**D. The Shear Resistance of Thin Fillet Welds.**

Effective parts of welds, (α = *<sup>p</sup> <sup>q</sup>* = 0.521):

$$l\_1 = h\_2 / \sin \theta\_2 = 80 / 0.6431 = 124.4 \text{ mm},$$

$$l\_2 = p\_{2,eff} = t\_w + 2r + 7t\_f \\ f\_{y0} / f\_{y2} = 6.5 + 2 \cdot 12 + 7 \cdot 11 \cdot 355 / 355 = 107.5 \text{ mm} \\ \therefore \, b\_2 = 60 \text{ mm}$$

Assumed *l*<sup>2</sup> = 60 mm.

$$\begin{aligned} b\_{j, \text{rad}} &= b\_2 - 2a\_w = 60 - 2 \cdot 3 = 54 \text{ mm}, \\ l\_3 = l\_{i, \text{rad}} / \sin \theta\_i &= (1 - \alpha) l\_i / \sin \theta\_i = (1 - 0.521) 60 / 0.7698 = 37.3 \text{ mm}, \\ l\_4 = p\_{1, \text{eff}} &= t\_w + 2r + 7t\_f \\ l\_5 &= 100 \text{ mm}, \end{aligned}$$

Assumed *l*<sup>4</sup> = 50 mm.

 $l\_5 = \frac{q}{(1 + \tan \theta\_2 / \tan \theta\_1) \cos \theta\_2} = \frac{40.6}{(1 + 0.8397 / 1.2062)0.7658} = 31.3 \text{ mm}/\text{s}$ 
$$l\_6 = b\_1 = 50 \text{ mm}.$$

Cross-section areas of effective lengths of welds. Assumed the thickness of welds *aw*1=*aw*2=*aw*3=*awb*=*aw*4=*aw*5=*aw*6= 3.0 mm,

$$A\_1 = l\_1 a\_{w1} = 124.4 \cdot 3 = 373.2 \text{ mm}^2, \; A\_2 = l\_2 a\_{w2} = 60 \cdot 3 = 180.0 \text{ mm}^2,$$

$$A\_3 = l\_3 a\_{w3} . 37.3 \cdot 3 = 111.9 \text{ mm}^2, \; A\_{2, \text{rad}} = l\_{2, \text{rad}} a\_{wb} . 54 \cdot 3 = 162.0 \text{ mm}^2,$$

$$A\_4 = l\_4 a\_{w4} = 50 \cdot 3 = 150.0 \text{ mm}^2, \; A\_5 = l\_5 a\_{w5} = 31.3 \cdot 3 = 93.9 \text{ mm}^2,$$

$$A\_6 = l\_6 a\_{w6} . 50 \cdot 3 = 150.0 \text{ mm}^2,$$


$$
\Delta K\_1 = \alpha K\_1 \sin \theta\_1 = 0.521 \cdot 103.2 \cdot 0.7698 = 41.4 \text{ kN/s}
$$

*red*Δ*K*<sup>2</sup> = *K*<sup>2</sup> sin θ<sup>2</sup> − α*K*<sup>1</sup> sin θ<sup>1</sup> =136.1 · 0.6431 − 0.521 · 103.2 · 0.7698 = 46.1 kN.

Loads in effective lengths parallel to the chord.

*P*1 - = (*K*<sup>2</sup> cos θ<sup>2</sup> + *K*<sup>1</sup> cos θ1)*A*1/ - *A* =(136.1 · 0.7658 + 103.2 · 0.6382)373.2/1462.2 = 43.4 kN, *P*2 - = (*K*<sup>2</sup> cos θ<sup>2</sup> + *K*<sup>1</sup> cos θ1)*A*2/ - *A* =(136.1 · 0.7658 + 103.2 · 0.6382)180.0/1462.2 = 20.9 kN, *P*3 - = (*K*<sup>2</sup> cos θ<sup>2</sup> + *K*<sup>1</sup> cos θ1)*A*3/ - *A* =(136.1 · 0.7658 + 103.2 · 0.6382)111.9/1462.2 = 13.0 kN, *P*4 - = (*K*<sup>2</sup> cos θ<sup>2</sup> + *K*<sup>1</sup> cos θ1)*A*4/ - *A* =(136.1 · 0.7658 + 103.2 · 0.6382)150.0/1462.2 = 17.4 kN, *Pb* - = (*K*<sup>2</sup> cos θ<sup>2</sup> + *K*<sup>1</sup> cos θ1)*Aj*,*red*/ - *A* =(136.1 · 0.7658 + 103.2 · 0.6382)162/1462.2 = 18.8 kN, *P*- <sup>5</sup> = Δ*K*<sup>1</sup> · *A*<sup>5</sup> · sin θ2/(2*A*<sup>5</sup> + *A*6) = 41.4 · 93.9 · 0.6431/(2 · 93.9 + 150) = 7.4 kN, *P*- <sup>6</sup> = Δ*K*<sup>1</sup> · *A*<sup>6</sup> · sin θ2/(2*A*<sup>5</sup> + *A*6) = 41.4 · 150 · 0.6431/(2 · 93.9 + 150) = 11.8 kN.

Loads in effective lengths perpendicular to the chord.

$$P\_1^{\prime \prime} = 0\_\prime$$

$$P\_2' = \operatorname{rd} \Delta K\_2 \cdot A\_2 / \left( A\_2 + A\_{2, \text{rd}} \right) = 46.1 \cdot 180 / (180 + 162) = 24.3 \text{ kN},$$

$$P\_3' = (1 - a) \Delta K\_1 \cdot A\_3 / (2A\_3 + A\_4) = (1 - 0.521) \cdot 41.4 \cdot 111.9 / (2 \cdot 111.9 + 150) = 5.9 \text{ kN},$$

$$P\_4'' = (1 - a) \Delta K\_1 \cdot A\_4 / (2A\_3 + A\_4) = (1 - 0.521) \cdot 41.4 \cdot 150 / (2 \cdot 111.9 + 150) = 8.0 \text{ kN},$$

$$P\_b'' = \operatorname{rd} \Delta K\_2 \cdot A\_{2 \operatorname{rd}} / \left( A\_2 + A\_{2 \operatorname{rd}} \right) = 46.1 \cdot 162 / (180 + 162) = 21.8 \text{ kN},$$

$$P\_5' = \Delta K\_1 \cdot A\_5 \cos \theta\_2 / \left( 2A\_5 + A\_6 \right) = 41.4 \cdot 93.9 \cdot 0.7658 / (2 \cdot 93.9 + 150) = 8.8 \text{ kN},$$

$$P\_6'' = \Delta K\_1 \cdot A\_6 \cos \theta\_2 / \left( 2A\_5 + A\_6 \right) = 41.4 \cdot 150 \cdot 0.7658 / (2 \cdot 93.9 + 150) = 14.1 \text{ kN}.$$

Stresses on the throat section of a fillet welds from the force component parallel to the chord: - longitudinal welds between the lower (overlapped) brace and the chord:

$$
\sigma' = 0,
$$

$$
\sigma'\_{\perp} = \tau'\_{\perp} = 0,
$$

$$
\tau'\_{II} = \frac{P'\_1}{a\_{w1}l\_1} = \frac{43.4 \cdot 10^3}{3.0 \cdot 124.4} = 116.3 \text{ MPa},
$$


$$\sigma' = \frac{P\_2'}{a\_{w2}l\_2} = \frac{20.9 \cdot 10^3}{3.0 \cdot 60} = 116.1 \text{ MPa}$$

$$\sigma'\_{\perp} = \sigma' \sin\frac{\theta\_2}{2} = 116.1 \cdot 0.3422 = 39.7 \text{ MPa}$$

$$\tau'\_{\perp} = \sigma' \cos\frac{\theta\_2}{2} = 116.1 \cdot 0.9396 = 109.1 \text{ MPa}$$

$$\tau'\_{ll} = 0,$$

.


$$
\sigma' = 0,
$$

$$
\sigma'\_{\perp} = \tau'\_{\perp} = 0,
$$

$$
\tau'\_{II} = \frac{P'\_3}{a\_{\text{B}3}l\_3} = \frac{13.0 \cdot 10^3}{3.0 \cdot 37.3} = 116.2 \text{ MPa},
$$


$$
\sigma' = \frac{P\_b'}{a\_{wb} b\_{j.rel}} = \frac{18.8 \cdot 10^3}{3.0 \cdot 54} = 116.0 \text{ MPa},
$$

$$
\sigma'\_{\perp} = \sigma' \cos \frac{\theta\_2}{2} = 116.0 \cdot 0.9396 = 109.0 \text{ MPa},
$$

$$
\tau'\_{\perp} = \sigma' \sin \frac{\theta\_2}{2} = 116.0 \cdot 0.3422 = 39.7 \text{ MPa},
$$

$$
\tau'\_{H} = 0,
$$


$$\sigma' = \frac{P\_4'}{a\_{w4}l\_4} = \frac{17.4 \cdot 10^3}{3.0 \cdot 50} = 116.0 \text{ MPa},$$

$$\sigma\_\perp' = \sigma' \sin\frac{\theta\_1}{2} = 116.0 \cdot 0.3849 = 44.6 \text{ MPa},$$

$$\tau\_\perp' = \sigma' \cos\frac{\theta\_1}{2} = 116.0 \cdot 0.9050 = 105.0 \text{ MPa},$$

$$\tau\_{II}' = 0,$$


$$
\sigma' = 0,
$$

$$
\sigma'\_{\perp} = \tau'\_{\perp} = 0,
$$

$$
\tau'\_{II} = \frac{P'\_5}{a\_{\text{tr}5}l\_5} = \frac{7.4 \cdot 10^3}{3.0 \cdot 31.3} = 78.8 \text{ MPa},
$$


$$\sigma' = \frac{P\_6'}{a\_{w6}l\_6} = \frac{11.8 \cdot 10^3}{3.0 \cdot 50} = 78.7 \text{ MPa}$$

$$\sigma\_\perp' = -\sigma' \cos \frac{\theta\_1 + \theta\_2}{2} = -78.7 \cdot 0.7049 = -55.5 \text{ MPa}$$

$$78.8 \cdot 0.7093 = 55.9 \text{ MPa}$$

$$\pi\_{II}' = 0$$

Stresses on the throat section of a fillet welds from the force component perpendicular to the chord:


$$P\_1'' = 0,$$

$$
\sigma\_\perp'' = \pi\_\perp'' = 0,$$

$$
\pi\_H'' = 0,$$


$$
\sigma'' = \frac{P\_2''}{a\_{w2}l\_2} = \frac{24.3 \cdot 10^3}{3.0 \cdot 60.0} = 135.0 \text{ MPa},
$$

$$
\sigma\_{\perp}'' = -\sigma'' \cos \frac{\theta\_2}{2} = -135.0 \cdot 0.9396 = -126.8 \text{ MPa},
$$

$$
\tau\_{\perp}'' = \sigma'' \sin \frac{\theta\_2}{2} = 135.0 \cdot 0.3422 = 46.2 \text{ MPa},
$$

$$
\tau\_{II}'' = 0,
$$


$$\sigma'' = \frac{P\_3''}{a\_{\text{m}3}l\_3} = \frac{5.9 \cdot 10^3}{3 \cdot 37.3} = 52.7 \text{ MPa}\_{\text{\textdegree}}$$

$$\sigma\_{\bot}'' = -\frac{\sigma''}{\sqrt{2}} = -\frac{52.7}{\sqrt{2}} = -37.2 \text{ MPa}\_{\text{\textdegree}}$$

$$\tau\_{\bot}'' = \frac{\sigma''}{\sqrt{2}} = \frac{52.7}{\sqrt{2}} = 37.2 \text{ MPa}\_{\text{\textdegree}}$$

$$\tau\_{II}'' = 0,$$


$$\begin{aligned} \sigma'' &= \frac{P''\_b}{a\_{wb} b\_{j.rel}} = \frac{21.8 \cdot 10^3}{3.0 \cdot 54} = 134.6 \text{ MPa}, \\\\ \sigma''\_{\perp} &= \sigma'' \cos \frac{\theta\_2}{2} = 134.6 \cdot 0.9396 = 126.4 \text{ MPa}, \\\\ \tau''\_{\perp} &= \sigma'' \sin \frac{\theta\_2}{2} = 134.6 \cdot 0.3422 = 46.1 \text{ MPa}, \\\\ & \qquad \tau''\_{II} = 0, \end{aligned}$$


$$
\sigma'' = \frac{P\_4'}{a\_{54}l\_4} = \frac{8.0 \cdot 10^3}{3.0 \cdot 50} = 53.3 \text{ MPa},
$$

$$
\sigma\_{\perp}'' = -\sigma'' \cos \frac{\theta\_1}{2} = -53.3 \cdot 0.9050 = -48.3 \text{ MPa},
$$

$$
\tau\_{\perp}'' = \sigma'' \sin \frac{\theta\_1}{2} = 53.3 \cdot 0.3849 = 20.5 \text{ MPa},
$$

$$
\tau\_{II}'' = 0,
$$

.


$$
\sigma'' = \frac{P\_5''}{a\_{w5}l\_5} = \frac{8.8 \cdot 10^3}{3.0 \cdot 31.3} = 93.7 \text{ MPa},
$$

$$
\sigma\_\perp'' = -\frac{\sigma''}{\sqrt{2}} = -\frac{93.7}{\sqrt{2}} = -66.3 \text{ MPa},
$$

$$
\tau\_\perp'' = \frac{\sigma''}{\sqrt{2}} = \frac{93.7}{\sqrt{2}} = 66.3 \text{ MPa},
$$

$$
\tau\_{ll}'' = 0,
$$


$$\sigma'' = \frac{P\_6'}{a\_{w6}l\_6} = \frac{14.1 \cdot 10^3}{3.0 \cdot 50} = 94.0 \text{ MPa},$$

$$\sigma\_\perp'' = \sigma'' \cos \frac{\theta\_1 + \theta\_2}{2} = 94 \cdot 0.7049 = 66.3 \text{ MPa},$$

$$\tau\_\perp'' = -\sigma'' \sin \frac{\theta\_1 + \theta\_2}{2} = -94 \cdot 0.7093 = -66.7 \text{ MPa},$$

$$\tau\_{II}' = 0$$

The normal and shear stresses in welds.


$$
\tau\_{\parallel} = \tau\_{\parallel}' + \tau\_{\parallel}'' = 116.3 + 0 = 116.3 \text{ MPa},
$$

$$
\sigma\_{\perp} = \sigma\_{\perp}' + \sigma\_{\perp}'' = 0 + 0 = 0 \text{ MPa},
$$

$$
\tau\_{\perp} = \tau\_{\perp}' + \tau\_{\perp}'' = 0 - 0 = 0 \text{ MPa},
$$


$$
\tau\_{\parallel} = \tau\_{\parallel}' + \tau\_{\parallel}' = 0 + 0 = 0,
$$

$$
\sigma\_{\perp} = \sigma\_{\perp}' + \sigma\_{\perp}' = 39.7 - 126.8 = -87.1 \text{ MPa},
$$

$$
\tau\_{\perp} = \tau\_{\perp}' + \tau\_{\perp}' = 109.1 + 46.2 = 155.3 \text{ MPa},
$$


$$
\tau\_{\parallel} = \tau\_{\parallel}' + \tau\_{\parallel}'' = 116.2 + 0 = 116.2 \text{ MPa},
$$

$$
\sigma\_{\perp} = \sigma\_{\perp}' + \sigma\_{\perp}'' = 0 \text{-37.2} = -37.2 \text{ MPa},
$$

$$
\tau\_{\perp} = \tau\_{\perp}' + \tau\_{\perp}'' = 0 + 37.2 = 37.2 \text{ MPa},
$$


$$
\tau\_{\parallel} = \tau\_{\parallel}' + \tau\_{\parallel}'' = 0 + 0 = 0,
$$

$$
\sigma\_{\perp} = \sigma\_{\perp}' + \sigma\_{\perp}'' = 109.0 + 126.4 = 235.4 \text{ MPa},
$$

$$
\tau\_{\perp} = \tau\_{\perp}' + \tau\_{\perp}'' = 39.7 + 46.1 = 85.8 \text{ MPa},
$$


$$
\tau\_{\parallel} = \tau\_{\parallel}' + \tau\_{\parallel}'' = 0 + 0 = 0 \text{ MPa}\_{\prime}
$$

$$
\sigma\_{\perp} = \sigma\_{\perp}' + \sigma\_{\perp}'' = 44.6 - 48.3 = -3.7 \text{ MPa}\_{\prime}
$$

$$
\tau\_{\perp} = \tau\_{\perp}' + \tau\_{\perp}'' = 105.0 + 20.5 = 125.5 \text{ MPa}\_{\prime}
$$


$$
\tau\_{\parallel} = \tau\_{\parallel}' + \tau\_{\parallel}'' = 78.8 + 0 = 78.8 \text{ MPa},
$$

$$
\sigma\_{\perp} = \sigma\_{\perp}' + \sigma\_{\perp}'' = 0 - 66.3 = -66.3 \text{ MPa},
$$

$$
\tau\_{\perp} = \tau\_{\perp}' + \tau\_{\perp}'' = 0 + 66.3 = 66.3 \text{ MPa},
$$


$$
\tau\_{\parallel} = \tau\_{\parallel}' + \tau\_{\parallel}'' = 0 + 0 = 0 \text{ MPa},
$$

$$
\sigma\_{\perp} = \sigma\_{\perp}' + \sigma\_{\perp}'' = -55.5 + 66.3 = 121.8 \text{ MPa},
$$

$$
\tau\_{\perp} = \tau\_{\perp}' + \tau\_{\perp}'' = 55.9 - 66.7 = -10.8 \text{ MPa}.
$$
