8.0 HORIZONTAL SHEAR STRESSES IN BEAMS
8.1 Introduction
In Chapter 7.0, we saw that when part of a beam is subjected to a constant bending moment and 0 Shear
Force, there will be only bending stresses in the beam. The shear stress will be 0 as shear stress is equal
to Shear Force divided by Area. As shear force is zero, shear stress will also be 0. In actual practice,
however, a beam is subjected to a bending moment which varies from section to section, and also the
shear force acting on the beam is not zero; and it also varies from section to section.
Due to these shear forces, the beam will be subjected to shear stresses. These shear stresses will be acting
across transverse sections of the beam. These transverse shear stresses will in turn produce
complimentary horizontal shear stresses which will be acting on longitudinal layers of the beam hence the
beam will be subjected to horizontal shear stresses.
8.2 Stress at a Section
Figure 9.1 (a) shows a simply supported beam carrying a uniformly distributed load. For a uniformly
distributed load, the shear force and bending moment will vary along the length of the beam. Consider
two sections AB and CD of this beam at a distance 𝑑𝑥 apart.
Figure 9.1
𝐿𝑒𝑡 𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐴𝐵: 𝐹 = 𝑆ℎ𝑒𝑎𝑟 𝐹𝑜𝑟𝑐𝑒; 𝑀 = 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡
𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐶𝐷: 𝐹 + 𝑑𝐹 = 𝑆ℎ𝑒𝑎𝑟 𝐹𝑜𝑟𝑐𝑒; 𝑀 + 𝑑𝑀 = 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡
, 𝐼 = 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑁𝑒𝑢𝑡𝑟𝑎𝑙 𝐴𝑥𝑖𝑠
It is required to find the shear stress at the section AB at a distance 𝑦 from the neutral axis. Figure 9.1 (c)
shows the cross-section of the beam. On the cross-section, let EF be a line at a distance 𝑦1 from the Neutral
Axis. Now consider the part of the beam above level EF and between sections AB and CD. This part of the
beam may be taken to constitute or consist of an infinite number of elemental cylinders each of area 𝑑𝐴
and length 𝑑𝑥. Consider one such elemental cylinder at distance 𝑦 from the Neutral Axis:
𝑑𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 (𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙)𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟.
𝑑𝑥 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑑𝑒𝑟.
𝑦 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑑𝑒𝑟 𝑓𝑟𝑜𝑚 𝑁𝑒𝑢𝑡𝑟𝑎𝑙 𝐴𝑥𝑖𝑠.
𝐿𝑒𝑡 𝜎 = 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑑𝑒𝑟 𝑜𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐴𝐵
𝑎𝑐𝑡𝑖𝑛𝑔 𝑛𝑜𝑟𝑚𝑎𝑙𝑙𝑦 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛.
𝜎 + 𝑑𝜎 = 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑟𝑒𝑠𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑜𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐶𝐷.
The bending stress at distance 𝑦 from the neutral axis is given by the equation:
𝑀 𝜎
=
𝐼 𝑦
𝑀𝑦
𝜎=
𝐼
For a given beam, the bending stress is a function of bending moment and the distance 𝑦 from the neutral
axis. Let us find the bending stress on the end of the elemental cylinder at the section AB and also at the
section CD. Therefore, the bending stress on the end of elemental cylinder on the section AB (where
bending moment is 𝑀) will be:
𝑀𝑦
𝜎=
𝐼
Similarly, bending stress on the end of elemental cylinder on the section CD (where bending moment is
𝑀 + 𝑑𝑀) will be:
(𝑀 + 𝑑𝑀)𝑦
𝜎 + 𝑑𝜎 =
𝐼
Now let us find the forces on the two ends of the elemental cylinder:
𝐹𝑜𝑟𝑐𝑒 𝑜𝑛 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑜𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐴𝐵 = 𝑠𝑡𝑟𝑒𝑠𝑠 × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
= 𝜎 × 𝑑𝐴
𝑀𝑦
= × 𝑑𝐴
𝐼
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝑡ℎ𝑒 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑜𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐶𝐷 = (𝜎 + 𝑑𝜎)𝑑𝐴
(𝑀 + 𝑑𝑀)𝑦
= × 𝑑𝐴
𝐼
At the two ends of the elemental cylinder, the forces are different. They are acting along the same line
but are in opposite directions. Hence, there will be an unbalanced force on the elemental cylinder.
, ∴ 𝑁𝑒𝑡 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑑𝑒𝑟
(𝑀 + 𝑑𝑀)𝑦 𝑀𝑦
=( × 𝑑𝐴) − ( × 𝑑𝐴)
𝐼 𝐼
𝑑𝑀 ⋅ 𝑦
= × 𝑑𝐴 … (𝑖)
𝐼
The total unbalanced force above level EF and between the two sections AB and CD may be found out by
considering all the elemental cylinders between sections AB and CD and above level EF (i.e. by integrating
the eqn (i))
∴ 𝑇𝑜𝑡𝑎𝑙 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑓𝑜𝑟𝑐𝑒
𝑑𝑀 ⋅ 𝑦
=∫ × 𝑑𝐴
𝐼
𝑑𝑀
= ∫ 𝑦 ⋅ 𝑑𝐴
𝐼
𝑑𝑀
= × 𝐴 × 𝑦̅ {𝑅𝑒𝑐𝑎𝑙𝑙: ∫ 𝑦 ⋅ 𝑑𝐴 = 𝐴 × 𝑦̅}
𝐼
𝑊ℎ𝑒𝑟𝑒: 𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑏𝑜𝑣𝑒 𝑙𝑒𝑣𝑒𝑙 𝐸𝐹 (𝑜𝑟 𝑎𝑏𝑜𝑣𝑒 𝑦1 )
= 𝑎𝑟𝑒𝑎 𝐸𝐹𝐺𝐻 (𝐹𝑖𝑔𝑢𝑟𝑒 9.1 (𝑐))
𝑦̅ = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐶𝑜𝐹 𝑜𝑓 𝑎𝑟𝑒𝑎 𝐴 𝑓𝑟𝑜𝑚 𝑁𝑒𝑢𝑡𝑟𝑎𝑙 𝐴𝑥𝑖𝑠
Due to the total unbalanced force acting on the part of the beam above level EF and between sections AB
and CD as shown in Figure 9.2 (a), the beam may fail due to shear. Hence, to prevent failure by shear, the
horizontal section of the beam at level EF must offer shear resistance. This shear resistance must at least
be equal (and opposite) to the total unbalanced force to avoid failure due to shear.
Figure 9.2
∴ 𝑆ℎ𝑒𝑎𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑜𝑟 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒)𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝐸𝐹
= 𝑇𝑜𝑡𝑎𝑙 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑓𝑜𝑟𝑐𝑒
𝑑𝑀
= × 𝐴 × 𝑦̅ … (𝑖𝑖)
𝐼
𝐿𝑒𝑡 𝜏 = 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠ℎ𝑒𝑎𝑟 𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝐸𝐹 (𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑠𝑠)
𝑏 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑏𝑒𝑎𝑚 𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝐸𝐹
∴ 𝐴𝑟𝑒𝑎 𝑜𝑛 𝑤ℎ𝑖𝑐ℎ 𝜏 𝑖𝑠 𝑎𝑐𝑡𝑖𝑛𝑔 = 𝑏 × 𝑑𝑥
8.1 Introduction
In Chapter 7.0, we saw that when part of a beam is subjected to a constant bending moment and 0 Shear
Force, there will be only bending stresses in the beam. The shear stress will be 0 as shear stress is equal
to Shear Force divided by Area. As shear force is zero, shear stress will also be 0. In actual practice,
however, a beam is subjected to a bending moment which varies from section to section, and also the
shear force acting on the beam is not zero; and it also varies from section to section.
Due to these shear forces, the beam will be subjected to shear stresses. These shear stresses will be acting
across transverse sections of the beam. These transverse shear stresses will in turn produce
complimentary horizontal shear stresses which will be acting on longitudinal layers of the beam hence the
beam will be subjected to horizontal shear stresses.
8.2 Stress at a Section
Figure 9.1 (a) shows a simply supported beam carrying a uniformly distributed load. For a uniformly
distributed load, the shear force and bending moment will vary along the length of the beam. Consider
two sections AB and CD of this beam at a distance 𝑑𝑥 apart.
Figure 9.1
𝐿𝑒𝑡 𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐴𝐵: 𝐹 = 𝑆ℎ𝑒𝑎𝑟 𝐹𝑜𝑟𝑐𝑒; 𝑀 = 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡
𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐶𝐷: 𝐹 + 𝑑𝐹 = 𝑆ℎ𝑒𝑎𝑟 𝐹𝑜𝑟𝑐𝑒; 𝑀 + 𝑑𝑀 = 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡
, 𝐼 = 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑁𝑒𝑢𝑡𝑟𝑎𝑙 𝐴𝑥𝑖𝑠
It is required to find the shear stress at the section AB at a distance 𝑦 from the neutral axis. Figure 9.1 (c)
shows the cross-section of the beam. On the cross-section, let EF be a line at a distance 𝑦1 from the Neutral
Axis. Now consider the part of the beam above level EF and between sections AB and CD. This part of the
beam may be taken to constitute or consist of an infinite number of elemental cylinders each of area 𝑑𝐴
and length 𝑑𝑥. Consider one such elemental cylinder at distance 𝑦 from the Neutral Axis:
𝑑𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 (𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙)𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟.
𝑑𝑥 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑑𝑒𝑟.
𝑦 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑑𝑒𝑟 𝑓𝑟𝑜𝑚 𝑁𝑒𝑢𝑡𝑟𝑎𝑙 𝐴𝑥𝑖𝑠.
𝐿𝑒𝑡 𝜎 = 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑑𝑒𝑟 𝑜𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐴𝐵
𝑎𝑐𝑡𝑖𝑛𝑔 𝑛𝑜𝑟𝑚𝑎𝑙𝑙𝑦 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛.
𝜎 + 𝑑𝜎 = 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑟𝑒𝑠𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑜𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐶𝐷.
The bending stress at distance 𝑦 from the neutral axis is given by the equation:
𝑀 𝜎
=
𝐼 𝑦
𝑀𝑦
𝜎=
𝐼
For a given beam, the bending stress is a function of bending moment and the distance 𝑦 from the neutral
axis. Let us find the bending stress on the end of the elemental cylinder at the section AB and also at the
section CD. Therefore, the bending stress on the end of elemental cylinder on the section AB (where
bending moment is 𝑀) will be:
𝑀𝑦
𝜎=
𝐼
Similarly, bending stress on the end of elemental cylinder on the section CD (where bending moment is
𝑀 + 𝑑𝑀) will be:
(𝑀 + 𝑑𝑀)𝑦
𝜎 + 𝑑𝜎 =
𝐼
Now let us find the forces on the two ends of the elemental cylinder:
𝐹𝑜𝑟𝑐𝑒 𝑜𝑛 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑜𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐴𝐵 = 𝑠𝑡𝑟𝑒𝑠𝑠 × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
= 𝜎 × 𝑑𝐴
𝑀𝑦
= × 𝑑𝐴
𝐼
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝑡ℎ𝑒 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑜𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐶𝐷 = (𝜎 + 𝑑𝜎)𝑑𝐴
(𝑀 + 𝑑𝑀)𝑦
= × 𝑑𝐴
𝐼
At the two ends of the elemental cylinder, the forces are different. They are acting along the same line
but are in opposite directions. Hence, there will be an unbalanced force on the elemental cylinder.
, ∴ 𝑁𝑒𝑡 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑑𝑒𝑟
(𝑀 + 𝑑𝑀)𝑦 𝑀𝑦
=( × 𝑑𝐴) − ( × 𝑑𝐴)
𝐼 𝐼
𝑑𝑀 ⋅ 𝑦
= × 𝑑𝐴 … (𝑖)
𝐼
The total unbalanced force above level EF and between the two sections AB and CD may be found out by
considering all the elemental cylinders between sections AB and CD and above level EF (i.e. by integrating
the eqn (i))
∴ 𝑇𝑜𝑡𝑎𝑙 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑓𝑜𝑟𝑐𝑒
𝑑𝑀 ⋅ 𝑦
=∫ × 𝑑𝐴
𝐼
𝑑𝑀
= ∫ 𝑦 ⋅ 𝑑𝐴
𝐼
𝑑𝑀
= × 𝐴 × 𝑦̅ {𝑅𝑒𝑐𝑎𝑙𝑙: ∫ 𝑦 ⋅ 𝑑𝐴 = 𝐴 × 𝑦̅}
𝐼
𝑊ℎ𝑒𝑟𝑒: 𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑏𝑜𝑣𝑒 𝑙𝑒𝑣𝑒𝑙 𝐸𝐹 (𝑜𝑟 𝑎𝑏𝑜𝑣𝑒 𝑦1 )
= 𝑎𝑟𝑒𝑎 𝐸𝐹𝐺𝐻 (𝐹𝑖𝑔𝑢𝑟𝑒 9.1 (𝑐))
𝑦̅ = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐶𝑜𝐹 𝑜𝑓 𝑎𝑟𝑒𝑎 𝐴 𝑓𝑟𝑜𝑚 𝑁𝑒𝑢𝑡𝑟𝑎𝑙 𝐴𝑥𝑖𝑠
Due to the total unbalanced force acting on the part of the beam above level EF and between sections AB
and CD as shown in Figure 9.2 (a), the beam may fail due to shear. Hence, to prevent failure by shear, the
horizontal section of the beam at level EF must offer shear resistance. This shear resistance must at least
be equal (and opposite) to the total unbalanced force to avoid failure due to shear.
Figure 9.2
∴ 𝑆ℎ𝑒𝑎𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑜𝑟 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒)𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝐸𝐹
= 𝑇𝑜𝑡𝑎𝑙 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑓𝑜𝑟𝑐𝑒
𝑑𝑀
= × 𝐴 × 𝑦̅ … (𝑖𝑖)
𝐼
𝐿𝑒𝑡 𝜏 = 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠ℎ𝑒𝑎𝑟 𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝐸𝐹 (𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑠𝑠)
𝑏 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑏𝑒𝑎𝑚 𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝐸𝐹
∴ 𝐴𝑟𝑒𝑎 𝑜𝑛 𝑤ℎ𝑖𝑐ℎ 𝜏 𝑖𝑠 𝑎𝑐𝑡𝑖𝑛𝑔 = 𝑏 × 𝑑𝑥
