SES - Block A - Eccentric loading. Step by step example

Structural Systems – Academic Year 2016/2017 Instructor: Maribel Castilla Heredia @maribelcastilla

Block A.

How to obtain normal stress due to eccentric axial loading

Step by step example.


Normal stress in a member subjected to an eccentric axial load

1. Find and mark on the cross section the point where te

eccentric axial load has been applied (cm).

2. Obtain the point where the neutral line interscts the Y

axis. Represent the neutral line on the figure (cm).

3. Normal stress on the centroid is 25 N/mm2. Find the

value of the axial load (kN).

4. Find the value and the point where the maximum positive

normal stress will appear (kN/cm2).

5. Find the value and the point where the maximum

negative normal stress will appear (kN/cm2).

6. The member is fix supported to the ground and free at

the other end. Represent on the 3D schematic the

neutral line, the area of the cross section subjected to

tension, the one subjected to compression, the expected

deformation and the point where the maximum normal

stress as an absolute value appears.

The cross section in the figure is subjected to an eccentric axial load which has been applied on line a-a’. Point

A belongs to the neutral line. Answer the following:



1. Find and mark on the cross section the point where the eccentric axial load

has been applied (cm).

We start at the general case of normal stress due to vending (acording to the sign

convention used througout this course). The cordinates of the point where the axial

force has been applied can be obtained after manipulating this formula:

yzx

z y

M zM yN

A I I

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm








If we substitute in the previous formula Mz and My with the moments that a point

load applied eccentrically would produce and use “P” as the value of the load that

has been applied, the formula would be this:

yzx

z y

M zM yN

A I I

y zx

z y

P e y P e zP

A I I

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm








If we substitute in the previous formula Mz and My with the moments that a point

load applied eccentrically would produce and use “P” as the value of the load that

has been applied, the formula would be this:

And obtaining P as a common factor:

yzx

z y

M zM yN

A I I

y zx

z y

P e y P e zP

A I I

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm1 y z

x

z y

e y e zP

A I I



ey and ez are the coordinates of the point where load P has been applied. The

subindexes indicate the axis related to each of these eccentricities. The following

schematic shows an example:

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm



ey and ez are the coordinates of the point where load P has been applied. The

subindexes indicate the axis related to each of these eccentricities. The following

schematic shows an example:

We’ve been told that normal stress at point A is zero (because it belongs to the

neutral axis). Substituting everything we know so far in the formula:

(0, 10)

0 ( 10)10

116 24318,67 8198,67

y zA

e eP

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm



If P where zero there wouldn’t be any problem to be solved, therefore the value to

be found verifies this:

0 ( 10)1 8198,670 7.067

116 24318,67 8198,67 1160

y zz

e ee cm

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm





We also know that P is on line a-a’, therefore the ey value can be obtained with the

geometrical relationship between them:

0 ( 10)1 8198,670 7.067

116 24318,67 8198,67 1160

y zz

e ee cm

1; 14,13

2z y ye e e cm

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm





We also know that P is on line a-a’, therefore the ey value can be obtained with the

geometrical relationship between them:

This is the point where the load has been applied:

0 ( 10)1 8198,670 7.067

116 24318,67 8198,67 1160

y zz

e ee cm

1; 14,13

2z y ye e e cm

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm



2. Obtain the point where the neutral line intersects the Y axis.

Represent the neutral line on the figure (cm).

Once the eccentricities have been obtained, we just have to substitute the “z”

coordinate with zero in the formula (because we are looking for a point that

belongs to the neutral line). The only unknown left will be the “y” coordinate we are

looking for.

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm








looking for.

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm

( ,0)

1 14,3 7,067 (0)0 ; 14.86

116 24318,67 8198,67EN y

yP y cm








looking for.

2

4

4

116

24318,67

8198,67

z

y

A cm

I cm

I cm

( ,0)

1 14,3 7,067 (0)0 ; 14.86

116 24318,67 8198,67EN y

yP y cm

Notice that the position of the neutral line doesn’t depend on the value

nor the sign of the axial load, just .



3. Normal stress on the centroid is 25 N/mm2. Find the value of the axial

load (kN).

In the general formula, let’s substitute all known values: eccentricity, coordinates of

the centroid (where normal stress is 25 N/mm2) and the normal stress itself and I’ll

be able to obtain “P”.

1 y zx

z y

e y e zP

A I I




load (kN).




1 y zx

z y

e y e zP

A I I

2(0,0) 2 4 4

3

1 14,3 0 7,067 02500

116 24318,67 8198,67

290 10 290

G

cm cmN Pcm cm cm cm

P N kN




load (kN).




1 y zx

z y

e y e zP

A I I

2(0,0) 2 4 4

3

1 14,3 0 7,067 02500

116 24318,67 8198,67

290 10 290

G

cm cmN Pcm cm cm cm

P N kN

4. Find the value and the point where the maximum positive normal stress will appear (kN/cm2).

The maximum positive normal stress (tensile) will take place in the furthest point from the neutral line in, the tension zone of

the cross section (shaded in blue) thus coordinates (+20,+10)




load (kN).




1 y zx

z y

e y e zP

A I I

2(0,0) 2 4 4

3

1 14,3 0 7,067 02500

116 24318,67 8198,67

290 10 290

G

cm cmN Pcm cm cm cm

P N kN

4. Find the value and the point where the maximum positive normal stress will appear (kN/cm2).

The maximum positive normal stress (tensile) will take place in the furthest point from the neutral line in, the tension zone of

the cross section (shaded in blue) thus coordinates (+20,+10)

2

(20,10) 2 4 4

1 14,3 ( 20 ) 7,067 ( 10 )290 8,360 /

116 24318,67 8198,67G

cm cm cm cmkN kN cm

cm cm cm



5, Normal stress on the centroid is 25 N/mm2. Find the value of the axial

load (kN).

The maximum negative normal stress (compression) will take place in the furthestpoint from the neutral line in, the compressed zone of the cross section (shaded inred) thus coordinates (-20,-10)

( 20, 10) 2 4 4

2

( 20, 10)

1 14,3 ( 20 ) 7,067 ( 10 )290

116 24318,67 8198,67

3,365 /

G

G

cm cm cm cmkN

cm cm cm

kN cm



6. The member is fix supported to the ground and free at the other end. Represent on the 3D schematic the neutral line,

the area of the cross section subjected to tension, the one subjected to compression, the expected deformation and the

point where the maximum normal stress as an absolute value appears.








The edge with the

greatest normal

stress as an

absolute value





Block A.

How to obtain normal stress due to eccentric axial loading

Step by step example.

Engineering

SES - Block A - Eccentric loading. Step by step example