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Concept:
Bending Moment Equation
MI=σy=ER
M=Iyσ=Zσ
For the constant bending moment: σ ∝ 1/Z
So, bending stress is inversely proportional to the Section Modulus
The strength of two beams of the same material can be compared by the section modulus values.
The beam is stronger when section modulus is more, the strength of the beam depends on section modulus.
The bending moment distribution in a beam is a function of distance x and it is given by M = (5x2 + 20x - 7) Nm. Find the shear stress at x = 2m if the shearing area is 2 mm2.
shearforce,F=dMdx
Calculation:
Bending moment (M) = (5x2 + 20x - 7)
Now,
∴ F = 10x + 20
(F)(x = 2 m) = 10(2) + 20
∴ F = 40 N
shear stress (τ)=FA=402
A circular cantilever beam is subjected to torsion at the tip. Because of tension, the angle of twist is found to be 0.5° as shown in the figure. The length of beam (in mm) is _______. Take (R = 100 mm, ∅ = 0.1°)
Arc length ‘S’ is given by,
S = Rθ
Also, in the case of torsion, we can write,
S = L∅
∴ L∅ = Rθ
L=Rθ∅
L=100×0.50.1
Maximum bending across shaft occurs at either top or bottom fiber and it is given by the bending formula:
σmax=MZ=2Mπ64d4×d2=64Mπd3
Maximum shear stress is given by the twisting formula:
τmax=TrI=(12)×(d2)π32d4=8Tπd3
σmaxτmax=64Mπd38Tπd3=8MT
A rectangle beam 60 mm wide and 150 mm deep is simply supported over a span of 4 m. If the beam is subjected to uniformly distributed load of 4.5 kN/m, then the maximum bending stress induced in the beam is _____ MPa.
Section modules of rectangular section, Z=db26
Given:
b = 60 mm, d = 150 mm
∴Z=bd26=60×15026=225×102mm3
The maximum bending moment at the center of a simply supported beam subjected to UDL is, M=wl28
w = 4.5 kN/m, l = 4 m = 4 × 103 mm
∴M=wl28=4.5×(4×103)28=9×106N−mm
∴σbmax=MZ=9×106225×103
∴ σmax = 40 MPa
Points to Remember:
i) For rectangular section Z=bd26
ii) Maximum Bending moment in a simply supported beam subjected to UDL in at center and M=wl28
iii) Maximum Bending Stress =MZ
A solid circular beam with a radius of 0.25 m and length of 2 m is subjected to a twisting moment of 20 kNm about the z-axis at the free end, which is the only load acting as shown in the figure. The shear stress component τxy at point ‘M’ in cross-section of the beam at a distance of 1 m from the fixed end is
For the given beam:
Let the point M is displaced from its original position under the action of external torque acting on the bar.
The torque so applied along the z-axis would displace the particle M in the y-z and x-z plane.
∴ The displacement in the y-z and x-z plane induces strain which is the result of stress in that plane.
As there is no displacement in the x-z plane, no strain and hence no stress is introduced in the x - y plane.
∴ The stress produced in the x-y plane is 0 (zero).
Using the flexural formula,
MI=ER=σby
And,
σb=M.yI
For the circular shaft, y=d2 and I=π64d4
σb=32Mπd3
Using torsion equation,
TJ=GθL=τr
τ=T.rJ
For the circular shaft, τ=τmax at r=d2 and J=π32d4
τ=16Tπd3
Using the given condition,
σb=1.5×τ
32Mπd3=1.5×16Tπd3
M=0.75T
Maximum principle stress is given as,
σ1=16πd3[M+M2+T2]
100×106=16π×0.083[0.75T+(0.75T)2+T2]
A cantilever beam is subjected to uniformly distributed torsion throughout its length. Find out the angle of twist at the free end. Where t = torsion per unit length (Nm/m)
Explanation:
Consider an element of length ‘dx’ at distance ‘x’ from fixed support of cantilever beam.
Torsion act at section x-x is,
Tx−x=t.x
The angle of twist,
dθ=Tx−xdxGJ
Integrating,
∫0θdθ=∫0LTx−xdxGJ
∫0θdθ=∫0Lt.xdxGJ
θ=[tx22GJ]0L
∴ θ=tL22GJ
TJ=Gθl=τr
Angleoftwist,θ=TlGJ
dB=0.5dA
lB=1.5lA
TA=TandTB=0.75T
θB=TBlBG×π32(d2)4
θB=0.75T×1.5lG×π32(d2)4=576TlGπd4
θA=TAlAG×π32d4
θA=TlG×π32d4=32TlGπd4
θBθA=576TlGπd432TlGπd4=18
i.e. strength in bending depends on geometrical parameters section modulus, Z=(Iy)
Z=Iy=bd312d2=bd26
Also from the geometry,
D2=b2+d2
∴d2=D2−b2 ...(1)
∴Z=b(D2−b2)6
Differentiating Z w.r.t. b and equating to zero, since rectangular cross-section beam of highest strength against bending is mentioned.
dZdb=(D2−3b2)6=0
∴D2=3b2 ... (2)
Put equation (2) in equation (1)
d2=3b2−b2
(db)2=2
∴(db)=2
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