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

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 = 𝑘𝛿 𝛿 = 𝑓𝑃

Where 𝑘 𝑎𝑛𝑑 𝑓 are constants of proportionality

𝑘 is the stiffness and it is defined as the force

required to produce a unit elongation 𝑘 = 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.

𝑘 =1

𝑓 𝑓 =

1

𝑘

Other terms for stiffness and flexibility are spring

constants and compliance respectively.

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

σ = εE ε =σ

E=

P

AE

From definition of strain

ε =δ

L

Equating and solving for deformation

δ =PL

AE

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, 𝑒𝑡𝑐, 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, 𝑒𝑡𝑐 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.

𝛿1 =𝑃1𝐿1

𝐴1𝐸1, 𝛿2 =

𝑃2𝐿2

𝐴2𝐸2 𝛿3 =

𝑃3𝐿3

𝐴3𝐸3 𝛿4 =

𝑃4𝐿4

𝐴4𝐸4…

In which 𝐿1 , 𝐿2 , 𝐿3 𝑎𝑛𝑑 𝐿4 are the length of the

segments and 𝐸𝐴 is the axial rigidity of the bar.

(4). Add 𝛿1, 𝛿2, 𝛿3 𝑎𝑛𝑑 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

BARS CONSISTING OF PRISMATIC SEGMENTS

δ = PiLi

AiEi

𝑛

𝑖=1

The stiffness and flexibility is given as

k =EA

L f =

L

EA

BARS WITH CONTINUOUSLY VARYING LOADS OR DIMENSIONS

δ =PL

EA

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

𝑑𝛿 =𝑁 𝑥 𝑑𝑥

𝐸𝐴 𝑥

Integrating over the length gives the elongation of the entire bar

𝑑𝛿

𝐿

0

= 𝑁 𝑥 𝑑𝑥

𝐸𝐴 𝑥

𝐿

0

Example 1:

Example 2:

Determine the deformation of the steel rod shown in fig

below under the given load [𝐸 = 200 × 109 𝑁 𝑚2

Example 4: A member 𝐴𝐵𝐶𝐷 is subjected to point loads

𝑃1 , 𝑃2 , 𝑃3 𝑎𝑛𝑑 𝑃4 as shown below. If 𝑃1 = 120 𝑘𝑁, 𝑃2 =

220 𝑘𝑁 𝑎𝑛𝑑 𝑃4 = 160 𝑘𝑁. Determine the net change in

length of the member. [Take 𝐸 = 200 𝐺 𝑁 𝑚2 ]

Example 5:

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

supported by two vertical bars 𝐵𝐷 𝑎𝑛𝑑 𝐵𝐶. Bar 𝐶𝐸 is pinned at both ends but bar

𝐵𝐷 is fixed to the foundation at its lower end. The distance from 𝐴 𝑡𝑜 𝐵 is

450 𝑚𝑚 𝑎𝑛𝑑 𝐵 𝑡𝑜 𝐶 𝑖𝑠 225 𝑚𝑚. Bars 𝐵𝐷 𝑎𝑛𝑑 𝐶𝐸 have lengths of 480 𝑚𝑚 and

600 𝑚𝑚 respectively, and their cross-sectional areas are

1020 𝑚𝑚2 𝑎𝑛𝑑 520 𝑚𝑚2 respectively. The bars are made of steel having

modulus of elasticity 𝐸 = 205 𝐺𝑃𝑎. Assuming the beam 𝐴𝐵𝐶 is rigid, find the

maximum allowable load 𝑃𝑚𝑎𝑥 if the displacement of 𝐴 is limited to 1.0 𝑚𝑚.

Taking moment about B summing forces in the vertical direction

gives.

𝐹𝐶𝐸 = 2𝑃 𝐹𝐵𝐷 = 3𝑃

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

Shortening of member BD

𝛿𝐵𝐷 =𝐹𝐵𝐷 ×𝐿𝐵𝐷

𝐸𝐴𝐵𝐷 =

3𝑃×480 𝑚𝑚

205𝐺𝑃𝑎×1020 𝑚𝑚 2 = 6.887𝑃 × 10−6𝑚𝑚

Where 𝑃 = 𝑁𝑒𝑤𝑡𝑜𝑛𝑠

Lengthening of member CE is

𝛿𝐶𝐸 =𝐹𝐶𝐸 × 𝐿𝐶𝐸

𝐸𝐴𝐶𝐸 =

2𝑃 × 600 𝑚𝑚

205𝐺𝑃𝑎 × 1020 𝑚𝑚2= 11.26𝑃 × 10−6𝑚𝑚

Using similar triangle from the displacement diagram

Consider triangle 𝐴′𝐴′′ 𝐶 ′ 𝑎𝑛𝑑 𝐵′𝐵′′ 𝐶 ′

𝐴′𝐴′′

𝐴′′ 𝐶 ′=

𝐵′𝐵′′

𝐵′′ 𝐶 ′ 𝑜𝑟

𝛿𝐴 + 𝛿𝐶𝐸

450 + 225=

𝛿𝐵𝐷 + 𝛿𝐶𝐸

225

Substituting for 𝛿𝐵𝐷 𝑎𝑛𝑑 𝛿𝐶𝐸 gives

𝛿𝐴 + 11.26𝑃 × 10−6

450 + 225=

6.887𝑃 × 10−6 + 11.26𝑃 × 10−6

225

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

gives 𝑃 as

𝑃 = 𝑃𝑚𝑎𝑥 = 23.200𝑁 𝑜𝑟 23.2 𝑘𝑁

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

𝑡𝑎𝑛𝛼 =𝐴′𝐴′′

𝐴′′ 𝐶 ′=

𝛿𝐴 + 𝛿𝐶𝐸

675

And 𝛿𝐶𝐸 = 0.261 𝑚𝑚 by substituting the value of P

𝑡𝑎𝑛𝛼 =1.0 + 1.261

675=

1.261

675= 0.001868

𝛼 = 0.11°

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

δL =4PL

πEd1d2