Two tie-bar plates of a bridge structure, 250 mm wide and 20 mm thick, are to be connected by a double-strap butt joint as shown in Fig. 8.61. The rivets and the plates are made of steel. The permissible stresses in tension, shear and compression are 80, 60 and 120 N/mm² respectively.
(i) Determine the diameter of the rivet by using the following empirical relationship,

Question

[latex] d=6 \sqrt{t} [/latex].

where t is the plate thickness.
(ii) Determine the number of rivets by equating the strength of the plate with the strength of the rivets. Assume that shear resistance of one rivet in double shear is 1.875 times its resistance in single shear.
(iii) Show the arrangement of rivets.
(iv) Determine the efficiency of the joint.
Assume the following relationships:
Margin (m) = 1.5d
Transverse pitch [latex] \left(p_{t}\right) [/latex] = 2d
Thickness of strap = 0.625t

Accepted Answer

Given w = 250 mm t = 20 mm.

[latex] \sigma_{t}=80 N / mm ^{2} \quad au=60 N / mm ^{2} \quad \sigma_{c}=120 N / mm ^{2} [/latex].

Step I Diameter of rivets
The diameter of the rivet is given by,

[latex] d=6 \sqrt{t}=6 \sqrt{20}=26.83 \quad ext { or } \quad 27 mm [/latex] (i).

Step II Number of rivets
The shear resistance of one rivet in double shear is given by,

[latex] P_{s}=1.875\left[\frac{\pi}{4} d^{2} au ight] [/latex].

[latex] =1.875\left[\frac{\pi}{4}(27)^{2}(60) ight]=64412.47 N [/latex] (a).

Crushing resistance of one rivet is given by,

[latex] P_{c}=d t \sigma_{c}=27(20)(120)=64800 N [/latex] (b).

[latex] ext { From (a) and (b), } P_{s}

Therefore, shear strength of the rivet is the criterion of design. It is assumed that rivets are so arranged that there is only one rivet in the outer row. The tensile strength of the plate in the outer row is given by,

[latex] P_{t}=(w-d) t \sigma_{t}=(250-27)(20)(80) [/latex].

= 356 800 N (c)
Equating the strength of the plate with the shear strength of n rivets

356 800 = n (64 412.47)
∴ n = 5.54 or 6 rivets (ii).

Step III Arrangement of rivets
The arrangement of six rivets in three rows is illustrated in Fig. 8.61. This type of arrangement in diamond shape is called Lozenge joint. The dimensions given in the figure have the following numerical values,
m = 1.5d = 1.5 (27) = 40.5 or 45 mm

[latex] p_{t}=2 d=2(27)=54 ext { or } 55 mm [/latex].

[latex] t_{1}=0.625 t=0.625(20)=12.5 mm [/latex].

From the figure,
m + p + p + m = 250 or 45 + 2p + 45 = 250
∴ p = 80 mm.

Step IV Efficiency of joint
In order to calculate the efficiency of the joint, we will proceed from the outer row of rivets to the inner row and fi nd out the weakest cross-section of the joint. Let us consider the failure behaviour of three cross-sections denoted by AA, BB and CC.
(i) The joint can fracture along the section-AA without affecting the rivets in the middle and inner rows.
(ii) The joint cannot fracture along the section- BB without shearing one rivet in double shear. This is the rivet in the outer row.
(iii) The joint cannot fracture along the section- CC, without shearing three rivets in double shear. These are the rivets in outer and middle rows.
Based on these assumptions the strength of the joint along three sections is calculated.
Along the section-AA,

[latex] ext { strength }=(w-d) t \sigma_{t}=(250-27)(20)(80) [/latex].

356 800 N (a)
Along the section-BB,

[latex] ext { strength }=(w-2 d) t \sigma_{t}+1 imes P_{s} [/latex].

= (250 – 2 × 27) (20) (80) + 64 412.47
= 378 012.47 N (b).

Along the section-CC,

[latex] ext { strength }=(w-3 d) t \sigma_{t}+3 imes P_{s} [/latex].

= (250 – 3 × 27)(20)(80)
+ 3 × 64 412.47
= 463 637.41 N (c).

[latex] ext { Shear resistance of all rivets }=6 imes P_{s} [/latex].

= 6 × 64 412.47 = 386 474.8 N (d)
From (a), (b), (c), and (d), the lowest strength of the joint is along the section-AA.

[latex] ext { Strength of solid plate }=w t \sigma_{t}=250(20)(80) [/latex].

= 400 000 N
Therefore

[latex] \eta=\frac{356800}{400000}=0.892 \quad ext { or } \quad 89.2 \% [/latex] (iv).

Question 8.20: Two tie-bar plates of a bridge structure, 250 mm wide and 20...

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