LECTURE

Third Edition

RODS: AXIAL LOADING AND DEFORMATION • A. J. Clark School of Engineering •Department of Civil and Environmental Engineering

3 Chapter 2.8

by Dr. Ibrahim A. Assakkaf SPRING 2003 ENES 220 – Mechanics of Materials Department of Civil and Environmental Engineering University of Maryland, College Park

Slide No. 1



Uniform Member – Consider a homogeneous rod as shown in the figure of the next viewgraph. – If the resultant axial stress σ = P/A dos not exceed the proportional limit of the material, then Hooke’s law can applied, that is σ = Eε

σ = axial stress, ε = axial strain E = modulus of elasticity

1

Slide No. 2

Uniform Member

L δ P

Slide No. 3



Uniform Member – From Hooke’s law, it follows that

σ = Eε (Hooke' s law) P = Eε A But ε =

δ

L

, therefore,

δ PL P = E ⇒δ = A L EA

2

Slide No. 4



Uniform Member – The deflection (deformation),δ, of the uniform member subjected to axial loading P is given by

PL δ= EA

(1)

Slide No. 5



Multiple Loads/Sizes – The expression for the deflection of the previous equation may be used only if the rod or the member is homogeneous (constant E) and has a uniform cross sectional area A, and is loaded at its ends. – If the member is loaded at other points, or if it consists of several portions of various cross sections, and materials, then

3

Slide No. 6



Multiple Loads/Sizes – It needs to be divided into components which satisfy individually the required conditions for application of the formula. – Denoting respectively by Pi, Li, Ai, and Ei, the internal force, length, cross-sectional area, and modulus of elasticity corresponding to component i,then n n PL δ = ∑δ i = ∑ i i i =1 i =1 Ei Ai

Slide No. 7



n

n

i =1

i =1

δ = ∑δ i = ∑

E1

E2

E3

L1

L2

L3

Pi Li Ei Ai

4

Slide No. 8

Multiple Loads/Sizes – The deformation of of various parts of a rod or uniform member can be given by n

n

i =1

i =1

δ = ∑δi = ∑

Pi Li Ei Ai

(2)

E1

E2

E3

L1

L2

L3

Slide No. 9



Example 4 – Determine the deformation of the steel rod shown under the given loads. Assume that the modulus of elasticity for all parts is – 29×106 psi 2 Area = 0.9 in

45 kips

30 kips

75 kips

Area = 0.3 in2 12 in

12 in

16 in

5

Slide No. 10



Example 4 (cont’d) – Analysis of internal forces 1

2 45 kips 75 kips

P3

3 30 kips

30 kips

→ + ∑ Fx = 0; − P3 + 30 = 0 ∴ P3 = 30 kips

Slide No. 11



Example 4 (cont’d) – Analysis of internal forces 1

2 45 kips 75 kips

P2

3

45 kips

30 kips

30 kips

→ + ∑ Fx = 0; − P2 − 45 + 30 = 0 ∴ P2 = −15 kips

6

Slide No. 12



Example 4 (cont’d) – Analysis of internal forces 2

1

3

45 kips 75 kips

30 kips

45 kips 75 kips

P1

30 kips

→ + ∑ Fx = 0; − P1 + 75 − 45 + 30 = 0 ∴ P1 = 60 kips

Slide No. 13



Example 4 (cont’d) – Deflection • Input parameters

L1 = 12 in

L2 = 12 in

L3 = 16 in

A1 = 0.9 in 2

A2 = 0.9 in 2

A3 = 0.3 in 2

From analysis of internal forces, P1 = 60 kips = 60,000 lb P2 = −15 kips = −15,000 lb P3 = 30 kips = 30,000 lb

7

Slide No. 14



Example 4 (cont’d) – Carrying the values into the deformation formula: Pi Li 1  P1 L1 P2 L2 P3 L3   =  + + E A E A A A i =1 i i 2 3   1 1  60,000(12) − 15,000(12) 30,000(16)  + +   29 × 106  0.9 0.9 0.3  δ = 0.0759 in 3

δ =∑

Slide No. 15



Relative Deformation A

δA

A δ B/ A = δ B − δ A =

L C

B

D

PL AE

δB

P

8

Slide No. 16



Relative Deformation – If the load P is applied at B, each of the three bars will deform. – Since the bars AC and AD are attached to the fixed supports at C and D, their common deformation is measured by the displacement δA at point A.

Slide No. 17



Relative Deformation – On the other hand, since both ends of bars AB move, the deformation of AB is measured by the difference between the displacements δA and δB of points A and B. – That is by relative displacement of B with respect to A, or

δ B/ A = δ B − δ A =

PL EA

(3)

9

Slide No. 18



Example 5 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.

Slide No. 19



Example 5 (cont’d) SOLUTION: • Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC. • Evaluate the deformation of links AB and DC or the displacements of B and D. • Work out the geometry to find the deflection at E given the deflections at B and D.

10

Slide No. 20



SOLUTION:

Example 5 (cont’d)

Free body: Bar BDE

∑MB = 0 0 = −(30 kN × 0.6 m ) + FCD × 0.2 m FCD = +90 kN tension

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

Slide No. 21



Example 5 (cont’d) Displacement of B:

δB = =

PL AE

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

= −514×10−6 m

δ B = 0.514 mm↑

11

Slide No. 22

Example 5 (cont’d) Displacement of D:

δD = =

PL AE

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

= 300 ×10−6 m

δ D = 0.300 mm ↓

Slide No. 23



Example 5 (cont’d)

Displacement of D: BB′ BH = DD′ HD 0.514 mm (200 mm ) − x = x 0.300 mm x = 73.7 mm EE ′ HE = DD′ HD

δE 0.300 mm

=

(400 + 73.7 )mm

δ E = 1.928 mm

73.7 mm

δ E = 1.928 mm ↓

12

Slide No. 24



Nonuniform Deformation – For cases in which the axial force or the cross-sectional area varies continuously along the length of the bar, then Eq. 1 – (PL / EA) is not valid. – Recall that in the case of variable cross section, the strain depends on the position of point Q, where it is computed from

ε=

dδ dx

Slide No. 25



Nonuniform Deformation

ε=

dδ dx

L

x

P

13

Slide No. 26



Nonuniform Deformation – Solving for dδ and substituting for ε

dδ = εdx – But ε = σ / E, and σ = P/A. therefore

dδ = εdx =

σ E

dx =

P dx EA

(4)

Slide No. 27



Nonuniform Deformation – The total deformation δ of the rod or bar is obtained by integrating Eq. 4 over the length L as L

δ =∫ 0

Px dx EAx

(5)

14

Slide No. 28



Example 6 – Determine the deflection of point a of a homogeneous circular cone of height h, density ρ, and modulus of elasticity E due to its own weight. a

h

Slide No. 29



Example 6 (cont’d) – Consider a slice of thickness dy – P = weight of above slice – = ρg (volume above) y 1 2  δy P = ρg  πr y  3   1 ρg  πr 2 y  Pdy ρg 3 dδ = ydy =  2 = EA E (πr ) 3E

r

h

15

Slide No. 30



Example 6 (cont’d) dδ = h

δ =∫

ρg 3E

ρg

3E 0

δ=

ρgh

ydy ydy =

ρg y 2 3E 2

h

0

2

6E

Slide No. 31



Normal Stresses in Tapered Bar – Consider the following tapered bar with a thickness t that is constant along the entire length of the bar.

h1

h(x)

h2

x

L

16

Slide No. 32



Normal Stresses in Tapered Bar h1

h(x)

x

h2

L

– The following relationship gives the height h of tapered bar as a function of the location x

hx = h1 + (h2 − h1 )

x L

(6)

Slide No. 33



Normal Stresses in Tapered Bar h1

h(x)

h2

x

L

– The area Ax at any location x along the length of the bar is given by

x  Ax = thx = t h1 + (h2 − h1 )  L 

(7)

17

Slide No. 34



Normal Stresses in Tapered Bar h1

h2

h(x)

x

L

– The normal stress σx as a function of x is given

σx =

P P = x Ax  t h1 + (h2 − h1 )  L 

(8)

Slide No. 35



Example 7 – Determine the normal stress as a function of x along the length of the tapered bar shown if – h1 = 2 in – h2 = 6 in – t = 3 in, and – L = 36 in x h1 h2 h(x) – P = 5,000 lb L

18

Slide No. 36



Example 7 (cont’d) – Applying Eq. 8, the normal stress as a function of x is given by P P = x Ax  t h1 + (h2 − h1 )  L  5000 5000 σx = = x x  32 + (6 − 2) )  6 + 3 36  

σx =

σx =

15000 18 + x

Slide No. 37



Example 7 (cont’d)

x (in)

– Max σ = 833.3 psi • At x =0

– Min σ = 277.8 psi • At x =36 in

h1

h(x) L

h2

x

0 3 6 9 12 15 18 21 24 27 30 33 36

σ (psi) 833.3 714.3 625.0 555.6 500.0 454.5 416.7 384.6 357.1 333.3 312.5 294.1 277.8

19

Slide No. 38



Deflection of Tapered Bar – Consider the following tapered bar with a thickness t that is constant along the entire length of the bar.

h1

dx

x

h2

h(x) L

Slide No. 39



Deflection of Tapered Bar – Recall Eq. 5

L

δ =∫ 0

– Substitute Eq. 7

Px dx EAx

– into Eq. 5, therefore L

PL 1 δ= dx ∫ Et 0 h1 L + (h2 − h1 )x

(9)

20

Slide No. 40 ENES 220 ©Assakkaf