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M, = (0.0781 -4.17+0.100 9.6) 1.90'= 4.64 kN-m

M, = (0.0331 • 4.17 + 0.0787 • 9.6)2,20’ = 4.32 kN-m

M, = (0,0462 • 4,17 + 0.0855 • 9.6)2,20’ = 4,91 kN-m

M, = - (0.105 • 4,17 + 0.119 - 9.6) [0.5(1,90 + 2.20))¹ = - 6.64 kN-m

Mc=- (0.079 - 4.17 + 0.111 • 9.6)2,20’ = -6.75 kN-m

Mi— = (0.0781 • 4,17 - 0.0263 - 9.6)1,90’ = 0.26 kN-m

M_„ = (0.0331 • 4.17 - 0,0461 • 9,6)2,20’ = - 1.47 kN-m

M— = (0.0462 • 4.17 - 0.0395 • 9,6)2,20’ = - 0,90 kN-m

M«— = - (0.105 • 4.1710.053 - 9.6) [0.5(1.90 + 2.20)]’ = - 3.98 kN-m

Mo— = - (0.079 - 4.17 + 0.040 • 9,6)2,20’ = - 3.45 kN-m

V_Ci— = - (0.474 - 4.17 + 0,576 - 9.6)200 = - 16.51 kN

1W = (0.500 - 4,17 + 0,591 • 9,6)2.20 = 17.07 kN

f,M' 13.3 1,0-0.07’    ’

fa-l-J-ty* =1-^1-2 0.075 = 0,078 <&„„ = Przekrój może być pojedynczo zbrojony, więc