Masonry Section Cracking Eccentricity

This solution details the calculation for the minimum eccentricity that causes cracking in a masonry section, given its specific properties like zero tensile strength.

Cracking Condition Analysis

A masonry section cracks when the induced tensile stress exceeds its tensile strength. Since the tensile stress capacity is stated as zero, cracking occurs as soon as any tensile stress develops at the extreme fiber due to an eccentric load.

Stress Distribution in Rectangular Section

Consider a rectangular masonry section with width b and height h. An axial load P is applied at an eccentricity e. The stress distribution across the section follows a linear pattern.

The general stress formula is:

$ \sigma = \frac{P}{A} \pm \frac{M}{Z} $

Where:

• P = Applied axial load
• A = Cross-sectional area ($A = b \times h$)
• M = Bending moment due to eccentricity ($M = P \times e$)
• Z = Section modulus. For bending about the axis parallel to the height h, $Z = \frac{hb^2}{6}$.

The stress at the extreme fiber on the tension side is given by subtracting the bending stress from the average stress:

$ \sigma_{tension} = \frac{P}{A} - \frac{Pe}{Z} $

Substituting the area and section modulus for a rectangle:

$ \sigma_{tension} = \frac{P}{bh} - \frac{Pe}{hb^2/6} $

Simplifying this expression yields:

$ \sigma_{tension} = \frac{P}{bh} - \frac{6Pe}{hb^2} $

Minimum Eccentricity Calculation for Cracking

Cracking initiates when the tensile stress $\sigma_{tension}$ becomes zero.

Set the tensile stress formula to zero:

$ \frac{P}{bh} - \frac{6Pe}{hb^2} = 0 $

Rearranging the terms to solve for the eccentricity e:

$ \frac{P}{bh} = \frac{6Pe}{hb^2} $

Assuming $P > 0$ and $h > 0$, we can cancel terms. After simplification, we get:

$ \frac{1}{b} = \frac{6e}{b^2} $

Solving for e gives the minimum eccentricity for cracking:

$ e_{min} = \frac{b}{6} $

Result for Specified Section Width

The given width of the masonry section is b = 600 mm.

Substitute the value of b into the formula for minimum eccentricity:

$ e_{min} = \frac{600 \text{ mm}}{6} $

$ e_{min} = 100 \text{ mm} $

The question asks for the answer rounded to one decimal place.

Therefore, the minimum eccentricity at which the section will crack is 100.0 mm.