# Analysis of Shear Walls-Worked Example

Shear walls are very stiff vertical structural elements in framed structures designed primarily to resist lateral forces (e.g. wind). In-order words, they form the lateral stability system of braced frames. As a building increases in height, the influence of wind forces, seismicity and other forms of lateral loads becomes very significant. Thus, one of the methods of ensuring lateral stability and avoiding excessive lateral deflection is by deploying shear walls across the width of the building to resist the lateral forces.

The fundamental of lateral stability was a topic discussed in a previous post. See Fundamentals of Lateral Stability. It’s a prerequisite to this post, you are encouraged to read the post before going ahead with the content of this post.

### Analysis of Shear Walls

Shear walls are analyzed as vertical cantilevers subjected to the lateral forces. It is always desired to ensure that the position of shear walls in the plan is symmetrical. This is to ensure that the centre of the stiffness (shear centre) of the structure coincides with the point of application of the lateral forces in order to avoid large torsional forces. However, in reality, this is not always possible as a result of architectural constraints. The walls must, therefore, be designed to withstand torsion due to eccentricity between the centre of the stiffness of the walls and the point of application of the loads.

The worked example presented in this post assumes symmetrical layout of shear walls. Where the layout is unsymmetrical, the effect of torsion must be considered. This is beyond the scope of this post, however, guidance can be found from the texts in the further reading section of the post.

### Derivation of Actions on Shear Walls

In addition to resisting the lateral forces which are primarily in the plane of the wall. Shear walls are also required to resist axial forces and moment outside the plane of the wall. In fact, the derivation of actions as well as the design of shear walls is very similar to that of a concrete column.

#### Axial Forces

The axial forces on the wall can be calculated using an elastic method of analysis. The tributary area method can be used or the vertical reaction from a subframe analysis.

#### Transverse Moment

Transverse moment, is the moment perpendicular to the plane of the wall. Its magnitude can be calculated from the analysis of a subframe.

#### In-plane Moment

This is the moment in the plane of the wall due to the applied lateral loads, its magnitude is determined by treating the wall as a vertical cantilever and taking moment about the foot of floors. It is very common to have several shear walls resisting the lateral actions. This lateral forces will be apportioned to the walls according to their stiffness’s. Thus a stiffer wall will attract more load.

For example, consider the shear walls shown in figure one, connected by floor slabs and subjected to uniformly distributed wind load wk.

Thus, since the walls are of equal heights, the load will be distributed between the walls in proportion to the second-moment area of the walls. i.e.

Wall\quad 1\quad { F }_{ 1 }=\left( \frac { { I }_{ 1 } }{ { I }_{ 1 }+{ I }_{ 2 } } \right) \cdot { F }_{ k }
Wall\quad 2\quad { F }_{ 2 }=\left( \frac { { I }_{ 2 } }{ { I }_{ 1 }+{ I }_{ 2 } } \right) \cdot { F }_{ k }

Where:

• F1 & F2 is the portion of wind action shared by wall one and wall two respectively.
• I1 & I2 is the second-moment area of wall one and wall two respectively.
• Fk is the wind action

### Worked Example

Figure 2 shows the typical floor layout of an 8 storey office building. Wall A is 200mm thick and provides vertical support to the 200mm thick flat slab and lateral stability to the building in the plane of the wall. The characteristic permanent and variable actions on the roof are 6.0kN/m2 and 1.5kN/m2 respectively. While on the floors the permanent and variable actions are 7.5kN/m2 and 3.0kN/m2. The net wind pressure in the plane of this wall is 2.5kN/m2 Derive the actions required to design this wall at mid-height and at the base. Take the storey height = 3@8=24m

How to derive wind load on buildings has been covered in a previous post See: Derivation of Wind Loads on Buildings

This example assumes that second-order effects are negligible. For practical purposes, second-order effects must be checked before applying the procedures described in this post

#### Actions

• On the roof, the permanent actions gk=6.0kN/m2 while the variable actions qk =1.5kN/m2
• On the floors, the permanent actions gk=6.0kN/m2 while the variable actions qk =3kN/m2
• Net wind pressure =2.5kN/m2
##### Actions from Roof

Permanent Actions

i.\quad Roof\quad load\quad =\left( 5 +\frac { 2.5 }{ 2 } \right) \left( \frac { 5 }{ 2 } +\frac { 2.5 }{ 2 } \right) \cdot 6.0=140.6kN
{ G }_{ k,roof }=140.6+75=215.6kN

Variable Actions

i.\quad Roof\quad load\quad =\left( 5+\frac { 2.5 }{ 2 } \right) \left( \frac { 5 }{ 2 } +\frac { 2.5 }{ 2 } \right) \cdot 1.5=35.16kN
{ Q }_{ k,roof }=35.16kN
##### Actions from floors

Permanent Actions

i.\quad Floors=\left(5 +\frac { 2.5 }{ 2 } \right) +\left( \frac { 5 }{ 2 } \right) \cdot 7.5=117.2kN
{ G }_{ k,floors }=117.2+46.88+75=239.08kN

Variable Actions

i.\quad Floors=\left( 5 +\frac { 2.5 }{ 2 } \right) \left( \frac { 5 }{ 2 } \right) \cdot 3=46.88kN
{ Q }_{ k,floors }=46.88+18.75=65.65kN

Imposed load reduction factors can be applied to the wall based on the numbers of storeys supported αn=1-n/10; where n=number of floors supported.

#### Axial Actions

Since we’re required to determine the design actions at mid height and base of the wall, we have

##### At Mid Height
{ G }_{ k }=933kN\quad { Q }_{ k }=162.4kN

{ N }_{ Ed }={ 1.35G }_{ k }+1.5{ Q }_{ k }
=(1.35\times 933)+(1.5\times 162.4)=1503.15kN

{ N }_{ Ed }={ 1.35G }_{ k }+1.5{ \psi Q }_{ k }
=(1.35\times 933)+(1.5\times 0.7\times 162.4)=1430.1kN
##### At Base
{ G }_{ k }=1889.3kN\quad { Q }_{ k }=296.6kN

{ N }_{ Ed }={ 1.35G }_{ k }+1.5{ Q }_{ k }
=(1.35\times 1889.3)+(1.5\times 296.6)=2995.5kN

{ N }_{ Ed }={ 1.35G }_{ k }+1.5{ \psi Q }_{ k }
=(1.35\times 1889.3)+(1.5\times 0.7\times 296.6)=2862kN

#### In-Plane Moments

By inspection, the walls are symmetrical about the plane of application of the wind actions. Therefore the shear centre of the wall will coincide with the point of application of the wind load, hence the effect of torsion can be neglected. Our in-plane moment is the overturning moment about the point at which it is desired.

The wind is being shared according to the stiffnesses of the walls

Lift\quad core\quad { I }_{ Lc }=\frac { 2.8\times { 2.8 }^{ 3 } }{ 12 } -\frac { 2.4\times { 2.4 }^{ 3 } }{ 12 } -\frac { 1.1\times { 0.2 }^{ 3 } }{ 12 } =2.36{ mm }^{ 4 }
{ I }_{ A }={ I }_{ B }=\frac { 0.2\times { 5 }^{ 3 } }{ 12 } =2.08{ mm }^{ 4 }
=\frac { 2.08 }{ 2.08+2.08+2.36 } =0.32

Therefore, the proportion of wind load taken by wall A is 32% of the total wind action. The wind action is converted to uniformly distributed load by multiplying the net wind pressure by the width of the building.

{ W }_{ k }=\frac { 32 }{ 100 } \times 2.5\times 37.5=30kN/m
##### Overturning Moment

At Mid height

M=30\times \frac { { 12 }^{ 2 } }{ 2 } =2160kN.m

{ M }_{ Ed,y }=1.5M=\left( 1.5\times 2160 \right) =3240kN.m

{ M }_{ Ed,y }=1.5\psi M=\left( 1.5\times 0.5\times 2160 \right) =1620kN.m

At Base

M=30\times \frac { { 24}^{ 2 } }{ 2 } =8640kN.m

{ M }_{ Ed,y }=1.5M=\left( 1.5\times 8640\right) =12960kN.m

{ M }_{ Ed,y }=1.5\psi M=\left( 1.5\times 0.5\times 8640 \right) =6480kN.m

The effect of global imperfection needs to be considered. This is done by applying horizontal notional loads in addition to the applied wind loads. This has been neglected in this post. However, for practical purposes, its effect must be considered. This has been covered in a previous post. See: Application of Notional Loads on Structures.

#### Transverse Moment

The out of plane moment will be determined using an equivalent frame consisting of the floor at a level and the walls above and below it. The slab frames into the wall and for the purpose of assessing the fixed end moment, the width of the slab can be determined as:

width=5+\frac { 2.5 }{ 2 } =6.25m

Figure 3 shows the subframe, note that the stabilizing effect of the stair has been ignored. This is conservative.

Wall\quad { k }_{ w }=\frac { I }{ l } =\frac { 5000\times { 200 }^{ 3 } }{ 12\times 3000 } =1.11\times { 10 }^{ 6 }
Slab\quad { k }_{ s }=\frac { I }{ l } =\frac { 6250\times { 200 }^{ 3 } }{ 2\times 12\times 5000 } =0.42\times { 10 }^{ 6 }
##### Moment

(FEM)=6.25\times \left[ \left( 1.35\times 7.5 \right) +\left( 1.5\times 3 \right) \right] \times \frac { { 5.0 }^{ 2 } }{ 12 } =190.43kN.m
{ M }_{ Ed,z }=\frac { 1.11 }{ \left( 2\times 1.11 \right) +0.42 } \times 190.43=80.1kN.m

(FEM)=6.25\times \left[ \left( 1.35\times 7.5 \right) +\left( 1.5\times 0.7\times 3 \right) \right] \times \frac { { 5.0 }^{ 2 } }{ 12 } =172.85kN.m
{ M }_{ Ed,z }=\frac { 1.11 }{ \left( 2\times 1.11 \right) +0.42 } \times 172.85=72.7kN.m

Table 2 & 3 gives a summary of the design actions on the wall. The Actions required to design this wall has been determined. By inspection, load case 2 prooves more onerous and therefore could be used to design the wall.

• P Bhatt., T.J MacGinley., B.S Choo- Reinforced Concrete Design to Eurocodes (4th ed)-CRC Press.
• The Concrete Centre (2009) Worked Examples to Eurocode 2: Volume 1 [Online] http://www.concretecentre.com/pdf/ Worked_Example_Extract_Slabs.pdf (Accessed: March 2014).

THANKYOU!!! WATCH OUT FOR THE DESIGN OF THE WALL

A dedicated, passion-driven and highly skilled engineer with extensive knowledge in research, construction and structural design of civil engineering structures to several codes of practices

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