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 s
2

L BC := 3m FB := 5kN FC := 3kN Solution:
+

kg wBC := 300 m kg wCA := 400 m L CA := 1.2m

↑Σ Fy = 0;

FA − wBC⋅ g ⋅ L BC − wCA⋅ g ⋅ LCA − 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

Problem 1-2 Determine the resultant internaltorque 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:

+

TA − TC = 0 T C := TA T C = 250 N⋅ m Ans

+

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 Band 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

+

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⋅ mAns

Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.

Problem 1-5 Determine the resultant internal loadings acting on the cross section through point D of member AB. Given: ME := 70N⋅ m a := 0.05m b := 0.3m

Solution: Segment AB: Support Reactions

+ ΣΜA=0;

−ME − By⋅ ( 2⋅ a + b) = 0 −ME By := By = −175 N 2a + b

At B:

⎛ 150 ⎞ Bx := By⋅ ⎜200 ⎝ ⎠
NB := −Bx ND + NB = 0 ND := −NB

Bx = −131.25 N VB := −By

+

Segment DB:

+

ΣF x=0;

ND = −131.25 N

Ans

+

ΣF y=0;

VD + VB = 0 VD := −VB VD = −175 N Ans

+ ΣΜD=0;

−MD − ME − By⋅ ( a + b) = 0 MD := −ME − By⋅ ( a + b) MD = −8.75 N⋅ m Ans

Problem 1-6 The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internal loadingsacting on the cross section at point D. Given: P := 5000N a := 0.8m b := 1.2m e := 0.6m Solution: θ := atan ⎜ φ := atan ⎜ c := 0.6m d := 1.6m

⎛ b⎞ ⎝ d⎠ ⎛ a + b⎞ − θ ⎝ d ⎠

θ = 36.87 deg φ = 14.47 deg

Member AB:

+ ΣΜA=0;

FBC⋅ sin ( φ ) ⋅ ( a + b) − P⋅ ( b) = 0 FBC := sin ( φ ) ⋅ ( a + b) P⋅ ( b)

FBC = 12.01 kN Segment BD: ΣF x=0; −ND − FBC⋅ cos ( φ ) − P⋅ cos ( θ ) = 0 ND := −FBC⋅ cos( φ ) − P⋅ cos ( θ ) ND = −15.63 kN Ans

+

+ ΣF y=0;

VD + FBC⋅ sin ( φ ) − P⋅ sin ( θ ) = 0 VD := −FBC⋅ sin ( φ ) + P⋅ sin ( θ ) VD = 0 kN Ans d−c

+ ΣΜD=0;

(FBC⋅ sin(φ) − P⋅ sin(θ))⋅ sin(θ) − MD = 0
MD := FBC⋅ sin ( φ ) − P⋅ sin ( θ ) ⋅ MD = 0 kN⋅ m Ans

(

)

sin ( θ )

d−c

Note: Member AB is the two-force member. Therefore the shear force and moment are zero. Problem 1-7 Solve Prob. 1-6 for the resultant internal loadings acting at point E. Given: P := 5000N a := 0.8m b := 1.2m e := 0.6m Solution: θ := atan ⎜ φ := atan ⎜ c := 0.6m d := 1.6m

⎛ b⎞ ⎝ d⎠ ⎛ a + b⎞ − θ ⎝ d ⎠

θ = 36.87 deg φ = 14.47 deg

Member AB:

+ ΣΜA=0;

FBC⋅ sin ( φ ) ⋅ ( a + b) − P⋅ ( b) = 0 FBC := sin ( φ ) ⋅ ( a + b) FBC = 12.01 kN P⋅ ( b)

Segment BE:

+

ΣF x=0;...
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