Question

Calculate the strength of joint of a zigzag double bolted lap joint as shown in figure. Assume Fe 410 grade plate and grade 4.6 bolts of diameter 20 mm and 8 mm thick plates.
$(n_s=0)$

Answer 1

Strength of 20 mm diameter bolt

(a) In single shear
$V_{dsb}=\frac{f_u}{\sqrt{3}{\gamma }_{m_b}}(n_n.A_{nb})$
$=\frac{400}{\sqrt{3}\times 1.25}\times (0.78\times \frac{\pi }{4}\times {20}^2)=45.3kN$

(b) In bearing on 8 mm plate

$V_{dpb}=\frac{2.5k_b.d.t.f_u}{{\gamma }_{m_1}}$
$k=min[\frac{50}{3\times 22},\frac{100}{3\times 22}-0.25,\frac{400}{410},1]=0.75$
$V_{dpb}=\frac{2.5\times 75\times 20\times 8\times 410}{1.25}=98.4kN$

Consider the 100 mm length of the joint shaded as shown in figure which represents typical condition. Line 1 - 1 has one bolt hole 22 mm in diameter. Line 2 - 2 bolt line 1 - 1 must fall with it. Hence, the resistance of plate in tearing along 2 - 2 = strength of net section of plate 2 - 2 plus one bolt strength. Hence for plate section 1 - 1 is critical

Net strength of plate
$=\frac{0.9\times (100-22)\times 8\times 410}{1.25}=184kN$

Strength of two bolts,
$=2\times 45.3=90.6kN$

Strength of joint = 90.6 kN