1 / 12

Strengths

Strengths . Chapter 9. 9-1 Intro. Dealing with relationship between the external loads applied to an elastic body and the intensity of the internal forces acting within the body Intensities of the internal resisting forces are called stresses Strengths deals with the deformations of the body

rayya
Download Presentation

Strengths

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Strengths Chapter 9

  2. 9-1 Intro • Dealing with relationship between the external loads applied to an elastic body and the intensity of the internal forces acting within the body • Intensities of the internal resisting forces are called stresses • Strengths deals with the deformations of the body • Deformation per unit length is called the strain • Subject of strength of materials involves strength – load carrying capacity based on stresses inside a member, stiffness deformation characteristics and stability the ability of a slender member to maintain its initial configuration without buckling while being subjected to compressive loading

  3. 9-2 Normal and shear stresses • Structural member is subjected to loads internal resisting forces are generated within the member so that the external forces can be balanced and the body can hold itself together • Intensities of internal forces per unit area are called stresses • Two types of stresses • Norma tresses – perpendicular to the area • Shear stresses by internal forces tangential to the area • Us customary system – pounds per square inch (psi, psf, or kips • SI newtons per square meter (N/m²) also designated pascal (Pa)

  4. 9-2 Normal and shear stresses • 1 kPa = 10³ Pa • 1 Mpa = 10(6) Pa • 1 Gpa = 10(9) Pa • Conversion factors • 1 psi = 6.895 kPa • 1 ksi = 6.895 Mpa • 1 psf = 47.88 Pa

  5. 9-3 Direct Normal Stresses • Bar uniform cross section is called a prismatic bar • Subject to equal and opposite pulling forces • A member subjected to axial loads is called an axially loaded member • Pulling forces to elongate stretch the bar = tension • Pushing forces contract or shorten said to be in compression • Uniform normal stress σ=p/a • σ = normal stress in the cross section • P = internal axial force at the section • A = the cross sectional area of the rod • Allowable axial load – members are limited stress level called allowable stress σ allow – which is the upper limit of stress not to be exceeded

  6. 9-3 Direct Normal Stresses • Required Area = minimum cross section area A of a member designed to carry a maximum axial load P without exceeding the stress • Internal Axial Force Diagram – variation of internal axial force along the length of a member can be depicted by an internal axial force diagram • Tensile force are positive • Compressive force as negative • Example 9-1 page 326 • Example 9-2 page 326 • Example 9-3 page 328 • Example 9-4 page 329

  7. 9-4 Direct Shear Stresses • Shear stress defined as intensity of internal force tangential to the area in question • Shear stress differs from normal stress because shear stress is parallel to the area on which it acts while normal stress is normal to the area • Horizontal force P applied to the protruded part tends to shear the part off the block along the shear plane abcd • Body resists the force P by developing resisting shear stresses in the shear plane • Resultant of the shear stresses must be equal to the applied force P

  8. 9-4 Direct Shear Stresses • Ƭ(avg) = p/A • Ƭ(avg) = the average shear stress • P = the internal resisting shear force tangent to the shear plane • A = the area of the shear plane • Direct shear stresses are found in bolts , rivets, pins, keys, etc. • Lap joint • Connecting two overlapping tension plates with a rivet – rivet is subjected to shear stress through section m-m • Shear stress in general is not uniformly distributed in the section • Since shear stress occurs only in one section of the rivet it is called a single shear

  9. 9-4 Direct Shear Stresses • Butt Joint – connects non overlapping tension plates using connecting plates • Ƭ(avg) = P/2A • Shear stresses occur in two sections of the rivet – rivet is said to be in double shear • Shaft Key –connects a gear to a shaft – moment M on the gear is transmitted to the shaft through the key – the key is subjected to forces • P=M/r • Average shear stress at section m-m of the key is Ƭ(avg) =P/A=M/r/bL =M/rbL • B is the width of the key , L is the length of the key, r is the radius of the shaft

  10. 9-5 Bearing Stresses • When one body presses against another bearing stress occur between the two bodies • Example 9-6 page 333 • Bearing stress in shaft key – occur between the key and the gear between the key and shaft • Example page 333 • Examples 9-5 page 334 • Examples 9-6 page 335 • Examples 9-7 page 336 • Examples 9-8 page 337

  11. 9-6 Stresses on Inclined Planes • Both normal and shear stresses exist • Force can be resolved into two components – normal component perpendicular to the inclined plane and tangential component parallel to the inclined plane • The normal component produces normal stress and tangential component produces shear stress • Examples 9-9 page 339 • Example 9-10 page 340

  12. 9-7 Stresses in thin walled pressure Vessels • Are leak proof containers subjected to internal pressure , boilers, fire extinguishers and compressed air tanks are examples • Examples 9-11 page 343 • Examples 9-12 page 343 • Examples 9-13 page 344

More Related