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

Stud Wall Design

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

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

Stud Wall Design Load bearing stud wall

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.

Stud Wall Design

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

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

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) = 866.7 and λ = 147.6

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

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

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

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 = 28.86 = 3800/28.86=131.67 = 14.43 =1900/14.43=131.67

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