BME Department of Mechanics, Materials and Structures Special Loadbearing Structures 2018/2019
T4. MASONRY STRUCTURES 1/4
T4/1 – Analysis of a barrel vault– simplified calculation
Exercise: Check the given masonry vault for symmetrical
loading!
Data:
𝑞𝑘 = 4 𝑘𝑁/𝑚2 (live load)
𝜌𝑚𝑎𝑠𝑜𝑛𝑟𝑦 = 17 𝑘𝑁/𝑚3 (specific weight of masonry wall)
𝜌𝑓𝑖𝑙𝑙 ≈ 𝜌𝑚𝑎𝑠𝑜𝑛𝑟𝑦(specific weight of fill of the extrados–
overestimated)
𝑓𝑑 = 1,0 𝑁/𝑚𝑚2 (compression strength of masonry)
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.
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.
Symmetrical loading
with fixed support at both ends 3-hinged structure
Barrel arch of each meter
circular arc
with fixed support at both ends
Asymmetrical loading
BME Department of Mechanics, Materials and Structures Special Loadbearing Structures 2018/2019
T4. MASONRY STRUCTURES 2/4
Estimation of the loads (on a 1 m wide section)
- live load:
𝑞𝑑 = 𝑞𝑘 ⋅ 1𝑚 ⋅ 𝛾𝑞 = 4 ⋅ 1 ⋅ 1.5 = 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.)
𝑔𝑑 = (𝜌𝑚𝑎𝑠𝑜𝑛𝑟𝑦 ⋅ 1𝑚 ⋅ 𝑡 + 𝜌𝑚𝑎𝑠𝑜𝑛𝑟𝑦 ⋅ 1𝑚 ⋅ ℎ𝑓𝑖𝑙𝑙) ⋅ 𝛾𝑔 =
= (17 ⋅ 1 ⋅ 0.15 + 17 ⋅ 1 ⋅ 0.25) ⋅ 1.35 = 9.18 𝑘𝑁/𝑚 - total load:
𝑝𝐸𝑑 = 𝑞𝑑 + 𝑔𝑑 = 6 + 9.18 = 15.18 𝑘𝑁/𝑚
Support reactions from the three-hinged structure
(Equilibrium equations of moments)
∑𝑀𝐴 = 0 → 𝑝𝐸𝑑𝐿2
2− 𝐵𝑉𝐿 = 0 → 𝐵𝑉 = 45.54 𝑘𝑁
∑𝐹𝑉 = 0 → 𝐴𝑉 + 𝐵𝑉 + 𝑝𝐸𝑑𝐿 = 0 → 𝐴𝑉 = 45.54 𝑘𝑁
∑𝑀𝐶𝑟𝑖𝑔ℎ𝑡
= 0 → 𝑝𝐸𝑑
𝐿
2
𝐿
4− 𝐵𝑉
𝐿
2+ 𝐵𝐻𝑓 = 0 → 𝐵𝐻 = 56.92 𝑘𝑁
∑𝑀𝐻 = 0 → 𝐴𝐻 = 𝐵𝐻 = 56.92 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.)
Data of the quarter point ‘D’
The radius of the arch axis from Pythagoras’ theorem:
2222
222
2,14,20,3
)()2/(
RRR
fRLR
35,4R m
The height at the quarter point (D):
(𝐿/4)2 + 𝑦2 = 𝑅2
08,45,135,4)4/( 2222 LRy m
The central angle at the quarter point:
29,20)/arccos( RyD
The height difference between points D and C:
27,0 yRh m
BME Department of Mechanics, Materials and Structures Special Loadbearing Structures 2018/2019
T4. MASONRY STRUCTURES 3/4
The internal forces at the quater point:
Equilirium equations of the right (separated) side:
∑𝐹𝑉𝑟𝑖𝑔ℎ𝑡
= 0 → −𝐶𝑉 − 𝐵𝑉 + 𝑝𝐸𝑑
𝐿
2= 0 → 𝐶𝑉 = 45.54 − 15.18 ⋅
6
2= 0
∑𝐹𝐻𝑟𝑖𝑔ℎ𝑡
= 0 → 𝐶𝐻 = 𝐵𝐻 = 56.92 𝑘𝑁
Equilibrium equations of the DC separated segments (see the figure):
∑𝐹𝑉𝐷𝐶 = 0 → −𝐷𝑉 − 𝐶𝑉 + 𝑝𝐸𝑑
𝐿
4= 0
𝑫𝑽 = 15.18 ⋅6
4= 𝟐𝟐. 𝟕𝟕 𝒌𝑵
∑𝐹𝐻𝐷𝐶 = 0 → 𝐷𝐻 − 𝐶𝐻 = 0
𝑫𝑯 = 𝐶𝐻 = 𝟓𝟔. 𝟗𝟐 𝒌𝑵
∑𝑀𝐷𝐷𝐶 = 0 → 𝑝𝐸𝑑
𝐿
4
𝐿
8− 𝐶𝐻ℎ + 𝑀𝐷 = 0
𝑴𝑫 = 56.92 ⋅ 0.27 − 15.18 ⋅6
4⋅
6
8= 𝟏. 𝟕 𝒌𝑵𝒎
The normal force at point D can be calculated from the internal forces. We calculate the components of the
internal forces parallel to the arch axis (see the figure):
11,6129,20sin77,2229,20cos92,56sincos DVDHD DDN kN
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 𝑓𝑑 in each points). The cross section is in equilibrium, therefore the resultant of the stresses
equals to 𝑁𝐷 and the moment of the stresses equal to 𝑀𝐷, therefore acting point of 𝑁𝐷 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.
𝜑𝐷
𝜑𝐷
𝐃𝐇
𝜑𝐷 𝐃𝐕
BME Department of Mechanics, Materials and Structures Special Loadbearing Structures 2018/2019
T4. MASONRY STRUCTURES 4/4
The eccentricity of the normal force at point D:
eD= MD/ND =1,7/61,11 = 0,028 m =28 mm
Note, that we consider an 1m long section of the barrel, therefore the width of the cross-section of the arch in
point D is 1m. The height of the compressed zone at point D:
942282
1502
2
De
hx mm
Calculation of the resistance of the cross-section assuming plastic stress state:
94100,110009410 33
dDRd, fbxN kN > NEd,D=61,11 → OK!
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.
with fixed support at both
Asymmetrical loading
BME Department of Mechanics, Materials and Structures Special Loadbearing Structures 2018/2019
T4. MASONRY STRUCTURES 5/4
T4/2 – 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.) Division of the vault into discrete elements.
2.) Determination 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. 2 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 (𝐴𝑉, 𝐴𝐻, 𝐵𝑉, 𝐵𝐻), plot the reaction forces into the figures
6.) These fix the point ‘O’ in the force diagram.
7.) Draw 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: http://web.mit.edu/masonry/mdejong/index.html