Lecture notes, lecture 15 - Axial loading PDF

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AXIAL LOADING...

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Chapter 10 Axial Loading

Objectives • To determine the deformation of axially loaded members

• To determine the support reactions when these reactions cannot be determined solely from the equation of equilibrium

• To analyze the effects of thermal stresses stresses

Chapter 10 – Axial Loading

10 10--2

Saint-Venant’s Principle Concentrated loads result in large stresses in the vicinity of the load application point. Stress and strain distributions become uniform at a relatively short distance from the load application points.

Saint-Venant’s Principle: Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points.

Chapter 10 – Axial Loading

10 10--3

Elastic Deformation of an Axially Loaded Member • The bar with cross-section which is gradually varies along its length L is subjected to concentrated loads at its ends and a variable external load distributed along its length • The normal stress will be uniformly distributed over the cross section →bar will deform uniformly • Differential element of length dx and cross-sectional A(x) is isolated from the bar at the arbitrary position x. The resultant internal axial force is represented as P(x). • The displacement of one end with respect to the other end is dδ. Chapter 10 – Axial Loading

10 10--4

Elastic Deformation of an Axially Loaded Member The stress and strain in the element are: σ=

P ( x) A( x )

and

ε=

dδ dx

From Hooke’s Law: σ = Eε

dδ =

→

P ( x) ⎛ dδ ⎞ = E⎜ ⎟ A( x) ⎝ dx ⎠

P( x) dx A (x )E

For the entire length L of the bar, the end displacement is L

δ =∫ 0

P( x) dx A( x ) E

Chapter 10 – Axial Loading

10 10--5

Deformation under Axial Loading

If a constant external force is applied at its end and the bar has a constant cross-sectional area A and the material is homogeneous (constant E), the end displacement is

With variations i ti iin loading, l di cross-section or material properties, δ =∑ i

δ =

Pi Li Ai E i

PL AE Chapter 10 – Axial Loading

10 10--6

Example 10.1 A composition steel bar shown in the figure is made from two segments, AB and BD, having cross-sectional areas of AAB = 600mm2 and ABD = 1200mm 2 . The modulus of elasticity E of the steel is 210 GPa. GPa Determine the vertical displacement of end A and the displacement of B relative to C.

Chapter 10 – Axial Loading

10 10--7

Example 10.1 Internal Force: Due to the application of the external loads, the internal axial forces in regions AB, BC, and CD will all be different. These forces are obtained be applying method of sections.

Chapter 10 – Axial Loading

10 10--8

Example 10.1 Di l Displacement: The vertical displacement of A relative to the fixed support D is: δA = ∑

PL + 75 × 103 (1000) + 35 × 103 ( 750) − 45 ×10 3 (500) + + = +0.61mm = AE 600( 210 × 10 3 ) 1200( 210 × 103 ) 1200( 210 ×103 )

Since the Si h result l iis positive, i i the h bar b elongates l andd so the h displacement at A is upward. The displacement of B relative to C is: δ B/ C

PBC LBC + 35× 10 3 (750) = + 0.104mm = = ABC E 1200( 210 ×103 )

Here B moves away from C, since the segment elongates

Chapter 10 – Axial Loading

10 10--9

Example 10.3

A rigid beam AB rests on the two short posts shown in the figure. AC is made of steel and has a diameter of 20 mm, and BD is i made d off aluminum l i andd has h a di diameter t off 40 mm. Determine the displacement of point F on AB if vertical load of 90 kN is applied over this point. Take Est = 200GPa and Eal = 70GPa Chapter 10 – Axial Loading

10 10--10

Example 10.3 Internal Force: The compressive forces acting at the top of each post are determined from the equilibrium of member AB. These forces are equal to the internal forces in each post.

Chapter 10 – Axial Loading

1010-11

Example 10.3 Displacement:

A diagram shows the centerline

The displacement of the top of each post is

displacements at point A, B and F on the beam. By proportion of

Post AC:

the shaded triangle, the displacement

− 60 ×103 (300) PAC L AC = δA = AAC E st π 2 20 (200 × 103 ) 4 = −0.286mm

of point F is therefore

Post BD: PBD L BD − 30 × 103 (300) = δB = ABD Eal π 2 × 3 40 (70 10 ) 4 = −0 .102mm

⎛ 400 ⎞ ⎟0.184mm 600 ⎝ ⎠

δ F = 0.102mm + ⎜ = 0.255mm(↓)

Chapter 10 – Axial Loading

10 10--12

Principle of Superposition The principle of superposition is often used to determine the stress or displacement at a point in a member when the member is subjected to a complicated loading. loading By subdividing the loading into components, the principle of superposition states that the resultant stress or displacement at the point can be determined by first finding the stress or displacement caused by its component load acting separately on the member. member The resultant stress or displacement is then determined by algebraically adding the contributions caused by each of of the the components.

Chapter 10 – Axial Loading

10 10--13

Principle of Superposition

Two conditions must be valid before the principle of superposition can be applied: • The loading must be linearly related to the stress or displacement that is to be determined. • The loading must not significantly change the original geometry or configuration of the member. Chapter 10 – Axial Loading

10 10--14

Statically Indeterminate Axially Loaded Member If a bar is fixed at both ends, two unknown axial reaction occur, and equations of equilibrium alone is not enough to solve these unknown. In this case, the bar is called statically indeterminate. For statically indeterminate problems, a compatibility or kinematic condition should be specified, in addition to equation of equilibrium. E ti off equilibrium: Equation ilib i

∑F

y

= 0 → FA + FB − P = 0

Compatibility or kinematic condition:

δ A /B = 0 Chapter 10 – Axial Loading

10 10--15

Statically Indeterminate Axially Loaded Member The compatibility condition can be expressed in terms of the applied loads by using load-displacement relationship, which depends on material behavior. Realizing that the internal force in segment AC is + FA and in segment CB is − F B , the compatibility equation can be written as: F A L AC F B L CB − =0 AE AE

Assuming that AE is constant, the two equations (equilibrium and compatibility) can be solved to obtain the two reactions. ⎛L ⎞ F A = P ⎜ CB ⎟ ⎝ L ⎠

⎛L ⎞ FB = P ⎜ AC ⎟ ⎝ L ⎠

Both of the results are positive, so the reaction are shown correctly on FBD. Chapter 10 – Axial Loading

10 10--16

Example 10.5

A steel rod in the figure has a diameter of 5 mm. mm It is attached to the fixed wall at A, and before it is loaded, there is a gap between the wall at B’ and the rod of 1mm. Determine the reactions at A and B’ if the rod is subjected to an axial force of P=20 kN as shown in the figure. Neglect the size of the collar at C. Take E st = 200GPa Chapter 10 – Axial Loading

10 10--17

Example 10.5 Check whether the gap is closed: Segment AC will be elongated due to the load P

PLAC + 20 ×103 (400) = δ A /C = AACE π 5 2 (200 ×10 3 ) 4 = +2.037mm

Equilibrium: The FDB of the bar is shown. The problem is statically indeterminate since there are two unknowns and only one equation of equilibrium.

Since δ A / C is larger than 1 mm, the gap will be closed due to the load P

∑F

x

= 0 → −FA − FB + 20kN = 0

Chapter 10 – Axial Loading

(1)

10 10--18

Example 10.5 δ B / A = 1mm =

Compatibility:

The loading causes point B to move to B’ with no further displacement. Therefore the compatibility condition for the rod is

δ B / A = 1mm

π 4

FA (400)

−

5 (200 × 10 ) 2

3

FA LAC FB LBC − AE AE π 4

FB (800)

= 1mm

5 (200 × 10 ) 2

3

FA (400) − FB (800) = 3927 N ⋅ m

This displacement can be expressed in terms of the unknown reactions by using the loading-displacement relationship applied to segments AC and CB.

(2)

Solving equations (1) and (2),

Chapter 10 – Axial Loading

FA = 16.6 kN FB = 3.39kN

1010-19

Thermal Stresses

A temperature changes results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports.

The thermal deformation of a member having a length L can be calculated as δ T = α Δ TL

α is coefficient of thermal expansion

Chapter 10 – Axial Loading

1010-20

Thermal Stresses There will be compressive stress in a bar with restrains at both ends when it is subjected to temperature increase. • Treat the additional support as redundant and apply the principal of superposition. PL δP = δT = αΔ TL AE • The thermal deformation and the d f deformation ti from f the th redundant d d t support must be compatible.

δ = δT + δP = 0 α ( ΔT )L +

PL =0 AE

Chapter 10 – Axial Loading

δ = δT + δ P = 0 P = − AEα ( ΔT )

σ=

P = − Eα (ΔT ) A 1010-21

Example 10.9 A steel bar shown in the figure is constrained t i d tto jjustt fit b etween t ttwo fixed supports when T1 =30 ˚C. If the temperature is raise to T2 = 60 ˚C, determine the avarage normal thermal stress developed in the bar. The linear coefficient of thermal expansion α of the steel is 12 ×10 −6 / ° C and the modulus of

elasticity E of the steel is 200 GPa.

Chapter 10 – Axial Loading

1010-22

Example 10.9 Equilibrium: The FBD of the bar is shown. Since there is no external load, the force at A is equal but opposite to the force acting at B; that is

∑F

y

= 0 → FA = FB = F

The problem is statically indeterminate since this force cannot be determined from equilibrium

Chapter 10 – Axial Loading

1010-23

Example 10.9 Compatibility:

Since δ A/ B =0, the thermal displacement δT at A that would occur is counteracted by the force F that would be required to push the bar δ F back to its original position. The compatibility condition at A becomes

δ A/ B = 0 = δ T − δ F Therefore, Applying the thermal and loaddisplacement relationships,

αΔ TL −

FL =0 AE

F = α ΔTAE = 12 × 10− 6 (60° − 30°)(10) 2 ( 200 × 103 ) = 7200 N

Chapter 10 – Axial Loading

1010-24

Example 10.9 From the magnitude of F, it should be apparent that changes in the temperature can cause large reactions forces in statically i d t indeterminate i t members. b Since F also represents the internal axial force within the bar, the average normal compressive stress is σ=

F 7200 = = 72 MPa A 10 2

Chapter 10 – Axial Loading

1010-25...