"7laha - Bending 1"
Smart structure problems for never stop thinking
Source: Mechanics of material by
Ferdinand p. Beer and E .Russell Johnston
SMART EXAMPLES
(1) The rectangular tube shown is extruded from an aluminum alloy for which σy =275 M pa, σ ultimate = 415 M pa and E=73 Gpa .Neglecting the effect of fillets .
(2)the corresponding radius of curvature of the tube.
Solution :
σall=σu/F.s =415/3 =138 Mpa
I=(80*(120^3))/12 –(68*(108^3))/12 =4.38*(10^-6)m^4.
NOTE THAT : σall<σy then the tube remains In elastic range.
C=t .006 m
M=I*σall/c =10.1 KN.m
Ρ=E*I/M =(73*4.38*1000)/10.1KN.m = 31.7 m
since we know the max stress is 1380Mpa .we can determine the max strain
єm = 138 Mpa/73 Gpa = 1.89*10^-3
Єm = C/ρ so ρ= 60/1.89*10^-3 = 31.7 m
a
Important note :THE MAX TENSILE STRESS BE AT THE FARTHEST LINE FROM THE MOMENT LINE AND NAX COMPRESSION STRESS AT THE CLOSED LINE FROM THE MOMENT LINE(STRUCTURE AXIS).
(2) A cast –iron machine part is acted upon by the 3KN.m couple shown Knowing that E=165 Gpa and neglecting the effect of fillets .
Determine:
(a)the max tensile and compressive stresses in the casting.
(b) the radius of curvature of the casting.
Solution :
we divide this body to two rectangular
Total area =(20*90)+(40*30) =0.3m^2
∑Y*A=(50*1800)+(20*1200)=0.114m so YG =38mm
Ix=∑(bh^3/12 +Ad^2) =(90*(20^3))/12*(90*20*12^2) )+(30*(40^3)/12
=(30*40*(18^2))
I=868*10^-9 m^4.
BENDING=COMPRESSION+TENSION
a
The main relations:
a
1- (єaxial =-Y/ρ) where y the distance between the structure axis and the point you want to calculate the strain .(to make our calculations simple and to our work be suitable to solve real problems we will add the term over what we studied (єmax =c/ρ ) where c =max value of y.
2- σx=-M*y/I where I = second moment of Area =b*h^3/12.(when we deal with rectangular),M=moment of bending.
where h always normal to the structure axis and b parallel to the structure axis. As previous σmax = -M*c/I .
3-The radius of curvature of the body after the moment (ρ=E*I/M).
As I think we can easily determine the Centroid of the circuit, triangle and rectangular. What about the compound body the matter is easy take this rule which passed you before: divide your body to known shapes.
YG (to the whole body)=∑y*A/∑A
YG (to the whole body)=∑y*A/∑A
SMART EXAMPLES
(1) The rectangular tube shown is extruded from an aluminum alloy for which σy =275 M pa, σ ultimate = 415 M pa and E=73 Gpa .Neglecting the effect of fillets .
DETERMINE:
(1) the bending moment M for which the factor of safety will be 3 .(2)the corresponding radius of curvature of the tube.
Solution :
σall=σu/F.s =415/3 =138 Mpa
I=(80*(120^3))/12 –(68*(108^3))/12 =4.38*(10^-6)m^4.
NOTE THAT : σall<σy then the tube remains In elastic range.
C=t .006 m
M=I*σall/c =10.1 KN.m
Ρ=E*I/M =(73*4.38*1000)/10.1KN.m = 31.7 m
Another solution :
since we know the max stress is 1380Mpa .we can determine the max strain
єm = 138 Mpa/73 Gpa = 1.89*10^-3
Єm = C/ρ so ρ= 60/1.89*10^-3 = 31.7 m
a
Important note :THE MAX TENSILE STRESS BE AT THE FARTHEST LINE FROM THE MOMENT LINE AND NAX COMPRESSION STRESS AT THE CLOSED LINE FROM THE MOMENT LINE(STRUCTURE AXIS).
(2) A cast –iron machine part is acted upon by the 3KN.m couple shown Knowing that E=165 Gpa and neglecting the effect of fillets .
Determine:
(a)the max tensile and compressive stresses in the casting.
(b) the radius of curvature of the casting.
we divide this body to two rectangular
Total area =(20*90)+(40*30) =0.3m^2
∑Y*A=(50*1800)+(20*1200)=0.114m so YG =38mm
Ix=∑(bh^3/12 +Ad^2) =(90*(20^3))/12*(90*20*12^2) )+(30*(40^3)/12
=(30*40*(18^2))
I=868*10^-9 m^4.
MAX tensile stress
σ a = M*C a /I =3KN.m *.022m/868*10^-9 =67 Mpa
Max Compressive stress =M*Cb/I =3 KN.m*.038 /868*10^-9 =-131.3 Mpa
σ a = M*C a /I =3KN.m *.022m/868*10^-9 =67 Mpa
Max Compressive stress =M*Cb/I =3 KN.m*.038 /868*10^-9 =-131.3 Mpa
NOTE THAT: THE APPLIED MOMENT BEND THE COUPLE TO UP WARDS SO THE CENTRE OF CURVATURE IS DOWNTHE SECTION.
Ρ=E*I/M =47.7 m
Ρ=E*I/M =47.7 m
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