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The L-shaped bar in Fig. 1 is applied with the point loads at different points and the single couple moment as shown. Simplif

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Gire dal F - SKN f5= 4 KN Fu= 4KN 25 Im 5 m 2m 2 m. im A klo Im 340 D 4.DE B. 4KN F6. F obvod = 2KN M =561.m F3 = 2KNAccording to the problem ₃ Efa Fx = 2005 50° - 2005 40 tusin 35 + 4 cosa5° 1.285 - 1.532 +2.294 + 3.625 Fa 5.672 KN tasty Fy

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