t4/1 analysis of a barrel vault simplified calculation courses/6... · the barrel vault behaves...

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



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



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



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



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

http://web.mit.edu/masonry/mdejong/index.html

t4/1 analysis of a barrel vault simplified calculation courses/6... · the barrel vault behaves...

Documents