Mechanic of Materials (Normal Stress)

Normal Stress*Normal Stress (1.1-1.5)MAE 314 Solid MechanicsYun Jing

Normal Stress

*PretestThis 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*Statics ReviewPins = no rxn moment

Normal Stress

Normal Stress*Statics ReviewAy and Cy can not be determined from these equations.

Normal Stress

Normal Stress*Statics ReviewSee section 1.2 in text for complete static analysis and review of method of joints. Bx

Normal Stress

Normal Stress*

Normal Stress

Normal Stress*Introduction to Normal StressMethods 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

Normal Stress*Introduction to Normal StressStress = Force per unit area

Normal Stress

Normal Stress*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 distributed over the cross-section of the bar. In reality F = resultant force over the end of the bar.

Introduction to Normal Stress

Normal Stress

Sign convention Tensile (member is in tension)Compressive (member is in compression)

Units (force/area)English: lb/in2 = psi kip/in2 = ksi

SI: N/m2 = Pa (Pascal) kN/m2 = kPa MPa, GPa, etc.

Normal Stress*Introduction to Normal Stress

Normal Stress

Normal Stress*Homogenous: material is the same throughout the barCross-section: section perpendicular to longitudinal axis of bar

Prismatic: cross-section does not change along axis of bar

Definitions and AssumptionsPrismaticNon-Prismatic

Normal Stress

Normal Stress*Uniaxial bar: a bar with only one axis

Normal Stress (): 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

Normal Stress*The uniform stress state does not apply near the ends of the bar.

Assume the distribution of normal stresses in an axially loaded member is uniform, except in the immediate vicinity of the points of application of the loads (Saint-Venants Principle).

Definitions and Assumptions

Normal Stress

Normal Stress*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

Normal Stress*Moments due to :

Set Mx = Mx and My = MyDefinitions and AssumptionsEquations for the centroid

Normal Stress

Normal Stress*Example ProblemCan the structure we used for our statics review safely support a 30 kN load? (Assume the entire structure is made of steel with a maximum allowable stress all=165 MPa.)Cross-section 30 mm x 50 mm

Normal Stress

Normal Stress*Example ProblemTwo cylindrical rods are welded together and loaded as shown. Find the normal stress at the midsection of each rod.

Normal Stress

Shearing and Bearing Stress*Shearing and Bearing Stress (1.6-1.8, 1.12)MAE 314 Solid MechanicsYun Jing

Normal Stress

Shearing and Bearing Stress*What is Shearing Stress?We learned about normal stress (), which acts perpendicular to the cross-section. Shear stress () 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

Shearing and Bearing Stress*Where Do Shearing Stresses Occur?Shearing stresses are commonly found in bolts, pins, and rivets.

Free Body Diagram of BoltBolt is in single shearForce P results in shearing stressForce F results in bearing stress (will discuss later)


Shearing and Bearing Stress*We do not assume is uniform over the cross-section, because this is not the case.

is the average shear stress.

The maximum value of may be considerably greater than ave, which is important for design purposes.Shear Stress Defined


Shearing and Bearing Stress*Double Shear

Free Body Diagram of BoltFree Body Diagram of Center of BoltBolt is in double shear


Shearing and Bearing Stress*Bearing StressBearing stress is a normal stress, not a shearing stress.

Thus,

where Ab = projected area where bearing pressure is applied P = bearing force

Single shear caseRead 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 connectionFrom 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.At the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is sBC = +159 MPa.

Pin Shearing Stresses

Pin Bearing Stresses

Shearing and Bearing Stress*Equilibrium of Shear StressesConsider an infinitesimal element of material. Apply a single shear stress, 1. Total shear force on surface is (1)bc. For equilibrium in the y-direction, apply 1 on (-) surface. For moment equilibrium about the z-axis, apply 2 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.

t1t2t2t1


Shearing and Bearing Stress*Equilibrium of Shear Stressest1t2t2t1What 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 stressesPositive face normal is in (+) x, y, or z directionNegative face - normal is in (-) x, y, or z direction

++--

FaceDirectionShear Stress++++----+-+-


Shearing and Bearing Stress*x = stress in x-direction applied in the plane normal to x-axis y = stress in y-direction applied in the plane normal to y-axis z = stress in z-direction applied in the plane normal to z-axisxy = stress in y-direction applied in the plane normal to x-axis xz = stress in z-direction applied in the plane normal to x-axis zy = stress in y-direction applied in the plane normal to z-axisAnd so on

Define General State of Stressyzx


Shearing and Bearing Stress*There are 9 components of stress: x, y, z, xy, xz, yx, yz, zx, zy

As shown previously, in order to maintain equilibrium: xy= yx, xz = zx, yz = zy

There are only 6 independent stress components.Define General State of Stressyzx


Shearing and Bearing Stress*Example ProblemA 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*Example ProblemLink 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.


Oblique Planes and Design Considerations*Oblique Planes and Design Considerations (1.11, 1.13)MAE 314 Solid MechanicsYun Jing

Normal Stress

Oblique Planes and Design Considerations*Stress on an Oblique PlaneWhat 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

Oblique Planes and Design Considerations*Stress on an Oblique PlaneAxial 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 direction must 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 is


Oblique Planes and Design Considerations*Stress on an Oblique PlaneWe 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.


Oblique Planes and Design Considerations*Stress on an Oblique PlaneSince and are functions of sine and cosine, we know the maximum and minimum values will occur at = 00, 450, and 900. At =900 =0At =450 =P/2A0At =00 =P/A0 (max)At =900 =0At =450 =P/2A0 (max)At =00 =0


Oblique Planes and Design ConsiderationsStress on an Oblique PlaneWhat does this mean in reality?


Oblique Planes and Design Considerations*Design ConsiderationsFrom 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 u and ultimate shear stress is denoted as u.


Oblique Planes and Design Considerations*Design ConsiderationsOften the allowable load is considerably smaller than the ultimate load.

It is a common design practice to use factor of safety.


Oblique Planes and Design Considerations*Example ProblemTwo 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*Example ProblemA 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.


*************Assuming pressure is uniform, then force is only applied on one pointBut the moments are the sameSo M \bar = M\bar y *P = M \bar

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Mechanic of Materials (Normal Stress)