2-Axial Loading.pdf

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

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Axial Loading Materials for this chapter are taken from : 1.

Ferdinand P. Beer, E. Russell Johnston,Jr, John T. Dewolf, David F. Mazurek “ Mechanics of Materials” 5th Edition in SI units

2.

R.C.Hibbeler “ Mechanics of Materials “ Seventh Edition

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 1

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Introduction Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced under loading. Statics analyses alone are not sufficient. Considering structures as deformable allows determination of member forces and reactions which are statically indeterminate. Determination of the stress distribution within a member also requires consideration of deformations in the member. Chapter 2 is concerned with deformation of a structural member under axial loading. Later chapters will deal with torsional and pure bending loads.

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 2

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Saint-Venant’s Principle 1.

Saint-Venant’s Principle states that both localized deformation and stress tend to “even out” at a distance sufficiently removed from these regions.

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 3

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Normal Strain under Axial Loading

P σ = = stress A

ε=

δ L

= normal strain

2P P = σ = 2A A

ε =

δ

L

MEC411 – MECHANICS OF MATERIALS

P σ= A 2δ δ = ε= 2L L Ch 2 - 4

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Stress-Strain Test

rutlandplastics.co.uk

www.tensilkut.com

deeshaimpex.com

Tensile Test Machines & Specimens MEC411 – MECHANICS OF MATERIALS

Ch 2 - 5

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Stress-Strain Test: Ductile Materials

necking rupture

Test specimen with tensile load elongation occurs until “necking” and rupture. MEC411 – MECHANICS OF MATERIALS

Ch 2 - 6

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Stress-Strain Test: Brittle Materials

Pottery, glass, and cast iron Material that breaks suddenly under stress at a point just beyond its elastic limit Brittle materials may also break suddenly when given a sharp knock

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 7

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Deformations of Members under Axial Loading From Hooke’s Law:

σ = Eε

ε=

σ E

=

P AE

From the definition of strain:

ε=

δ L

Equating and solving for the deformation,

PL δ = AE With variations in loading, cross-section or material properties,

PL δ =∑ i i

i Ai Ei

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 8

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Example 2.1

75 k

45 k 30 k

STEPS: Divide the rod into components at the 120 mm

120 mm

200 mm

load application points.

E = 200 GPa

Apply a free-body analysis on each

D = 100mm d = 50mm

component to determine the internal force

Determine the deformation of the steel rod shown under the given

Evaluate the total of the component deflections.

loads.

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 9

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

SOLUTION: Apply free-body analysis to each component to Divide the rod into three

determine internal forces,

components: A

B

75 k

C

P1 = 30kN D 30 k

P2 = −15kN P3 = 60kN

45 k P1

30 k

Evaluate total deflection,

45 k 30 k P2

δ =∑ i

75 k

45 k 30 k

P3

PL 1  PL P L P L  i i =  1 1+ 2 2 + 3 3 Ai Ei E  A1 A2 A3 

 ( 30 × 103 ) 0.12 ( −15 × 103 ) 0.12 ( 60 × 103 ) 0.2  1 = + +   200 × 109  7.854 × 10−3 7.854 × 10−3 1.96 × 10−3  = 3.1758 × 10−5 m

L1 = L2 = 120mm

L3 = 200mm

A1 = A2 = 7.854 × 10-3 m 2 A3 = 1.96 × 10−3 m 2

δ = 3.1758 × 10−5 m

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 10

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Example 2.2

The rigid bar BDE is supported by two links AB and CD. Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2. Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600 mm2). For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E.

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 11

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Solution

Displacement of B:

Free body: Bar BDE

δB =

PL AE

( − 60 ×103 N )(0.3 m ) = (500 ×10-6 m2 )(70 ×109 Pa ) = −514 ×10 − 6 m

δ B = 0.514 mm ↑ Displacement of D:

∑MB = 0 0 = −(30 kN × 0.6 m ) + FCD × 0.2 m

δD =

FCD = +90 kN tension

∑ MD = 0 0 = −(30 kN × 0.4 m ) − FAB × 0.2 m FAB = −60 kN compression

PL AE

( 90 ×103 N )(0.4 m ) = (600 ×10-6 m2 )(200 ×109 Pa ) = 300 ×10 − 6 m

δ D = 0.300 mm ↓ MEC411 – MECHANICS OF MATERIALS

Ch 2 - 12

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Displacement of D:

BB′ BH = DD′ HD

0.514 mm ( 200 mm) − x = 0.300 mm x x = 73.7 mm

EE′ HE = DD′ HD

δE

400 + 73.7) mm ( =

0.300 mm 73.7 mm δE =1.928 mm

δ E = 1.928mm↓

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 13

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Statically Indeterminate Problems

Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate.

A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium.

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 14

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Statically Indeterminate Problems Redundant reactions are replaced with unknown loads which along with the other loads

must

produce

compatible

deformations. Deformations due to actual loads and redundant

reactions

are

determined

separately and then added or superposed.

δ = δL +δR = 0

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 15

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Example 2.3

Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied.

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 16

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Example 2.3 STEPS: Consider the reaction at B as redundant, release the bar from that support, and solve for the displacement at B due to the applied loads. Solve for the displacement at B due to the redundant reaction at B. Require that the displacements due to the loads and due to the redundant reaction be compatible, i.e., require that their sum be zero. Solve for the reaction at A due to applied loads and the reaction found at B.

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 17

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Solve for the displacement at B due to the applied loads with the redundant constraint released,

P1 = 0 P2 = P3 = 600×103 N P4 = 900×103 N A1 = A2 = 400×10−6 m2

A3 = A4 = 250×10−6 m2

L1 = L2 = L3 = L4 = 0.150 m Pi Li 1.125×109 δL = ∑ = E i Ai Ei Solve for the displacement at B due to the redundant constraint,

P1 = P2 = − RB A1 = 400 ×10 − 6 m 2 L1 = L2 = 0.300 m

(

A2 = 250 × 10− 6 m 2

)

Pi Li 1.95 × 103 RB δR = ∑ =− A E E i i i MEC411 – MECHANICS OF MATERIALS

Ch 2 - 18

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Require that the displacements due to the loads and due to the redundant reaction be compatible,

δ = δL +δR = 0

(

)

1.125×109 1.95×103 RB δ= − =0 E E RB = 577 ×103 N = 577 kN Find the reaction at A due to the loads and the reaction at B

∑ Fy = 0 = R A − 300 kN − 600 kN + 577 kN R A = 323 kN

RA = 323 kN RB = 577 kN

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 19

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Example 2.4 The A-36 steel rod shown has a diameter of 5 mm. It is attached to the fixed wall at A, and before it is loaded there is a gap between the wall at and the rod of 1 mm. Determine the reactions at A and B’.

Solution Consider the support at B’ as redundant and using principle of superposition,

+→

MEC411 – MECHANICS OF MATERIALS

0.001 = δ p − δ B

Ch 2 - 20

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Example 2.4 By substituting into compatibility equation,

Thus,

( )

 20 103  ( 0.4 ) PL   δ P = AC = AE π ( 0.0025 )  200 109    = 0.002037 m

( )

δB =

FB (1.2 ) FB LAB = AE π ( 0.0025 )  200 109   

( )

(

(

)

0.001 = 0.002037 − 0.3056 10−6 FB

( )

FB = 3.39 103 = 3.39 kN (Ans) From the free-body diagram,

)

= 0.3056 10−6 FB

+→

∑F

x

= 0;

− FA + 20 − 3.39 = 0 FA = 16.6 kN (Ans)

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 21

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Problems Involving Temperature Changes A temperature change 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. Treat the additional support as redundant and apply the principle of superposition.

PL AE α = thermalexpansion coef.

δT = α ( ∆T ) L ; δ P =

The thermal deformation and the deformation from the redundant support must be compatible.

δ = δT + δ P = 0 PL α ( ∆T ) L + =0 AE

MEC411 – MECHANICS OF MATERIALS

δ = δT + δ P = 0 P = AEα ( ∆T )

σ=

P = Eα ( ∆T ) A Ch 2 - 22

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Example 2.5

A steel rod is stretched between two rigid walls and carries a tensile load of 5000 N at 20°C. If the allowable stress is not to exceed 130 MPa at -20°C, what is the minimum diameter of the rod? Assume α = 11.7 µm/(m—°C) and E = 200 GPa.

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 23

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Example 2.5 Apply the total deformation formula

Get the diameter considering

and obtain the area of the rod.

formula for area.

δ = −δ T + δ st σL PL = −α L ( ∆T ) +

1 2 π d = 137.36 4 d = 13.22 mm Ans.

E

AE

σ = −α E ( ∆T ) +

(

P A

)

130 = − 11.7 ×10−6 ( 200000 )( −20 − 20 ) +

5000 A

A = 137.36 mm 2

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 24

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Example 2.6 The rigid bar is fixed to the top of the three posts made of A-36 steel and 2014-T6 aluminum. The posts each have a length of 250 mm when no load is applied to the bar, and

the

temperature

is

T1

=

20°C.

Determine the force supported by each post if the bar is subjected to a uniform distributed load of 150 kN/m and the temperature is raised to T2 = 80°C.

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 25

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Example 2.6 Solution From free-body diagram we have

+↑

∑F

y

= 0;

( )

2 Fst + Fal − 90 103 = 0

(1)

The top of each post is displaced by an equal amount and hence,

( + ↓)

δ st = δ al

The final position of the top of each post is equal to its displacement caused by the temperature increase and internal axial compressive force.

( + ↓) ( + ↓)

δ st = − (δ st )T + (δ st ) F δ al = − (δ al )T + (δ al ) F

(2) MEC411 – MECHANICS OF MATERIALS

Ch 2 - 26

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Applying Eq. 2 gives

− (δ st )T + (δ st )F = −(δ st )T + (δ al )F With reference from the material properties, we have

[ ( )]

− 12 10 −6 (80 − 20)(0.25) +

Fst (0.25) Fal (0.25) −6 ( )( ) = − 23 10 80 − 20 0 . 25 + π (0.02)2 200 109 π (0.03)2 73.1 109

[ ( )]

[ ( )]

( )

Fst = 1.216 Fal − 165.9 103

[ ( )]

(3)

Solving Eqs. 1 and 3 simultaneously yields

Fst = −16.4 kN and Fal = 123 kN (Ans)

MEC411 – MECHANICS OF MATERIALS

Ch 2 - 27

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Supplementary Problem 2 1. An aluminium tube is fastened between a steel rod and a bronze rod as shown. Axial loads are applied at the positions indicated. Find the value of P that will not exceed a maximum overall deformation of 2 mm or a stress in the steel of 140 MPa, in the aluminium of 80 MPa, or in bronze of 120 MPa. Assume that the assembly is suitably braced to prevent buckling. 0.6 m

1.0 m

0.8 m 2P

3P P Bronze A = 450 mm2 E = 83 GPa

4P Aluminium A = 600 mm2 E = 70 GPa

Steel A = 300 mm2 E = 200 GPa D

2.

The assembly consists of three titanium (Ti-6A14V) rods and a rigid bar AC. The cross-sectional area of each rod is given in the figure. If a force of 30 kN is applied to the ring F, determine the horizontal displacement of point F.

C

ACD = 600 mm2 0.6 m AEF = 1200 mm2

0.3 m F

E 0.3 m

AAB = 900 mm2 B

MEC411 – MECHANICS OF MATERIALS

1.2 m

1.8 m

A

Ch 2 - 28

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Supplementary Problem 2 500 mm

3. The composite bar in Fig. 2.3 is stress-free before the axial loads P1 and P2 are applied. Assuming that the walls are rigid, calculate the stress in each material if P1 = 150 kN and P2 = 90 kN.

4.

P1

Aluminium A = 900 mm2 E = 70 GPa

250 mm

350 mm

P2

Steel A = 2000 mm2 E = 200 GPa

Bronze A = 1200 mm2 E = 83 GPa

A rigid block of mass M is supported by three symmetrically spaced rods as shown. Each copper rod has an area of 900 mm2; E = 120 GPa; and the allowable stress is 70 MPa. The steel rod has an area of 1200 mm2; E = 200 GPa; and the allowable stress is 140 MPa. Determine the largest mass M which can be supported. MEC411 – MECHANICS OF MATERIALS

Ch 2 - 29

CHAPTER 2 AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS

Supplementary Problem 2 5.

6.

The assembly has the diameters and material make-up indicated. If it fits securely between its fixed supports when the temperature is T1 = 20°C, determine the average normal stress in each material when the temperature reaches T2 = 40°C. The two circular rod segments, one of aluminum and the other of copper, are fixed to the rigid walls such that there is a gap of 0.2 mm between them when T1 = 15°C. What larger temperature T2 is required in order to just close the gap? Each rod has a diameter of 30 mm . Determine the average normal stress in each rod if T2 = 95°C.

1.2 m

1.8 m

0.9 m

A

D B Aluminium E = 70 GPa Ø = 300 mm

C Bronze E = 83 GPa Ø = 200 mm

Stainless Steel E = 200 GPa Ø = 100 mm

0.2 mm

Copper

Aluminium

100 mm

200 mm

Copper E = 126 GPa α =17(10-6)/oC

MEC411 – MECHANICS OF MATERIALS

Aluminium E = 70 GPa α = 24 (10-6)/oC

Ch 2 - 30

FACULTY OF MECHANICAL ENGINEERING

CHAPTER 2 AXIAL LOADING

DIVISION OF ENGINEERING MECHANICS

Supplementary Problem 2 7. Four steel bars jointly support a mass of 15 Mg as shown in Fig. 2.7. Each bar has a cross-sectional area of 600

mm2.

Find the

load

bar

after

carried

by

each

a

A

B

C

D

60o 15o

15o

temperature rise of 50°C. Assume α = 11.7 µm/(m—°C) and E = 200 GPa.

MEC411 – MECHANICS OF MATERIALS

15 Mg

Ch 2 - 31