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MECHANICS OF ENGINEERING ENGINEERING MATERIAL MATERIALS S Top opic ic 10: 10: Elementary Elementary Plast Plastic icit ity y – Plasti lastic c Bending Bendi ng Plasti lastic c Col Collapse lapse
Contents 1.Plastic bending of beams: plastic moment 2.Plastic collapse of beams 3.Plastic torsion of shafts
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MECHANICS OF ENGINEERING ENGINEERING MATERIAL MATERIALS S 1. Plastic bending of beams: plastic moment Beam Pure bending assumption: 1. Beam Beam mater material ial is in in a condi condition tion of simpl simplee tensio tension n or compre compressio ssion n 2. Any cross cross-sec -sectio tion n of the beam beam will will rema remain in plan planee during during bending bending
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MECHANICS OF ENGINEERING ENGINEERING MATERIAL MATERIALS S Fundamental undamental concepts con cepts of pla pl astic sti c beha behavi viour our
Stress-strain curv e for mild mil d steel steel σ
uts
σ
yu
Plastic (slope = 0)
σ
y
Failure
Strain hardening (slope ≈ 0.04E)
Elastic (slope = E)
0.0012
0.014
≈ 0.2
Strain
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MECHANICS OF ENGINEERING ENGINEERING MATERIAL MATERIALS S Elastic – plastic stress – strain relationship
nonlinear
linear
a real strain – hardening material
semi-idealized, hardening linearly from yield
strain occurs inertial
strain hardening ignore Linear elastic non – hardening plastic relationship
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MECHANICS OF ENGINEERING ENGINEERING MATERIAL MATERIALS S
idealized elastoplastic material
which doesn’t exist
Member has a rectangular cross section of width b and depth d= 2c
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MECHANICS OF ENGINEERING MATERIALS Rectangular section
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MECHANICS OF ENGINEERING MATERIALS Case (a) : Normal stress σx dose not exceed the yield strength
σy
- Hook’s law applies
σ x ≤ σ Y
σ m
=
Mc I
<
σ Y
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MECHANICS OF ENGINEERING MATERIALS Case (b) : Bending moment increase,
σ
m eventually
reaches the value
σ m σ y
=
=
σy
σ Y σ m
=
M Y c I
I M Y = σ Y c = maximum elastic moment
For rectangular cross section: 1 3 b( 2c ) I M Y = σ Y = 12 σ Y c c
=
2 3
2
bc σ Y = σ Y
bd 2 6
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MECHANICS OF ENGINEERING MATERIALS Case (c) : • If the moment is increased beyond the maximum elastic moment, plastic zones develop around an elastic core. • The total bending moment is obtained by consideration of both the plastic stress near the top and bottom of the beam and the elastic stress in the core of the beam (elastic – plastic bending moment) h
M = M 1 + M 2 + M 3 + M 4 M 3
=
M 2
M 4
=
M 1
M = F ave ⋅ d = σ ave ⋅ A ⋅ d d − h 1 h M 1 = σ 1ave ⋅ A1 ⋅ d 1 = σ Y ⋅ bh ⋅ c − = bσ Y h( ) = bσ Y h( d − h ) 2 2 2 M 2
=
σ 2 ave ⋅ A2 ⋅ d 2
=
σ Y 2
⋅
b(c − h ) ⋅
2 3
(c − h) =
1 3
bσ Y (c − h)
2
=
1 12
bσ Y (d − 2h)
2
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MECHANICS OF ENGINEERING MATERIALS M = M 1 + M 2 + M 3 + M 4 M = 2 ⋅
1 2
bσ Y h(d − h ) + 2 ⋅
1 12
bσ Y (d − 2 h ) = bσ Y h(d − h ) + 2
h h σ Y bd 2 = 1 + 2 1 − 6 d d At a distance (d/2-h) from the neutral axis, the stress has just reached the value of σ y; then, if R is the radius of curvature,
The relationship between M and 1/R is linear up to value of M =MY. Beyond this point the relationship is nonlinear and the slope decrease in depth, h, of the plastic state
1 6
bσ Y (d − 2 h )
2
h
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MECHANICS OF ENGINEERING MATERIALS Case (d) :
In the limit as the moment is increased further, the elastic core thickness goes to zero, corresponding to a fully plastic deformation.
h=
d 2
M p =
3 2
σ
bd 2
Y
3 = M Y 2
6
= σ Y =
bd 2 4
fully plastic moment
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MECHANICS OF ENGINEERING MATERIALS
Z plastic =
M p M Y
For rectangular section,
= plastic section modulus
Z plastic =
bd 2 4
Z plastic is simply a function of the geometry of the cross section and is often available in standard tables of data.
M p = σ Y Z plastic
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MECHANICS OF ENGINEERING MATERIALS I section
Elastic condition:
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MECHANICS OF ENGINEERING MATERIALS In the fully plastic condition: The fully plastic moment is:
The shape factor is:
In an I-beam, 100 mm x 300 mm, with flanges and we 14 m and 98 mm thick respectively
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MECHANICS OF ENGINEERING MATERIALS Asymmetrical section If the cross-section is asymmetrical about the axis of bending, then the position of the neutral axis must be determined
- YY is the axis of symmetry - ZZ passes through the centroid of the section
-In the fully plastic condition, the beam is bent about the neutral axis NN. -A1 and A2 are the area of the cross section above and below NN respectively -Since there can be no longitudinal resultant force in the beam during bending without end load
A1σ y = A2σ y A1 = A2 =0.5 A, where A is the total area of the cross - section
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In the fully plastic condition, the neutral axis divides the cross-section into two equal areas and the stress diagram is shown in the above Figure • If C1 is the centroid of the area A1, • C2 the centroid of the area A2 • h the distance between C1 and C2 Then the fully plastic moment is given by
1
1
M b = Aσ y h = σ y Ah 2 2
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MECHANICS OF ENGINEERING MATERIALS Example : The flange and web of the T-bar section are each 12 mm thick, the flange width is 100 mm, and the overall depth of the section is 100 mm. The centroid of the section is at a distance of 70.6 mm from the bottom of the web, and the second moment of area I z of the section about a line through the centroid and parallel to the flange is 20.3x10 6 mm4. Determine the value of the shape factor.
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MECHANICS OF ENGINEERING MATERIALS Solution:
100 Let m be the distance of the neutral axis NN from the top of the flange, then
100
70.6
If n is the distance of the centroid of area A2 from the bottom of the web, then
(100 × 0.7 × 88.35) + (12 × 88 × 44) = [(100 × 0.7 ) + (12 × 88)]n n = 46.8mm
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MECHANICS OF ENGINEERING MATERIALS Therefore h, the distance between C 1 and C2 is:
So that
The shape factor is:
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MECHANICS OF ENGINEERING MATERIALS
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MECHANICS OF ENGINEERING MATERIALS Asymmetric sections do not yield simultaneously at the top and bottom of the section. The neutral axis moves as yield spreads through the section from the centroid before yield, to the axis that bisects the cross sectional area.
Can you identify the position of the plastic neutral axis?
20 mm abo ve the base. Calcul ate the fully plastic moment of th e section? Take y = 275 N/mm 2
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MECHANICS OF ENGINEERING MATERIALS C = T = area × σ y =
4000 × 275 = 1100kN
M p
=
C × 50 + T × 10
= 1100 × 60 =
M p Z p = M p / σ y
=
6
=
66kNm
Z p × σ y
66 × 10 / 275 = 240 × 10
Plastic Section Modulus
3
mm
3
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MECHANICS OF ENGINEERING MATERIALS 2. Plastic collapse of beams The fully plastic ending moment developed in the preceding section was due to the application of pure bending. A beam would therefore become fully plastic at all cross-section along the whole length once M p was reached. However, in practice pure bending rarely occurs and the bending moment distribution varies depending on the loading conditions.
The point of maximum bending moment along the beam will be the first crosssection which becomes fully plastic as the load magnitude increases. Cross-sections adjacent to the fully plastic section will have commenced yielding to various depths.
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MECHANICS OF ENGINEERING MATERIALS Progression of Yield Yone Leading to Fully Plastic Hinge and Collapse •
Stresses reach Yield Magnitude at extreme fibres
•
Yield Zones spreads towards Neutral axis
•
Yield Zones join, are now spread through entire x-section
•
Plastic Hinge causes structural collapse
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MECHANICS OF ENGINEERING MATERIALS Elastic and plastic zones in beams:
•
A simply supported beam carrying a central concentrated load • The boundary between elastic and plastic material is parabolic in shape
•
•
a simply supported beam carrying a distributed loading Plastic zones is triangular in shape
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MECHANICS OF ENGINEERING MATERIALS When a cross-section reaches the fully plastic state it can not carry any higher loading and the beam forms a plastic hinge at that cross-section.
•When one or more plastic hinge occur such that the beam becomes a mechanism then this situation is described as plastic collapse.
•the maximum bending moment is at the center and is M =
WL 4
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MECHANICS OF ENGINEERING MATERIALS
Figure 15.8 (a)
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MECHANICS OF ENGINEERING MATERIALS
A cantilever beam simply supported at one end, fully fixed at the other end and carrying a concentrated load at mid-span
Plastic
elastic
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MECHANICS OF ENGINEERING MATERIALS Free body diagram of the beam:
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Take moments about B: while Thus
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If the deformation is elastic:
Therefore:
Use singularity function method
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MECHANICS OF ENGINEERING MATERIALS
A cantilever beam fully fixed at both ends and carrying a uniformly distributed load
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Plastic hinges will form at A, B and C.
Take a free body diagram of AB: - Moment equilibrium about A
M = 2
M p =
w p L
16
wL2
16
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MECHANICS OF ENGINEERING MATERIALS
M p
=
w p L2 16
Elastic moment Use singularity function method wL2 M = 12
M Y =
wY L2 12
Therefore:
w p wY
=
4 M p 3 M Y
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MECHANICS OF ENGINEERING MATERIALS Example: The beam illustrated below is made of I-section mild steel having a shape factor of 1.15 and a yield stress of 240 MN/m 2. Using a load factor against collapse of 2 find the required section modulus.
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MECHANICS OF ENGINEERING MATERIALS Solution:
10kN
30kN
MA R A
R C
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MECHANICS OF ENGINEERING MATERIALS 10kN
30kN
MA R A
R C 10kN
30kN
MA R A M
M
R C
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MECHANICS OF ENGINEERING MATERIALS M
Take a free body diagram of AB: - Moment equilibrium about B
Take a free body diagram of AC: - Moment equilibrium about C
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MECHANICS OF ENGINEERING MATERIALS
Z =
2 M shape factor × σ Y
Therefore the required section modulus is 0.362 m 3.
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MECHANICS OF ENGINEERING MATERIALS • Determine the Plastic Collapse Load for the mild steel beam shown below (σy=300MPa)
Pc 20 30
300 mm
100 mm 400 mm
5
5
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MECHANICS OF ENGINEERING MATERIALS Plastic Modulus, Zzz A2 20 5 z
z 25
A1 5
C
• Locate zz such that there is equal area above and below Neutral Axis 5C=5(20)+5(25-C) C=22.5 mm
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MECHANICS OF ENGINEERING MATERIALS Stress Distribution
20
=300 MPa
σy
2.96
5 2.5
F
4.54 z
z 11.25
15.793
22.5 F 5 Section Properties
=300 MPa
σy
Plastic Stress Distribution
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MECHANICS OF ENGINEERING MATERIALS Fully Plastic Moment, Mp
M p = F(d) F=σy(A1)=300MPa(22.5)(5)N =33750 N M p = F(d) = 33750(15.793) = 533000 N.mm
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MECHANICS OF ENGINEERING MATERIALS Plastic Collapse Mechanism Pc
A
C B Hinge, Ma
Hinge
M b
•2 Plastic Hinges are required for the collapse •The order in which they occur does not matter •Find Pc for Ma = M b = M p
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MECHANICS OF ENGINEERING MATERIALS Statics - Equilibrium - Right Side M p R c 100
M p -R c(100)=0 R c=M p/400 =533000/100 =5330N
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MECHANICS OF ENGINEERING MATERIALS Whole Structure Pc
M p R a
R c 300
100
R a+R c=Pc R a=Pc-R c= Pc-5330 R a(400)-Pc(100)-M p=0 (Pc-5330)400-100Pc-533000=0 Pc=8880 N
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MECHANICS OF ENGINEERING MATERIALS Plastic torsion of shafts: plastic torque T J
=
τ c
=
Gθ L
When the shaft has a torque applied in the elastic zone, Max. shear stress occurs when c = the outer radius
T =
τπ 2
r 3
When the shear stress at the surface of the shaft has reached the value τy
T y =
τ yπ 2
3
r
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MECHANICS OF ENGINEERING MATERIALS When the torque increase beyond Ty then plasticity occurs in fibres at the surface of the shaft The torque carried by the elastic core is
T 1 =
τ yπ 2
r y
3
Where r y is the interface radius between elastics and plastic material, and that carried by the plastic zone is r o
2π τ y r o3 − r y2 T 2 = 2π r τ y dr = 3 r y
∫
(
2
)
The total torque, T, is
T 1 + T 2 =
1 2
τ yπ r + 3 y
2π 3
τ y (r o3 − r y3 )
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MECHANICS OF ENGINEERING MATERIALS Therefore 3 r 2 3 y T = π r o τ y 1 − 3 4r o 3
When fully plastic condition is reached τ = τy and r y = 0.0
T p
2 =
T p T y
3
π r o3τ y
=
The angle of twist
θ y
=
τ y L Gr o
4 3
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MECHANICS OF ENGINEERING MATERIALS At the elastic – plastic condition
θ =
τ y L Gr y
Since we assume the radii remain straight, then the outer plastic region has same angle of twist
θ y θ
=
r y r o
It is evident that as the shaft approaches the fully plastic state, the angle of twist tends to infinity
1 θ y T = π r τ 1 − 4 θ 3 2
3 o y
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MECHANICS OF ENGINEERING MATERIALS Example: A mild steel shear coupling in a metal - working process is 40 mm in diameter and 250 mm in length. It is subjected to an overload torque of 1800 Nm which is known to have caused shear yield in the shaft. Determine the radial depth to which plasticity has penetrated and the angle of twist. τ y = 120MN/m2 , G = 80GN/m2. Solution 3 r 2 3 y T = π r o τ y 1 − 3 4r 3 o
3 r 2 y 3 6 1800 = π × 0.02 ×120 ×10 1 − 4 × 0.023 3 r y = 15mm
Hence the depth of plastic deformation is 5 mm
