Normal Stress 1
Normal Stress (1.1-1.5)
MAE 314 – Solid Mechanics
Yun Jing
2
Pretest
This is a structure which was designed to support a 30kN load, it consists of a boom AB and of a rod BC. The boom and rod are connected by a pin at B and are supported by pins and brackets at A and C. (1) Is there a reaction moment at A? why? (2) What is the reaction force in the vertical direction at A? (3) What is the internal force in AB? (4) What is the internal force in BC?
Normal Stress 3
Statics Review
Pins = no rxn moment
Normal Stress 4
Statics Review
• 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
• Solve for reactions at A & C:
Normal Stress 5
Statics Review
• Results: kN30kN40kN40 yx CCA
0
m8.00
y
yB
A
AM
• Consider a free-body diagram for the boom:
kN30yC
substitute into the structure equilibrium equation
See section 1.2 in text for complete static analysis and review of method of joints.
Bx
Normal Stress 6
Normal Stress 7
Introduction to Normal Stress Methods of statics allow us to determine forces and moments in a structure, but how do we determine if a load can be safely supported? Factors: material, size, etc. Need a new concept….Stress
Normal Stress 8
Introduction to Normal Stress
Stress = Force per unit areaA
F
Normal Stress 9
If stress varies over a cross-section, the resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the axis. Thus, we can write the stress at a point as We assume the force F is evenly distributedover the cross-section of the bar. In realityF = resultant force over the end of the bar.
Introduction to Normal Stress
A
FdA
A
F
A
lim0
Sign convention Tensile (member is in tension)Compressive (member is in compression)Units (force/area)English: lb/in2 = psi kip/in2 = ksiSI: N/m2 = Pa (Pascal) kN/m2 = kPa MPa, GPa, etc.
Normal Stress 10
Introduction to Normal Stress
00
Tensile
Compressive
Normal Stress 11
Homogenous: material is the same throughout the bar Cross-section: section perpendicular to longitudinal axis of bar
Prismatic: cross-section does not change along axis of bar
Definitions and Assumptions
F’ F
A
Prismatic Non-Prismatic
Normal Stress 12
Uniaxial bar: a bar with only one axis Normal Stress (σ): stress acting perpendicular to the cross-section. Deformation of the bar is uniform throughout. (Uniform Stress State) Stress is measured far from the point of application. Loads must act through the centroid of the cross-section.
Definitions and Assumptions
Normal Stress 13
The uniform stress state does not apply near the ends of the bar. Assume the distribution of normal stresses in an axially loadedmember is uniform, except in theimmediate vicinity of the points ofapplication of the loads (Saint-Venant’s Principle).
Definitions and Assumptions
“Uniform” Stress
Normal Stress 14
How do we know all loads must act through the centroid of the cross-section? Let us represent P, the resultant force, by a uniform stress over the cross-section (so that they are statically equivalent).
Definitions and Assumptions
Normal Stress 15
Moments due to σ:
Set Mx = Mx and My = My
Definitions and Assumptions
A
y
A
x
dAxM
dAyM
AA
AA
xdAA
dAxP
x
ydAA
dAyP
y
11
11
Equations for the centroidMM xx
MM yy
Normal Stress 16
Example Problem Can the structure we used for our statics review safely support a30 kN load? (Assume the entire structure is made of steel with a maximum allowable stress σall=165 MPa.)
Cross-section 30 mm x 50 mm
Normal Stress 17
Example Problem Two cylindrical rods are welded together and loaded as shown. Find the normal stress at the midsection of each rod.
mmd
mmd
30
50
2
1
Shearing and Bearing Stress 18
Shearing and Bearing Stress (1.6-1.8, 1.12)
MAE 314 – Solid Mechanics
Yun Jing
Shearing and Bearing Stress 19
What is Shearing Stress?
We learned about normal stress (σ), which acts perpendicular to the cross-section. Shear stress (τ) acts tangential to the surface of a material element.
Normal stress results in a volume change.
Shear stress results in a shape change.
Shearing and Bearing Stress 20
Where Do Shearing Stresses Occur?
Shearing stresses are commonly found in bolts, pins, and rivets.
Free Body Diagram of Bolt
Bolt is in “single” shear
Force P results in shearing stressForce F results in bearing stress (will discuss later)
Shearing and Bearing Stress 21
We do not assume τ is uniform over the cross-section, because this is not the case. τ is the average shear stress.
The maximum value of τ may be considerably greater than τave, which is important for design purposes.
Shear Stress Defined
A
F
A
Pave
Shearing and Bearing Stress 22
Double Shear
Free Body Diagram of Bolt Free Body Diagram of Center of Bolt
A
F
A
F
A
Pave 2
2
Bolt is in “double” shear
Shearing and Bearing Stress 23
Bearing Stress
Bearing stress is a normal stress, not a shearing stress.
Thus,
where Ab = projected area where bearing pressure is
appliedP = bearing force
td
P
A
P
bb
Single shear case
Read section 1.8 in text for a detailed stress analysis of a structure.
Would like to determine the stresses in the members and connections of the structure shown.
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)
Rod & Boom Normal StressesThe 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 = +159 MPa.
Pin Shearing StressesThe 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
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
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
Shearing and Bearing Stress 29
Equilibrium of Shear Stresses Consider an infinitesimal element of material. Apply a single shear stress, τ1. Total shear force on surface is (τ1)bc. For equilibrium in the y-direction, applyτ1 on (-) surface. For moment equilibrium about the z-axis,apply τ2 on top and bottom surfaces. Moment equilibrium equation about z-axis: Thus, a shear stress must be balanced by three other stresses for the element to be in equilibrium.
1
2
2
1
bacabc )()( 21
21
Shearing and Bearing Stress 30
Equilibrium of Shear Stresses
1
2
2
1
Face DirectionShear Stress
+ + +
+ - -
- - +
- + -
What does this tell us? Shear stresses on opposite (parallel) faces of an element are equal in magnitude and opposite in direction. Shear stress on adjacent (perpendicular) faces of an element are equal in magnitude and both point towards or away from each other.
Sign convention for shear stresses Positive face – normal is in (+) x, y, or z direction Negative face - normal is in (-) x, y, or z direction +
+
-
-
Shearing and Bearing Stress 31
σx = stress in x-direction applied in the plane normal to x-axis σ y = stress in y-direction applied in the plane normal to y-axis σ z = stress in z-direction applied in the plane normal to z-axis τxy = stress in y-direction applied in the plane normal to x-axis τ xz = stress in z-direction applied in the plane normal to x-axis τ zy = stress in y-direction applied in the plane normal to z-axis And so on…
Define General State of Stressy
zx
Shearing and Bearing Stress 32
There are 9 components of stress: σx, σ y, σ z, τxy, τ xz, τ yx, τ yz, τ zx, τ zy As shown previously, in order to maintain equilibrium:τ xy= τ yx, τ xz = τ zx, τ yz = τ zy There are only 6 independent stress components.
Define General State of Stressy
zx
Shearing and Bearing Stress 33
Example Problem
A load P = 10 kips is applied to a rod supported as shown by a plate with a 0.6 in. diameter hole. Determine the shear stress in the rod and the plate.
Shearing and Bearing Stress 34
Example Problem Link AB is used to support the end of a horizontal beam. If link AB is subject to a 10 kips compressive force determine the normal and bearing stress in the link and the shear stress in each of the pins.
ind
int
inb
1
41
2
Oblique Planes and Design Considerations 35
Oblique Planes and Design Considerations (1.11, 1.13)
MAE 314 – Solid Mechanics
Yun Jing
Oblique Planes and Design Considerations 36
Stress on an Oblique Plane What have we learned so far?
Axial forces in a two-force member cause normal stresses.
Transverse forces exerted on bolts and pins cause shearing stresses.
Oblique Planes and Design Considerations 37
Stress on an Oblique Plane Axial forces cause both normal and shearing stresses on planes which are not perpendicular to the axis. Consider an inclined section of a uniaxial bar. The resultant force in the axial directionmust equal P to satisfy equilibrium. The force can be resolved into components perpendicular to the section, F, and parallel to the section, V. The area of the section iscosPF sinPV
cos/cos 00 AAAA
Oblique Planes and Design Considerations 38
Stress on an Oblique Plane We can formulate the average normal stress on the section as The average shear stress on the section is
Thus, a normal force applied to a bar on an inclined section produces a combination of shear and normal stresses.
2
00
coscos/
cos
A
P
A
P
A
F
cossincos/
sin
00 A
P
A
P
A
V
Oblique Planes and Design Considerations 39
Stress on an Oblique Plane Since σ and τ are functions of sine and cosine, we know the maximum and minimum values will occur at θ = 00, 450, and 900.
At θ=±900 σ=0
At θ=±450 σ=P/2A0
At θ=00 σ=P/A0 (max)
At θ=±900 τ=0
At θ=±450 τ=P/2A0 (max)
At θ=00 τ=0
2
0
cosA
P
cossin0A
P
Oblique Planes and Design Considerations
Stress on an Oblique Plane What does this mean in reality?
Oblique Planes and Design Considerations 41
Design Considerations From a design perspective, it is important to know the largest load which a material can hold before failing. This load is called the ultimate load, Pu. Ultimate normal stress is denoted as σu and ultimate shear stress is denoted as τu.
A
Puu
A
Puu
Oblique Planes and Design Considerations 42
Design Considerations Often the allowable load is considerably smaller than the
ultimate load.
It is a common design practice to use factor of safety.
all
u
P
P
loadallowable
loadultimateSF ..
all
u
stressallowable
stressultimateSF
..all
u
stressallowable
stressultimateSF
..
Oblique Planes and Design Considerations 43
Example Problem Two wooden members are spliced as shown. If the maximum allowable tensile stress in the splice is 75 psi, determine the largest load that can be safely supported and the shearing stress in the splice.
Oblique Planes and Design Considerations 44
Example Problem A load is supported by a steel pin inserted into a hanging wooden piece. Given the information below, determine the load P if an overall factor of safety of 3.2 is desired.
mmd
mmc
mmb
MPa
MPa
tensioninMPa
steelu
woodu
woodu
12
55
40
145
5.7
)(60
_
_
_