2.Timber Tutorial Solutions Tension and Compression PDF

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Timber Engineering Tutorial solutions – Tension and compression Tension members 1. Evaluate the capacity of a 190 x 35 mm MGP10 seasoned timber tension member 2.8m long. It will be used as an internal principal member in a roof structure for a grandstand in Brisbane. Evaluate the capacity to resist  Dead load only  A wind load combination The member is connected through 4 M12 bolts each end (2 rows of 2). All bolt holes are 13mm diameter.

13mm dia holes

N*t

N*t 2.8m long 190 x 35mm MGP10

Question 1 solution 190 x 35 MGP10 N*t = Nd,t = Ø k1 k4 k6 [f’tAt] Ø:

Table 2.1 Machine graded, primary structural member  Ø = 0.7

k1 :

(a) Dead load only  Table 2.3  k1 = 0.57 (b) Wind load combination  Table 2.3  k1 = 1.0 [k1 is the value for the shortest duration load in a load combination]

k4:

Seasoned timber in an interior environment  Clause 2.4.2  k4 = 1.0

k6:

Brisbane is south of 25˚S  Clause 2.4.3  k6 = 1.0

f’t

Table H.3.1 1  f’t = 7.1 MPa

At = 190 x 35 – 2 x 35 x 13 = 5740 mm2 Hence (a) (b)

Dead load only N*t

= 0.7 x 0.57 x 1 x 1 x 7.1 x 5740 / 1000 = 16.3kN

Wind load comb. N*t

= 16.3 X 1.0/0.57 = 28.5kN

2. You are selecting a 3m long tension member that will support an internal timber mezzanine floor by hanging it from a portal frame above. M16 bolts along the centre-line of the member will be used to connect to steel brackets at both top and bottom. The mezzanine will support office accommodation in a commercial warehouse in Ballarat (Vic.). Only four such tension members support the entire floor of 96 m 2. The loadings in the tension member are as follows: 51 kN dead load per tension member 96 kN live load per tension member a) Factor the loads to find long-term and short-term load combinations for the strength limit state b) Select the critical design load c) Evaluate the k factors and  for this application d) Select an appropriate cross-sectional area and actual member size assuming F17 hardwood is used

Question 2 solution G = 51kN, Q = 96 kN (a)

(b)

Long term load combination

1.2G + 1.5 x ΨlQ = 1.2 x 51 + 1.5 x 0.4 x 96 = 119 kN

Or permanent action(50+ years)

1.35G = 1.35 x 51 = 68.9 kN

Short term combination:

1.2G + 1.5Q = 1.2 x 5.1 + 1.5 x 96 = 205 kN

Critical design load will depend on k1 factor as all others will be the same for each combination, so take k1 to LHS of equation, i.e. N*t / k1 = Ø k4 k6 [f’tAt] (RHS is now a constant for all load combinations)

According to table G1 For Long term load combination 1.2G + 1.5 x ΨlQ: k1 = 0.57  119/0.57=209KN For 1.35G k1 = 0.57  68.9/0.57 = 120 For Short term 1.2G+1.5Q (distributed floor live load) , k1 = 0.8  205/0.8 = 256 ← critical

(c)

Nd,t = Ø k1 k4 k6 [f’tAt] Ø: Principal member, assume visual graded  Ø = 0.85 k1: = 0.8 k4 : Internal application k4 = 1 k6 : Ballarat  k6 = 1

(d)

Nd,t ≥ N*t  Nd,t ≥ 205 kN/(0.85) = 241 kN 241 x 103 = 0.8 x1x1xf’tAt  f’tAt = 301 kN

Try F17 hardwood, f’t = 25 MPa (Table H2.1)  At = 301 x 103/25 = 12050 mm2 Try 300x50 F17 hardwood (assume unseasoned) M16 = 17mm diameter hole An = (297 x 47 – 17 x 47 ) = 13160 mm2 ØNd,t = 0.85 x 0.94 x 1 x 1 x 25 x 13160 = 263 kN › 205 kN OK  Adopt 300 x 50 F17 hardwood

Compression members 3. Evaluate the compression capacity of the member used in Question 1 for a wind load combination. It will prove difficult to provide any lateral bracing between the ends.

Question 3 solution 190 x 35 MGP10 under wind load, compression Nd,c = Ø k1 k4 k6 k12 [f’cAc] Ø: Table 2.1  MPG10  Ø = 0.7

for primary member

k1: Table 2.3, wind gust  k1 = 1.0 k4 = k6 = 1 as for Question 1 k12:

Member length = 2800 No lateral bracing either direction End connections = 4 bolts each end Table 3.2  g13 = 0.75 [or 0.7 could be argued] Since no restraint either direction, minor axis buckling will obviously be critical S4 =

g13l 0.75 x2800 = = 60 35 b

c: Table 3.3

MPG10  c = 0.96 cS = 0.96 x 60 = 57.6 > 20 k12 = 200/(57.6)2 = 0.06 f’c = 18 MPa (Table H3.1) Ac = Agross for filled bolt holes  N*c = Nd,c = 0.7 x 1.0 x 1 x1x 0.06 [18 x 190 x 35]

= 5.0 kN

4. A multi-residential building to be built in Hobart has a timber framed structure and the walls at the internal stairwell have a substantial machinery load as well as some live load. The frame will be designed in the first instance for vertical loads and would then be checked for combined actions (not part of this question). Select a suitable cross-section for the studs using unseasoned F7 oregon (Douglas Fir).

The studs will have a height floor to floor of 3.3 m and noggings at mid-height. The double stud wall frame will have studs at 450 mm centres in each of the overlapping frames. The figure above shows one frame only. This will be clad on one side only with decorative panelling. The loads given here are for just one stud in one wall in the stair well. The four compression load cases are given below.

Factored strength limit state load combinations on a single stud are as follows:  dead load plus machinery weight (permanent) 6.4 kN  dead load plus construction loads (medium term) 5.3 kN  dead load plus machinery weight plus operation and maintenance loads (long-term) 7.2 kN  dead load plus crowd loading (short-term) 8.2 kN

Question 4 solution: Nd,c = Ø k1 k4 k6 k12 [f’cAc] Ø:

Unseasoned F7, unknown origin  assume visual grading. Studs normally secondary members, but these are supporting machine loading  primary members  Ø = 0.70 (Table 2.1)

k1

Factored loads provided, find critical combination - permanent = 6.4 kN, k1 = 0.57  N*/k1 = 11.2 critical - medium term = 5.3 kN, k1 = 0.94  N*/k1 = 5.6 - long term = 7.2 kN, k1 = 0.80  N*/k1 = 9.0 - short term = 8.2 kN, k1 = 0.94  N*/k1 = 8.7

k4

Unseasoned, to be used indoor  k4 = 1

k6

Hobart not hot!  k6 = 1

k12

Need to select a trial stud size Try 175 x 50 F7 oregon Major axis buckling – no intermediate restraint Table 3.2  g13 = studs in light framing = 0.9 0.9 x3300 = 17.3  S3 = 172 Minor axis buckling – noggings at mid-height  Lay = 3300/2 = 1650 mm 1650 S4 = = 35.1  critical 47 c = 1.08 (F7, unseasoned, Table 3.3)  k12 =

200 (1.08x 35.1) 2

= 0.139

f’c = 13 MPa (Table H.2.1, F7 grade) Ac = 172 x 47 mm2  Nd,c =

0.7 x 0.57 x1x1x1x 0.139 x [13 x 172 x 47] =

5.83 kN < 6.4 kN  no good, need bigger stud

Try 150 x 75 Only change will be to k12 S3 =

0.9 x3300 = 20.2 147

S4 =

1650 = 22.9  still critical 72

k12 =

200 = 0.323 (1.08x 22.9) 2

Nd,c = 0.7 x 0.57 x 1 x 1 x 1 x 0.323 x [13 x 147 x 72] = 17.73 kN > 6.4kN OK Thicker stud much more effective since weak axis buckling governed design. Since lots of capacity, try 100 x 75 0.9 x3300 = 30.6, S4 = 22.9  major axis critical 97

S3

=

k12

= 200/(1.08 x 30.6)2 = 0.183

Nd,c

= 0.7 x 0.57 x1 x 1 x 0.183 x [13 x 97 x 72] = 6.63kN  OK

 Adopt 100 x 75 F7 oregon stud

5

An isolated column which supports a second storey dining room will be made by nailing two pieces of 35 mm thick MGP 12 timber to give a total column thickness of 70 mm. The column height will be 2.8 m and, because it is an isolated column, no bracing will be possible in either direction. In this case g 13 can be taken as 0.9. The column will be used for an interior column in a McDonalds restaurant in Perth. Unfactored design compression loads are as follows:  dead load (permanent) 2.76 kN  live load (crowd - 5 days) 11 kN Find strength limit state load combinations representing:  dead load plus an estimate of permanent live load  dead load plus crowd live load Select a member size to be used to make the column from two pieces of timber nailed side by side. Determine  and k factors and check that your chosen size has adequate final capacity.

Question 5 solution N*c = Nd,c = Ø k1 k4 k6 k12 [f’cAc] Ø:

MGP12 machine graded, primary member since isolated column supporting floor, Table 2.1  Ø = 0.7

k1:

Permanent combination 1.2G + 1.5ΨlQ = 1.2 x 2.76 + 1.5 x 0.4 x 11 = 9.9 kN k1 = 0.57  N*/k1 = 17.4 Crowd loading = 5 days duration (§ 2.4.1.1(c)) 1.2G + 1.5Q = 1.2 x 2.76 + 1.5 x 11 = 19.8 kN k1 = 0.94  N*/k1 = 21.1  critical combination

k4:

Seasoned timber, used indoors  k4 = 1

k6

Perth  k6 = 1

k12 No bracing possible either direction  weak axis buckling will govern g13 = 0.9

L = 2800

b = 2 x 35mm  S4 = 0.9 x 2800/(2 x 35) = 36 MGP12  c =0.98 (Table 3.3)

k12 =

200 = 0.16 (0.98x 36) 2

f’c

=

24 MPa for MGP12 (Table H3.1)

Ac

=

d x 70mm



19800 = 0.7 x 0.94 x 1x1x 0.0.16 x 24 x d x 70 d ≥ 112mm

 Adopt 2-120 x 35 MGP12 studs, nail laminated together to form column...