The document analyzes maximum shear stress in parallel and transverse fillet welds under various conditions, including the required angles for maximum stress. It derives equations for shear stress, calculates allowable loads per millimeter for both types of welds, and concludes that the strength of transverse fillet welds is 1.17 times greater than that of parallel welds. The findings emphasize the importance of understanding shear stress implications in mechanical design.

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

4. 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)

5. 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)

6. 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,
τmax = P(Sin45°+Cos45°)/hl
=1.414P/hl=P/(1/4.414)hl=P/0.707hl
Or τmax =P/0.707hl
The above equation is same as Eq. (8.5). Substituting (l=1mm) in
Eq.(8.13), the allowable load Pall per mm length of the weld is give by
Pall = 0.707hτmax = 0.707(8)(94)
=531.66 N/mm

7. 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.

8. 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 Ps
and the normal force is Pn.
Considering equilibrium of vertical force, [Fig. (b)and(c)],
2P=2PsSinθ+2PnCos θ
P= PsSinθ+PnCos θ (a)
Since the resultant of Ps and Pn is vertical, their horizontal
components must be equal and opposite.
Therefore,
PsCosθ = Pn Sinθ
Or,
Pn = Ps Cosθ/Sin θ (b)
Substituting Eq. (b) in Eq. (a).
P=Ps sin θ+ (Ps cos θ cos θ/sin θ) (c)

9. Multiplying both sides of the above equation by sin θ.
P sin θ = Ps sin² θ+Ps cos² θ
= sin² θ+ cos² θ
Or,
Ps = P sinθ (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/(Sinθ+PnCos 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=(Ps/t’l)

10. From Eq. (d) and (e)
τ = P sinθ(Sinθ+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/dθ=cos θ dv/dθ=(cos θ-sin θ

11. 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 2θ+cos 2θ (h)
From (g) and (h)
sin 2θ+cos 2θ
sin 2θ=-cos 2θ
tan 2θ=-1
2θ= 135°
Or,
θ=67.5° (8.15)

12. 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,
τmax= Psin(67.5°)[sin(67.5°)+cos(67.5°)]/hl
τmax= 1.21P/hl (8.16)
Substituting (l=1mm) in Eq. (8.16), the allowable load Pall per
mm length of transverse filet weld is given by,
Pall= hτmax/1.21
Or Pall= 0.8284hτmax (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).
Pall= 0.8284hτmax
622.96 N/mm

13. 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,
Pall for transverse weld/Pall for parallel weld= 0.8284hτmax/
0.707hτmax = 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.