Design a tie bar if trial thickness of 30 cm was found safe. The or details are:
Width of lane = 3.5 m.
Allowable working stress in concrete = 1250 kg/cm²
Unit weight of concrete = 2400 kg/m³
Coefficient of friction = 1.5
Allowable bond stress in plain bars = 17.5 kg/cm²
Allowable tensile stress in plain bars = 1250 kg/cm².

There are various steps to design the tie bar:
Step 1: Area of steel per m width of joint
\(A_s=\frac{fWbh×10^{-2}}{s_s}\)
Where AS is the area of steel per metre of joint width to resist the frictional force in the slab
f is the coefficient of friction = 1.5
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 = 30 cm
SS is the allowable tensile stress in plain bars = 1250 kg/cm²
Substituting these in the equation,
\(A_s=\frac{1.5×2400×3.5×30×10^{-2}}{1250}\)=3.02 cm²/m
Step 2: Spacing of tie bar
Assuming the diameter of the plain bar as 1.2 cm, the cross-sectional area can be found as:
\(A=\frac{?}{4}×d^2=\frac{?}{4}×1.2^2=1.13 cm^2\)
The spacing of tie bar = \(\frac{A}{A_s} =\frac{1.13}{3.02}\)=0.3742 m≈37.42 cm
Therefore, the spacing to be provided is 38 cm.
Step 3: Length of the tie bar
L=\(\frac{2S_s A}{S_b P}\)
Where L is the length of the tie bar
S_S is the allowable tensile stress in plain bars = 1250 kg/cm²
A is the cross-sectional area of tie bar = 1.13 cm²
P is the perimeter of the tie bar = πd=π×1.2=3.77 cm
S_b is the allowable bond stress in plain bars = 17.5 kg/cm²
After substituting these in the equation,
\(L_t=\frac{2×1250×1.13}{17.5×3.77}\)=42.82 cm
The obtained length has to be increased by 10 cm to account for the loss of bond due to painting and an additional 5 cm as tolerance in placement. Therefore, the length of the tie bar would be
L = 42.82 + 10 + 5 = 57.82 ≈58 cm