Unreinforced Clay Masonry in Compression

The following example demonstrates how the compression capacity of an unreinforced clay masonry wall can be calculated based on design information provided by the engineer.

Import Modules

We start by importing the Clay() object.

from toms_structures.unreinforced_masonry import Clay

Create Masonry Object

We next define the masonry properties. Note, toms-structures is intended to be unit specific, entering units other than those specified by a particular parameter may lead to incorrect results.

fuc = 20  # unconfined compressive strength in MPa
mortar_class = 3  # mortar class
height = 3000  # height in mm
length = 1000  # height in mm
thickness = 110  # thickness in mm
bedding_type = True  # Whether the masonry is face-shell bedded or fully bedded

With the properties defined, the Clay() object can be created.

wall = Clay(
    fuc=fuc,
    length=length,
    thickness=thickness,
    height=height,
    bedding_type=bedding_type,
    mortar_class=mortar_class,
)

Properties
==========
length: 1000 mm
height: 3000 mm
thickness: 110 mm
bedding_type: Full bedding
mortar class: M3
fuc: 20 MPa
fmt: 0.2 MPa
Joint thickness tj: 10 mm
Masonry unit height hu: 76 mm

Compute compressive capacity

Next we will compute the compressive capacity of the wall in accordance with the simplified method given in Cl 7.3.3 of AS 3700:2018. To do this, there are 3 parameters required which correspond to options given in Cl 7.3.3. For this example, we assume the wall is laterally supported along its top edge, with no engaged piers, and a concrete slab over.

simple_av = 1  # Member is laterally supported along top edge
kt = 1  # There are no engaged piers
compression_load_type = 1  # Concrete slab over.

We can now use the compression_capacity() method to calculate the simplified compression capacity

wall.compression_capacity(
    simple_av=simple_av,
    kt=kt,
    compression_load_type=compression_load_type,
    verbose=True,
)

Basic Compressive Capacity, refer Cl 7.3.2(2) AS3700
====================================================
bedding_type: Full
km: 1.4
kh: 1.0, based on a masonry unit height of 76 mm and a joint thickness of 10 mm
fmb: 6.26 MPa
fm: 6.26 MPa
bedded area Ab: 110000 mm2
grouted area Ag: 0 mm2
phi_compression: 0.75
basic_compressive_capacity = 516.45 KN

Compresion Capacity, refer Cl 7.3.3.3 AS3700
============================================
Buckling capacity
-----------------
Srs = 1 * 3000 / 1 * 110
Srs = 27.27 (Simplified slenderness ratio Cl 7.3.3.3)
Load type: Concrete slab over
k = min(0.67 - 0.02 * (27.27 - 14), 0.67)
k = 0.4
Simple compression capacity kFo: 206.58 KN

{'Simple': 206.58}

Compute refined compressive capacity

It is often necessary to use the refined methods for calculating compressive capacity given in Cl 7.3.4 of AS3700:2018.

refined_av = 0.75  # wall laterally and rotationally restrained top and bottom.
refined_ah = 0  # There are no return walls.
kt = 1  # There are no engaged piers
e1 = 0  # Assume no eccentricity.
e2 = 0  # Assume no eccentricity.

We can now use the refined_compression() method to calculate the refined compressive capacity.

wall.refined_compression(
    refined_av=refined_av, refined_ah=refined_ah, kt=kt, e1=e1, e2=e2
)

Basic Compressive Capacity, refer Cl 7.3.2(2) AS3700
====================================================
bedding_type: Full
km: 1.4
kh: 1.0, based on a masonry unit height of 76 mm and a joint thickness of 10 mm
fmb: 6.26 MPa
fm: 6.26 MPa
bedded area Ab: 110000 mm2
grouted area Ag: 0 mm2
phi_compression: 0.75
basic_compressive_capacity = 516.45 KN

Refined Compression Capacity, refer Cl 7.3 AS3700
=================================================
effective length of wall used in calculation: 1000 mm

Crushing capacity
-----------------
End eccentricity, e1: 5.5 mm, e2: 5.5 mm, refer AS3700 Cl 7.3.4.4
k (crushing): 0.900
crushing_compressive_capacity = 464.81 kN

Buckling capacity
-----------------
av: 0.75
ah: 0
kt: 1
Sr (vertical): 20.45
Sr (horizontal) = inf
Horizontal:
k for lateral instability = 0.5 * (1 + 5.5 / 5.5) * (  (1 - 2.083 * 5.5 / 110)  - (0.025 - 0.037 * 5.5 / 110) * (1.33 * inf - 8) ) + 0.5 * (1 - 0.6 * 5.5 / 110) * (1 - 5.5 / 5.5) * (   1.18 - 0.03 * inf )
k for lateral instability: 0.0
Vertical:
k for lateral instability = 0.5 * (1 + 5.5 / 5.5) * (  (1 - 2.083 * 5.5 / 110)  - (0.025 - 0.037 * 5.5 / 110) * (1.33 * 20.45 - 8) ) + 0.5 * (1 - 0.6 * 5.5 / 110) * (1 - 5.5 / 5.5) * (   1.18 - 0.03 * 20.45 )
k for lateral instability: 0.45
k (buckling): 0.45
Effective length: 1000.0 mm
kFo = 232.4 kN

{'Crushing': 464.81, 'Buckling': 232.4}

By computing the refined capacity, we were able to achieve a higher compression capacity than the simplified method alone.