Statics and Mechanics of Materials

Statics and Mechanics of Materials Chapter 4-1 Internal force, normal and shearing Stress

Department of Mechanical Engineering

Outlines



Internal Forces - cutting plane  

Result of mutual attraction (or repulsion) between molecules on both sides of the cutting plane These result is distributed over the entire surface of the cutting plane


Internal Forces - cutting plane  Each

part of the body satisfies the equilibrium equation  The resultant of the internal forces R must be in equilibrium with other applied forces in the body part  Stress is the intensity of the R  So either body part can be used to determine the internal forces


Internal Force – cutting plane  If

the cutting plane is perpendicular to the bar axis  the internal forces, internal stress, and the resultant will be perpendicular  in normal direction  If the cutting plane is not perpendicular  the resultant will still be perpendicular, but it has normal and tangential components


Normal stress – axial loading  Axial

loading = the loading/force is collinear with the axis of the bar  Stress = intensity of the internal force  Generally speaking,

 Or

symbolically,


Normal stress – some notes  Generally,

the stress is not uniformly distributed over the area  For many applications, it’d be assumed that it is uniformly distributed  Cross area changes under loading  Engineering stress uses initial cross sectional area  True stress uses the deformed area Department of Mechanical Engineering

Shearing stresses in connections  Loads

are transmitted to individual members through connections that use rivets, bolts, pins, nails, or welds

 Single

shear  Double shear  Punching shear  Bearing stress Department of Mechanical Engineering

Single shear and double shear

Single shear

Double shear


Punching shear  Example:

Shear stress developed due to action of punch in forming a rivet hole

P  As P

P


Bearing stress  Bearing

stress = compressive normal stress  While the amount of the force = compression load, the area depends on the mode of the contact  Examples: – Between the head of the bolt and the top plate (a) – Between the surfaces of the shanks and hole which they pass (b)

F b  Ab Department of Mechanical Engineering

Units of Stress  Dimension

= FL-2

 USCS;

– psi (pounds per square inch), – ksi (kilo pounds per square inch), – ksi = 1000 psi  SI;

– Pa (Pascal = N/m2), – kPa (kilo Pa) = 1000 Pa, or – MPa (mega Pa) = 106 Pa Department of Mechanical Engineering

Example Problem 4-1  The

cross-sectional area = 3 in2.  Determine the axial stress in the bar on a cross section; – 20” to the right of A – 20” to the right of B – 20” to the right of C  First

thing to do; to determine the internal force on the section  use cutting plane


Example Problem 4-4  The

column experiences compression  Determine the bearing stress on the surface between the bearing plate and the column


Example Problem 4-4  FBD

of the bearing plate

Compression developed in the timber beam do

di

Compression developed in the column

F b  Ab

Ab 

Cross section of the column



 d 4

2 o

 d i2




Problem 4-12  Average

punching shear stress in the collar  Average bearing stress between the collar and the plate

Plate

Bearing stress Punching shear stress Department of Mechanical Engineering

Please read and practice example problems 4-3, 4-4 and 4-5



Maximum and minimum stresses normal stress  when  = 0 (or 180) P  max  A  Maximum shearing stress  when  = 45 (or 135) (opposite directions)   P  Maximum

max

2A

stress = 0, when  = 90  Note: maximum stresses don’t appear on the same angle  Minimum


Example Problem 4-7  Given:

─ A = 200x100 mm2 ─ AB = 12.00 MPa ─  = 36o  Questions:

– P? – AB=? – Max normal and shearing stresses


Example Problem 4-7: Answer  Follow

the solution in the book or use P (1  cos 2 ) 2A  n 2A P  694.7 kN (1  cos 2 )

n 



 And



P sin 2  16.52MPa 2A

the maximum are P  34.7 MPa A P   17.37 MPa 2A

 max   max
