(Jun 2011)
Example 13
A bar with a 30 mm diameter and 80 mm length is subjected to a tensile force of 100kN. The ultimate stress is 230 MN/m2. As a result of this force, the bar elongated by 0.0585 mm and the diameter be comes 29.994 mm. Determine:
i) The tensile stress
ii) The tensile strain in x-x direction
iii) The tensile strain in y-y direction
iv) Modulus of elasticity
v) Strain energy
vi) Safety factor
vii) Poisson’s ratio
Satu bar bergarispusat 30 mm dan panjang 80 mm dikenakan beban kerja sebanyak 100 kN. Disebabkan daya ini, bar telah memanjang sebanyak 0.0585 mm dan diameternya berubah ke 29.994 mm. Diketahui, tegasan muktamad bagi bar ini ialah 230 MN/m2. Kirakan:
i) Tegasan
ii) Terikan paksi
iii) Terikan sisi
iv) Modulus Young
v) Tenaga Terikan
vi) Faktor keselamatan
vii) Nisbah poisson
Penyelesaian:
Data diberi:
D0 = 30 mm
L0 = 80 mm
D1 = 29.994mm
DL = 0.0585 mm
P = 100 kN
smuktamad =230 MN/m2
i) Tegasan
P
s = --------
A
π D2
Ü A = ---------------
4
π (30 x 10-3)2
= ------------------
4
= 7.069 x 10-4 m2 ##
P
Ü s = --------
A
100 x 103
= -------------------------
7.069 x 10-4
= 141.463 MN/m2 @ MPa ##
ii) Terikan paksi
Lf –Lo
epaksi = -----------
Lo
0.0585 x 10-3
= ----------------
80 x 10-3
= 7.3125 x 10-4 M ##
iii) Terikan sisi
-(Do –Df )
esisi = ------------
Do
-(30 x 10-3) –(29.994 x 10-3 )
= -------------------------------------
30 x 10-3
= 2 x 10-4 M ##
iv) Modulus Young
s
Ü E = -------
epaksi
141.463 x 106
= ---------------------
7.3125 x 10-4
= 193.454 x109 N/m2##
v) Tenaga terikan
1
U = --- PDL
2
1
= ---- (100 x 103 ) ( 5.85 x 10-5 )
2
= 2.925J##
vi) Faktor keselamatan jika tegasan muktamad 432 MPa
smax/muktamad
F.K = ---------------
skerja
230 x 106
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