Figure 1-1(a)

Solution:

The given section is symmetric about vertical YY axis but not symmetric about horizontal XX axis as shown in figure 1-1(b). The horizontal parts of the section are known as flange and the vertical part is known as web or stem as given in the following figure 1-1(c).

Figure 1-1(b)

We will calculate the center of gravity or centroid (O) as x′ with reference to an extreme left edge of the flange , y′ with reference to the bottom edge of the flange. Let the centroid O is at a distance of y′ from the bottom edge of bottom flange. We divide the whole section into horizontal (top flange and bottom flange) and vertical (web) parts as shown in figure 1-1(c) and figure 1-1(d).

Figure 1-1(c)

Figure 1-1(d)

Take the moment of all the rectangular areas about the bottome edge of the section to calculate the position of XX axis. Let the distance of centroid of A1 is y1 from the bottom edge, distance of centroid of A2 is y2 from the bottom edge and the distance of centroid of A3 is y3 from the bottom edge as shown in figure 1-1(e).

Figure 1-1(e)

We can easily calculate the values of y1, y2 and y3 as;

y1 = 1+10+0.5 = 11.5 cm from the bottom edge

y2 = 1+5 = 6 cm from the bottom edge

y3 =0.5 cm from the bottom edge

y′ = {(A1) x (y1) + (A2) x (y2) + A3 x (y3)}/(A1 + A2 + A3)

Therefore y′ = (5 x 11.5 + 10 x 6 + 10 x 0.5)/(5+10+10)

= (57.5 +60 + 5)/35 = 122.5/25 = 4.9 m from bottom edge of the bottom flange.

x′ = 5 cm from the left edge of the section as the centroid of each rectangular are A1, A2, A3 is at a distance of 5 cm from the extreme left edge as shown in the figure. We can calculate as given below.

x′ = {(A1) x (x1) + (A2) x (x2) + A3 x (x3)}/(A1 + A2 + A3)

Therefore x′ = (5 x 5 + 10 x 5 + 10 x 5)/(5+10+10) = 5 cm from the extreme left edge as the reference

Moment of inertia is also known as second moment of area. The formula for Moment of Inertia (M.I.) for rectangular section of breadth B and depth D is given by the following;

About x-axis of the rectangle, I_xx = BD³/12 = (side parallel to x-axis of rectangle) x (cube of the side perpendicular to x-axis of rectangle)/12

About yy-axis of rectangle, I_yy = DB³/12 = (side parallel to y-axis of rectangle) x (cube of the side perpendicular to y-axis of rectangle)/12. Refer to the following figure 1-1(f)

Figure 1-1(f)

We are using xx and yy axes for the individual area and XX and YY axes for the whole section.

Now we calculate the moment of inertia (M.I.) about YY-axis of the given section. This is easy because we find that the YY axis of the section is coinciding with the centroid of each rectangular area.

I_yy for area A1 = (I_yy)₁ = 1 x 5³/12 = 10.416 cm⁴

I_yy for area A2 = (I_yy)₂ = 10 x 1³/12 = 0.8333 cm⁴

I_yy for area A3 = (I_yy)₃ = 1 x 10³/12 = 83.333 cm⁴

M.I. about YY axis of the whole section,

I_YY = I_yy for area A1 + I_yy for area A2 + I_yy for area A3

= 10.416 cm⁴ + 0.8333 cm⁴ + 83.333 cm⁴ = 94.582 cm⁴

Now we will calculate the M.I. of the given section about XX-axis of the section. We find that the xx-axis of each the rectangular areas A1, A2 and A3 is not coinciding with the XX-axis of the given section. Therefore in this situation we will use parallel axes theorem which states that

I_AB = I_CD + A x h²

Where I_AB = M.I. about an axis AB; I_CD = M.I. about an axis CD which is parallel to axis AB; A = area of the section; h = distance between the two parallel axes AB and CD.

First we calculate M.I. of each area (A1, A2, A3) about its own xx-axis.

I_xx for A1 = (I_xx)₁ = 5 x 1³/12 = 0.416 cm⁴

I_xx for A2 = (I_xx)₂ = 1 x 10³/12 = 83.333 cm⁴

I_xx for A3 = (I_xx)₃ = 10 x 1³/12 = 0.833 cm⁴

Now we transform these values of I_xx about the XX-axis of the whole section by using parallel axes theorem as given below. In this case we will take CD as the xx-axis of the rectangular area and AB as the XX-axis of the whole section.

I_XX for A1 =(I_XX)₁ = (I_xx)₁ + A1 x h1²

I_XX for A2 =(I_XX)₂ = (I_xx)₂ + A2 x h2²

I_XX for A3 =(I_XX)₃ = ( I_xx)₃ + A3 x h3²

Where, h1 = distance between xx-axis of the area A1 and XX-axis of the whole section

= y1 - y′ = 11.5 - 4.9 = 6.6 cm

h2 = distance between xx-axis of the area A2 and XX-axis of the whole section

= y2 - y′ = 6 - 4.9 = 1.1 cm

h3 = distance between xx-axis of the area A3 and XX-axis of the whole section

= y′ - y3 = 4.9 - 0.5 = 4.4 cm

Now we substitute these values to get ;

(I_XX)₁ = 0.416 + 5 x 6.6² = 218.216 cm⁴

(I_XX)₂ = 83.333 + 10 x 1.1² = 95.433 cm⁴

(I_XX)₃ = 0.833 + 10 x 4.4² = 194.433 cm⁴

Therefore I_XX for the whole section is calculated as below

I_XX = (I_XX)₁ + (I_XX)₂ + (I_XX)₃

I_XX = 218.216 + 95.433 + 194.433 = 508.082 cm⁴

Least value of M.I. is 94.582 cm⁴

You can also use our online calculator for moment of inertia of plane sections

Example 5-2 SFD and BMD for Simple beam

Example 5-3 SFD and BMD for Overhanging beam