The thickness of a PCC pavement is 26 cm and width of lane is 3.5 m. Find length of a tie bar system along longitudinal joints if allowable working stress in concrete is 1250 kg/cm², unit weight of concrete is 2400 kg/m³, coefficient of friction is 1.4, allowable bond stress in deformed bars is 24.6 kg/cm² and allowable tensile stress in deformed bars is 2000 kg/cm².

The procedure given by IRC to find the spacing of a tie bar system is carried out as below.
Step 1: Cross-sectional area of the tie bar
\(A_s=\frac{fWbh×10^{-2}}{s_s}\)
Where A_s is the cross-sectional area
f is the coefficient of friction = 1.4
W is the unit weight of concrete = 2400 kg/m³
b is the width of the slab = 3.5 m
h is the thickness of the slab = 26 cm
S_sis the allowable tensile stress in deformed bars = 2000 kg/cm²
Substituting these in the equation,
\(A_s=\frac{1.4×2400×3.5×26×10^{-2}}{2000}\)=1.53 cm²
Step 2: Length of the tie bar
\(L_t=\frac{S_s d}{2S_b}\)
Where L_t is the length of the tie bar
S_s is the allowable tensile stress in deformed bars = 2000 kg/cm²
d is the diameter of the tie bar and it is assumed as 1.2 cm
S_b is the allowable bond stress in deformed bars = 24.6 kg/cm²
After substituting these in the equation,
\(L_t=\frac{2000×1.2}{2×24.6}=48.78?50\) cm
Step 3: Spacing of the tie bar
No. of bars = \(\frac{Total \,area}{Area \,of \,one \,bar}=\frac{1.53}{\frac{?}{4}×1.2^2}\)=1.35
Spacing = \(\frac{1 \,m \,slab \,length}{No. \, of \,bars}=\frac{100 cm}{1.35}=74.07?70\) cm