Transcript
Page 1: Axially Loaded Members

AXIALLY LOADED MEMBERS

Fig.1-1: Spring subjected to an axial load P

Two major things that will happen when the spring is

loaded

It gets longer if loaded in tension

It shortens when compressed

SPRINGS

Fig: 1-2: Elongation of axially loaded spring

Natural length of the spring is called

unstressed, relaxed or free length

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The spring increases by 𝛿 under the action

of force 𝑃 and the final length is 𝐿 + 𝛿

If material is linearly elastic, load and

elongation will be proportional

𝑃 = π‘˜π›Ώ 𝛿 = 𝑓𝑃

Where π‘˜ π‘Žπ‘›π‘‘ 𝑓 are constants of proportionality

π‘˜ is the stiffness and it is defined as the force

required to produce a unit elongation π‘˜ = 𝑃 𝛿

𝑓 is the flexibility and it is defined as the

elongation produced by a load of unit value

𝑓 = 𝛿 𝑃

NOTE:

The stiffness and flexibility of a spring is reciprocal

of each other.

π‘˜ =1

𝑓 𝑓 =

1

π‘˜

Other terms for stiffness and flexibility are spring

constants and compliance respectively.

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ELONGATION OF PRISMATIC BAR:

Fig.1-3: Elongation of prismatic bar

A prismatic bar is a structural member having a straight longitudinal axis and constant cross section throughout its length. From Hooke’s Law

Οƒ = Ξ΅E Ξ΅ =Οƒ

E=

P

AE

From definition of strain

Ξ΅ =Ξ΄

L

Equating and solving for deformation

Ξ΄ =PL

AE

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BARS WITH INTERMEDIATE AXIAL LOADING

SOLUTION PROCEDURE:

(1). Identify the segments of the bar under study. (2). Determine the internal axial forces 𝐹1, 𝐹2 , 𝐹3, 𝑒𝑑𝑐, In all segments from the free-body diagrams of the bar under study. Note that the internal axial forces are denoted by the letter 𝐹1, 𝐹2 , 𝐹3, 𝑒𝑑𝑐 to distinguish them from the external loads P. By summing forces in the vertical or horizontal directions, the expressions for the axial forces can be obtained. (3). Determine the changes in the length of the segments.

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𝛿1 =𝑃1𝐿1

𝐴1𝐸1, 𝛿2 =

𝑃2𝐿2

𝐴2𝐸2 𝛿3 =

𝑃3𝐿3

𝐴3𝐸3 𝛿4 =

𝑃4𝐿4

𝐴4𝐸4…

In which 𝐿1 , 𝐿2 , 𝐿3 π‘Žπ‘›π‘‘ 𝐿4 are the length of the

segments and 𝐸𝐴 is the axial rigidity of the bar.

(4). Add 𝛿1, 𝛿2, 𝛿3 π‘Žπ‘›π‘‘ 𝛿4 to obtain 𝛿, the change

in length of the entire bar;

𝛿 = 𝛿𝑖

3

𝑖=1

= 𝛿1 + 𝛿2 + 𝛿3 + 𝛿4 …

The changes in lengths must be added

algebraically, with elongation being positive and

shortenings negative.

ALTERNATIVE

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BARS CONSISTING OF PRISMATIC SEGMENTS

Ξ΄ = PiLi

AiEi

𝑛

𝑖=1

The stiffness and flexibility is given as

k =EA

L f =

L

EA

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BARS WITH CONTINUOUSLY VARYING LOADS OR DIMENSIONS

Ξ΄ =PL

EA

Substituting N x for P, dx for L and A x for A

𝑑𝛿 =𝑁 π‘₯ 𝑑π‘₯

𝐸𝐴 π‘₯

Integrating over the length gives the elongation of the entire bar

𝑑𝛿

𝐿

0

= 𝑁 π‘₯ 𝑑π‘₯

𝐸𝐴 π‘₯

𝐿

0

Example 1:

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Example 2:

Determine the deformation of the steel rod shown in fig

below under the given load [𝐸 = 200 Γ— 109 𝑁 π‘š2

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Example 4: A member 𝐴𝐡𝐢𝐷 is subjected to point loads

𝑃1 , 𝑃2 , 𝑃3 π‘Žπ‘›π‘‘ 𝑃4 as shown below. If 𝑃1 = 120 π‘˜π‘, 𝑃2 =

220 π‘˜π‘ π‘Žπ‘›π‘‘ 𝑃4 = 160 π‘˜π‘. Determine the net change in

length of the member. [Take 𝐸 = 200 𝐺 𝑁 π‘š2 ]

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Example 5:

The device shown in the figure below consists of a horizontal beam 𝐴𝐡𝐢

supported by two vertical bars 𝐡𝐷 π‘Žπ‘›π‘‘ 𝐡𝐢. Bar 𝐢𝐸 is pinned at both ends but bar

𝐡𝐷 is fixed to the foundation at its lower end. The distance from 𝐴 π‘‘π‘œ 𝐡 is

450 π‘šπ‘š π‘Žπ‘›π‘‘ 𝐡 π‘‘π‘œ 𝐢 𝑖𝑠 225 π‘šπ‘š. Bars 𝐡𝐷 π‘Žπ‘›π‘‘ 𝐢𝐸 have lengths of 480 π‘šπ‘š and

600 π‘šπ‘š respectively, and their cross-sectional areas are

1020 π‘šπ‘š2 π‘Žπ‘›π‘‘ 520 π‘šπ‘š2 respectively. The bars are made of steel having

modulus of elasticity 𝐸 = 205 πΊπ‘ƒπ‘Ž. Assuming the beam 𝐴𝐡𝐢 is rigid, find the

maximum allowable load π‘ƒπ‘šπ‘Žπ‘₯ if the displacement of 𝐴 is limited to 1.0 π‘šπ‘š.

Taking moment about B summing forces in the vertical direction

gives.

𝐹𝐢𝐸 = 2𝑃 𝐹𝐡𝐷 = 3𝑃

Member CE is in tension and member BD is in compression.

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Shortening of member BD

𝛿𝐡𝐷 =𝐹𝐡𝐷 ×𝐿𝐡𝐷

𝐸𝐴𝐡𝐷 =

3𝑃×480 π‘šπ‘š

205πΊπ‘ƒπ‘ŽΓ—1020 π‘šπ‘š 2 = 6.887𝑃 Γ— 10βˆ’6π‘šπ‘š

Where 𝑃 = π‘π‘’π‘€π‘‘π‘œπ‘›π‘ 

Lengthening of member CE is

𝛿𝐢𝐸 =𝐹𝐢𝐸 Γ— 𝐿𝐢𝐸

𝐸𝐴𝐢𝐸 =

2𝑃 Γ— 600 π‘šπ‘š

205πΊπ‘ƒπ‘Ž Γ— 1020 π‘šπ‘š2= 11.26𝑃 Γ— 10βˆ’6π‘šπ‘š

Using similar triangle from the displacement diagram

Consider triangle 𝐴′𝐴′′ 𝐢 β€² π‘Žπ‘›π‘‘ 𝐡′𝐡′′ 𝐢 β€²

𝐴′𝐴′′

𝐴′′ 𝐢 β€²=

𝐡′𝐡′′

𝐡′′ 𝐢 β€² π‘œπ‘Ÿ

𝛿𝐴 + 𝛿𝐢𝐸

450 + 225=

𝛿𝐡𝐷 + 𝛿𝐢𝐸

225

Substituting for 𝛿𝐡𝐷 π‘Žπ‘›π‘‘ 𝛿𝐢𝐸 gives

𝛿𝐴 + 11.26𝑃 Γ— 10βˆ’6

450 + 225=

6.887𝑃 Γ— 10βˆ’6 + 11.26𝑃 Γ— 10βˆ’6

225

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Substituting 𝛿𝐴 = 1.0 i.e. its limiting value and solve

gives 𝑃 as

𝑃 = π‘ƒπ‘šπ‘Žπ‘₯ = 23.200𝑁 π‘œπ‘Ÿ 23.2 π‘˜π‘

Note: When the load reaches this value, the downward

displacement of at point A is 1.0 mm

Angle of displacement:

From the displacement diagram, the angle of rotation

π‘‘π‘Žπ‘›π›Ό =𝐴′𝐴′′

𝐴′′ 𝐢 β€²=

𝛿𝐴 + 𝛿𝐢𝐸

675

And 𝛿𝐢𝐸 = 0.261 π‘šπ‘š by substituting the value of P

π‘‘π‘Žπ‘›π›Ό =1.0 + 1.261

675=

1.261

675= 0.001868

𝛼 = 0.11Β°

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Question 6

Consider a circular bar that tapers uniformly from diameterd1at the bigger end to diameter d2 at the Smaller end, and subjected to axial tensile load P as

shown in fig. below:

Show that the total elongation of the tapering bar

Ξ΄L =4PL

Ο€Ed1d2


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