Exercise 1:

1- Define the following

a-     Stress, σ

b-    Strain, ε

c-     Modulus of Elasticity (Young’s Modulus), E

d-    Modulus of Rigidity (Shear Modulus), G

e-     Second Moment of Area, I

f-      Polar Moment of Area, J

g-     Bending moment Diagram, BMD

h-    Torque, T

i-       Moment, M

j-       Friction force, Ff

k-     Centroid of area

l-       Factor of safety (Safety factor), SF

m-  Example of:

i.            Energy transmission elements

ii.            Energy storage elements

iii.            Locating (fixing, joining) elements

iv.            Friction reduction elements

n-    The differences between:

i.             Tensile and Compression Stresses, σt σc

ii.            Shear and Torsional Stresses, τsτt

iii.            Yield and Ultimate Stresses, σyσu

iv.            (Pa) and (MPa)

v.            Simple beam and Cantilever

o-    Find the following:

i.            Area (A) and Second moment of area (I) of a rectangle (B,H)

ii.            Area (A) and Second/Polar moment of area (I, J) of a circle (R)

 2- Find the point and value of maximum stress on the beam, (circular cross sectional area (d = 20 mm)) . Where: F = 150 N, F1 = 100 N, I = 2000 mm4 L = 400 mm, L1 = 100 mm, L2 = 250 mm Solution Hint: Using ΣFy = 0, ΣMo = 0, to find R1 = R2 = Draw: BMD and obtain point and value of Mmax Calculate: σb (using: M, y, I)

3- A Force of 2000 N, is applied on a tip of a beam of rectangular cross section (10 x 4 mm) and with a length of 120 mm. Chose suitable steel, taking SF = 6.

a- show the force direction on the element and the reaction due to this force to induce the following stress:

 i-                   Tensile ii-                 Compression iii-               Bending iv-               Torsion, J = b h/[3 (b2 + h2)] v-                 Shear And find the stress due to the force applied:   Solution; i-                   Tensile stress, σt = F/A = 2000/(10x4) = 50 MPa σt = σall =50 MPa σu = σall x SF = 50 x 6 = 300 MPa From chart 9, chose steel EN1A (420 MPa)