Answer:
Bore = 7 cm
stroke = 6.36 cm
compression ratio = 10.007
Explanation:
Given data:
Cubic capacity of the engine, V = 245 cc
Clearance volume, v = 27.2 cc
over square-ratio = 1.1
thus,
D/L = 1.1
where,
D is the bore
L is the stroke
Now,
V =
or
V =
on substituting the values, we have
245 =
or
D = 7.00 cm
Now,
we have
D/L = 1.1
thus,
L = D/1.1
L = 7/1.1
or
L= 6.36 cm
Now,
the compression ratio is given as:
on substituting the values, we get
or
Compression ratio = 10.007
Answer:
The maximum length of the specimen before the deformation was 358 mm or 0.358 m.
Explanation:
The specific deformation ε for the material is:
(1)
Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:
(2)
Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:
So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:
Then de specific defotmation:
Finally the maximum specimen lenght for a elongation 0f 0.45 mm is: