Exercicios resolvidos hibbeler cap.1
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Problem 1-1 Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the column has a mass of 200 kg/m. ( a) Given: g := 9.81 m s2
L BC := 3m FB := 5kN FC := 3kN Solution:
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kg wBC := 300 m kg wCA := 400 m L CA := 1.2m
↑Σ Fy = 0;
FA − wBC⋅ g ⋅ L BC − wCA⋅ g ⋅ L CA − FB − 2FC = 0 FA := wBC⋅ g ⋅ LBC + wCA⋅ g ⋅ LCA + FB + 2FC FA = 24.5 kN Ans
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)
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( b) Given:
g := 9.81 L := 3m
m s
2
w := 200
kg m
F1 := 6kN F2 := 4.5kN
FB := 8kN Solution:
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↑Σ Fy = 0;
FA − ( w⋅ L) ⋅ g − FB − 2F1 − 2F2 = 0 FA := ( w⋅ L ) ⋅ g + FB + 2F1 + 2F2 FA = 34.89 kN Ans
Problem 1-2 Determine the resultant internal torque acting on the cross sections through points C and D of the shaft. The shaft is fixed at B. Given: T A := 250N⋅ m T CD := 400N⋅ m T DB := 300N⋅ m Solution: Equations of equilibrium:
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TA − TC = 0 T C := TA T C = 250 N⋅ m Ans
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T A − TCD + TD = 0 T D := T CD − T A T D = 150 N⋅ m Ans
Problem 1-3 Determine the resultant internal torque acting on the cross sections through points B and C. Given: T D := 500N⋅ m T BC := 350N⋅ m T AB := 600N⋅ m Solution: Equations of equilibrium: Σ Mx = 0; T B + T BC − T D = 0 T B := −TBC + TD T B = 150 N⋅ m Σ Mx = 0; TC − TD = 0 T C := TD T C = 500 N⋅ m Ans Ans
Problem 1-4 A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A. Given: P := 80N θ := 30deg a := 0.3m Solution: Equations of equilibrium: NA − P⋅ cos ( φ − θ ) = 0 NA := P⋅ cos ( φ − θ ) NA = 77.27 N Ans φ := 45deg b := 0.1m
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ΣF x'=0;
+ ΣF y'=0;
VA − P⋅ sin ( φ − θ ) = 0 VA := P⋅ sin ( φ − θ ) VA = 20.71 N Ans
+ ΣΜA=0;
MA + P⋅ cos ( φ ) ⋅ a cos ( θ ) − P⋅ sin ( φ ) ⋅ ( b + a sin ( θ ) ) = 0 MA := −P⋅ cos ( φ ) ⋅ a cos ( θ ) + P⋅ sin ( φ ) ⋅ ( b + a sin ( θ ) ) MA = −0.555 N⋅ m