T4/1 Analysis of a barrel vault simplified calculation

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1 T4. MASONRY STRUCTURES 1/4 T4/1 Analysis of a barrel vault simplified calculation Exercise: Check the given masonry vault for symmetrical loading! ata: q k = 4 kn/m (live load) ρ masonry = 17 kn/m 3 (specific weight of masonry wall) ρ fill ρ masonry (specific weight of fill of the extrados overestimated) f d = 1,0 N/mm (compression strength of masonry) Barrel arch of each meter circular arc A 1.0 m wide section of the barrel vault is considered. The barrel vault behaves like a series of arches positioned next to each other. The schematic statical model is a three-times statically indeterminate structure. However, due to the loads and small movements at the supports, the arch is cracked at the crown (at the keystone) and at the springing line. In the simplified calculation, hinges are assumed at these points, which results in a statically determinate, three-hinged structure. The axis of the three-hinged structure is assumed to coincide with the arch axis. Symmetrical loading with fixed support at both ends 3-hinged structure This simplified method is not suitable for asymmetrical loading, where 4 cracks evolve leading to a four-hinged mechanism. Complex cases can be calculated with a finite element software or thrust line analysis. Asymmetrical loading with fixed support at both ends

2 T4. MASONRY STRUCTURES /4 Estimation of the loads (on a 1 m wide section) - live load: q d = q k 1m γ q = = 6.0 kn/m - dead loads (self-weight of the arch and the fill): They can be estimated as uniformly distributed loads (In contrary to the load on the figure! This estimation can be used as the arch is sufficiently flat.) g d = (ρ masonry 1m t + ρ masonry 1m h fill ) γ g = = ( ) 1.35 = 9.18 kn/m - total load: p Ed = q d + g d = = kn/m Support reactions from the three-hinged structure (Equilibrium equations of moments) M A = 0 p EdL B V L = 0 B V = kn F V = 0 A V + B V + p Ed L = 0 A V = kn M right L L C = 0 p Ed 4 B L V + B Hf = 0 B H = 56.9 kn M H = 0 A H = B H = 56.9 kn Strength check We examine the position of the thrust line, which is the path of the compression forces in the structure. In a cross-section, the location of the thrust line is the location of the normal force (the eccentricity of the normal force is e = M/N). The stability requirements of a structure are fulfilled if all of the points of the thrust line is inside the cross-section and if the maximal stress of each cross-section is less than the strength of the material. The most dangerous cross-sections of the structure are located at points where the eccentricity is large. These points are near the points of maximum bending moment, approximately at the quarter point (have a look at the bending moment diagrams of the first page). (In case of asymmetrical loading the maximal moment emerges near the midpoint of the heavily loaded side. The thrust line is approximately the inverse of the moment diagram.) ata of the quarter point The radius of the arch axis from Pythagoras theorem: R ( L / ) ( R f ) R 3,0 R R 4,35 m,4r 1, The height at the quarter point (): (L/4) + y = R y R ( L/ 4) 4,35 1,5 4,08 m The central angle at the quarter point: arccos( y / R) 0, 9 The height difference between points and C: h R y 0,7 m

3 T4. MASONRY STRUCTURES 3/4 The internal forces at the quater point: H φ φ φ V Equilirium equations of the right (separated) side: F V right = 0 C V B V + p Ed L = 0 C V = = 0 F H right = 0 C H = B H = 56.9 kn Equilibrium equations of the C separated segments (see the figure): F V C = 0 V C V + p Ed L 4 = 0 V = =. 77 kn 4 F H C = 0 H C H = 0 H = C H = kn M C L L = 0 p Ed 4 8 C Hh + M = 0 M = = 1. 7 knm 8 The normal force at point can be calculated from the internal forces. We calculate the components of the internal forces parallel to the arch axis (see the figure): N cos sin 56,9 cos 0,9,77 sin 0,9 61,11 kn H V Plastic analysis We assume, that the cross section is cracked on the tension side and in a plastic stress state on the compression side (=the stress is f d in each points). The cross section is in equilibrium, therefore the resultant of the stresses equals to N and the moment of the stresses equal to M, therefore acting point of N is at the centroid of the compressed zone. To calculate the resistance of the cross-section, we need to obtain the actual size of the compressed zone.

4 T4. MASONRY STRUCTURES 4/4 The eccentricity of the normal force at point : e= M/N =1,7/61,11 = 0,08 m =8 mm Note, that we consider an 1m long section of the barrel, therefore the width of the cross-section of the arch in point is 1m. The height of the compressed zone at point : h 150 x e 8 94 mm Calculation of the resistance of the cross-section assuming plastic stress state: 3 3 N x b f , kn > NEd,=61,11 OK! Rd, d Additional notes: The analysis in case of asymmetrical loading is more complex (4 hinged mechanism) but at the same time, it is much more dangerous. The most critical cross section is at the quarter point of the less loaded side. Asymmetrical loading with fixed support at both

5 T4. MASONRY STRUCTURES 5/4 T4/ Construction of the thrust line The construction of the thrust line is based on the principles of grafostatics: force polygons are redacted which fulfil the equilibrium equations = if the force polygon is closed, it means the force system is in equilibrium (eg. vector triangle). Two figures are prepared, the so-called force diagram and the form diagram. The two diagrams are affine pairs of each other. The force diagram expresses the equilibrium of the form diagram. The steps of the construction: 1.) ivision of the vault into discrete elements..) etermination of the loads of each element. Then the loads are substituted by the resultants (concentrated forces) along their line of action. 3.) Three data must be fixed for the explicit drawing of the thrust line (otherwise there would be multiple solutions). In the present example the expected location of the thrust line is given at the springings and at the keystone (red circles). It would be also possible to fix e.g. points and 1 steep. First, we draw the force diagram. 4.) The force vectors from the loads are plotted under each other (their direction and size matters!). Name the vectors in the figure by numbers! 5.) Use the calculated support reactions (A V, A H, B V, B H ), plot the reaction forces into the figures 6.) These fix the point O in the force diagram. 7.) raw straight lines from point O to the ends of each load vector. Name them by letters. These are the segments of the thrust line. The last step is the drawing of the form diagram: 8.) First, draw the vector of the support reaction A from the support until it crosses the load vector of the first element. Then, continue from this point with a straight line parallel to the vector a. Continue the drawing in the same manner. If the construction is carried out correctly, the chain of the vectors should arrive at support B. The resulting polygon is the thrust line. Note: the shape of the thrust line is opposing to the moment diagram. In this problem, the moment diagram is concave down, therefore the thrust line is concave up. This comes from the fact, that the moment diagram is plotted on the tension side of the axis, while the thrust line is the path of the compression force. More information:

TYPES OF STRUCUTRES. HD in Civil Engineering Page 1-1

TYPES OF STRUCUTRES. HD in Civil Engineering Page 1-1 E2027 Structural nalysis I TYPES OF STRUUTRES H in ivil Engineering Page 1-1 E2027 Structural nalysis I SUPPORTS Pin or Hinge Support pin or hinge support is represented by the symbol H or H V V Prevented: