'236 (a) For the axial loading shown, determine the change in height and the change
PROBLEM 2- in qume of the brass cylinder shown. (b) Solve parts assuming that [he loading is
hydrostatic with q. - In;- a: - - 70 MPa.
1 " SOLUTION
he: ['3 MM 'i 0
Bulk Mndulns: Veiurne changes are prndueed uni].-r h'ji nermal stresses.
Fur example, ennsider an element leaded with unljr nnrmal stresses (prin
eipal stresses) as shuwn in Fig. 2.16m). The change in 1Irdlume ean he
shuwn tn he [fer small strains).
Generalized Heekes Law: As neted preyieusiy. Heekes ialsr fer ene
dimensien er fer the eentiilzien ef missial stress and strain fer elastie mate-
rials is giyen by e = E s. Using the principle ef sttp-erpensitieli1 the gener-
alised Heeke's law fer a thre
A 100- by 150-mm rectangular plate QABC is deformed into
the shape shown by the dashed lined in the gure. All
dimensions shown in the figure are in mm. Assuming the .1025
strain state to be uniform, determine at point 0 the strain
components an, a, and 73
In many situations, phvsical constraints prevent strain from occurring in a given direction. For example,
.933 = 0, in the case shown, where iongitudinal movement ofthe long prism is prevented at everv point.
Plane sections perpendicular to the longitudin
1'. cfw_a For the axial loading shown, determine the change in
height and the change in volume ofthe brass cylinder shown.
cfw_in Solve part or, assuming that the loading is hydrostatic with
on = a = 033 = 70 MPa.
Answer: (a) 20.0246 mm; 2143.91 mms.
5. Two blocks of rubber with a modulus of rigidity G = 10 MPa are
bonded to rigid supports and to plate A. Knowing that b = 200 mm
and c = 125 mm, determine the largest allowable load P and the
smallest allowable thickness o of the blocks if the shearing
5. Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensiie
and compressive stresses in portion BC of the beam.
10 mm 10 mm
_L A 14704
L _| I 250 mm
50 mm 15] mm 150 mm
Answer: amp = -102.4 MPa, (Th
The state of stress in a structure given is given as follows:
on =y2 +43:2 y2) (332:1:2 +322
cry,=Jr.2+.s(y2 x2) txy=f(x,y) 1:13 =33 =0
Determine f(x,y) so that the structure may be in equilibrium in the absence of body forces.
Answer: fcfw_x,y) = 20.xy
At room temperature cfw_20 Cl, a 0.5-mm gap exists
between the ends of the rods shown. At a later time
when the temperature has reached 140 C, determine
cfw_a the normal stress in the aluminum rod, cfw_b the change
in length of the aluminum rod. Hint: The
Consider a cubic element subjected to a uniform hydrostatic
pressure, Co as shown in the figure. Find the change in the
volume. Hint: You may use the superposition principle to
nd the individual effects of the stress components, am, am
and all, on the
1. The deformation of a body is dened bv the displacement components
u1 = 0.2!1 112 = 0.5X1 + 0.2.32 113 = 0
Consider a unit square material of OABC, where the initial positions (Xi, X2 of the material points (1.4,
B, and C are given bv cfw_0,0, (1,0), (1
M.S. AUTOMOTIVE ENGINEERING
Investigation of Aerodynamic Influence with Truck Platooning on Single Lane
LECTURER: Assoc. Professor Kunt ATALIK
2. OVERVIEW OF AERODYNAMIC FEATURES