5 Bending stress tutorial solutions PDF

Preview

Summary

answers...

Description

EG22010 : Bending Stresses 1.

I yy = (0.06 * 0.013)/12 + 2 ( 0.01 * 0.083 ) / 12 = 858.3 x 10-9 m4 I xx = (0.01 * 0.063)/12 + 2 [ ( 0.08 * 0.013 ) / 12 + 0.08 * 0.01 * 0.0352 ] = 2153 x 10-9 m4 M = σI / y M y = 80 x 106 * 858.3 x 10-9 / 0.04 = 1717 Nm M x = 80 x 106 * 2153 x 10-9 x 10-9 / 0.04 = 4306 Nm

2.

y centroid = Σ Ay / Σ A y centroid = [ (60*10*30) + (70*10*65) ] / [ (60*10) + (70*10) ] = = 48.8mm from base of t I xx = [ 0.01*0.063/12 +0.01*0.06*0.01882 ]+[ 0.07*0.013/12 +0.07*0.01*0.01622] = 582 x 10-9 m4 σmax tension σ = My / I centroid 48.8mm

σ max tension = 1000 * 0.0212 / 582 x 10-9 = 36.45 MPa σ max compression = 1000 * 0.0488 / 582 x 10-9 = 83.84 MPa

3.

Will first need to find the maximum bending moment. Find restraints ΣF v = 0, A v + Bv - 100 * 3 = 0 ΣM = 0, Bv * 2 – 3 * 100 * 1.5 = 0 B v = 225 N, A v = 75 N Consider AB : 100 N/m M z

Q

75N

Q = 75 – 100z N M = 75z – 100z (z/2) = 75z – 50z2 Nm

σmax compression

Consider AC : 100 N/m M Q

z 75N

225N (Bv)

Q = 75 + 225 – 100z = 300 – 100z N M = 75z – 100z (z/2) + 225(z-2) = 75z – 50z2 + 225z – 450 Nm M = -50z2 + 300z – 450 Nm

Shear Force (N) 75

-125

Bending Moment (Nm)

Here maximum +ve bending moment occurs when SF = 0 75 - 100z = 0, z = 0.75m BM = 75 x 0.75 – 50 x 0.752 = 28.12 Nm Here the top surface is in compression the bottom in tension.

28.12

0.75m

-50 Here maximum -ve bending moment occurs at z = 2 BM = (50 x 22) + (300 x 2) – 450 = -50 Nm Here the top surface is in tension the bottom in compression. Beam will bend like this - but is the maximum tensile stress on the top surface here…….

….or the bottom surface here ???

Similarly for compression, hence work out tensile and compressive stresses for +ve & -ve maximum bending moments at both positions along the beam and select the worst cases. Maximum positive BM occurs at z = 0.75m, M = 28.12Nm, This puts the top surface in compression σ max comp = 28.12 * 0.0212 / 582 e –9 = 1.02 MPa σ max tens = 28.12 * 0.0488 / 582 e –9 = 2.36 MPa Maximum negative BM occurs at z = 2m, M = 50Nm, This puts the top surface in tension σ max tens = 50 * 0.0212 / 582 e –9 = 1.82 MPa σ max comp = 50 * 0.0488 / 582 e –9 = 4.19 MPa Therefore: The maximum tensile stress occurs on the bottom surface at z = 0.75m and is of 2.36 MPa. The maximum compressive stress occurs on the bottom surface at z = 2m and is of 4.19 MPa.

4.

A = 0.022 – 0.012 = 3 x 10-4 m2 I = BD3 / 12 – bd3 / 12 I = 0.024 / 12 – 0.014 / 12 = 1.25e-8 m4 Bending stress : σ = My/I = 10 * 0.01 / 1.25e-8 = ±8.00 MPa Direct stress : σ = F / A = 1000 / 3 e-4 = 3.33 MPa σmax tens = 3.33+8.00 = 11.33 z (m) 0.02

To find neutral axis use similar triangles : σ z -0.02

z / 4.67 = (0.04 –z) / 11.33 z = 0.012m from bottom of beam.

σmax comp = 3.33-8.00 = -4.67 Bendstress_tut_sol...