Ultimate Strength for Short, Circular Members when Controlled by Tension Formula

GoUltimate Strength for Short, Circular Members when Governed by Compression Formula

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Axial Load Capacity is defined as the maximum load along the direction of the drive train. Check FAQs

P u = 0.85 \cdot f' c \cdot (D^{2}) \cdot Φ \cdot (\sqrt{({((0.85 \cdot \frac{e}{D}) - 0.38)}^{2}) + (Rho' \cdot m \cdot \frac{D b}{2.5 \cdot D})} - ((0.85 \cdot \frac{e}{D}) - 0.38))

P_u - Axial Load Capacity?f'_c - 28-Day Compressive Strength of Concrete?D - Overall Diameter of Section?Φ - Resistance Factor?e - Eccentricity of Column?Rho' - Area Ratio of Gross Area to Steel Area?m - Force Ratio of Strengths of Reinforcements?D_b - Bar Diameter?

Ultimate Strength for Short, Circular Members when Controlled by Tension Example

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Here is how the Ultimate Strength for Short, Circular Members when Controlled by Tension equation looks like with Values.

Here is how the Ultimate Strength for Short, Circular Members when Controlled by Tension equation looks like with Units.

Here is how the Ultimate Strength for Short, Circular Members when Controlled by Tension equation looks like.

1.3E+6 = 0.85 \cdot 55 \cdot (250^{2}) \cdot 0.85 \cdot (\sqrt{({((0.85 \cdot \frac{35}{250}) - 0.38)}^{2}) + (0.9 \cdot 0.4 \cdot \frac{12}{2.5 \cdot 250})} - ((0.85 \cdot \frac{35}{250}) - 0.38))

1.3E+6N = 0.85 \cdot 55MPa \cdot ({250mm}^{2}) \cdot 0.85 \cdot (\sqrt{({((0.85 \cdot \frac{35mm}{250mm}) - 0.38)}^{2}) + (0.9 \cdot 0.4 \cdot \frac{12mm}{2.5 \cdot 250mm})} - ((0.85 \cdot \frac{35mm}{250mm}) - 0.38))

1.3E+6 = 0.85 \cdot 55 \cdot (250^{2}) \cdot 0.85 \cdot (\sqrt{({((0.85 \cdot \frac{35}{250}) - 0.38)}^{2}) + (0.9 \cdot 0.4 \cdot \frac{12}{2.5 \cdot 250})} - ((0.85 \cdot \frac{35}{250}) - 0.38))

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Ultimate Strength for Short, Circular Members when Controlled by Tension Solution

Follow our step by step solution on how to calculate Ultimate Strength for Short, Circular Members when Controlled by Tension?

FIRST Step Consider the formula

P u = 0.85 \cdot f' c \cdot (D^{2}) \cdot Φ \cdot (\sqrt{({((0.85 \cdot \frac{e}{D}) - 0.38)}^{2}) + (Rho' \cdot m \cdot \frac{D b}{2.5 \cdot D})} - ((0.85 \cdot \frac{e}{D}) - 0.38))

Next Step Substitute values of Variables

∴

P u = 0.85 \cdot 55MPa \cdot ({250mm}^{2}) \cdot 0.85 \cdot (\sqrt{({((0.85 \cdot \frac{35mm}{250mm}) - 0.38)}^{2}) + (0.9 \cdot 0.4 \cdot \frac{12mm}{2.5 \cdot 250mm})} - ((0.85 \cdot \frac{35mm}{250mm}) - 0.38))

Next Step Prepare to Evaluate

∴

P u = 0.85 \cdot 55 \cdot (250^{2}) \cdot 0.85 \cdot (\sqrt{({((0.85 \cdot \frac{35}{250}) - 0.38)}^{2}) + (0.9 \cdot 0.4 \cdot \frac{12}{2.5 \cdot 250})} - ((0.85 \cdot \frac{35}{250}) - 0.38))

Next Step Evaluate

∴

P u = 1328527.74780593N

LAST Step Rounding Answer

∴

P u = 1.3E+6N

Ultimate Strength for Short, Circular Members when Controlled by Tension Formula Elements

Variables

Functions

Axial Load Capacity

Axial Load Capacity is defined as the maximum load along the direction of the drive train.

Symbol: P_u

Measurement: ForceUnit: N