mekanika bahan session 01 - 02
TRANSCRIPT
Stress and StrainSession 1 - 2
Course : S1014 - Mechanics of MaterialsYear : July 2013
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EQUILIBRIUM OF A DEFORMABLE BODY
External loads• Surface forces
– Area of contact– Concentrated force– Linear distributed force– Centroid C (or geometric center)
• Body force (e.g., weight)
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Support reactions• for 2D problems
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Equations of equilibrium• For equilibrium
– balance of forces– balance of moments
• Draw a free-body diagram to account for all forces acting on the body
• Apply the two equations to achieve equilibrium state
∑ F = 0∑ MO = 0
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Internal resultant loadings• Define resultant force (FR) and moment (MRo) in 3D:
– Normal force, N– Shear force, V– Torsional moment or torque, T– Bending moment, M
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Internal resultant loadings• For coplanar loadings:
– Normal force, N– Shear force, V– Bending moment, M
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Internal resultant loadings• For coplanar loadings:
– Apply ∑ Fx = 0 to solve for N– Apply ∑ Fy = 0 to solve for V– Apply ∑ MO = 0 to solve for M
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Procedure for Analysis• Method of sections
1. Choose segment to analyze2. Determine Support Reactions3. Draw free-body diagram for whole body4. Apply equations of equilibrium
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Procedure for analysis• Free-body diagram
1. Keep all external loadings in exact locations before “sectioning”2. Indicate unknown resultants, N, V, M, and T at the section,
normally at centroid C of sectioned area3. Coplanar system of forces only include N, V, and M4. Establish x, y, z coordinate axes with origin at centroid
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Procedure for analysis• Equations of equilibrium
1. Sum moments at section, about each coordinate axes where resultants act
2. This will eliminate unknown forces N and V, with direct solution for M (and T)
3. Resultant force with negative value implies that assumed direction is opposite to that shown on free-body diagram
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EXAMPLE 1.1Determine resultant loadings acting on cross section at C of beam.
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Support Reactions• Consider segment CBFree-Body Diagram:Keep distributed loading exactly where it is on segment
CB after “cutting” the section. Replace it with a single resultant force, F.
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Intensity (w) of loading at C (by proportion)w/6 m = (270 N/m)/9 m
w = 180 N/m
F = ½ (180 N/m)(6 m) = 540 N
F acts 1/3(6 m) = 2 m from C.
Free-Body Diagram:
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Equilibrium equations:
∑ Fx = 0;
∑ Fy = 0;
∑ Mc = 0;
− Nc = 0Nc = 0
Vc − 540 N = 0Vc = 540 N
−Mc − 504 N (2 m) = 0Mc = −1080 N·m
+
+
+
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Equilibrium equations:
Negative sign of Mc means it acts in the opposite direction to that shown below
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What is Stress ?Much Work with limited time High Stress
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What is Stress ?
Less Work with long time Low Stress
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What is Stress ?stress is according to strength and failure of solids. The stress field is the distribution of internal "tractions" that balance a given set of external tractions and body forces
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Stress stress is according to strength and failure of solids. The stress field is the distribution of internal "tractions" that balance a given set of external tractions and body forces
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Stress
Look at the external traction T that represents the force per unit area acting at a given location on the body's surface.
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Stress Traction T is a bound vector, which means T cannot slide along its line of action or translate to another location and keep the same meaning. In other words, a traction vector cannot be fully described unless both the force and the surface where the force acts on has been specified. Given both DF and Ds, the traction T can be defined as
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Stress The internal traction within a solid, or stress, can be defined in a similar manner.
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Stress Suppose an arbitrary slice is made across the solid shown in the above figure, leading to the free body diagram shown at right.
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Stress Surface tractions would appear on the exposed surface, similar in form to the external tractions applied to the body's exterior surface.
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Stress
The stress at point P can be defined using the same equation as was used for T.
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Stress
Stress therefore can be interpreted as internal tractions that act on a defined internal datum plane. One cannot measure the stress without first specifying the datum plane.
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Stress Surface tractions, or stresses acting on an internal datum plane, are typically decomposed into three mutually orthogonal components. One component is normal to the surface and represents direct stress. The other two components are tangential to the surface and represent shear stresses.
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Stress What is the distinction between normal and tangential tractions, or equivalently, direct and shear stresses?
Direct stresses tend to change the volume of the material (e.g. hydrostatic pressure) and are resisted by the body's bulk modulus (which depends on the Young's modulus and Poisson ratio).
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Stress What is the distinction between normal and tangential tractions, or equivalently, direct and shear stresses?
Shear stresses tend to deform the material without changing its volume, and are resisted by the body's shear modulus.
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STRESS
Concept of stress• To obtain distribution of force acting over a sectioned area • Assumptions of material:
1. It is continuous (uniform distribution of matter)
2. It is cohesive (all portions are connected together)
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Concept of stress• Consider ΔA in figure below• Small finite force, ΔF acts on ΔA• As ΔA → 0, Δ F → 0• But stress (ΔF / ΔA) → finite limit (∞)
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Normal stress• Intensity of force, or force per unit area, acting normal
to ΔA• Symbol used for normal stress, is σ (sigma)
Tensile stress: normal force “pulls” or “stretches” the area element ΔACompressive stress: normal force “pushes” or “compresses” area element ΔA
σz =
limΔA →0
ΔFz
ΔA
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Shear stress• Intensity of force, or force per unit area, acting
tangent to ΔA• Symbol used for normal stress is τ (tau)
τzx =
limΔA →0
ΔFx
ΔA
τzy =
limΔA →0
ΔFy
ΔA
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General state of stress• Figure shows the state of stress acting
around a chosen point in a body
Units (SI system)• Newtons per square meter (N/m2) or a
pascal (1 Pa = 1 N/m2)• kPa = 103 N/m2 (kilo-pascal)• MPa = 106 N/m2 (mega-pascal)• GPa = 109 N/m2 (giga-pascal)
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AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Examples of axially loaded bar• Usually long and slender structural members• Truss members, hangers, bolts• Prismatic means all the cross sections are the same
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Assumptions
1. Uniform deformation: Bar remains straight before and after load is applied, and cross section remains flat or plane during deformation
2. In order for uniform deformation, force P be applied along centroidal axis of cross section
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Average normal stress distribution
σ = average normal stress at any point on cross sectional area
P = internal resultant normal forceA = x-sectional area of the bar
FRz = ∑ Fxz ∫ dF = ∫A σ dA
P = σ A
+
PAσ =
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Equilibrium• Consider vertical equilibrium of the element
∑ Fz = 0 σ (ΔA) − σ’ (ΔA) = 0σ = σ’
Above analysis applies to members subjected to tension or compression.
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Maximum average normal stress
• For problems where internal force P and x-sectional A were constant along the longitudinal axis of the bar, normal stress σ = P/A is also constant
• If the bar is subjected to several external loads along its axis, change in x-sectional area may occur
• Thus, it is important to find the maximum average normal stress
• To determine that, we need to find the location where ratio P/A is a maximum
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Maximum average normal stress
• Draw an axial or normal force diagram (plot of P vs. its position x along bar’s length)
• Sign convention: – P is positive (+) if it causes tension in the member– P is negative (−) if it causes compression
• Identify the maximum average normal stress from the plot
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Procedure for AnalysisAverage normal stress• Use equation of σ = P/A for x-sectional area of a member when
section subjected to internal resultant force P
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Procedure for AnalysisAxially loaded members• Internal Loading: • Section member perpendicular to its longitudinal axis at pt
where normal stress is to be determined• Draw free-body diagram• Use equation of force equilibrium to obtain internal axial force
P at the section
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Procedure for AnalysisAxially loaded members• Average Normal Stress: • Determine member’s x-sectional area at the section• Compute average normal stress σ = P/A
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EXAMPLE
Bar width = 35 mm, thickness = 10 mm
Determine max. average normal stress in bar when subjected to loading shown.
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EXAMPLE 1.6 (SOLN)Internal loading
Normal force diagramBy inspection, largest loading area is BC, where PBC = 30 kN
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Average normal stress
σBC =PBC
A
30(103) N(0.035 m)(0.010 m)
= = 85.7 MPa
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Specific weight γst = 80 kN/m3
Determine average compressive stress acting at points A and B.
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Internal loadingBased on free-body diagram, weight of segment AB determined from
Wst = γstVst
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Average normal stress
+ ∑ Fz = 0; P − Wst = 0
P − (80 kN/m3)(0.8 m)(0.2 m)2 = 0
P = 8.042 kN
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Average compressive stressCross-sectional area at section is:
A = (0.2)m2
8.042 kN
(0.2 m)2
P
A=σ =
σ = 64.0 kN/m2
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AVERAGE SHEAR STRESS
• Shear stress is the stress component that act in the plane of the sectioned area.
• Consider a force F acting to the bar• For rigid supports, and F is large enough, bar will deform
and fail along the planes identified by AB and CD• Free-body diagram indicates that shear force, V = F/2 be
applied at both sections to ensure equilibrium
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Average shear stress over each section is:
τavg = average shear stress at section, assumed to be same at each pt on the section
V = internal resultant shear force at section determined from equations of equilibrium
A = area of section
PAτavg =
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• Case discussed above is example of simple or direct shear• Caused by the direct action of applied load F• Occurs in various types of simple connections, e.g., bolts,
pins, welded material
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Single shear• Steel and wood joints shown below are examples of single-
shear connections, also known as lap joints. • Since we assume members are thin, there are no moments
caused by F
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Single shear• For equilibrium, x-sectional area of bolt and bonding surface
between the two members are subjected to single shear force, V = F
• The average shear stress equation can be applied to determine average shear stress acting on colored section in (d).
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Double shear • The joints shown below are examples of double-shear
connections, often called double lap joints.• For equilibrium, x-sectional area of bolt and bonding surface
between two members subjected to double shear force, V = F/2
• Apply average shear stress equation to determine average shear stress acting on colored section in (d).
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Procedure for analysisInternal shear1. Section member at the pt where the τavg is to be determined
2. Draw free-body diagram3. Calculate the internal shear force VAverage shear stress1. Determine sectioned area A2. Compute average shear stress τavg = V/A
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Depth and thickness = 40 mm
Determine average normal stress and average shear stress acting along (a) section planes a-a, and (b) section plane b-b.
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Part (a)Internal loadingBased on free-body diagram, Resultant loading of axial force, P = 800 N
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Part (a)Average stress
Average normal stress, σ
σ = PA
800 N(0.04 m)(0.04 m)
= 500 kPa =
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Part (a)Internal loadingNo shear stress on section, since shear force at section is zero.
τavg = 0
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Part (b)Internal loading
+
∑ Fx = 0; − 800 N + N sin 60° + V cos 60° = 0+
∑ Fy = 0; V sin 60° − N cos 60° = 0
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Part (b)Internal loading
Or directly using x’, y’ axes,
∑ Fx’ = 0;
∑ Fy’ = 0;
+
+
N − 800 N cos 30° = 0
V − 800 N sin 30° = 0
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Part (b)Average normal stress
σ = NA
692.8 N(0.04 m)(0.04 m/sin 60°) = 375 kPa =
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Part (b)Average shear stress
τavg = VA
400 N(0.04 m)(0.04 m/sin 60°) = 217 kPa =
Stress distribution shown below
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ALLOWABLE STRESS• When designing a structural member or mechanical
element, the stress in it must be restricted to safe level• Choose an allowable load that is less than the load the
member can fully support• One method used is the factor of safety (F.S.)
F.S. = Ffail
Fallow
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• If load applied is linearly related to stress developed within member, then F.S. can also be expressed as:
F.S. = σfail
σallow
F.S. = τfail
τallow
In all the equations, F.S. is chosen to be greater than 1, to avoid potential for failureSpecific values will depend on types of material used and its intended purpose
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DESIGN OF SIMPLE CONNECTIONS• To determine area of section subjected to a normal force, use
A = Pσallow
A = Vτallow
• To determine area of section subjected to a shear force, use
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Cross-sectional area of a tension member
Condition: The force has a line of action that passes through the centroid of the cross section.
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Cross-sectional area of a connecter subjected to shear
Assumption: If bolt is loose or clamping force of bolt is unknown, assume frictional force between plates to be negligible.
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Assumptions: 1. (σb)allow of concrete <
(σb)allow of base plate
2. Bearing stress is uniformly distributed between plate and concrete
Required area to resist bearing• Bearing stress is normal stress produced by the compression of
one surface against another.
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• Although actual shear-stress distribution along rod difficult to determine, we assume it is uniform.
• Thus use A = V / τallow to calculate l, provided d and τallow is known.
Required area to resist shear caused by axial load
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Procedure for analysisWhen using average normal stress and shear stress equations, consider first the section over which the critical stress is actingInternal Loading1. Section member through x-sectional area2. Draw a free-body diagram of segment of member3. Use equations of equilibrium to determine internal resultant force
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Procedure for AnalysisRequired Area• Based on known allowable stress, calculate required area
needed to sustain load from A = P/τallow or A = V/τallow
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EXAMPLE
The two members pinned together at B. If the pins have an allowable shear stress of τallow = 90 MPa, and allowable tensile stress of rod CB is (σt)allow = 115 MPa
Determine to nearest mm the smallest diameter of pins A and B and the diameter of rod CB necessary to support the load.
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Draw free-body diagram:
σ = PA
800 N(0.04 m)(0.04 m)
= 500 kPa =
No shear stress on section, since shear force at section is zero
τavg = 0
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EXAMPLE 1.13 (SOLN)Diameter of pins:
dA = 6.3 mm
AA =VA
Tallow
2.84 kN90 103 kPa = = 31.56 10−6 m2 = (dA
2/4)
dB = 9.7 mm
AB =VB
Tallow
6.67 kN90 103 kPa = = 74.11 10−6 m2 = (dB
2/4)
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Diameter of pins:
dA = 7 mm dB = 10 mm
Choose a size larger to nearest millimeter.
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Diameter of rod:
dBC = 8.59 mm
ABC = P(σt)allow
6.67 kN115 103 kPa = = 58 10−6 m2 = (dBC
2/4)
dBC = 9 mm
Choose a size larger to nearest millimeter.
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What is Strain?A propotional
dimensional change ( intensity or degree of distortion )
What is Strain measure?a total elongation per unit length of material due to some applied stress.
oL
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What are the types of strain ?
1.Elastic Strain2.Plastic Deformatio
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Elastic StrainTransitory dimensional change that exists only while the initiating stress is applied and dissapears immediately upon removal of the stress.
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Elastic Strain
The applied stresses cause the atom are displaced the same amount and still maintain their relative geometic. When streesses are removed, all the atom return to their original positions and no permanent deformation occurs
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ELASTIC CURVE
The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called the elastic curve, which is characterized by the deflection and slope along the curve. E.g.
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Moment-curvature relationship:Sign convention:
Moment-curvature relationship:Sign convention:
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Consider a segment of width dx, the strain in are ds, located at a position y from the neutral axis is ε = (ds’ – ds)/ds. However, ds = dx = ρdθ and ds’ = (ρ-y) dθ, and so ε = [(ρ – y) dθ – ρdθ ] / (ρdθ), or
1 ρ
= – ε y
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Comparing with the Hooke’s Law ε = σ / E and the flexure formula σ = -My/IWe have
1 ρ = M
EI or
1 ρ
= – σ Ey
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Plastic Deformation
a dimentional change that does not dissapear when the initiating stress is removed. It is usually accompanied by some elastic strain.
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Elastic Strain & Plastic DeformationThe phenomenon of elastic strain & plastic Deformation in a material are called elasticity & Plasticity respectively
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Elastic Strain & Plastic DeformationMost of Metal material At room temperature they have some elasticity, which manifests itself as soon as the slightest stress is applied. Usually, they are also posses some plasticity , but this may not become apparent until the stress has been raised appreciablty.
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Elastic Strain & Plastic DeformationMost of Metal material
The magnitude of Plastic strain, when it does appear , is likely to be much greater than that of the elastic strain for a given stress increment
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