The composite beam is acted upon by 5 kNm couple shown Knowi

The composite beam is acted upon by 5 kN-m couple shown. Knowing that E = 190 GPa and solve (a) the neutral axis for the distance(mm) above the bottom as given, (b) the moment of inertia (mm^8) relative to the neutral axis for the composite beam, (c) the maximum tensile, (d) compressive stresses (MPa), and (e) the radius of curvature (m) of this beam

Solution

Data given:

M (couple acting on beam) = 5 kN-m

E(modulus) = 190 GPa ;                                                 10@9 mean 10 power 9, 10@-3 mean 10 power -3.                               

Cross Section of beam is as shown below:

                                                                                                                            X

The moment acting on the beam lies on the plane of the beam and with other bending theory assumptions, such as plane sections remain plane before and after bending and Isotropic and homogenous material.

The flexural formulae is written as:

(M/ I ) = (s / y) = (E/R)

Where M is bending moment N.m

I is moment of inertia of the section

s is Bending stress.

Y distance of the layer from neutral axis.

E is young’s modulus of the material.

R is radius of curvature of the beam.

Neutral axis is nothing but the centroid of the section of the beam since the composite beam of same E = 190 GPa.

Centroid of the above I section is given by:

The y -coordinate is : yc (from x-axis as shown above) = ((60*10*40)+(20*25*22.5)*(10*40*5))/(600+500+400) = 24.83 mm from bottom of the section.(X-axis shown).

Therefore Neutral axis passes through yc.

The moment of Inertia is given by:

The moment of inertia of a rectangle about its centroidal axis is

(B*D*D*D)/12.

By using transformation theorem from centroidal axis to neutral axis we get the total moment of inertia of the section as:

= ((60*10*10*10)/12)+(60*10*(40-24.83)(40-24.83))+((20*25*25*25)/12)+(20*25*(24.83-22.5)*(24.83-22.5))+((40*10*10*10)/12)+(10*40*(24.83-5)*(24.83-5))

= 5000+138077.34+26041.66+2714.45+3333.33+157291.56

= 332458.34 mm4 is the moment of inertia about neutral axis.

M/I =E/R

Radius of curvature R = EI/M = (190 * 10@9 * 332458.34 * 10@ -12)/5000          

                                                = 12.633 metres.

The stress in the beam:

The stress at neutral axis is zero.

The stress is given by:

s/y = E/R Þ s = Ey/R.

As the moment induces hogging to the beam the top layers of the beam are under tension and the bottom layers are under compression

s (at y = -24.83) = (190 * 10@9 * 24.83*10@-3)/12.633 = 373.442MPa (compressive)

s (at y = 20.17) = (190 * 10@9 * 20.17*10@-3)/12.633 = 303.356 MPa (tensile).