Beam bending

Design model #274130

What's It?

[σ] =  MPa

[σ] =  MPa

[σ] =  MPa

[σ] =  MPa for d/D=

[σ] =  MPa for h/b=



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Beam bending, SFD & BMD diagram

Change distributed load to equivalent force

Q1 = 6·2 = 12кН

Find support reactions using equilibrium equations

Σ MA = + P · 2 + Q1 · 3 - M - RE · 6= + 12 · 2 + 12 · 3 - 10 - RE · 6=0

Σ ME = - P · 4 - Q1 · 3 - M + RA · 6= - 12 · 4 - 12 · 3 - 10 + RA · 6=0

Support reactions

RA = 15.67kN.

RE = 8.333kN.

Shear force and bending moment equations

For segment AB: (0 ≤ z1 ≤ 2 м )

Q(z1) = + RA = + 15.67 = 15.667 kN

M(z1) = + RA · z = + 15.67 · z

M(0) = 0 kNm

M(2) = 31.333 kNm

For segment BC: (2 ≤ z2 ≤ 4 м )

Q(z2) = + RA - P - q1·(z - 2) = + 15.67 - 12 - 6·(z - 2)

Q(2) = 3.667 kN

Q(4) = -8.333 kN

M(z2) = + RA · z - P·(z - 2) - q1·(z - 2)2/2 = + 15.67 · z - 12·(z - 2) - 6·(z - 2)2/2

M(2) = 31.333 kNm

M(4) = 26.667 kNm

Shear force equal 0 at z = 0.611 м, so this point has extremum on BMD

M(0.611) = 32.5 kNm

For segment CD: (4 ≤ z3 ≤ 5 м )

Q(z3) = + RA - P - Q1 = + 15.67 - 12 - 12 = -8.333 kN

M(z3) = + RA · z - P·(z - 2) - Q1·(z - 3) = + 15.67 · z - 12·(z - 2) - 12·(z - 3)

M(4) = 26.667 kNm

M(5) = 18.333 kNm

For segment DE: (5 ≤ z4 ≤ 6 м )

Q(z4) = + RA - P - Q1 = + 15.67 - 12 - 12 = -8.333 kN

M(z4) = + RA · z - P·(z - 2) - Q1·(z - 3) - M = + 15.67 · z - 12·(z - 2) - 12·(z - 3) - 10

M(5) = 8.333 kNm

M(6) = 0 kNm

Solve section sizes using maximal moment Mmax = 32.5 kNm.

Strength condition for bending σ = Mmax / W ≤ [σ]

Wmin=Mmax / [σ]


Select I-beam with allowable stress [σ] = 160 MPa
Wmin=32500 / 160 = 203.125 cm3
Select I-beam #22 with resistance moment W = 231.82 cm3 and area A = 30.6 cm2
Maximal normal stress is
σmax = Mmax/Wx = 32500/231.82 = 140.19 MPa
Maximal shear stress (at axis) is
τmax = Qmax×Sx/b×Ix = 15700×114.96×10-6/0.0054×2550×10-8 = 13.107×106 Pa = 13.107 MPa
Shear stress between Flange and Web is
τmax = Qmax×Sx'/b×Ix = 15700×101.11×10-6/0.0054×2550×10-8 = 11.528×106 Pa = 11.528 MPa,
where Static moment of Flange is
Sx'=b×t×(h-t)/2=11×0.87×(22-0.87)/2=101.11 cm3.
Normal and shear stresses in I-beams section:

Select round section.
Wmin=32500/160=203 cm3
Resistance moment of round section
W=π×d3 / 32
d3=32×W / π = 32×203 / π = 2069
Diameter of the section d=12.7 cm
Section area
A=π×d2/4=π×12.72/4=126.61 cm2
Maximal normal stress is
σmax = 32×Mmax/π×d3 = 32×32500/π×12.73 = 161.69 MPa
Maximal shear stress
τmax = 4Qmax/3A = 4×15700/3×126.61×100 = 1.653 MPa
Normal and shear stresses in round section:

Select tube with α = d/D = 0.9
Wmin=32500 / 160=203 cm3
Resistance moment of tube is
W=π×D3 ×(1-α4)/32
D3=32×W / π×(1-α4) = 32×203 / π×(1-0.94)=6016
Diameter D=18.2 cm
Area A=π×D2(1-α2)/4=π×18.22(1-0.92)/4=49.4 cm2
Maximal normal stress is
σmax = 32×Mmax/π×D3×(1-α4) = 32×32500/π×18.23×(1-0.94) = 159.76 MPa
Maximal shear stress
τmax = Qmax×Sx/b×Ix, где b=D-d
Static moment of half section
Sx=2R3/3-2r3/3=(D3-d3)/12=(18.23-(18.2×0.9)3)/12=136.15 cm3
Moment of Inertia
Ix=π×D4×(1-α4)/64=π×18.24×(1-0.94)/64=1851.26 cm4
τmax = 15700×136.15×10-6/(18.2-0.9×18.2)×0.01×1851.26-8=0.063×106 Pa = 0.063 MPa
Normal and shear stress diagrams in tube section:

Select square section.
Wmin=32500 / 160=203 cm3
Resistance moment of square section
W=a3/6
Size of square a= 10.7 cm
Area A=a2=10.72=114.49 cm2

Select rectangle section for h / b=2
Wmin=32500 / 160 = 203 cm3
Resistance moment of rectangle
W=b×h2 / 6 = b3 × 22 / 6 = b3×0.67
b3=203 / 0.67=303
Width b=6.7 cm, Height h=b×2=6.7×2=13.4 cm
Area A=b×h=6.7×13.4=89.78 cm2
Maximal normal stress is
σmax = 6×Mmax/b×h2 = 6×32500/6.7×13.42 = 162.09 MPa
Maximal shear stress for rectangle
τmax = 3Qmax/2A = 3×15700/2×89.78×100 = 2.623 MPa
Normal and shear stress diagrams in rectangle section:

Section angle and deflection equations using DOUBLE INTEGRATION METHOD

For segment AB: (0 ≤ z1 ≤ 2 м )

EJ×φ(z) = EJ×φ0 + RA·z2/2

EJ×v(z) = EJ×v0 + EJ×φ0×z + RA·z2/2

For segment BC: (2 ≤ z2 ≤ 4 м )

EJ×φ(z) = EJ×φ0 + RA·z2/2 - P·(z - 2)2/2 - q1·(z - 2)3/6

EJ×v(z) = EJ×v0 + EJ×φ0×z + RA·z2/2 - P·(z - 2)2/2 - q1·(z - 2)3/6

For segment CD: (4 ≤ z3 ≤ 5 м )

EJ×φ(z) = EJ×φ0 + RA·z2/2 - P·(z - 2)2/2 - q1·(z - 2)3/6 + q1·(z - 4)3/6

EJ×v(z) = EJ×v0 + EJ×φ0×z + RA·z2/2 - P·(z - 2)2/2 - q1·(z - 2)3/6 + q1·(z - 4)3/6

For segment DE: (5 ≤ z4 ≤ 6 м )

EJ×φ(z) = EJ×φ0 + RA·z2/2 - P·(z - 2)2/2 - q1·(z - 2)3/6 + q1·(z - 4)3/6 - M· (z - 5)

EJ×v(z) = EJ×v0 + EJ×φ0×z + RA·z2/2 - P·(z - 2)2/2 - q1·(z - 2)3/6 + q1·(z - 4)3/6 - M· (z - 5)

Calculate initial conditions using support conditions:

- angle of rotation φ0 = -61.83 kNm2

- deflection v0 = 0 kNm3

Angles of rotation and deflections for all sections

For segment AB

EJ×φ(0) = -61.83 kNm2

EJ×v(0) = 0 kNm3

EJ×φ(0.5) = -59.88 kNm2

EJ×v(0.5) = -30.59 kNm3

EJ×φ(1) = -54 kNm2

EJ×v(1) = -59.22 kNm3

EJ×φ(1.5) = -44.21 kNm2

EJ×v(1.5) = -83.94 kNm3

EJ×φ(2) = -30.5 kNm2

EJ×v(2) = -102.8 kNm3

For segment BC

EJ×φ(2) = -30.5 kNm2

EJ×v(2) = -102.8 kNm3

EJ×φ(2.5) = -14.5 kNm2

EJ×v(2.5) = -114.1 kNm3

EJ×φ(3) = 1.667 kNm2

EJ×v(3) = -117.3 kNm3

EJ×φ(3.5) = 17.25 kNm2

EJ×v(3.5) = -112.5 kNm3

EJ×φ(4) = 31.5 kNm2

EJ×v(4) = -100.2 kNm3

For segment CD

EJ×φ(4) = 31.5 kNm2

EJ×v(4) = -100.2 kNm3

EJ×φ(4.25) = 37.91 kNm2

EJ×v(4.25) = -91.54 kNm3

EJ×φ(4.5) = 43.79 kNm2

EJ×v(4.5) = -81.31 kNm3

EJ×φ(4.75) = 49.16 kNm2

EJ×v(4.75) = -69.68 kNm3

EJ×φ(5) = 54 kNm2

EJ×v(5) = -56.78 kNm3

For segment DE

EJ×φ(5) = 54 kNm2

EJ×v(5) = -56.78 kNm3

EJ×φ(5.25) = 55.82 kNm2

EJ×v(5.25) = -43.04 kNm3

EJ×φ(5.5) = 57.13 kNm2

EJ×v(5.5) = -28.91 kNm3

EJ×φ(5.75) = 57.91 kNm2

EJ×v(5.75) = -14.52 kNm3

EJ×φ(6) = 58.17 kNm2

EJ×v(6) = 0 kNm3


Examples
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