Hi, i was wonderwing why with the normal stress at B the moment Mx * Cy is minus. can you explain how to set up the correct equation for the normal stress. Because often i had a correct answer but the minus and pluses different.

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Hi, i was wonderwing why with the normal stress at B the moment Mx * Cy is minus. can you explain how to set up the correct equation for the normal stress. Because often i had a correct answer but the minus and pluses different.

 

 
 
 
 
Determine the normal stress (A) at point A using the relation:
0A = +²+M₁²₁ (4)
Here, P is the vertical load acting, A is the cross section area, I, is the moment of inertia about x axis, I, is the
moment of inertia about y axis, c, is the distance from point A to the origin along y axis, c, is the distance
from point A to the origin along x axis, M, is the moment along x axis, and My moment alongy axis.
Substitute -800 x 10³ N for P, 0.09 m² for A, 35 x 10³ Nm for M, 0.675 x 10-³ m² for I, 0.675 x 10-³ m²
for Iy, 20 x 10³ N-m for My, 0.15 m for cy, and 0.15 m for c, in Equation (4).
-800(10³)
GA =
0.09
(35x10³)x0.15
0.675 (10³)
20x10²x0.15
0.675(10-³)
= 3.33333 MPa (T)
Hence, the normal stress (A) at corner A of the column is 3.33 MPa (T).
Determine the normal stress (8) at point B using the relation:
Mycy
(5)
08 = +
Here, P is the vertical load acting, A is the cross section area, I, is the moment of inertia about x axis, I, is the
moment of inertia about y axis, cy is the distance from point A to the origin along y axis, c, is the distance
from point A to the origin along x axis, M, is the moment along x axis, and My moment alongy axis.
Substitute -800 x 10³ N for P, 0.09 m² for A, 35 x 10³ Nm for Mx, 0.675 x 10-³ m² for I, 0.675 x 10-³ m²
for Iy, 20 × 10³ N-m for My, 0.15 m for cy, and 0.15 m for cx in Equation (5).
(35x10³)x0.15
0.675 (10-³)
--800(10³)
0.09
20x10²x0.15
0.675 (10³)
= -12.22 MPa
= 12.22 MPa (C)
Hence, the normal stress (8) at corner B of the column is 12.2 MPa (C).
OB =
Transcribed Image Text:Determine the normal stress (A) at point A using the relation: 0A = +²+M₁²₁ (4) Here, P is the vertical load acting, A is the cross section area, I, is the moment of inertia about x axis, I, is the moment of inertia about y axis, c, is the distance from point A to the origin along y axis, c, is the distance from point A to the origin along x axis, M, is the moment along x axis, and My moment alongy axis. Substitute -800 x 10³ N for P, 0.09 m² for A, 35 x 10³ Nm for M, 0.675 x 10-³ m² for I, 0.675 x 10-³ m² for Iy, 20 x 10³ N-m for My, 0.15 m for cy, and 0.15 m for c, in Equation (4). -800(10³) GA = 0.09 (35x10³)x0.15 0.675 (10³) 20x10²x0.15 0.675(10-³) = 3.33333 MPa (T) Hence, the normal stress (A) at corner A of the column is 3.33 MPa (T). Determine the normal stress (8) at point B using the relation: Mycy (5) 08 = + Here, P is the vertical load acting, A is the cross section area, I, is the moment of inertia about x axis, I, is the moment of inertia about y axis, cy is the distance from point A to the origin along y axis, c, is the distance from point A to the origin along x axis, M, is the moment along x axis, and My moment alongy axis. Substitute -800 x 10³ N for P, 0.09 m² for A, 35 x 10³ Nm for Mx, 0.675 x 10-³ m² for I, 0.675 x 10-³ m² for Iy, 20 × 10³ N-m for My, 0.15 m for cy, and 0.15 m for cx in Equation (5). (35x10³)x0.15 0.675 (10-³) --800(10³) 0.09 20x10²x0.15 0.675 (10³) = -12.22 MPa = 12.22 MPa (C) Hence, the normal stress (8) at corner B of the column is 12.2 MPa (C). OB =
Show the diagram of direction of force and moment as shown in Figure 1.
300 KN
Mx
Z
500 KN
-800 P
P = -800 KN
My
J
Figure 1
Let the negative sign of the force indicates the direction of the force acting downward M, and net moment
along y at My.
Calculate the resultant force acting the center of block due to vertical loads.
Σ F, = (FR),
-500-300 = P
Find the value of moment acting on block along x-axis.
300 (0.05)-500 (0.1) = Mx
15 - 50 = M,
M₂ = -35 kN - m
Transcribed Image Text:Show the diagram of direction of force and moment as shown in Figure 1. 300 KN Mx Z 500 KN -800 P P = -800 KN My J Figure 1 Let the negative sign of the force indicates the direction of the force acting downward M, and net moment along y at My. Calculate the resultant force acting the center of block due to vertical loads. Σ F, = (FR), -500-300 = P Find the value of moment acting on block along x-axis. 300 (0.05)-500 (0.1) = Mx 15 - 50 = M, M₂ = -35 kN - m
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