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 s
2

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

(

)

(

)

(

)

(

)

( b) Given:

g := 9.81 L := 3m

m s
2

w := 200

kg m

F1 := 6kN F2 := 4.5kN

FB := 8kN Solution:
+

↑Σ Fy = 0;

FA − ( w⋅ L) ⋅ g − FB − 2F1 − 2F2 = 0 FA := ( w⋅ L ) ⋅ g + FB + 2F1 + 2F2 FA = 34.89 kN Ans …exibir mais conteúdo…

Therefore the shear force and moment are zero.

Problem 1-8 The boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist and load weigh 1500 N, determine the resultant internal loadings in the crane on cross sections through points A, B, and C. Given: P := 1500N a := 2.1m c := 0.6m Solution: Equations of Equilibrium: For point A w := 750 b := 1.5m d := 2.4m N m

e := 0.9m

+
+

ΣF x=0; ΣF y=0;

NA := 0

Ans

VA − w⋅ e − P = 0 VA := w⋅ e + P −MA − ( w⋅ e) ⋅ ( 0.5 ⋅ e) − P⋅ ( e) = 0 MA := −( w⋅ e) ⋅ ( 0.5 ⋅ e) − P⋅ ( e)

VA = 2.17 kN

Ans

+ ΣΜA=0;

MA = −1.654 kN⋅ m

Ans

Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B

+
+

ΣF x=0; ΣF y=0;

NB := 0

Ans

VB − w⋅ ( d + e) − P = 0 VB := w⋅ ( d + e) + P

VB = 3.98 kN

Ans

+ ΣΜB=0;

−MB − [ w⋅ ( d + e) ] ⋅ [ 0.5 ⋅ ( d + e) ] − P⋅ ( d + e) = 0 MB := −[ w⋅ ( d + e) ] ⋅ [ 0.5 ⋅ ( d + e) ] − P⋅ ( d + e) MB = −9.034 kN⋅ m

Ans

Note: Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C

+
+

ΣF x=0; ΣF y=0;

VC := 0

Ans

−NC − w⋅ (