq. 506.4-13 & ---)kc shall not be taken less than 0.35 nor greater than 0.76 Tension Flange Yielding (NSCP Section 506.4.4)a. For , the limit state of TFY does not apply.b. For ,(NSCP Eq. 506.4-14 & ---)

Step 4.4: (NSCP Section 506.5)If (slender web) bent about their major axis.

1. Solve for the limiting lengths Lp and Lr:(NSCP Eq. 506.4-7 & ---)(NSCP Eq. 506.5-5 & ---)

2. Determine the bending strength reduction factor Rpg:

3. Mn shall be the lower value obtained according to the following limit states: Compression Flange Yielding (NSCP Section 506.5.1)(NSCP Eq. 506.5-1 & ---) Lateral-Torsional Buckling (NSCP Section 506.5.2)(NSCP Eq. 506.5-2 & ---)a. For , the limit state of LTB does not apply.b. For , (NSCP Eq. 506.5-3 & ---)c. For , `(NSCP Eq. 506.5-4 & ---) Compression Flange Local Buckling (NSCP Section 506.5.3)(NSCP Eq. 506.5-7 & ---)a. For ,(NSCP Eq. 506.5-8 & ---)b. For,(NSCP Eq. 506.5-9 & ---)kc shall not be taken less than 0.35 nor greater than 0.76 Tension Flange Yielding (NSCP Section 506.5.4)a. For , the limit state of TFY does not apply.b. For ,(NSCP Eq. 506.5-10 & ---)

Step 4.5: (NSCP Section 506.5)For I-Shaped members bent about their minor axis.1. Mn shall be the lower value obtained according to the following limit states:

Yielding (NSCP Section 506.6.1)(NSCP Eq. 506.6-1 & ---) Flange Local Buckling (NSCP Section 506.6.2)a. For, the limit state of yielding shall apply.b. For ,(NSCP Eq. 506.6-2 & ---)c. For,(NSCP Eq. 506.6-3 & ---)(NSCP Eq. 506.6-4 & ---)

Step 5: (NSCP Section 506.1)For LRFD, check if the section is capable of resisting the design loads, the section must satisfy the equation:

For ASD, check if the section is capable of resisting the service loads, the section must satisfy the equation:

For all provisions of Section 506, and .

= width-thickness ratiof = width-thickness ratio of flangep = upper limit for compact category r = upper limit for non-compact categoryw = width-thickness ratio of webpf = flange upper limit for compact category rf = flange upper limit for non-compact categorypw = web upper limit for compact categoryrw = web upper limit for non-compact categorybf = flange widthCb = lateral-torsional buckling modification factorE = modulus of elasticity of steelFy = specified minimum yield stress of the type of steel being usedh = web heightho = distance between the flange centroidsIy = moment of inertia taken about the y-axisJ = torsional constantLb = length between points that are either braced against lateral displacement of compression flange or braced against twist of the cross-sectionLp = limiting laterally unbraced length for the limit state of yieldingLr = limiting laterally unbraced length for the inelastic lateral-torsional bucklingMA = absolute value of moment at quarter point of the unbraced segmentMB = absolute value of moment at centerline of the unbraced segmentMC = absolute value of moment at third quarter point of the unbraced segmentMmax = absolute value of maximum moment in the unbraced segmentMn = nominal moment capacityMp = plastic momentRm = cross-section monosymmetry parameterSx = elastic section modulus taken about the x-axisSy = elastic section modulus taken about the y-axisrx= radius of gyration with respect to x-axisry= radius of gyration with respect to y-axistf = flange thicknesstw = web thicknessZx = plastic section modulus about the x-axisZy = plastic section modulus about the y-axis

DESIGN OF MEMBERS FOR FLEXURE(NSCP 5TH Edition 2001)

Step 1:Determine the following: Ma for ASD Load Combinations (SEI/ASCE 7, Section 2.4)

Step 2:Assume steel section from AISC Manual. Determine its design load due to self-weight and add the corresponding values to the dead load of ASD load combinations.

Step 3:(NSCP Table 502-1)From the assumed section, classify cross-sectional shapes as compact, partially compact, or non-compact, depending on the values of the width-thickness ratios of the flange and web. For I-shaped sections: For compact sections, the following conditions must be satisfy:a. Its flanges must be continuously connected to the web.b. The section must have a flange width-thickness ratio of its compression elements:

c. The section must have a depth to web thickness ratio:

For partially compact sections, the following conditions must be satisfy:a. The section satisfy the requirements for compact sections except that their flanges are non-compact:

For non-compact sections, the following conditions must be satisfy:a. The section must not qualify as compact shapes and must have a flange width-thickness ratio of its compression elements:

Determine the lateral-torsional buckling modification factor for non-uniform moment diagrams when both ends of the unsupported segment are braced, Cb.For doubly symmetric members, Cb is permitted to be conservatively taken as 1.0 for all cases. For cantilevers or overhangs where the free end is unbraced, Cb=1.0.

Step 4:(NSCP Table 502-1)From the assumed section, determine the value of Lb, Lc and Lu which will be considered in the computations. Lb is the laterally unsupported length of the compression flange. Lc shall be the smaller value of L1 or L2, and Lu shall be the larger value of L1 or L2.

Step 5:Determine the value of the allowable bending stress, Fb depending on the given parameters that the section has satisfy. Laterally Supported Beamsa. For compact sections bending about the strong axis (x-axis) and .

b. For partially-compact sections bending about the strong axis (x-axis) and .

c. For non-compact sections bending about the strong axis (x-axis) and .

Laterally Unsupported Beamsa. If

b. If (2 Cases)Case 1:Fb is the larger value obtained in the following equations:

The allowable bending stress shall not exceed 0.6Fy ()Case 2:Fb is the larger value obtained in the following equations:

The allowable bending stress shall not exceed 0.6Fy () Sections bending about its weak axisa. For compact sections:

b. For partially-compact sections:

c. For non-compact sections bending about the strong axis (x-axis) and .

Step 6:Solve for the actual bending stress and compare to the allowable bending stress. The section must satisfy the equation:

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Steps for Steel Design