Steel Design

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Last updated 6:37 PM on 8/14/26
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28 Terms

1
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Two types of Connection

  1. Bolted

  2. Welded

2
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Tension Based on Gross Area of Plate

ft=PAg;ft=0.6Fyf_{t}=\frac{P}{A_{g}};f_{t}=0.6F_{y} (ASD Factor of Safety = 1.67)

P=0.6FyAgP=0.6F_{y}A_{g}

3
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Tension Based on Net Rupture

ft=PAn;ft=0.5Fuf_{t}=\frac{P}{A_{n}};f_{t}=0.5F_{u}

P=0.5FuAeP=0.5F_{u}A_{e}

4
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Bearing Failure of Plates (LRFD)

fp=PA;fp=1,2Fuf_{p}=\frac{P}{A};f_{p}=1,2F_{u}

T=1.2FuApT=1.2F_{u}A_{p}

5
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Deformation at Bolt Hole and Tear Out Failure (ASD)

Rn=2.4dtFuR_{n}=2.4\cdot d\cdot t\cdot F_{u}

Rn=1.2LctFuR_{n}=1.2\cdot L_{c}\cdot t\cdot F_{u}

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Allowable Tensile Stresses

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Relationship of Reduction Factor and Factor of Safety

1.5ϕ=Ωt\frac{1.5}{\phi}=\Omega_{t}

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Effective Net Area

Ae=A_{e}= lesser of:

(1) 0.85Ag0.85A_{g}

(2) UAnUA_{n}

Where U → Shear Lag Factor

9
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Shear Lag Factor, U

For Plates: 1.0

For Angles:

U=1xLU=1-\frac{\overline{x}}{L}

L = Length of Connection

<p>For Plates: 1.0</p><p></p><p>For Angles:</p><p>$$U=1-\frac{\overline{x}}{L}$$ </p><p></p><p>L = Length of Connection </p>
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Cochrane’s Equation for Staggered Bolts

Anet+s24gA_{net}+\frac{s^2}{4g}

Where S (stagger or pitch) = Horizontal Distance

g (gage) = Vertical Distance

<p> $$A_{net}+\frac{s^2}{4g}$$ </p><p></p><p>Where S (stagger or pitch) = Horizontal Distance</p><p>g (gage) = Vertical Distance</p>
11
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Block Shear Strength

P=0.5FuAt+0.3FuAvP=0.5F_{u}A_{t}+0.3F_{u}A_{v}

12
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Out of Plane Eccentric Load

Causes Bending

<p>Causes Bending</p>
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In-Plane Eccentric Load

Causes Torsion

fv=TρJf_{v}=\frac{T\rho}{J}

<p>Causes Torsion</p><p></p><p>$$f_{v}=\frac{T\rho}{J}$$ </p>
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Stress due to Out of Plane Eccentric Load

ft=McAy2f_{t}=\frac{Mc}{A\sum_{}^{}y^2}

ft=McIf_{t}=\frac{Mc}{I}

For welds, don’t neglect the first term I=bh312+Ad2I=\frac{bh^3}{12}+Ad^2

15
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Stress Due to Welds

Pweld=σweldAweldσweld=0.3FuP_{weld}=\sigma_{weld}A_{weld}\to\sigma_{weld}=0.3F_{u}

Aweld=0.707tLA_{weld}=0.707tL

16
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Conversion of Electrode Welds

EXX Electrode → Where XX is in Ksi

Convert XX to mPa

1mPa = 145psi

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Kips to Newton Conversion

1 kip = 4,448.22 Newtons (N)

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Torsional Shearing Stress in Welds

fvt=TρJf_{vt}=\frac{T\rho}{J}

J=L(L212+x2+y2)J=\sum^{}L\left(\frac{L^2}{12}+x^2+y^2\right)

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Bending Stress in Beams

fb=McI=MSS=Icf_{b}=\frac{Mc}{I}=\frac{M}{S}\to S=\frac{I}{c}

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Shear Stress in Beams (General)

fv=VQIbf_{v}=\frac{VQ}{Ib}

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Average Web Shear Stress (Shear Stress in Web)

fv=Vdtwf_{v}=\frac{V}{d_{tw}}

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Unit Weight of A36 Steel

γ=77kNm3\gamma=77\frac{kN}{m^3}

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Allowable Bending Stress

Fb=0.66FyF_{b}=0.66Fy

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Allowable Shear Stress

FV=0.4FyF_{V}=0.4F_{y} (Allowable Web Shear Stress)

25
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Unless Otherwise Specified

Beam connected to Another Beam → Simply Supported

Beam connected to Column → Fixed

26
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Purlins Design (Max Moment for 1 Sagrod at Midpoint)

Mmax=w(l2)28M_{\max}=\frac{w\left(\frac{l}{2}\right)^2}{8}

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Interaction Equation for Purlins

fbxFbx+fbyFbyRatio\frac{f_{bx}}{F_{bx}}+\frac{f_{by}}{F_{by}}\to Ratio

28
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Interaction Equation for Purlins (Loads Acting on the Top Flange)

fbxFbx+2fbyFbyRatio\frac{f_{bx}}{F_{bx}}+\frac{2f_{by}}{F_{by}}\to Ratio

or fby=MySy2f_{by}=\frac{M_{y}}{\frac{S_{y}}{2}}