Download - 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
The spring increases by πΏ under the action
of force π and the final length is πΏ + πΏ
If material is linearly elastic, load and
elongation will be proportional
π = ππΏ πΏ = ππ
Where π πππ π are constants of proportionality
π is the stiffness and it is defined as the force
required to produce a unit elongation π = π πΏ
π is the flexibility and it is defined as the
elongation produced by a load of unit value
π = πΏ π
NOTE:
The stiffness and flexibility of a spring is reciprocal
of each other.
π =1
π π =
1
π
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 Hookeβs Law
Ο = Ξ΅E Ξ΅ =Ο
E=
P
AE
From definition of strain
Ξ΅ =Ξ΄
L
Equating and solving for deformation
Ξ΄ =PL
AE
BARS WITH INTERMEDIATE AXIAL LOADING
SOLUTION PROCEDURE:
(1). Identify the segments of the bar under study. (2). Determine the internal axial forces πΉ1, πΉ2 , πΉ3, ππ‘π, In all segments from the free-body diagrams of the bar under study. Note that the internal axial forces are denoted by the letter πΉ1, πΉ2 , πΉ3, ππ‘π 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.
πΏ1 =π1πΏ1
π΄1πΈ1, πΏ2 =
π2πΏ2
π΄2πΈ2 πΏ3 =
π3πΏ3
π΄3πΈ3 πΏ4 =
π4πΏ4
π΄4πΈ4β¦
In which πΏ1 , πΏ2 , πΏ3 πππ πΏ4 are the length of the
segments and πΈπ΄ is the axial rigidity of the bar.
(4). Add πΏ1, πΏ2, πΏ3 πππ πΏ4 to obtain πΏ, the change
in length of the entire bar;
πΏ = πΏπ
3
π=1
= πΏ1 + πΏ2 + πΏ3 + πΏ4 β¦
The changes in lengths must be added
algebraically, with elongation being positive and
shortenings negative.
ALTERNATIVE
BARS CONSISTING OF PRISMATIC SEGMENTS
Ξ΄ = PiLi
AiEi
π
π=1
The stiffness and flexibility is given as
k =EA
L f =
L
EA
BARS WITH CONTINUOUSLY VARYING LOADS OR DIMENSIONS
Ξ΄ =PL
EA
Substituting N x for P, dx for L and A x for A
ππΏ =π π₯ ππ₯
πΈπ΄ π₯
Integrating over the length gives the elongation of the entire bar
ππΏ
πΏ
0
= π π₯ ππ₯
πΈπ΄ π₯
πΏ
0
Example 1:
Example 2:
Determine the deformation of the steel rod shown in fig
below under the given load [πΈ = 200 Γ 109 π π2
Example 4: A member π΄π΅πΆπ· is subjected to point loads
π1 , π2 , π3 πππ π4 as shown below. If π1 = 120 ππ, π2 =
220 ππ πππ π4 = 160 ππ. Determine the net change in
length of the member. [Take πΈ = 200 πΊ π π2 ]
Example 5:
The device shown in the figure below consists of a horizontal beam π΄π΅πΆ
supported by two vertical bars π΅π· πππ π΅πΆ. Bar πΆπΈ is pinned at both ends but bar
π΅π· is fixed to the foundation at its lower end. The distance from π΄ π‘π π΅ is
450 ππ πππ π΅ π‘π πΆ ππ 225 ππ. Bars π΅π· πππ πΆπΈ have lengths of 480 ππ and
600 ππ respectively, and their cross-sectional areas are
1020 ππ2 πππ 520 ππ2 respectively. The bars are made of steel having
modulus of elasticity πΈ = 205 πΊππ. Assuming the beam π΄π΅πΆ is rigid, find the
maximum allowable load ππππ₯ if the displacement of π΄ is limited to 1.0 ππ.
Taking moment about B summing forces in the vertical direction
gives.
πΉπΆπΈ = 2π πΉπ΅π· = 3π
Member CE is in tension and member BD is in compression.
Shortening of member BD
πΏπ΅π· =πΉπ΅π· ΓπΏπ΅π·
πΈπ΄π΅π· =
3πΓ480 ππ
205πΊππΓ1020 ππ 2 = 6.887π Γ 10β6ππ
Where π = πππ€π‘πππ
Lengthening of member CE is
πΏπΆπΈ =πΉπΆπΈ Γ πΏπΆπΈ
πΈπ΄πΆπΈ =
2π Γ 600 ππ
205πΊππ Γ 1020 ππ2= 11.26π Γ 10β6ππ
Using similar triangle from the displacement diagram
Consider triangle π΄β²π΄β²β² πΆ β² πππ π΅β²π΅β²β² πΆ β²
π΄β²π΄β²β²
π΄β²β² πΆ β²=
π΅β²π΅β²β²
π΅β²β² πΆ β² ππ
πΏπ΄ + πΏπΆπΈ
450 + 225=
πΏπ΅π· + πΏπΆπΈ
225
Substituting for πΏπ΅π· πππ πΏπΆπΈ gives
πΏπ΄ + 11.26π Γ 10β6
450 + 225=
6.887π Γ 10β6 + 11.26π Γ 10β6
225
Substituting πΏπ΄ = 1.0 i.e. its limiting value and solve
gives π as
π = ππππ₯ = 23.200π ππ 23.2 ππ
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‘πππΌ =π΄β²π΄β²β²
π΄β²β² πΆ β²=
πΏπ΄ + πΏπΆπΈ
675
And πΏπΆπΈ = 0.261 ππ by substituting the value of P
π‘πππΌ =1.0 + 1.261
675=
1.261
675= 0.001868
πΌ = 0.11Β°
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
ΟEd1d2