(1) The document analyzes the stress resultants at critical points A and B of a vertical post due to loads P1, P2, and bending moments M1 and M2. (2) At point A, the stresses are 360 psi compression due to P1, M1, and M2. At point B, the stresses are 179 psi compression and 60 psi shear due to P1, M1, and P2. (3) The maximum shear stress is calculated to be 180 psi at point A and 60 psi at point B.

1. 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.

2. 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

3. 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

4. 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