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**Design Strength**

**28.1.1 Butt Joints**

Butt joints generally fail in tension.

Module 4 Lesson 28 Fig.28.1

**Fig.28.1.**

The design strength of butt joint in tension is,

\[{T_d}={\sigma _t}{l_e}{t_e}\]

where,

\[{t_e}={\rm{effective throat dimension of the weld}}\]

\[{l_e}={\rm{effective length of the weld}}\]

**28.1.2 Fillet Joints**

Fillet joints generally fail in shear.

**Fig. 28.2.**

The design strength of fillet joint in shear is,

\[{V_d}={\sigma _S}{l_e}{t_e}\]

where,

\[{\sigma _S}={\rm{ allowable shear stress of the weld}}\]

\[{t_e}={\rm{effective throat dimension of the weld}}\]

\[{l_e}={\rm{effective length of the weld}}\]

**28.1.3 Circular fillet weld subjected to torsion**

**Fig. 28.3.**

Maximum shear stress occurs at the throat area and is given by,

\[{\tau _{\max }}={{T\left( {0.5d + {t_e}} \right)} \over J}\]

where,

\[d={\rm{outer diameter of the shaft }}\]

\[J={\rm{polar moment of area of the throat section}}={\pi\over {32}}\left[ {{{\left( {d + 2{t_e}} \right)}^4} - {d^4}} \right]\]

Now the maximum shear stress \[{\tau _{\max }}\] should not be more than the allowable shear stress of the weld. Therefore at the limiting case,

\[{\sigma _s}={\tau _{\max }}={{T\left( {0.5d + {t_e}} \right)} \over J}\]

**Example 1**

Find the length of the fillet weld required for the following connection. Both plates are 10mm thick. Assume allowable shear stress of the weld is 70MPa.

**Solution**

Effective length of the fillet weld \[{l_e}=2l\]

Effective throat dimension \[{t_e}={t \over {\sqrt 2 }}\] [ t is the thickness of the plate]

Therefore, design strength of the weld,

\[{V_d}={\sigma _S}{l_e}{t_e}=70 \times {10^6} \times 2l \times {{10 \times {{10}^{ - 3}}} \over {\sqrt 2 }}=19.8 \times {10^5}l\]

Equating V_{d} with the applied load,

\[19.8 \times {10^5}l = 75 \times {10^3} \Rightarrow 37.87{\rm{ mm}}\]

**Example 2**

Two plates of thickness 16mm and 12mm are to be connected by a groove weld. The joint is subjected to a tensile load of 300kN. Assume allowable tensile stress of the weld is 250MPa. Determine the length of the weld required for the following cases.

(*i*) Single V groove joint (Figure 28.5a)

(*ii*) Double V groove joint (Figure 28.5b)

**Fig. 28.5.**

**Solution**

(*i*) case 1

Effective throat dimension \[{t_e}={5 \over 8} \times {\rm{thickness of thinner plate}}={5 \over 8} \times {\rm{12}}=7.5{\rm{ mm}}\]

\[{T_d}={\sigma _t}{l_e}{t_e}\]

\[\Rightarrow 300 \times {10^3} = 250 \times {10^6} \times {l_e} \times 7.5 \times {10^{ - 3}}\]

\[\Rightarrow {l_e}=160{\rm{ mm}}\]

(*ii*) case 2

Effective throat dimension \[{t_e}={\rm{thickness of thinner plate}}=12{\rm{ mm}}\]

\[{T_d}={\sigma _t}{l_e}{t_e}\]

\[\Rightarrow 300 \times {10^3}=250 \times {10^6} \times {l_e} \times 12 \times {10^{ - 3}}\]

\[\Rightarrow {l_e} = 100{\rm{ mm}}\]

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