Determine the normal stresses, Mechanical Engineering

Determine the normal stresses:

A short hollow pier 1.6 m × 1.6 m outsides as well as 1.0 m × 1.0 m intersides supports a vertical load of 2000 kN at a point located on a diagonal 0.5 m from the vertical axis of the pier. Determine the normal stresses at the 4 corners of the section of the pier, neglecting its self weight.

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Figure

Solution

Figure illustrated the section of the pier. At point P, the load of 2000 kN is applied on the pier.

Area of cross-section A = 1.62 - 1.02 = 1.56 m2.

Section modulus, Z xx= Z yy 1.60 - 1.00 × (2/60) = 0.5875 m3

Eccentricity about XX-axis = Eccentricity around YY-axis

= 0.50 sin 45o = 0.353 m

Bending moment around XX-axis = Bending moment around YY-axis

= (2000 × 0.353)

= 706 kNm

Direct stress, f0  = P/ A =2000/1.56 = 1282.05 kN/m2  (compressive)

Bending stress around XX-axis = Bending stress around YY-axis

=± M/ Z =±     706 / 0.5785 = ± 1220.4 kN/m2

∴          Resultant stresses at corners,

f  = (P/ A) ± M xx/ Z xx ± M yy/ Z yy

Stress at corner A = 1282.05 + 1220.4 + 1220.4

                              = 3722.85 kN/m2 (compressive)

 Stress at corner B = 1282.05 - 1220.4 + 1220.4

                              = 1282.05 kN/m2 (compressive)

Stress at corner C = 1282.05 - 1220.4 - 1220.4

                           = - 1158.75 kN/m2 (tensile)

Stress at corner D = 1282.05 + 1220.4 - 1220.4

                              = 1282.05 kN/m2 (compressive)

Posted Date: 1/21/2013 7:16:25 AM | Location : United States







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