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CHAPTER 5:Axially Loaded Members

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OUTLINE

-Elastic deformation of an axially loaded member

-Statically indeterminate axially loaded member

-Thermal stress problems

-Combined problems

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*Rectangular bar will deform elastically when the bar is subjected to P*Localized deformation tends to decrease and becomes uniform.

SaintSaint--VenantVenant’’ss principleprinciple

[Reference # 3: 17 – 132] 4

Elastic deformation of an axially Elastic deformation of an axially loaded memberloaded member

The load P will deform the element into the shape indicated by the dashed outline. The stress and strain in the element are

)()(xAxP

=σdxdδε =and

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For problems do not exceed the proportional limit,

or

εσ E=

⎟⎠⎞

⎜⎝⎛=dxdE

xAxP δ)()(

ExAdxxPd)()(

=δ

∫=L

ExAdxxP

0 )()(δ

AEPL

=δ

(Hooke’s law)

∑= AEPLδ

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Several different forces or cross-sectional area or E changes

orAEPL

=δ ∑= AEPLδConstant load

and cross-sectional area

= displacement of one point relative to another pointL = distance between the pointsP = internal axial forceA = cross-sectional areaE = modulus of elasticity for the material

δ

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Sign conventionSign convention

Both the force and displacement will be considered to be positive if they cause tensionand elongation, respectively.

Negative force and displacement will cause compression and contraction, respectively.

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ExampleExample

The composite A-36 steel bar shown in below is made from two segments, AB and BD, having cross-section areas of AAB = 600 mm2 and AAD = 1200 mm2. Determine the vertical displacement of end A and the displacement of B relative to C. [E = 210(103) MPa]

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SolutionSolution

∑ ++

+==

]/)10)(210(1200[)10)(75.0](35[

]/)10)(210(600[)10)(1](75[

232

6

232

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mkNmmmkN

mkNmmmkN

AEPL

Aδ

mmmkNmm

mkN 61.0]/)10)(210(1200[

)10)(5.0](45[232

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+=−

+

mmmkNmm

mkNEALP

BC

BCBCCB 104.0

]/)10)(210(1200[)10)(75.0](35[

232

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/ +=+

==δ

(A is upward)

(B moves away from C)10

Example1. Determine the load Prequired to displace the roller downward 0.2 mm.

Homework # 12: Reference # 3 Problem 4-13

P

1.6 m

1.2 m

1.6 m

E=200GPa, A=400mm2

for all truss members

2. Determine the displacement of its end A when it is subjected to the distributed loading.

1.5 m

A

w = 500x1/3 N/mmx

E=250GPa, A=1800mm2

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Principle of superpositionPrinciple of superposition

The resultant stress or displacement is determined by algebraically adding the contributions caused by each of the components.

[Reference # 3: Page 133 – 147] 12

Statically indeterminate axially Statically indeterminate axially loaded memberloaded member

0=−+ PFF AB

Statically indeterminate: the equilibrium equation is not sufficient to determine the reactions.

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Compatibility conditionCompatibility conditionSince the end supports are fixed,

0/ =BAδ

AEPL

=δFrom,

0=−AELF

AELF CBBACA

⎟⎠⎞

⎜⎝⎛=LL

PF CBA ⎟

⎠⎞

⎜⎝⎛=LL

PF ACBand

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Flexibility or force methodFlexibility or force method

No displacement at B Displacement at B when redundant force at B is removed.

Displacement at B when only the redundant force at B is applied.

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BP δδ −=0

AELF

AEPL BAC −=0

⎟⎠⎞

⎜⎝⎛=LL

PF ACB

0=−+⎟⎠⎞

⎜⎝⎛ PFLL

P AAC

⎟⎠⎞

⎜⎝⎛=LL

PF CBA

Compatibility conditionCompatibility conditionNo displacement occurs at B,

Results are the same!

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Example

The A-36 steel column is encased in high-strength concrete as shown. If an axial force of 300 kN is applied to the column, determine the required area of the steel so that the force is shared equally between the steel and concrete. How far does the column shorten? It has an original length of 3.0 m. EA-36= 210 GPa, Econc = 29 GPa.

3 m

225 mm

300 kN

400 mm


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ExampleThe A-36 steel pipe has a aluminum core. It is subjected to a tensile force of 200 kN. Determine the average normal stress in the aluminum and the steel due to this loading. The pipe has an outer diameter of 80 mm and an inner diameter of 70mm. Eal = 68.9 GPa, Est = 200 GPa

200 kN200 kN

400 mm

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Example

The stainless steel post A is surrounded by a red brass tube B. Both rest on the rigid surface. If a force of 25 kN is applied to the rigid cap, determine the required diameter d of the steel post so that the load is shared equally between the post and tube. Est= 200 GPa, Ebr = 100 GPa.


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ExampleThe column is constructed from high-strength concrete and four A-36 steel reinforcing rods. If it is subjected to an axial force of 800 kN, determine the required diameter of each rod so that one-fourth of the load is carried by the steel and three-fourths by the concrete (Est = 200 GPa, Ec = 25GPa )

800 kN300 mm300 mm

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ExampleThe bolt AB has a diameter of 20 mm and passes through a sleeve that has an inner diameter and of 40 mm and an outer diameter of 50 mm . The bolt and sleeve are made of A-36 steel and are secured to the rigid brackets as shown. If the bolt length is 220 mm and the sleeve length is 200 mm , determine the tension in the bolt when a force of 50 kN is applied to the brackets.

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Solution


∑Fx = 0 Pb+Ps = 50 x 103 … (1)

PbLb / AbE = PsLs / AsEPb(220x10-3)/π(102)(10-6)(20x109) = Ps(200x10-3)/π(252-202)(10-6)(20x109)

Pb = 0.4040Ps … (2)Pb = 14.4 kN Ps = 35.61 kN #

δb = δs

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Thermal stress problemsThermal stress problemsA change in temperature can cause a material to change its dimensions.

TLT Δ=αδ= the algebraic change in length of the member= a property of the material (linear coefficient of thermal expansion)= the algebraic change in temperature of the member

= the original length of the member

Tδα

TΔL

[Reference # 3: Page 148 – 155]

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ExampleThree bars each made of different materials are connected together and placed between two walls when the temperature is T1 = 12°C. Determine the force exerted on the (rigid) supports when the temperature becomes T2 = 18°C. The material properties and cross-sectional area of each bar are given in the figure.

Steel Brass CopperEst = 200 GPa Ebr = 100 GPa Ecu = 120 GPaαst = 12(10-6)/°C αbr = 21(10-6)/°C αcu = 17(10-6)/°C

300 mm

Ast = 200 mm2

100 mm200 mm

Abr = 450 mm2Acu = 515 mm2

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ExampleA thermo gate consists of a 6061-T6-aluminum plate AB and an Am-1004-T61-magnesium plate CD, each having a width of 15 mm and fixed supported at their ends. If the gap between them is 1.5 mm when the temperature is T1 = 25°C, determine the temperature required to just close the gap. Also, what is the axial force in each plate if the temperature becomes T2 = 100°C? Assume bending or buckling will not occur. Eal = 68.9 GPa, EAm = 44.7 GPa.

600 mm

A B

400 mm

C D10 mm10 mm

1.5 mm

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ExampleThe A-36 steel pipe having a cross-sectional area of 300 mm2 is connected to fixed supports and carries a liquid that causes the pipe to be subjected to a temperature drop of ΔT = (-x3/2)°C, where x is in millimeters. Determine the maximum and minimum normal strain. Est = 200 GPa.

2500 mmx

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ExampleThe rigid block has a weight of 400 kN and is to be supported by posts A and B , which is made of A-36 steel, and the post C, which is made of C83400 red brass. If all the posts have the same original length before they are loaded, determine the average normal stress developed in each post when post C is heated so that its temperature is increased by 20°C. Each post has a cross-sectional area of 5000 mm2. Ebr = 100 GPa, αbr = 17(10-6)/°C, Est = 200 GPa, αst = 12(10-6)/°C

A BC

1 m 1 mHomework # 16: Reference # 3 Problem 4-90

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