Example 12
Satu rod 50 mm panjang dan berdiameter 12mm mengalami pemanjangan 0.0435 mm dan perubahan diameter 11.996 mm apabila dikenakan daya tegangan 20 kN. Kirakan
i) Tegasan
ii) Terikan sisi
iii) Terikan paksi
iv) Modulus Young
v) Nisbah poisson
vi) Faktor keselamatan jika tegasan muktamad 432 MPa
A rod is 50 mm long and 12 mm diameter extension having a diameter of 0.0435 mm and 11.996 mm change when subjected to tensile force 20 kN. Calculate:
i) The stress
ii) The strain in the side
iii) The axial strain
iv) Young’s Modulus
v) Poisson’s ratio
vi) Safety factor if the final stress 432 MPa
Penyelesaian:
Data diberi:
L = 50 mm
D1 = 12mm
DL = 0.0435 mm
D0 = 11.996 mm
P = 20 kN
i) Tegasan
P
s = --------
A
π D2
Ü A = ---------------
4
π (12 x 10-3)2
= ------------------
4
= 1.1309x 10-4 m2 ##
P
Ü s = --------
A
20 x 103
= -------------------------
1.1309x 10-4
= 176.83 MN/m2 @ MPa ##
ii) Terikan sisi
-(Do –Df )
esisi = ------------
Do
-(11.996 x 10-3) –(12 x 10-3 )
= -------------------------------------
11.996 x 10-3
= 3.3344 x 10-4 M ##
iii) Terikan paksi
Lf –Lo
epaksi = -----------
Lo
0.0435 x 10-3
= ----------------
50 x 10-3
= 8.7 x 10-4 M ##
iv) Modulus Young
s
Ü E = -------
epaksi
176.83 x 106
= ---------------------
8.7 x 10-4
= 203.25 x109 N/m2##
v) Nisbah poisson
esisi
n = ----------------
epaksi
3.3344 x 10-4
= ----------------
8.7 x 10-4
= 0.383##
vi) Faktor keselamatan jika tegasan muktamad 432 MPa
smax/muktamad
F.K = ---------------
skerja
432 x 106
= ----------------
176.83 x 106
= 2.44##
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