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1. Stress
CHAPTER OUTLINE
1. Introduction
2. Equilibrium of a deformable body
3. Stress
4. Average normal stress in an axially loaded bar
5. Average shear stress
6. Allowable stress
7. Design of simple connections
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1. Stress
A branch of mechanics• It studies the relationship of
– External loads applied to a deformable body, and– The intensity of internal forces acting within the body
• Deals with the behavior of solid bodies subjected to various types of loading
• Study body’s stability when external forces are applied to it.
• A thorough understanding of mechanical behavior is essential for the safe design of all structures.
1.1 INTRODUCTION
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1. Stress
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1. Stress
1.1 INTRODUCTION
Historical development• The historical development of mechanics of materials is a
fascinating blend of both theory and experiment.• Leonardo da Vinci (1452–1519) and Galileo Galilei (1564–
1642) performed experiments to determine the strength of wires, bars, and beams.
• In recent times, with advanced mathematical and computer techniques, more complex problems can be solved.
• As a result, this subject expanded into more advanced mechanics such as theory of elasticity and plasticity
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1. Stress
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1. Stress
• The main objective of the course is to provide the future engineer with the means of analyzing and designing various machines and load bearing structures.
• Both the analysis and design of a given structure involve the determination of stresses and deformations. This chapter is devoted to the concept of stress.
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1. Stress
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
External loads• Surface forces – caused by the
direct contact of one body with the surface of another.– Area of contact– Concentrated force– Linear distributed force– Centroid C (or geometric
center)• Body force (e.g., weight) – one
body exerts a force on another without direct physical contact.
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1. Stress
Support reactions• for 2D problems
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
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
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
Internal resultant loadings
• Define resultant force (FR) and moment (MRo) acting within the body (3D):– Normal force, N– Shear force, V– Torsional moment or torque, T– Bending moment, M
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
Internal resultant loadings• For coplanar loadings:
– Normal force, N– Shear force, V– Bending moment, M
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
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
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
Procedure for Analysis• Method of sections
1. Choose segment to analyze
2. Determine Support Reactions
3. Draw free-body diagram for whole body
4. Apply equations of equilibrium
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
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 area
3. Coplanar system of forces only include N, V, and M
4. Establish x, y, z coordinate axes with origin at centroid
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
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
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
EXAMPLE 1.1
Determine resultant loadings acting on cross section at C of beam.
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1. Stress
EXAMPLE 1.1 (SOLN)
Support Reactions• Consider segment CB
Free-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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1. Stress
EXAMPLE 1.1 (SOLN)
Intensity (w) of loading at C (by proportion)w/6 m = (270 N/m)/9 m
w = 180 N/m
F = area of triangle = ½ (180 N/m)(6 m) = 540 N
F acts at centroid = 1/3(6 m) = 2 m from C.
Free-Body Diagram:
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1. Stress
EXAMPLE 1.1 (SOLN)
Equilibrium equations:
∑ Fx = 0;
∑ Fy = 0;
∑ Mc = 0;
− Nc = 0Nc = 0
Vc − 540 N = 0Vc = 540 N
−Mc − 540 N (2 m) = 0Mc = −1080 N·m
+
+
+
Resultant moment at point C. Alternatively at point B;
+ ∑ MB = 0; −Mc + 540 N (4 m) - 540N(6m) = 0Mc = −1080 N·m
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1. Stress
EXAMPLE 1.1 (SOLN)
Equilibrium equations:
Negative sign of Mc means it acts in the opposite direction to that shown below
We cannot use segment AC unless the values for Va, Na and Ma are given in the question.
Va = 1215N ; Na = 0 ; Ma = 3645Nm
Exercise: Try calculate Mc, Vc and Nc using segment AC. Should get the same answer.
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1. Stress
EXAMPLE 1.1 (SOLN)
Equilibrium equations:
∑ Fx = 0;
∑ Fy = 0;
Nc = 0
-Vc + 1215 N-135N-540N = 0Vc = 540 N
Mc + 3645Nm – 135N(1m)-540N(1.5m) – 540N(3m) = 0Mc = −1080 N·m
+
+
∑ MA = 0;+
Mc + 540N(1.5m) + 135N(2m) – 1215N(3m) +3645Nm = 0
Mc = −1080 N·m (same answer)
∑ MC = 0;+
Alternatively resultant moment at point C:
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1. Stress
Review of Statics
• The structure is designed to support a 30 kN load
• Perform a static analysis to determine the internal force in each structural member and the reaction forces at the supports (point A & C).
• The structure consists of a beam and rod joined by pins (zero moment connections) at the junctions and supports
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1. Stress
Structure Free-Body Diagram
• Structure is detached from supports and the loads and reaction forces are indicated
• Ay and Cy can not be determined from
these equations
kN30
0kN300
kN40
0
kN40
m8.0kN30m6.00
yy
yyy
xx
xxx
x
xC
CA
CAF
AC
CAF
A
AM
• Conditions for static equilibrium:
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1. Stress
Component Free-Body Diagram
• In addition to the complete structure, each component must satisfy the conditions for static equilibrium
• Results: kN30kN40kN40 yx CCA
Reaction forces are directed along boom and rod
0
m8.00
y
yB
A
AM• Consider a free-body diagram of the boom;
kN30yC
substitute into the structure equilibrium equation
(boom)
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1. Stress
Method of Joints• The boom and rod are 2-force members,
i.e., the members are subjected to only two forces which are applied at member ends
kN50kN40
3
kN30
54
0
BCAB
BCAB
B
FF
FF
F
• Joints must satisfy the conditions for static equilibrium which may be expressed in the form of a force triangle:
• For equilibrium, the forces must be parallel to to an axis between the force application points, equal in magnitude, and in opposite directions
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1. Stress
Stress Analysis
• Conclusion: the strength of member BC is adequate
MPa 165all
• From the material properties for steel, the allowable stress is
Can the structure safely support the 30 kN load?
• At any section through member BC, the internal force is 50 kN with a force intensity or stress ofRod Diameter, dBC = 20 mm
Rod Area, ABC = π(dBC/2)2
• From a statics analysisFAB = 40 kN (compression)
FBC = 50 kN (tension)
3
-6 2
50 10 N39.79 MPa
125.7 10 mBC
P
A
159MPa
3.14x10-4
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1. Stress
1.3 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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1. Stress
1.3 STRESS
Concept of stress• Consider ΔA in figure below• Small finite force, ΔF acts on ΔA
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1. Stress
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 ΔA
• Compressive stress: normal force “pushes” or “compresses” area element ΔA
1.3 STRESS
σz =lim
ΔA →0
ΔFz
ΔA
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1. Stress
Shear stress• Intensity of force, or force per unit area, acting
tangent to ΔA• Symbol used for normal stress is τ (tau)
1.3 STRESS
τzx =lim
ΔA →0
ΔFx
ΔA
τzy =lim
ΔA →0
ΔFy
ΔA
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1. Stress
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1. Stress
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1. Stress
1.4 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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1. Stress
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
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
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
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
FRz = ∑ Fxz ∫ dF = ∫A σ dA
P = σ A
+
PA
σ =
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1. Stress
Equilibrium• Consider vertical equilibrium of the element• When the bar is stretched by the force P, the resulting
stresses are tensile stresses• If the force P cause the bar to be compressed, we
obtain compressive stresses
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
• Sign convention for normal stresses is:• (+) for tensile stresses and• (-) for compressive stresses
• Because the normal stress σ is obtained by dividing the axial force by the cross–sectional area, it has units of force per unit of area.
• In SI units:• Force is expressed in newtons (N) and area in
square meters (m2). A N/m2 is a pascals (Pa).
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
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
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
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
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
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
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
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
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
Procedure for AnalysisAxially loaded members• Average Normal Stress: • Determine member’s x-sectional area at the
section• Compute average normal stress σ = P/A
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
EXAMPLE 1.6
Bar width = 35 mm, thickness = 10 mm
Determine max. average normal stress in bar when subjected to loading shown.
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1. Stress
EXAMPLE 1.6 (SOLN)
Internal loading
Normal force diagram
By inspection, largest loading area is BC, where PBC = 30 kN
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1. Stress
EXAMPLE 1.6 (SOLN)
Average normal stress
σBC =PBC
A
30(103) N
(0.035 m)(0.010 m)= = 85.7 MPa
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1. Stress
1.5 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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1. Stress
1.5 AVERAGE SHEAR STRESS
Average shear stress over each section is:
τavg = average shear stress at section, assumed to be same at each pt on the section
P = V = internal resultant shear force at section determined from equations of equilibrium
A = area of one section2V = F
V = F/2
=PAτavg =
F2A
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1. Stress
1.5 AVERAGE SHEAR STRESS
• 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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1. Stress
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
1.5 AVERAGE SHEAR STRESS
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51
1. Stress
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 coloured section in (d).
1.5 AVERAGE SHEAR STRESS
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52
1. Stress
1.5 AVERAGE SHEAR STRESS
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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53
1. Stress
Shearing Stress Examples
A
F
A
Pave
Single Shear
A
F
A
P
2ave
Double Shear
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54
1. Stress
1.5 AVERAGE SHEAR STRESS
Procedure for analysis
Internal shear
1. Section member at the pt where the τavg is to be determined
2. Draw free-body diagram
3. Calculate the internal shear force V
Average shear stress
1. Determine sectioned area A
2. Compute average shear stress τavg = V/A
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55
1. Stress
EXAMPLE 1.10
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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56
1. Stress
EXAMPLE 1.10 (SOLN)
Part (a)Internal loadingBased on free-body diagram, Resultant loading of axial force, P = 800 N
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57
1. Stress
EXAMPLE 1.10 (SOLN)
Part (a)Average stress
Average normal stress, σ
σ = P
A 800 N
(0.04 m)(0.04 m)= 500 kPa =
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58
1. Stress
EXAMPLE 1.10 (SOLN)
Part (a)Internal loadingNo shear stress on section, since shear force at section is zero.
τavg = 0
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59
1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Internal loading
+
∑ Fx = 0; − 800 N + N sin 60° + V cos 60° = 0+
∑ Fy = 0; V sin 60° − N cos 60° = 0
Solve and we can get N = 692.8N and V = 400N
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60
1. Stress
EXAMPLE 1.10 (SOLN)
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
Solve and we can get N = 692.8N and V = 400N
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61
1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Internal loading
Or using a force triangle,
V N
800N
cos 60o = V/800
V = 400N
sin 60o = N/800
N = 692.8N
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62
1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Average normal stress
σ = N
A 692.8 N
(0.04 m)(0.04 m/sin 60°) = 375 kPa =
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63
1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Average shear stress
τavg = V
A
400 N
(0.04 m)(0.04 m/sin 60°) = 217 kPa =
Stress distribution shown below
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64
1. Stress
EXAMPLE 1.11 (SOLN)
ave
2
For the rod;
V
A5000N
63.7MPa0.005m
indicated as abcd
2rA
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65
1. Stress
ave
For the strut;
V
A2500N
3.12MPa0.04 0.02m m
EXAMPLE 1.11 (SOLN)
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66
1. Stress
• Would like to determine the stresses in the members and connections of the structure shown.
Stress Analysis & Design Example
• Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection
• From a statics analysis:FAB = 40 kN
(compression) FBC = 50 kN (tension)
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67
1. Stress
Rod & Boom Normal Stresses• The rod is in tension with an axial force of 50 kN.
• The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa.
• The minimum area sections at the boom ends are unstressed since the boom is in compression.
MPa167m10300
1050
m10300mm25mm40mm20
26
3
,
26
N
A
P
A
endBC
• At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline,
• At the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is BC = + 39.79 MPa.
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1. Stress
Pin Shearing Stresses
• The cross-sectional area for pins at A, B, and C,
262
2 m104912
mm25
rA
MPa102m10491
N105026
3
,
A
PaveC
• The force on the pin at C is equal to the force exerted by the rod BC,
• The pin at A is in double shear with a total force equal to the force exerted by the boom AB,
MPa7.40m10491
kN2026,
A
PaveA
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1. Stress
• Divide the pin at B into sections to determine the section with the largest shear force,
(largest) kN25
kN15
G
E
P
P
MPa9.50m10491
kN2526,
A
PGaveB
• Evaluate the corresponding average shearing stress,
Pin Shearing Stresses
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1. Stress
Bearing Stress in Connections
• Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces of the members they connect.
dt
P
A
Pb
• Corresponding average force intensity is called the bearing stress,
• The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin.
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71
1. Stress
Pin Bearing Stresses
• To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,
MPa3.53mm25mm30
kN40
td
Pb
• To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm,
MPa0.32mm25mm50
kN40
td
Pb
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1. Stress
1.6 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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1. Stress
1.6 ALLOWABLE STRESS
• 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 failure.
• Specific values will depend on types of material used and its intended purpose.
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1. Stress
Factor of safety considerations:• uncertainty in material properties • uncertainty of loadings• uncertainty of analyses• number of loading cycles• types of failure• maintenance requirements and deterioration effects• importance of member to structures integrity• risk to life and property• influence on machine function
1.6 ALLOWABLE STRESS
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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
• To determine area of section subjected to a normal force, use
A = P
σallow
A = V
τallow
• To determine area of section subjected to a shear force, use
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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
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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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
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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1. Stress
Assumptions:
1. (σb)allow of concrete < (σb)allow of base plate
2. Bearing stress is uniformly distributed between plate and concrete
1.7 DESIGN OF SIMPLE CONNECTIONS
Required area to resist bearing• Bearing stress is normal stress produced by the
compression of one surface against another.
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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
• 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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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
Procedure for analysis
When using average normal stress and shear stress equations, consider first the section over which the critical stress is acting
Internal Loading
1. Section member through x-sectional area
2. Draw a free-body diagram of segment of member
3. Use equations of equilibrium to determine internal resultant force
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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
Procedure for Analysis
Required 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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82
1. Stress
EXAMPLE 1.13
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 MPaDetermine 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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1. Stress
EXAMPLE 1.13 (SOLN)
Draw free-body diagram:
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84
1. Stress
EXAMPLE 1.13 (SOLN)
Diameter of pins:
dA = 6.3 mm
AA =VA
Tallow
2.84 kN
90 103 kPa = = 31.56 10−6 m2 = (dA2/4)
dB = 9.7 mm
AB =VB
Tallow
6.67 kN
90 103 kPa = = 74.11 10−6 m2 = (dB2/4)
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85
1. Stress
EXAMPLE 1.13 (SOLN)
Diameter of pins:
dA = 7 mm dB = 10 mm
Choose a size larger to nearest millimeter.
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86
1. Stress
EXAMPLE 1.13 (SOLN)
Diameter of rod:
dBC = 8.59 mm
ABC = P
(σt)allow
6.67 kN
115 103 kPa = = 58 10−6 m2 = (dBC
2/4)
dBC = 9 mm
Choose a size larger to nearest millimeter.
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1. Stress
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1. Stress
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1. Stress
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90
1. Stress
CHAPTER REVIEW
• Internal loadings consist of
1. Normal force, N
2. Shear force, V
3. Bending moments, M
4. Torsional moments, T• Get the resultants using
1. method of sections
2. Equations of equilibrium
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91
1. Stress
CHAPTER REVIEW
• Assumptions for a uniform normal stress distribution over x-section of member (σ = P/A)
1. Member made from homogeneous isotropic material
2. Subjected to a series of external axial loads that,
3. The loads must pass through centroid of cross-section
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92
1. Stress
CHAPTER REVIEW
• Determine average shear stress by using τ = V/A equation
– V is the resultant shear force on cross-sectional area A
– Formula is used mostly to find average shear stress in fasteners or in parts for connections
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1. Stress
CHAPTER REVIEW
• Design of any simple connection requires that– Average stress along any cross-section not
exceed a factor of safety (F.S.) or– Allowable value of σallow or τallow– These values are reported in codes or
standards and are deemed safe on basis of experiments or through experience