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Shear Force in Web Formula

GoShear Force at Junction of Top of Web Formula

GoMaximum Shear Force in I Section Formula

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Shear Force on Beam is the force which causes shear deformation to occur in the shear plane. Check FAQs

F s = \frac{I \cdot b \cdot 𝜏 beam}{\frac{B \cdot (D^{2} - d^{2})}{8} + \frac{b}{2} \cdot (\frac{d^{2}}{4} - y^{2})}

F_s - Shear Force on Beam?I - Moment of Inertia of Area of Section?b - Thickness of Beam Web?𝜏_beam - Shear Stress in Beam?B - Width of Beam Section?D - Outer Depth of I section?d - Inner Depth of I Section?y - Distance from Neutral Axis?

Shear Force in Web Example

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Here is how the Shear Force in Web equation looks like with Values.

Here is how the Shear Force in Web equation looks like with Units.

Here is how the Shear Force in Web equation looks like.

0.0699 = \frac{0.0017 \cdot 7 \cdot 6}{\frac{100 \cdot (9000^{2} - 450^{2})}{8} + \frac{7}{2} \cdot (\frac{450^{2}}{4} - 5^{2})}

0.0699kN = \frac{0.0017m⁴ \cdot 7mm \cdot 6MPa}{\frac{100mm \cdot ({9000mm}^{2} - {450mm}^{2})}{8} + \frac{7mm}{2} \cdot (\frac{{450mm}^{2}}{4} - {5mm}^{2})}

0.0699 = \frac{0.0017 \cdot 7 \cdot 6}{\frac{100 \cdot (9000^{2} - 450^{2})}{8} + \frac{7}{2} \cdot (\frac{450^{2}}{4} - 5^{2})}

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Home » Physics » Mechanical » Strength of Materials » Shear Force in Web

Shear Force in Web Solution

Follow our step by step solution on how to calculate Shear Force in Web?

FIRST Step Consider the formula

F s = \frac{I \cdot b \cdot 𝜏 beam}{\frac{B \cdot (D^{2} - d^{2})}{8} + \frac{b}{2} \cdot (\frac{d^{2}}{4} - y^{2})}

Next Step Substitute values of Variables

∴

F s = \frac{0.0017m⁴ \cdot 7mm \cdot 6MPa}{\frac{100mm \cdot ({9000mm}^{2} - {450mm}^{2})}{8} + \frac{7mm}{2} \cdot (\frac{{450mm}^{2}}{4} - {5mm}^{2})}

Next Step Convert Units

∴

F s = \frac{0.0017m⁴ \cdot 0.007m \cdot 6E+6Pa}{\frac{0.1m \cdot ({9m}^{2} - {0.45m}^{2})}{8} + \frac{0.007m}{2} \cdot (\frac{{0.45m}^{2}}{4} - {0.005m}^{2})}

Next Step Prepare to Evaluate

∴

F s = \frac{0.0017 \cdot 0.007 \cdot 6E+6}{\frac{0.1 \cdot (9^{2} - {0.45}^{2})}{8} + \frac{0.007}{2} \cdot (\frac{{0.45}^{2}}{4} - {0.005}^{2})}

Next Step Evaluate

∴

F s = 69.8512991960517N

Next Step Convert to Output's Unit

∴

F s = 0.0698512991960517kN

LAST Step Rounding Answer

∴

F s = 0.0699kN

Shear Force in Web Formula Elements

Variables

Shear Force on Beam

Shear Force on Beam is the force which causes shear deformation to occur in the shear plane.

Symbol: F_s

Measurement: ForceUnit: kN

Note: Value can be positive or negative.

Moment of Inertia of Area of Section

Moment of Inertia of Area of Section is the second moment of the area of the section about the neutral axis.

Symbol: I

Measurement: Second Moment of AreaUnit: m⁴