Section 6

Axially Loaded Structural MembersThis section will introduce how to solve problems of axially loaded members such as stepped and tapered rods loaded in tension. The concept of strain energy will also be introduced.

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Contents• Axially Loaded Structural Members• Deformation of Axially Loaded Members• Deformation in Members with Varying Cross- Sections• Example: Two-Step Steel Rod• Solution: Two-Step Steel Rod• Two-Step Steel Rod: FBD• Solution: Two-Step Steel Rod• Axial Member with Tapered Cross-Section• Axial Member with Tapered Cross-Section (Circular Cross-Section)• Axial Member with Tapered Cross-Section (Rectangular Cross-Section)• Example: Flat Bar of Rectangular Cross-Section • Strain Energy in Tension and Compression• Strain Energy – Axial Loading • Strain Energy – Shear Loading• Example: Strain Energy• Credits & Notices

Axially Loaded Structural Members

• In previous lectures we have looked at stress and strain behaviour of materials and the deformation caused by axial loads

• We will now look at the deformation caused by axial loads in structures with a stepped or tapered geometry

• This will then lead into the work done by the external loads and how strain energy is calculated.

Deformation of Axially Loaded Members

The term EA/L is the stiffness of the member so we can rewrite theequation as

( )PL PLEEA A

The deformation of a structural member, with known geometry, andsubjected to an axial load can be determined by using the equation fromthe notes in Section 3

Parameter k is called stiffness (sometimes or spring constant).The reciprocal of the stiffness, k-1, is called compliance.

PL PEA k

Deformation in Members with Varying Cross- Sections

Individual elongations 1 and 2 are added algebraically togive an overall elongation of the entire system as

1 1 2 21 2

1 21 2tot

PL P LE A E A

Axially loaded stepped member

L2 L1

PPB

B AC

(P2) (P1)RCX

RCY

MCZ

(Pi, Li, Ei and Ai are local values e.g. Pi is internal force)

Deformation in Members with Varying Cross- Sections

For an n number of levels

1 2 11 1

n ni i

tot n n ii i i i

PLE A

Axially loaded n level step bar

Rx

RY

MZ

Where Pi is the internal axial force in member i (i.e. not external load) and Li, Ei and Ai are all local values)

Pn … 4 3 2 1

Ln Li L4 L3 L2 L1

Example: Two-Step Steel Rod

The two-step steel rod is subjected to the three external loads shown. The large and small sections of the rod have a diameter of 30 mm and 15 mm respectively. Young’s modulus for steel is 210 GPa. Calculate the elongation of the rod.

Axially loaded two-step rod

300

200kN300kN

B AC

RCX

RCY

MCZ

300 400

500kN

Solution: Two-Step Steel Rod

First Step: We need to determine the actual loads experienced by the rod in each part of the assembly

How do we do this? How many free body diagrams?

Axially loaded two-step bar

300

200kN300kN

B AC

RCX

RCY

MCZ

300 400

500kN

Two-Step Steel Rod: FBD

Red lines show cuts to establish internal forces

2 13

200kNP1 Internal force P1 = 200kN (tension)

200kN300kNP2 P2 = -100kN (compression)

200kN300kN500kNP3 P3 = 400kN (tension)

Cut 1 1

Cut 2 2

Cut 3 3

2 13

200kN300kN

B AC

RCX

RCY

MCZ

500kN

Solution: Two-Step Steel Rod

33 31 1 2 2

1 1 2 3

2 2 2

1

1 200 400 100 300 400 300210 7.5 15 15

1 1422.2 133.3 533.3 2.16 0.20 0.812102.76

tot ii

P LPL P LE A A A

mm

(Units used are kN and mm)

300

200kN300kN

B AC

RCX

RCY

MCZ

300 400

500kN

Axial Member with Tapered Cross-Section

Elongation of differential element of length is and is calculated as follows:

d

Elongation over entire length of bar given by integration (In limit )

do o

i

x dP xEA

x dxP dx

EA

l l

x

L

AB

x

d1d2di

P

Axial Member with Tapered Cross-Section(Circular Cross-Section)

x

L

AB

x

d1d2di

P

2 1 2id d d dx

l l

Force equilibrium at any cross section shows P is constant along length

Taper is linear so diameter d of rod at distance x is di

2 2

1 224 4

ii

d d dA x d

ll

Area A of rod at distance x is Ai

Axial Member with Tapered Cross-Section(Circular Cross-Section)

x

L

AB

x

d1d2di P

21 2

2

11 11 2

2 2 11 2 1 20

1 2

1 2 2 1 1 2 1 2 1 2

4d

4 4

4 1 1 4 4 (For unifo

o o oi

d dP Pdx x d dxEA E

d dP Px d d dE d d E d d

d dP P PE d d d d E d d d d E d d

l l l

l

ll

l lll

l l l1 2rm bar )d d

Elongation over entire length

Axial Member with Tapered Cross-Section(Rectangular Cross-Section)

A flat bar of rectangular cross section is subjected to the tensile forces P shown. It has a length L and a constant thickness t with Young’s modulus E. The width of the bar varies linearly from b1 at left end to b2 at right end.

Determine the elongation of the bar.

x

L

x

b1b2biP P

Example: Flat Bar of Rectangular Cross-Section

1 2 1

2 11

Breadth of bar at distance is

Area A of bar at distance is

. . .

i

i

i

i i

b x bb b b b

x

x Ab bA t b t x b

l

l

x

L

x

b1b2biP P

Example: Flat Bar of Rectangular Cross-Section

1

2 11

2 11

2 1

2

2 1 1

Using d

.

ln .

ln

o o

o

o

P dxEA x

b bP t x b dxE

b bP x bE t b b

bPE t b b b

l l

l

l

l

ll

l

x

L

x

b1b2biP P

Strain Energy in Tension and Compression

External loads cause deformation in structures. The deformation requiresan energy input or external work.

The work done W by the external load P to deform the bar by an elongation of is equal to the area below the load displacement graph.

P

P

2PW

Strain Energy in Tension and Compression

P

P

2PW

Strain energy U is stored internally in the bar during the loading process.If the bar behaves elastically, it is called elastic strain energy.

For energy conservation: Internal strain energy = external work W required to deform barU = W=P/2 Units for strain energy: Nm or Joule (J).

Strain Energy – Axial Loading

2

2

2 2

Internal strain energy

=2

Substituting = =2

Substituting = 2

Substituting = 2 2

In terms of energy density - - energy per unit volume of the material

2

PU

PL P LUEA EA

EA EAP UL L

EA P kk U ORL k

uU PuAL AL

2 2

2 2 2EOR

E

Strain Energy – Shear Loading

2 2

Internal shear strain energy

=2

In terms of energy density - - energy per unit volume of the material

2 2 2

VU

uGu

G

Example: Strain Energy

A straight bar of length L and diameter d with a Young’s modulus E has one end fixed and is subjected to the axial tensile force P at the free end. Its stored strain energy is Uo.

Calculate the amount of strain energy stored in the stepped bar below which is made of the same material as straight bar and compare the internal energies.

L/4

L

2d 2dd P

Example: Strain Energy

2

0

2 22 2 2

1

1 0

Internal energy in straight bar =2

Internal energy in stepped bar 3

3 74 4=2 4 2 32 8 32

7 The stepped bar has less than 1/2 energy of straight bar16

P LUEA

L LP PP L P L P LU

E A EA EA EA EA

U U

L/4

L

2d 2dd P

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Credits