Engineering Mechanics - Statics Episode 2 Part 2 potx

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Positive means Tension, Negative means Compression

− 600 400

− 1414 1600

The truss, used to support a balcony, is subjected to the loading shown Approximate each joint

as a pin and determine the force in each member State whether the members are in tension or

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Joint D: ( FDC FAD − ) cos ( ) θ + FDE = 0

− 800 0 1131 1600

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− 3.13 9.37

− 9.37

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− 6.25 18.75

− 18.75

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Positive means tension, Negative means compression.

− 4.167

− 13.125

− 1.458

− 3.125

− 0

Determine the force in each member of the truss in terms of the external loading and state if the

members are in tension or compression.

Solution:

ΣMA = 0; − a C P + y2a − P a = 0

Cy = P Joint C:

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The maximum allowable tensile force in the members of the truss is Tmax, and the maximum

allowable compressive force is Cmax Determine the maximum magnitude P of the two loads

that can be applied to the truss.

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− 1.374 1.374 1.667

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− 471

− 667 667 0 943

− 471

Negative means compression.

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− 1179

− 1167 1167 500 1650

− 471

Negative means compression.

Problem 6-12

Determine the force in each member of

the truss and state if the members are in

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− sin ( ) β − FCDsin ( ) α − FEC = 0

Joint E − FFE + FED = 0

FEC P2 − = 0

Joint D − FCD cos ( ) α − FED = 0

FCDsin ( ) α = 0

455

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Negative means Compression

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− sin ( ) β − FCDsin ( ) α − FEC = 0

Joint E − FFE + FED = 0

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FAB = 1 lb FAF = 1 lb FBC = 1 lb

FBF = 1 lb FFC = 1 lb FFE = 1 lb

FED = 1 lb FEC = 1 lb FCD = 1 lb

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− 233.3 47.1

− 112.7 112.7 300.0 377.1

FAB = 1 lb FAF = 1 lb FBC = 1 lb

FBF = 1 lb FFC = 1 lb FFE = 1lb

FED = 1 lb FEC = 1 lb FCD = 1 lb

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Joint B FBC FAB cos − ( ) φ = 0

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− 266.7 188.6 56.4 56.4 0.0 188.6

Problem 6-16

Determine the force in each member of the truss in terms of the load P and state if the members

are in tension or compression.

Solution:

Support reactions:

ΣME = 0; Ax 3 2 d − P d = 0 Ax = 2P 3

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The maximum allowable tensile force in the members of the truss is Tmax and the maximum

allowable compressive force is Cmax Determine the maximum magnitude of the load P that can

be applied to the truss.

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− 0.373

− 0.333 0.373

− 1.202 0

Determine the force in each member of the truss and state if the members are in tension or

compression Hint: The horizontal force component at A must be zero Why?

465

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Engineering Mechanics - Statics Chapter 6

− 692.82

compression Hint: The resultant force at the pin E acts along member ED Why?

Units Used:

kN = 10 3 N

Units Used:

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− 3 3

− 2.704 6.31

Problem 6-20

Each member of the truss is uniform and has a mass density ρ Determine the approximate force

in each member due to the weight of the truss State if the members are in tension or

compression Solve the problem by assuming the weight of each member can be represented as

a vertical force, half of which is applied at each end of the member.

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− 389 314

− 736 1204

Problem 6-21

Determine the force in each member of the truss in terms of the external loading and state if the

members are in tension or compression.

Solution:

Joint B:

+ ↑ Σ Fy = 0 ; FBA sin 2 ( ) θ − P = 0

FBA = P csc 2 ( ) θ ( ) C +

→ Σ Fx = 0; FBAcos 2 ( ) θ − FBC = 0

FBC = Pcot 2 ( ) θ ( ) C

Joint C:

+

→ Σ Fx = 0; P cot 2 ( ) θ + P + FCDcos 2 ( ) θ − FCA cos ( ) θ = 0

+↑Σ F = 0; FCDsin 2 ( ) θ − FCA sin ( ) θ = 0

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→ Σ Fx = 0; FDA − ⎡⎣ cot 2 ( ) θ + 1 ⎤⎦ ⎡⎣ cos 2 ( ) θ ⎤⎦ P = 0

FDA = ⎡⎣ cot 2 ( ) θ + 1 ⎤⎦ ⎡⎣ cos 2 ( ) θ ⎤⎦ P ( ) C

Problem 6-22

The maximum allowable tensile force in the members of the truss is Tmax, and the maximum

allowable compressive force is Cmax Determine the maximum magnitude P of the two loads that

can be applied to the truss

Initial guesses (assume all bars are in tension) Use a unit load

for P and then scale the answer later.

FBA = 1 kN FBC = 1 kN FCA = 1 kN

FCD = 1 kN FDA = 1 kN P = 1 kN

471

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− 2.732 1.577

− 0.789

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Engineering Mechanics - Statics Chapter 6

The Fink truss supports the loads shown Determine the force in each member and state if the

members are in tension or compression Approximate each joint as a pin.

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Problem 6-24

Determine the force in each member of the double scissors truss in terms of the load P and state

if the members are in tension or compression.

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compression Hint: The vertical component of force at C must equal zero Why?

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− 8 18.5 19.799

− 4.5

Negative means Compresson.

Problem 6-26

Each member of the truss is uniform and has a mass density ρ Remove the external loads F1

and F2 and determine the approximate force in each member due to the weight of the truss.

State if the members are in tension or compression Solve the problem by assuming the weight

of each member can be represented as a vertical force, half of which is applied at each end of

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Guesses FAB = 1 N FAE = 1 N FBE = 1 N

FBC = 1 N FBD = 1 N FDE = 1 N Given

Joint E FDE FAE − = 0

FBE WAE WBE

− 857 1045

− 216 118

Tiêu đề	Statics Episode 2 Part 2 potx
Trường học	Pearson Education, Inc.
Chuyên ngành	Engineering Mechanics
Thể loại	Giáo trình
Năm xuất bản	2007
Thành phố	Upper Saddle River

Định dạng
Số trang	40
Dung lượng	439,07 KB