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How does ideCAD calculate steel members' shear strength according to AISC 360-16?


Tip
  • The shear strength of steel elements is calculated automatically according to AISC 360-16.


Symbols

E = Modulus of elasticity of steel = 29,000 ksi (200 000 MPa)
Fy = Specified minimum yield stress
h = clear distance between flanges less the fillet at each flange, in. (mm)
Vn = Nominal shear strength, kips (N)
tw = thickness of web, in. (mm)
λw = Limiting width-to-thickness parameter


In design per share, it is assumed that shear force is completely carried by the web, no interaction between bending and shear occurs, yield stress in shear is,

Mathinline
body$$ \normalsize f_y \cong 0.60 F_y $$

image-20240126-073334.png
  • The design shear strength =

    Mathinline
    body$$ \normalsize \phi_vV_n $$

  • The design shear strength =

    Mathinline
    body$$ \normalsize V_n/\Omega_v $$

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body$$ \normalsize \phi_v=0.90 \;\; (LRFD) \;\;\;\;\;\;\;\;\;\; \Omega_v=1.67 \;\; (ASD) $$

The nominal shear strength, Vn, shall be determined depending on the slenderness defined by (h/tw) according to the limit state of shear yielding and shear buckling.

  • If

    Mathinline
    body--uriencoded--$$ \normalsize \lambda_w=\dfrac%7Bh%7D%7Bt_w%7D \le \lambda_%7Bpv%7D=2.24 \sqrt%7B \dfrac%7BE%7D%7BF_y%7D %7D $$
    satisfies, shear yielding controls.

  • If

    Mathinline
    body--uriencoded--$$ \normalsize \lambda_w=\dfrac%7Bh%7D%7Bt_w%7D > \lambda_%7Bpv%7D=2.24 \sqrt%7B \dfrac%7BE%7D%7BF_y%7D %7D $$
    satisfies, shear buckling controls.