GATE Past Year Questions: Bolted, Riveted & Welded Joint | Design of Machine Elements - Mechanical Engineering PDF Download

P = .707 t × L_e × τ_S.

Given shear strength of material = 200 MPa
FOS = 200/58.94 = 3.4

For plates, σ = P/A

P = 167kN

For rivet,

∴ P_s = 100 × d × t × 3 = 15 kN
and P_c = 150 × d × t × 3 = 22.5 kN
∴ p = max (15, 22.5) = 22.5 k N

Given : Number of rivets = 4,
Diameter of each rivet , d = 6 × 10^-3 m Direct load on the bracket = 1000 N since there is no eccentricity
∴ Load on each rivet = 1000/4 = 250 N

d/P = 0.25

Pre-tensioning increases the stiffness of the bolt.

P_P = P/n = P/3 = 2kN
r_A = r_C = 50, (P_S)_A = (P_S)_C
r_B = 0 ⇒ (P_S)_B = 0

Q_A = 180°, Q_C = 0°
Resultant shear force
(R_C) = P_P + (P_S)_C.
R_C = 2 + 15 = 17 KN

σ_y = 200 MPa
P = 19 × 10³ N

M10

Resultant load on P = F_sec – F_prim
= 20 – 2 = 18 kN

Let F₁ and F₂ resist the 4 kN eccentric oad. The primary shear load is due to a direct shear stress of 4kN, which is equally divided amongst the two bolts, i.e. 2 kN,
∴ F_primary on P = 2 kN

The secondary load is to resist the moment produced by the eccentric load 4 kN about the C.
G. of the joint
Moment M = 4 k N × 2000 mm
= 8000
Nm Now this moment must be balanced by a moment produced by equal and opposite forces i.e. secondary loads on the bolts i .e. F × 400 = 800 N m
⇒ F = 20 k N

Shear load at A and C


(τ_max)_A,C = 242.5MPa
(τ_max)B = 42.44MPa

Tightening force = 2200 N
Pitch = 4 mm = 0.004 m

= 1.4 N-m