MAXIMUM SHEAR STRESS IN PARALLEL WELD AND TRANSVERSE FILLET WELD

MDID is nearly all heart in mechanical engineer and this is for them .
this will discbribe abou maximum stress in parallel and transverse fillet weld.

1. GUJRAT POWER ENGINEERING & RESEARCH INSTITUTE NAME :- THAKKAR VIJAYKUMAR B. ENROLLMENT NO:- 161043119012 SUB:- MACHINE DESIGN & INDUSTRIAL DRAFTING GUIDED BY:- PARIN PATEL SIR

2. MAXIMUM SHEAR STRESS IN PARALLEL FILLET WELD A double parallel fillet weld of equal legs subjected to a force of (2P) is shown in Fig(a). It is required to find out the inclination ( θ ) of the plane in the weld, where shear stress is included and also, the magnitude of the maximum shear stress. The effect of bending is to be neglected. The free body diagram of forces acting on the vertical plate with two welds cut symmetrically is shown in Fig(b). The symbol x (cross) indicates a force perpendicular to the plane of paper, which goes away from the observer. The symbol .(dot) indicates a force perpendicular to the plane of paper, which is towards the observer. The welds are cut at an angle θ with the horizontal. t ’ is the width of plane that is inclined at angle θ with the horizontal

3. In the triangle ABC (Fig c) AB=BC=h ∴ ∠ECD=45 ° = DE  |  BC = BC=EC+BE = BE+DE (DE=EC) = BD Cos θ +BD Sin θ = BD (Cos θ +Sin θ ) Or. h=t’(Sin θ +Cos θ ) Therefore, t’= h/(Sin θ +Cos θ ) (8.11)

4. The area of the weld in the plate inclined at angle θ with horizontal is ( t’l ). Therefore, the shear stress in this plane is given by, τ = P/ t’l Substituting Eq. (8.11) in the above expression, P(Sin θ +Cos θ ) τ = hl In order to find out the plane with maximum shear stress, differentiate τ with respect to θ and set the derivative equal to zero. ∂τ/∂ θ = P/hl (Sin θ +Cos θ ) = 0 Cos θ -Sin θ = 0 Cos θ =Sin θ Tan θ = 1 Therefore, θ =45 ° (8.12)

5. The condition for plane with maximum shear stress is ( θ =45 ° ). Substituting this value θ in Eq. (a), the maximum shear stress is given by, τ max = P(Sin45 °+ Cos45 °) /hl =1.414P/hl=P/(1/4.414)hl=P/0.707hl Or τ max =P/0.707hl The above equation is same as Eq. (8.5). Substituting (l=1mm) in Eq.(8.13), the allowable load P all per mm length of the weld is give by P all = 0.707h τ max = 0.707(8)(94) =531.66 N/mm

6. MAXIMUM SHEAR STRESS IN TRANSVERSE FILLET WELD A double shear transverse fillet weld of equal legs is subjected to force (2P) as shown in Fig. 8.15(a). It is required to find out the inclination ( θ ) of the plane in the weld where the maximum shear stress is incuded and also, the magnitude of the maximum shear stress. The effect of bending is to be negleted .

7. The free body diagram of force acting on the vertical plate with two symmetrical cut welds is shown in Fig.(b). The shear force is P s and the normal force is P n . Considering equilibrium of vertical force, [Fig. (b)and(c)], 2P=2P s Sin θ +2P n Cos θ P= P s Sin θ + P n Cos θ (a) Since the resultant of P s and P n is vertical, their horizontal components must be equal and opposite. Therefore, P s Cos θ = P n Sin θ Or, P n = P s Cos θ /Sin θ (b) Substituting Eq. (b) in Eq. (a). P=P s sin θ + (P s cos θ cos θ /sin θ ) (c)

8. Multiplying both sides of the above equation by sin θ . P sin θ = P s sin² θ +P s cos² θ = sin² θ + cos² θ Or, P s = P sin θ (d) From Eq. (8.11), the width t’ of the plane in the weld that is inclined at angle θ with the horizontal, is ( t’l ). Therefore, the shear stress in this plane is given by, t’= h/(Sin θ + P n Cos θ ) (e) The area of the weld, in a plane that is inclined at angle θ with horizontal, is ( t’l ). Therefore, the shear stress in this plane is given by, τ =(P s / t’l )

9. From Eq. (d) and (e) τ = P sin θ (Sin θ +Cos θ )/hl (f) In order to find out the plane with the maximum shear stress, differentiate τ with respect to θ and set the differentiate equal to zero. ∂τ/∂ θ =0 (p/hl)(∂/∂ θ )[sin θ (sin θ + cos θ )] = 0 Or, ∂/∂ θ [sin θ (sin θ + cos θ )] = 0 (g) Since, d/ dx ( uv )=u( dv / dx )+v( du+dx ) x= θ u=sin θ v=(sin θ + cos θ ) du/d θ = cos θ dv /d θ =( cos θ -sin θ

10. Substituting, ∂/∂ θ [sin θ (sin θ + cos θ )] = sin θ ( cos θ -sin θ )+(sin θ + cos θ ) cos θ = 2sin θ cos θ +(cos² θ -sin² θ ) = sin 2 θ + cos 2 θ (h) From (g) and (h) sin 2 θ + cos 2 θ sin 2 θ =- cos 2 θ tan 2 θ =-1 2 θ = 135 ° Or, θ =67.5 ° (8.15)

11. The condition or the plane with the maximum shear stress is ( θ =67.5 ° ) Substituting the above value θ in Eq. (f), the maximum shear stress is given by, τ max = Psin (67.5 ° )[sin(67.5 ° )+ cos (67.5 ° )]/hl τ max = 1.21P/hl (8.16) Substituting (l=1mm) in Eq. (8.16), the allowable load P all per mm length of transverse filet weld is given by, P all = h τ max /1.21 Or P all = 0.8284h τ max (8.17) Suppose it is required to find out allowable load per mm length of transverse fillet weld for the permissible shear stress of 94 N/mm² and the leg dimension of 8 mm. From eq. (8.17). P all = 0.8284h τ max 622.96 N/mm

12. It is observed from Eq. (8.14) and (8.17) that the allowable load for a transverse fillet weld is more than that of a parallel weld. Or, P all for transverse weld/P all for parallel weld= 0.8284h τ max / 0.707h τ max = 1.17 The strength of transverse fillet weld is 1.17 times of the strength of parallel fillet weld. As mentioned in section 8.8, many times transverse fillet weld is designed by using the same equations of parallel fillet weld. Such design is on the safer side the additional advantage of simple calculation.