TUNNELLING IMPACTS | TECHNICAL
A1
σv Y Y t X σH M(θ) RY(θ)
AA(θ) RX(θ)
X σH HH
Far left, figure A1: External loading on the masonry sewer
Left, figure A2:
Consideration of ¼ of the masonry sewer
A2
σv
APPENDIX A:
THRUST LINE ANALYSIS OF A CIRCULAR MASONRY SEWER Background The thrust line analysis aims to confirm if the line of thrust can be developed within an arch ring and equilibrate the given external loading (see Figure A.1) without exceeding the masonry compressive strength. If this is the case, the arch ring will not collapse under this loading in accordance with the lower bound theorem of limit analysis. The following calculation is mainly based on the methodology documented in CIRIA C671 (2009).
Assumptions The self-weight of the brickwork is ignored for simplicity. Also, a quarter of the masonry sewer is considered in the thrust line analysis because of symmetry (see Figure A.2). The tensile strength of the masonry is considered to be negligible for assessment purposes.
Input parameters re rc
t External radius of the sewer
Radius along centerline of the sewer Sewer thickness
σm Masonry compressive strength σH σV KT
/σV )
Calculations Thrust at crown is: HH = σH
× re
Horizontal and vertical force components at a given section located by the angle (θ) are: RX(θ) = HH – σH RY(θ) = σV
× (1 - cosθ) × re × re × sinθ
The minimum depth from the intrados/extrados of the sewer at which the thrust line can be located (dmin
) without exceeding the masonry compressive strength is:
External horizontal pressure at sewer axis level External vertical pressure at sewer axis level
Ratio of external horizontal pressure to external vertical pressure (σH
dmin
= 2 3
HH σm
which is based on the compressive stress distribution shown on Figure A.3 where
1 3 h = dmin
This defines the locations of the ULS intrados and extrados envelopes around the arch. Within the bounds of the ULS envelopes, a further check is undertaken to confirm if the thrust line can be developed within the ‘middle-third’ of the arch thickness (i.e. dmin
= t/3) so that there will be no tension within
the arch. Bending moment at a given section located by the angle (θ) is given by:
M(θ) = HH ×
+ – σv × rc – × re × (sinθ)2 {[re + re × cosθ
[ + – σH × × (1 – cosθ)
Axial force at a given section located by the angle (θ) is given by: AA (θ)= RX (θ)× cosθ + RY (θ) × sinθ
Eccentricity of the thrust line (i.e. offset from the centerline of the arch) at a given section located by the angle (θ) is given by:
ecc(θ)= M(θ) AA(θ)
Worked example Thrust line analysis of a 2m internal diameter masonry sewer with an axis depth of 8m: Step 1 Calculation of input parameters re rc t
External radius of the sewer = 1.6m
Radius along centerline of the masonry arch = 1.3m Sewer thickness = 0.6m
February 2022 | 31
re 2
2 – rc × cosθ × re {[
t 2
– dmin + (1 – cosθ)× rc
{ [ [ {[
re
re
rc
θ
rc
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