The timber beam shown is supported by bearing plates at A an

The timber beam shown is supported by bearing plates at A and B, and the beam supports a load P on a bearing plate at C. The ultimate bearing stress far the beam perpendicular to the gram is 6.03 MPa and parallel to the gram 26.5 MPa. If the factor of safety is 3.5 and the load on the beam is limited by bearing stress, what is the allowable load?

Solution

Say, reaction at bearing A = R_a Newton

Reaction force at bearing B = R_b Newton

Therefore,

by equilibrium of force R_a + R_b = P

Taking moment about P,

R_a x 1.6 = R_b x 1.1

Or, R_a = R_b x 0.6875

R_b + R_b x 0.6875 = P

Or, R_b = P/ 1.6875

Or, R_a = 0.6875 x P/ 1.6875 = 0.40741 x P

Area of Bearing A, a_A = 100 x 150 mm² = 15000 mm²

Area of Bearing B, a_B = 150 x 150 mm² = 22500 mm²

Area of Bearing C, a_C = 150 x 150 mm² = 22500 mm²

The grains are aligned along the length of the beam.

Hence, bearing strength of the beam at the corresponding to bearing location A, B and C = 6.03 MPa

Therefore, allowable bearing stress at the bearing points = Ultimate bearing stress / F.O.S

= 6.03 / 3.5 = 1.723 MPa

Considering failure of bearing A,

R_a / a_A = 1.723

Or, 0.40741 x P / 15000 = 1.723

Or, P = 63.432 KN

Considering failure of bearing B,

R_b / a_B = 1.723

Or, (P/ 1.6875) / 22500 = 1.723

Or, P = 65.42 KN

Considering failure of bearing C,

R_c / a_C = 1.723

Or, P / 22500 = 1.723

Or, P = 38.768 KN

Therefore allowable load = min of P above = min of [63.432, 65.42, 38.768] KN

= 38.768 KN