The actual design of steel structure construction is now popular with bolts and base bolts for steel column bonds, with columns that are welded directly into the base. This article presents the process of making this associated calculation table for convenience in the daily design practice for the daily design for Structural engineers.

This type of foot calculation is based on the calculation process of the AISC (American Institute of Steel Construction):

 

Choose the length of the base plate L (each side from the column an ​​additional 100-200m segment enough to arrange for 1 row of bolts: l = h+(200 ÷ 400) mm.

 

Calculate the width of the base:

 

$$B=\frac{N}{\psi{R}_{b.loc}}+\frac{6M}{\psi{R}_{b.loc}L^2}$$

 

Calculate the stress of $\sigma_ {max}, \sigma_{min}$ must be less than $r_{b.loc}$ according to the selected size of B and L:

 

$$\sigma_{max,min}=\frac{N}{BL}\pm\frac{6M}{BL^2}$$

 

The positive direction is compressive stress

$R_ {b.loc}$: The local calculation strength strength of the foundation concrete under the base, the design engineer is determined according to TCVN 5574: 2012: $ r_ {b.loc} = \alpha\varphi_br_b $ $

 

$\alpha$: Nail dependence factor, $\alpha = 1$ with label less than 350#, $\alpha = 13.5R_ {bt}/r_b$ with label above 350

 

$psi$: The factor depends on the distribution of compressed load on concrete, when present of the pole moment, the pressure of the compressed base on the foundation concrete is irregular: $\psi = 0.75$. In case the compression pressure is evenly distributed, $\psi = 0.1$

 

$\varphi_b=\sqrt[3]{\frac{A_m}{A_{bd}}}$

 

$A_ {bd}$: The area of ​​the base

 

$A_m$: foundation area/ concrete pedestal

Flexural design of Base plate

Calculate the eccentricity e = m/n and consider 2 basic design cases:

when $e\leq{L}/6$:

The stress chart has a trapezoid shape (the entire compressive base). Let x be the coordinates running from the point of $\sigma_ {max}$ in the direction of L, the stress value at x is:

 

$$\sigma_x=\sigma_{max}-\left(\frac{\sigma_{max}-\sigma_{min}}{L}\right)x$$

 

Because the stress of $\sigma_x$ is a bending load for the base, the integral of the equation is 2 times the equation $\sigma_x$ will be caused by the moment on the base (on the long unit):

 

$$M_b=\sigma_{max}\frac{x^2}2-\left(\frac{\sigma_{max}-\sigma_{min}}{L}\right)\frac{x^3}6$$

 

At $x = 0.5 (L-0.95H)$, the maximum maximum torque value. The minimum sole thickness is:

$$t_{bd}=\sqrt{\frac{6M_b}{f\gamma_c}}$$

control $t_{bd}\geqslant{12}mm$$

when $e<l/6$:

The stress under the base will have 2 parts: tensile stress and compression stress. To favor safety, considering the maximum compression stress by the compressive strength of $\sigma_ {max} =\psi r_ {b.loc}$ and the tensile stress in the bolt reaches the tensile calculation intensity of the bolt of $f_ {tb}$.

 

Call the deformation of the bolt and the concrete is $\varepsilon_ {bl}, \varepsilon_b$ (when the deformed base): $\varepsilon_ {BL} = \frac {f_{tb}}E, \varepsilon_b = \frac {R_ {b.loc}} {E_b}$

 

$E$: elastic module of steel

 

$E_ b$: elastic module of nail concrete

 

Call y is the length of the compressive stress distribution, according to the same geometric conditions:

 

$$\frac{\varepsilon_{bl}}{\varepsilon_b}=\frac{L-y-c}{y}$$

 

$c$: distance from the bolt center to the outer edge of the base

 

$$\implies y=\frac{n\psi R_{b.loc}(L-c)}{f_{tb}+nR_{b.loc}}$$

 

$n=E/E_b$

 

The equation of compressive stress chart at point X is:

 

$$\sigma_x=\psi R_{b.loc}-\psi R_{b.loc}\frac xy$$

 

Integrated 2 times the bending moment equation in the base:

 

$$M_b=\psi R_{b.loc}\frac{x^2}2-\psi R_{b.loc}\frac{x^3}{6y}$$

 

The bending moment also reached the maximum value when $x = 0.5 (L-0.95H)$, calculating the smallest sole thickness similar to the above case:

 

$$t_{bd}=\sqrt{\frac{6M_b}{f\gamma_c}}$$

The structural design school has a large wind pressure, the compression force due to static roof load is smaller than the wind spitting due to the wind, it is necessary to calculate the thickness of the base according to the conditions of the plucking. At that time, the internal force of the column was $n_k, m_k$. To be pulled, usually arrange at least 8 bolts (4 bolts on each side of the column, symmetrical through the column as shown).

 

Pulling force on one side: $t = \frac {m_k} {l_b}+ \frac {n_K} 2 $ 2

 

Moment causes bending to the base: $m_b = T.L_1/8$

 

$L_b$: The distance between the two focus of the bolt area (on both sides)

 

$L_1 = P-0.5\phi$

 

$P$: The distance of the two symmetrical bolts through the column wings

 

$\phi$: bolt diameter

 

The minimum sole thickness is also calculated: $t_ {bd} = \sqrt {\frac {6m_b} {f\gamma_c}}$

 

$b$: column width

Calculation of anchor bolt

Compressed stress chart force: $r = 0.5byr_{b.loc}$

 

Pulling force on bolts: $t = r-n$

 

The total area of ​​anchor bolts on one side of the column: $a_ {bl} = t/f_ {tb}$

 

Check the cutting of anchor bolt:

 

Cutting capacity of all bolts: $[Q] = AF_ {vb}\gamma_b$

 

$A$: The area of ​​the bolt body (no thread)

Welding road bearing vertical, cutting force, moment bending simultaneously, checking under the conditions of welding stress stress:

 

$$\tau_{tđ}=\sqrt{\left(\frac M{W_{wf}+\frac N{A_{wf}}}\right)^2+\left(\frac Q{A_{wf}}\right)^2}\leqslant f_{wf}$$

 

$A_{wf}$: Total area of ​​column welding

 

$I_{wf}, W_ {wf}$: Moment inertia and the static moment of the total welding line with the axis of the column

Structural engineers can make a spreadsheet as attached.

Reference:

“Design of industrial steel frame” – MSc. Hoang Van Quang, MSc. Tran Manh Dung, MSc. Nguyen Quoc Cuong – Science & Technology Publishing House.