Bài giảng Axial Load SBVL

Bài giảng SBVL1 Axial Load được rút gọn với những công thức và ví dụ chính có thể xem để hiểu và ôn lưu ý đây là bài giảng bằng tiếng anh cho các khóa CLC tuy nhiên ai đọc cũng có thể hiểu mong mọi người thành công

Axial Load

Normal stress

Normal strain

Hooke’s Law

The two aluminum rods AB and AC have diameters of 10 mm and 8 mm, respectively Determine the largest vertical force P that can be

supported The allowable tensile stress for the aluminum is σallow = 150 MPa

Elastic Deformation σ = N ( x)

A( x)

ϵ = dδδ dδx

Hooke’s Law

σ = E ( x) ϵ dδδ= N ( x) dδx

E ( x ) A( x )

For the entire length L of the bar, δ= ∫

0

L

N ( x) dδx

E ( x ) A( x )

How to determine δ = displacement of one

point on the bar relative to the other point

P = constant, AE = constant

General loads, AE = constant on L i

δ= ∑

i=1

n SN L i

( AE )i

SN L i is area of N on L i

Sign Convention

Segments AB and CD of the assembly are solid circular rods, and segment BC is a tube If the assembly is made of 6061-T6 aluminum

(E=68.9 GPa), determine the displacement of end D with respect to

end A

Work of a

Force If no energy is lost: the Strain Energy

external work done by the loads converted → converted into internal work called strain energy which is caused by the action of either normal

or shear stress

Strain energy of normal stress

The work done by dFz

The strain energy

Axial Load

External Work and Strain Energy

Castigliano’s Theorem

Conservation

of Energy

δ j=∑

i =1

n

N i(∂N i

∂P )

L i

(AE) i

1

2L

2 AE

If A=100 mm2, E = 200 Gpa, determine the

horizontal displacement at point B

N1=P /√3 N2=−2 P /√3 N3=P

L1=1m L2=2 m L3=√3 m

A1=A A2=A A3=A

E1=E E2=E E3=E

Method 1: Conservation

of Energy

Method 2: Castigliano’s Theorem

1

2 Pδ= 1 2 AE (L1N12+L2N22+L3N32)

δ=(3+√3) P

AE δ=4.732 mm→

∂N1/ ∂P=1/√3 ∂N2/∂P=−2/√3 ∂ N3/∂P=1

δ= 1

AE(

P

√3×

1

√3×1+ 2 P√3 ×

2

√3×2+ P×1×√3)

δ=(3+√3) P

AE δ=4.732 mm→

Statically indeterminate axially loaded member

Principle of superposition

Determine the force developed in each bar Bars

AB and EF each have a cross- sectional area of 50 mm2 , and bar CD has a cross-sectional area of 30 mm2

Force Method for statically

conditions

The equations of equilibrium are not sufficient

to determine all the reactions on a member

→ converted statically indeterminate problem

N1

N2 N

3

X 1 = N4 i=1

i=2

i=3

11X1 p 0

   

0 1

1

0, ,

1,0

N f X P









   

 

2 3

1

3

0 1 1

1

i i

i i

i i

p i

i i

N

N N

L EA







 





11

,

p

i i

X N f X P





Compatibility conditions

Review problems

The assembly shown in Fig.a consists of an

aluminum tube AB having a cross-sectional area of

400 mm2 A steel rod having a diameter of 10 mm is

attached to a rigid collar and passes through the

tube If a tensile load of 80 kN is applied to the rod,

determine the displacement of the end C of the

rod Take Est = 200 Gpa, Eal = 70 GPa

Rigid beam AB rests on the two short posts shown in Fig a

AC is made of steel and has a diameter of 20 mm, and BD is made of aluminum and has a diameter of 40 mm Determine the displacement of point F on AB if a vertical load of 90 kN

is applied over this point Take Est = 200 GPa, Eal = 70 GPa

The rigid bar is supported by the

pin-connected rod CB that has a

cross-sectional area of 14 mm 2 Determine the

vertical deflection of the bar at D when

the distributed load is applied

Review problems

Determine the support reactions at the rigid supports A

and C The material has a modulus of elasticity of E

Determine the vertical displacement at C