Two flat plates subjected to a tensile force P are connected together by means of double-strap butt joint as shown in Fig. 8.60. The force P is 250 kN and the width of the plate w is 200 mm. The rivets and plates are made of the same steel and the permissible stresses in tension, compression

Question

Two flat plates subjected to a tensile force P are connected together by means of double-strap butt joint as shown in Fig. 8.60. The force P is 250 kN and the width of the plate w is 200 mm. The rivets and plates are made of the same steel and the permissible stresses in tension, compression and shear are 70, 100 and 60 N/mm² respectively. Calculate:

(i) the diameter of the rivets;
(ii) the thickness of the plates;
(iii) the dimensions of the seam, viz., p, $p_{t}$ and m; and
(iv) the efficiency of the joint.

(i) the diameter of the rivets;
(ii) the thickness of the plates;
(iii) the dimensions of the seam, viz., p, [latex] p_{t} [/latex] and m; and
(iv) the efficiency of the joint.

Accepted Answer

Given P = 250 kN w = 200 mm

[latex] \sigma_{t}=70 N / mm ^{2} \quad au=60 N / mm ^{2} \quad \sigma_{c}=100 N / mm ^{2} [/latex].

Step I Diameter of rivets
There are five rivets subjected to double shear.
From Eq. (8.36),

[latex] P_{s}=2\left[\frac{\pi}{4} d^{2} au n ight] \quad ext { (for double shear) } [/latex] (8.36).

[latex] P_{s}=2\left[\frac{\pi}{4} d^{2} au n ight] [/latex].

[latex] ext { or } \quad\left(250 imes 10^{3} ight)=2\left[\frac{\pi}{4} d^{2}(60)(5) ight] [/latex].

∴ d = 23.03 or 25 mm (i).

Step II Thickness of the plates
In case of tension in the plate, the first thought is to investigate the section-XX passing through the centres of the three holes as shown in Fig. 8.60.
This section has minimum cross-sectional area.
However, if the main plate is to break at this section, the two-hole section, denoted by YY, must also fail before the joint will break. This failure at the section-YY may be due to shear failure in two rivets or crushing failure in the plate at two holes.
On the other hand, the plate might fail in tension at the two-hole section without affecting the three-hole section. Therefore, strength equations are written for two-hole sections rather than three-hole sections. Or,

[latex] (w-2 d) t \sigma_{t}=P [/latex].

[latex] ext { or }(200-2 imes 25) t(70)=250 imes 10^{3} [/latex].

t = 23.81 or 25 mm (ii)
Step III Pitch of rivets
The pitch of the rivets is given by

[latex] p=\frac{ ext { width of plate }}{ ext { number of rivets }}=\frac{200}{3}=66.67 mm [/latex].

or p = 65 mm
As mentioned in Section 8.19, the dimensions of the seam are as follows

m = 1.5d = 1.5(25) = 37.5 or 40 mm

[latex] p_{t}=0.6 p=0.6(65)=39 ext { or } 40 mm [/latex] (iii).

Step IV Efficiency of joint
From Eqs (8.36), (8.37) and (8.38)

[latex] P_{s}=2\left[\frac{\pi}{4} d^{2} au n ight] \quad ext { (for double shear) } [/latex] (8.36).

[latex] P_{t}=(p-d) t \sigma_{t} [/latex] (8.37).

[latex] P_{c}=d t \sigma_{c} n [/latex] (8.38).

[latex] P_{s}=2\left[\frac{\pi}{4} d^{2} au n ight]=2\left[\frac{\pi}{4}(25)^{2}(60)(5) ight] [/latex].

= 294 524.31 N (a).

[latex] P_{t}=(w-2 d) t \sigma_{t}=(200-2 imes 25)(25)(70) [/latex].

= 262 500 N (b).

[latex] P_{c}=d t \sigma_{c} n=25(25)(100)(5)=312500 N [/latex] (c).

From (a), (b) and (c), the lowest strength is 262 500 N. The strength of the solid plate is given by,

[latex] P=w t \sigma_{t}=200(25)(70)=350000 N [/latex].

Therefore, the efficiency of the joint is given by,

[latex] \eta=\frac{262500}{350000}=0.75 \quad ext { or } \quad 75 \% [/latex] (iv).

Question 8.19: Two flat plates subjected to a tensile force P are connected...

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