Prove shear stirrup spacing or Ld formula

Proof of $L_d=\dfrac{0.87f_y\,\phi}{4\,\tau_{bd}}$

Tension in a bar of diameter $\phi$ at design stress $0.87f_y$:

$$T=0.87f_y\times\frac{\pi\phi^2}{4}$$

Resisting bond force along embedded length $L_d$ at average design bond stress $\tau_{bd}$ on the surface area $\pi\phi L_d$:

$$F=\tau_{bd}\,\pi\phi\,L_d$$

Equating $T=F$:

$$0.87f_y\frac{\pi\phi^2}{4}=\tau_{bd}\pi\phi L_d\ \Rightarrow\ L_d=\frac{0.87f_y\,\phi}{4\,\tau_{bd}}$$

Proof of $S_v=\dfrac{0.87f_yA_{sv}d}{V_s}$

Consider a diagonal crack at 45$^\circ$ in a beam of effective depth $d$. The horizontal length of the crack is about $d$ (for a vertical stirrup the crack crosses $d/s_v$ stirrups).

        crack at 45 deg
   ____________/_________
  |           /  |  |  |
  |          /   |  |  |  stirrups, spacing Sv
  |_________/____|__|__|
  |<--- d --->|

Number of stirrups crossing the crack $=\dfrac{d}{S_v}$. Each stirrup (two legs, total area $A_{sv}$) carries design tension $0.87f_yA_{sv}$ at yield. The vertical component of the force carried by the stirrups equals the shear $V_s$ assigned to them (vertical stirrups):

$$V_s=0.87f_yA_{sv}\times\frac{d}{S_v}\ \Rightarrow\ \boxed{S_v=\frac{0.87f_yA_{sv}d}{V_s}}$$

where $V_s=V_u-\tau_cbd$ is the shear to be carried by the stirrups.