Generally, shear is not a problem in steel design, because the webs of rolled shapes are capable to resist large shearing forces. The shear might play important role under some conditions:
Tall buildings: if columns in the upper floors are offset with the respect to the columns below them. A horizontal beam than carries weight of the upper beam and a great shear might be present.
From mechanics of material, shear is a stress or the forces resisting shear or attempts to shear a member. Since shear is a stress, its average value over the cross-section is $\tau = V/A$. But most of the shear is carried close to normal axis. At any point along the $y$ axis:
$$ \begin{equation} \tau(y) = \frac{V S_x}{I_x b} \label{shear:eq1} \end{equation} $$where $V$ is shear force (load), $S_x$ is statical moment of portion area at considered point against centroidal axis, $I_x$ moment of inertia of the whole section, $b$ is thickness/width of the section at the considered point (discontinuities of width of sections cause discontinuities of the shear forces along the height).
The assessment according to the Eurocode is
$$ \begin{equation} V_{Ed} \le V_{c, Rd}, \end{equation} $$where $V_{Ed}$ is the design shear force from load and $V_{c, Rd}$ is design shear resistance. For $V_{c, Rd}$ is either $V_{pl,Rd}$ taken if plastic design is considered, or $V_{c,Rd}$ when elastic design is requested.
assuming no torsion is involved, $A_V$ is the shear area. $A_{V}$ should be taken according to the code (complicated): for commonly used profiles the parts of sections parallel to the force can be taken.
If no shear buckling needs to be considered (chapter 5 of EN 1993-1-5), then for verifying the design elastic shear resistance $V_{c,Rd}$, the assessment is being made by comparing stresses at the critical point of the cross-section:
$$ \frac{\tau_{Ed}}{f_y / (\sqrt{3}\cdot \gamma_{M0})} \le 1.0 $$where $\tau_{Ed}$ is determined for $V_{Ed}$ from $(\ref{shear:eq1})$. For I- or H- sections the shear stress in the web might be taken as
$$ \begin{equation} \tau_{Ed} = \frac{V_{Ed}}{A_w} \end{equation} $$if area of flanges $A_f = b_f t_f$ is large compared to area $A_w = h_w t_w$ of the web: $A_f / A_w \ge 0.6$.