1. Mahesh Dhakal mkdhakal@hotmail.com
Stud Wall Design
Mahesh Dhakal

2. Stud Wall Design

3. Stud Wall Design
In timber frame housing the load bearing walls are normally constructed using stud walls
BS 5268: Part 6: Code of Practice for Timber Frame Walls; Section 6.1: Dwellings not exceeding four storeys
Consist of vertical timber members, commonly referred to as studs
Studs are held in position by top and bottom plate or sole

4. Stud Wall Design
The most common stud sizes are 100 × 50, 47, 38 mm and 75 × 50, 47, 38 mm
The studs are usually placed at 400 or 600 mm centres
As C/C spacing is normally less than 610 mm, the load-sharing factor K8 will apply to the design of stud walls
Frame is covered by a cladding material such as plasterboard for aesthetic reason but also provide lateral restraint to the studs about y-y axis
Horizontal noggings acts as bracing and provide lateral restraint against buckling
In addition to noggings, additional bracing may be provided if required

5. Stud Wall Design
Load bearing stud wall

6. Stud Wall Design Design example:
A stud wall panel has an overall height of 3.75 m including top and bottom rails and vertical studs at 600 mm centres with nogging pieces at mid-height. Assuming that the studs, rail framing and nogging pieces comprise 44 × 100 mm section of strength class C22, calculate the maximum uniformly distributed long term total load the panel is able to support.

7. Stud Wall Design

8. Stud Wall Design Effective height
Lex = coefficient × L = 1.0 × 3750 = 3750 mm
Ley = coefficient × L /2 = 1.0 × 3750/2 = 1875 mm
Radius of gyration

9. Stud Wall Design Slenderness ratio
Important: where two values of λ are possible the larger value must always be used to find σc,adm,||

10. Stud Wall Design For timber of strength class C22,
Grade stress and modulus of elasticity values are (Table 8):
Grade stress, σc,g,|| = 7.5 N/mm2 & Modulus of Elasticity (Emin) = 6500 N/mm2
Modification factors, K2 = 1.0 (assuming service class 1&2); K3 = 1.0 and K8 = 1.1 (due to load sharing)
Emin /σc,|| = Emin/(σc,g,|| K2 K3) = 6500/(7.5*1*1) = and λ = 147.6

11. Table 22 (page 36) for K12 By interpolation, K12 = 0.212
Stud Wall Design
Table 22 (page 36) for K12
By interpolation, K12 = 0.212

12. Stud Wall Design Permissible compressive stress:
σc,adm,|| = σc,g,||* K2* K3*K8*K12 = 7.5 ×1.0 x 1.0 × 1.1 × 0.212
= 1.75 N/mm2
Axial load capacity of stud:
PC = σc,adm,|| * Area = 1.75 × (44 × 100) = 7.7 kN
Hence, uniformly distributed load capacity of stud wall panel is
7.7/0.6 = 12.8 KN/m

13. Stud Wall Design
Design a stud wall of length 4.2 m and height 3.8 m, using timber of
strength class C24 to support a long term uniformly distributed load of 14 kN/m.
Solution:
1) Find applied stress (σc,a,|| )
UDL = 14KN/m
Assuming that vertical studs are placed at a spacing of 600mm C/C
Therefore, total load = 14KN/m * 0.6 m = 8.4 KN
Assuming a stud size of 100 mm x 50 mm
Therefore, applied stress = σc,a,|| = 8.4x103 /(100X50) =1.68 N/mm2

14. Stud Wall Design 2) Find permissible stress (σc,adm,|| )
σc,adm,|| = σc,g,||*K2* K3*K8*K12 = 7.9*1.0*1.0*1.1* K12
Lex= 1.0 *height = 3.8 m = 3800 mm
Ley = (½)*height = 3.8 m/2 = 1900 mm
= = 3800/28.86=131.67
= =1900/14.43=131.67

15. Stud Wall Design = 7200/(7.9*1.0) = 911.40
K12 from interpolation (table 22) = 0.267
Therefore,
σc,adm,|| = σc,g,||* K3*K8*K12 = 7.9 * 1.0*1.1* K12 =7.9*1.0*1.1*0.267
=2.32 N/mm2
Since,
σc,a,|| (1.68 N/mm2) < σc,adm,|| (2.32 N/mm2) Hence OK
Note: the header spans 0.6m and should also be checked as a beam for given loading