Data Source & Methodology

This calculator determines the design loads for a masonry lintel based on the "triangular prism" load distribution method, commonly known as **arching action**. This methodology assumes the load from the masonry above the opening is carried by a 45-degree triangle of material, transferring the load to the adjacent masonry rather than directly onto the lintel. This is the standard, accepted method in structural masonry design.

• Authoritative Source: The Masonry Society (TMS)
• Reference: TMS 402/602-22: Building Code Requirements and Specification for Masonry Structures.

All calculations are based strictly on the formulas derived from this "arching action" principle. This tool checks if the available height of masonry ($H_{avail}$) is sufficient to develop the full arch ($H_{req} = L/2$). If not, it conservatively calculates the load based on the full rectangular block of masonry below the available height.

The Formulas Explained (Arching Action)

This calculator determines the maximum bending moment ($M_{max}$) and shear force ($V_{max}$) that the lintel must resist. It assumes the lintel is a simply supported beam.

1. Check for Arching Action: The calculator first determines the required height $H_{req}$ for a full 45° arch to form.

It then compares this to the available height $H_{avail}$.

Case 1: Full Arching ($H_{avail} \ge H_{req}$) The lintel supports the triangular prism of masonry.

Case 2: No Arching ($H_{avail} < H_{req}$) The lintel supports the full rectangular block of masonry up to $H_{avail}$.

2. Calculate Total Loads (Moment & Shear): The calculator sums the loads from the masonry ($M_m, V_m$) and any other superimposed Uniformly Distributed Loads (UDL) like self-weight ($w_l$) and floor loads ($w_s$).

The final design values are the sum of the masonry and UDL components:

(Note: For Case 1, $M_m = W_m \cdot L / 6$ and $V_m = W_m / 2$. For Case 2, $M_m = (w_{m,UDL} \cdot L^2) / 8$ and $V_m = (w_{m,UDL} \cdot L) / 2$.)

Glossary of Variables

• $L$ (Clear Span): The width of the opening the lintel must span.
• $H_{avail}$ (Available Height): The height of continuous masonry above the lintel, uninterrupted by openings or floor/roof lines.
• $w_m$ (Masonry Weight): The area weight of the masonry wall (e.g., psf or N/m²).
• $w_s$ (Superimposed UDL): Additional Uniformly Distributed Loads from floors, roofs, etc. (e.g., plf or N/m).
• $w_l$ (Lintel Self-Weight): The weight of the lintel beam itself, as a UDL.
• $M_{max}$ (Max Bending Moment): The maximum rotational force the lintel must resist, used to design for bending/deflection.
• $V_{max}$ (Max Shear Force): The maximum vertical force the lintel must resist, used to design for shear and bearing.

How It Works: A Step-by-Step Example

Let's find the design loads for a 6-foot wide opening in a standard 40 psf block wall, with 4 feet of masonry above it and a 100 plf floor load.

• Units: Imperial
• Clear Span ($L$): 6 ft
• Available Height ($H_{avail}$): 4 ft
• Masonry Weight ($w_m$): 40 psf
• Superimposed UDL ($w_s$): 100 plf
• Lintel Self-Weight ($w_l$): 15 plf (assumed)

1. Check Arching:
   $H_{req} = L / 2 = 6 \text{ ft} / 2 = 3 \text{ ft}$.
   Since $H_{avail} (4 \text{ ft}) \ge H_{req} (3 \text{ ft})$, full arching action is assumed.
2. Calculate Masonry Load (Triangular):
   $W_m = (L^2 \cdot w_m) / 4 = (6^2 \cdot 40) / 4 = 360 \text{ lbf}$ (Total)
3. Calculate Masonry Moment & Shear:
   $M_m = (W_m \cdot L) / 6 = (360 \cdot 6) / 6 = 360 \text{ lbf-ft}$
   $V_m = W_m / 2 = 360 / 2 = 180 \text{ lbf}$
4. Calculate Other UDL Loads:
   $w_{UDL} = w_s + w_l = 100 \text{ plf} + 15 \text{ plf} = 115 \text{ plf}$
   $M_{UDL} = (w_{UDL} \cdot L^2) / 8 = (115 \cdot 6^2) / 8 = 517.5 \text{ lbf-ft}$
   $V_{UDL} = (w_{UDL} \cdot L) / 2 = (115 \cdot 6) / 2 = 345 \text{ lbf}$
5. Calculate Totals:
   $M_{max} = M_m + M_{UDL} = 360 + 517.5 = 877.5 \text{ lbf-ft}$
   $V_{max} = V_m + V_{UDL} = 180 + 345 = 525 \text{ lbf}$
6. Final Result: $M_{max} = 0.88 \text{ k-ft}$ ; $V_{max} = 0.53 \text{ kips}$

Frequently Asked Questions (FAQ)

H_{req} = \frac{L}{2}

W_m = (\frac{1}{2} \cdot L \cdot H_{req}) \cdot w_m = \frac{L^2 \cdot w_m}{4}

w_{m,UDL} = H_{avail} \cdot w_m

w_{UDL} = w_s + w_l

M_{UDL} = \frac{w_{UDL} \cdot L^2}{8} \quad ; \quad V_{UDL} = \frac{w_{UDL} \cdot L}{2}

M_{max} = M_m + M_{UDL}

$ H_{req} = \frac{L}{2} $

$ W_m = (\frac{1}{2} \cdot L \cdot H_{req}) \cdot w_m = \frac{L^2 \cdot w_m}{4} $

$ w_{m,UDL} = H_{avail} \cdot w_m $

$ w_{UDL} = w_s + w_l $ $ M_{UDL} = \frac{w_{UDL} \cdot L^2}{8} \quad ; \quad V_{UDL} = \frac{w_{UDL} \cdot L}{2} $