Professor Terje Haukaas University of British Columbia, Vancouver Size Effect Models. = 1 F(τ ) (1) = ( 1 F(τ )) n (2) p s

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1 Size Effect Models Weibull Theory This theory is based on a weaest lin approach to material strength, and essentially simplifies the problem to the calibration of two parameters m and. To understand the theory, consider a structural member that consists of n small elements. If any of the small elements fail, then the member fails. Denote the strength of each small element by τ and assume that the probability distribution of this strength, i.e., F(τ) is nown. Then, the survival probability of each element, p s,n is p s,n F(τ ) () If the small elements are statistically independent then the survival probability for the entire member is For large n, series expansion yields p s ( F(τ )) n (2) p s e n F(τ ) (3) Consequently, the failure probability of the member, for large n, is e n F(τ ) (4) According to Weibull, the CDF for the strength of each small element is F(τ ) (5) for τ>, where > is called the shape parameter and m> is called the scale parameter. Eq. (5) is intended as a convenient approximation of the lower tail of the unnown actual probability distribution of the strength. Substitution of Eq. (5) into (4) yields e n (6) If the structural member is divided into infinitely many infinitesimally small elements then the failure probability reads: e d (7) Under the assumption that the structural member has unit volume and uniform stress, the failure probability becomes: e m (8) Size Effect Models Updated February 22, 24 Page

2 where is the strength of the unit reference volume. Eq. (8) is the standard Weibull distribution. Suppose this reference strength is nown. The strength of another structural member non-unit volume and/or non-uniform stress distribution is obtained by equating its failure probability with the unit-volume reference member: e d e m Hence, a design chec can be of the form or equivalently of the form τ d τ d ( ) (9) () τ d. () To arrive at the O86 code equation for shear capacity of large glulam beams, multiply Eq. () by the constant W f on both sides W f τ d W f (2) The shear stress distribution over a rectangular cross-section is 3 τ(x, y, z) 2 A 6 z2 A d 2 Hence, the volume integral in Eq. (2) reads τ d b d 2 d A 6 z2 A d 2 (x) (3) dz (x) dx Substitution into Eq. (2) yields 3.3 A A (x) dx 3.3 Z A (x) dx (4) Size Effect Models Updated February 22, 24 Page 2

3 A 3.3 W f In contrast, the O86 code formula is where the factor C v is r φ F v.48 A g K N C v Z.8 Z W f (5) (x) dx ( ).48 A g (.) C v Z.8.9 f v K D K H K Sv K T (6) W f C v.825 W f G W f 6 5 dx where G is substituted by the integral of the shear force diagram because the code specifies.2 (7) G l a 5 A ( B + 4 C ) (8) which is Simpson s integration rule without the factor /6. It is also noted that W f is the total load, Z is the beam volume, and A g is the cross-sectional area. In summary, the O86 code equation is φ F v A.6 W f 5 dx Compared this with the theoretically derived Eq. (5)..2 Z.8 W f (9) Possible Simpler Implementation τ d ( ) (2) τ d A 3.3 A (x) dx.27 A (x) dx (2) (22) Size Effect Models Updated February 22, 24 Page 3

4 Determination of Shape and Scale Parameters The shape parameter,, and the scale parameter, m, are determined from tested beams with nown volume, shear force diagram, and shear failure load. To calculate and m the stress at a reference location in the beam due to the failure load is first computed. The maximum shear stress in the cross-section is typically selected, which for rectangular cross-sections read τ M 3 2A Next, this reference stress enters a generic equation that expresses the stress at any location in the beam: Substitution into Eq. (9) yields (23) τ(x, y, z) τ M θ(x, y, z) (24) (τ M θ) d ( ) τ M θ d (25) Factorize out the shear force from the integral and introduce I(), and equivalently β: τ M ( I() ) β() ( ) In a plot with ln(τ M ) along the abscissa axis versus ln() along the ordinate axis the slope of the function is /. To ease the regression, the median τ M for each beam configuration is usually plotted. Knowing, β, and the median values of τ M, the corresponding media values of are computed from Eq. (25). This gives the median value of the Weibulldistributed random variable, which corresponds to the inverse CDF value at.5 probability of exceedance (26) Solving for m yields:.5 e m τ median m (27) τ median ( ln(.5) ) / (28) Regression results from Foschi and Barrett for.m 3 reference volume is 5.53 and m2,54n/m 2. The values for and m are substituted into the Weibull equation, i.e., Eq. (8), to obtain reference strength values for different exceedance probabilities: For example: (p) m( ln( p) ) / (29) Size Effect Models Updated February 22, 24 Page 4

5 τ.5 τ.5 (2,54N/m 2 ) ln(.5) (2,54N/m 2 ) ln(.5) ( ) /5.53 2,377N/m 2 (3) ( ) /5.53,484N/m 2 References Weibull Foschi and Barrett Size Effect Models Updated February 22, 24 Page 5