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Diameter of side crankshaft at juncture of crankweb for max torque Formula

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Diameter of Crankshaft at Crank-web Joint is the distance measured through the center of the crankshaft around it's circumference at the juncture of crank web and crankshaft. Check FAQs

d = {(\frac{16}{π \cdot τ} \cdot \sqrt{{M h}^{2} + {M v}^{2} + {(P t \cdot r)}^{2}})}^{\frac{1}{3}}

d - Diameter of Crankshaft at Crank-web Joint?τ - Shear Stress in Shaft at Crank-web Joint?M_h - Horizontal Bending Moment at Crank-web Joint?M_v - Vertical Bending Moment at Crank-web Joint?P_t - Tangential Force at Crankpin?r - Distance Between Crank Pin and Crankshaft?π - Archimedes' constant?

Diameter of side crankshaft at juncture of crankweb for max torque Example

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Here is how the Diameter of side crankshaft at juncture of crankweb for max torque equation looks like with Values.

Here is how the Diameter of side crankshaft at juncture of crankweb for max torque equation looks like with Units.

Here is how the Diameter of side crankshaft at juncture of crankweb for max torque equation looks like.

30.4493 = {(\frac{16}{3.1416 \cdot 57.382} \cdot \sqrt{29800^{2} + {316.625}^{2} + {(80 \cdot 75)}^{2}})}^{\frac{1}{3}}

30.4493mm = {(\frac{16}{3.1416 \cdot 57.382N/mm²} \cdot \sqrt{{29800N*mm}^{2} + {316.625N*m}^{2} + {(80N \cdot 75mm)}^{2}})}^{\frac{1}{3}}

30.4493 = {(\frac{16}{3.1416 \cdot 57.382} \cdot \sqrt{29800^{2} + {316.625}^{2} + {(80 \cdot 75)}^{2}})}^{\frac{1}{3}}

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Diameter of side crankshaft at juncture of crankweb for max torque Solution

Follow our step by step solution on how to calculate Diameter of side crankshaft at juncture of crankweb for max torque?

FIRST Step Consider the formula

d = {(\frac{16}{π \cdot τ} \cdot \sqrt{{M h}^{2} + {M v}^{2} + {(P t \cdot r)}^{2}})}^{\frac{1}{3}}

Next Step Substitute values of Variables

∴

d = {(\frac{16}{π \cdot 57.382N/mm²} \cdot \sqrt{{29800N*mm}^{2} + {316.625N*m}^{2} + {(80N \cdot 75mm)}^{2}})}^{\frac{1}{3}}

Next Step Substitute values of Constants

∴

d = {(\frac{16}{3.1416 \cdot 57.382N/mm²} \cdot \sqrt{{29800N*mm}^{2} + {316.625N*m}^{2} + {(80N \cdot 75mm)}^{2}})}^{\frac{1}{3}}

Next Step Convert Units

∴

d = {(\frac{16}{3.1416 \cdot 5.7E+7Pa} \cdot \sqrt{{29.8N*m}^{2} + {316.625N*m}^{2} + {(80N \cdot 0.075m)}^{2}})}^{\frac{1}{3}}

Next Step Prepare to Evaluate

∴

d = {(\frac{16}{3.1416 \cdot 5.7E+7} \cdot \sqrt{{29.8}^{2} + {316.625}^{2} + {(80 \cdot 0.075)}^{2}})}^{\frac{1}{3}}

Next Step Evaluate

∴

d = 0.0304493004760827m

Next Step Convert to Output's Unit

∴

d = 30.4493004760827mm

LAST Step Rounding Answer

∴

d = 30.4493mm

Diameter of side crankshaft at juncture of crankweb for max torque Formula Elements

Variables

Constants

Functions

Diameter of Crankshaft at Crank-web Joint

Diameter of Crankshaft at Crank-web Joint is the distance measured through the center of the crankshaft around it's circumference at the juncture of crank web and crankshaft.

Symbol: d

Measurement: LengthUnit: mm