Stress Resultants

The force P1 acting on the pad plate is statically equivalent to a force P1 and a moment

M1 = P1d acting on the centroid of the cross section of the vertical post. The load P2 is also shown.

The stress resultant at the base of the post due to the loads P1 and P2 and the moment M1 are as

follows:

(A) An axial compressive force P1 = 550 kg = 1212.542 lb. (Vertical Load)

(B) A bending moment M1 produced by the force P1: M1 = P1d = (1212.542lb) (8in) = 9700.336lb-in

(C) A shear force P2 = 165 kg= 363.76 lb.

(D) A bending moment M2 produced by the force P2:

M2 = P2h = (363.76lb) (17.95in) = 6529.492lb.in (Horizontal load)

P1

d

P2 M1=P1d

h

Pipe support

A B

These stress resultants shows that both M1 and M2 produce maximum compressive stresses at point A

and the shear force produces maximum shear stresses at point B. Therefore, A and B are the critical

points where the stresses should be determined.

P1

10 mm THK PAD PLATE M1=P1d

M2=P2h

P2

A B

b=6in

Cross section of the angle support

t= 3in

Stresses at points A and B

(A) The axial force P 1 produces uniform compressive stresses throughout the post. These stresses are

σP1 = P 1 / A where A is the cross section area of the post

A = b² – (b – 2t) ² = 4t (b-t) = 4 (3in) (6in – 3in) = 9in²

σP1 = P 1 / A = 1212.542 lb. / 9in² = 135psi

(B) The bending moment M 1 produces compressive stresses σM1 at points A and B. These stresses are

obtained from the flexure formula

σM1 = M 1 (b / 2) / I

Where, Ι is the moment of inertia of the cross section.

The moment of inertia is Ι = [b⁴ - (b -2t) ⁴] / 12 = [(6in) ⁴ – (0in) ⁴] / 12 = 108in⁴

Thus, σM1 = M 1 b / 2Ι = (9700.336lb.in) (6in) / (2) (108in⁴) = 44psi

(C) The shear force P2 produces a shear stress at point B but not at point A.

We know that an approximate value of the shear stress can be obtained by dividing the shear force by

the web area.

τP2 = P2 / Aweb =P2 / (2t (b – 2t)) =363.76 lb. / (2) (3in) (6in–6in) = 60psi

The stress τp2 acts at point B in the direction shown in the above figure. We can calculate the shear

stress τP2 from the more accurate formula. The result of this calculation is τP2 = 60psi, which shows

that the shear stress obtained from the approximate formula, is satisfactory.

D) The bending moment M2 produces a compressive stress at point A but no stress at point B. The

stress at A is σM2 = M2 b / 2Ι = (6529.492lb.in) (6in) / (2) (108in⁴) = 181psi. This stress is also shown in

the above figure.

Stress Elements each element is oriented so that the y-axis is vertical (i.e. parallel to the longitudinal

axis of the post) and the x-axis is horizontal axis Point A: The only stress in point A is a compressive

stress σa in the y direction

σA = σP1 + σM1 + σM2 y

σA = 135psi + 44psi + 181psi = 360psi (compression) σA = σP1 + σM1 + σM2

Thus, this element is in uniaxial stress.

Principal Stresses and Maximum Shear Stress

σx = 0 x

σy = - σa = - 360psi

τxy = 0

Since the element is in uniaxial stress,

σ1 = σx and σ2 = σy = - 360psi

And the maximum in-plane shear stress is

τmax = (σ1 - σ2) / 2 = ½ (360psi) = 180psi

The maximum out-of-plane shear stress has the same magnitude.

A

0

σB = σP1 + σM1

Point B: Here the compressive stress in the y direction is

σB = σP1 + σM1

σB = 135psi + 44psi = 179psi (compression) X

And the shear stress is

τB = τP2 = 60psi

The shear stress acts leftward on the top face and downward on the x face of the element.

Principal Stresses and Maximum Shear Stress

σx =0

σy = - σB = - 179psi 𝜎1,2 = (𝜎𝑥 + 𝜎𝑦) ± √(𝜎𝑥−𝜎𝑦)²

2+ (𝜏𝑥𝑦)²

τxy = - τP2 = - 60psi

Substituting σ1, 2 = - 89.5psi +/- 107.75psi

σ1 = 18 psi and σ2 = - 197psi

The maximum in-plane shear stresses can be obtained from the equation because the principal stresses

have opposite signs; the maximum in-plane shear stresses are larger than the maximum out-of-plane

shear stresses.

𝜏𝑚𝑎𝑥= √(𝜎𝑥

2) ² + (𝜏𝑥𝑦)²

Then, τmax= 60 psi. = 4.21 KG/cm²

This is the maximum Shear stress acting on the pad plate is 60 psi for 10 THK PAD PLATE 300 SQ & 180

SQ.

B

0