1. General characteristics of the building

• Building length 32 m

• Building width 21,6 m

• Building height H=1,5m+28x3m=85,5m

• Localization: Szczecin (50 m.a.s.l.)

• Category of terrain: Exposed to the open sea (0 category)

1. Moment of inertia

$$\sum_{\mathbf{i}}^{\mathbf{n}}{\mathbf{I}_{\mathbf{i}}\mathbf{=}\mathbf{I}_{\mathbf{1}}\mathbf{+}\mathbf{I}_{\mathbf{2}}\mathbf{+ .. +}\mathbf{I}_{\mathbf{n}}}$$

• I₁=I₇

nr	h	b	Iy	x2	P	z	Iy	x2
	[cm]	[cm]	[cm^4]	[cm^4]	[cm^2]	[cm]	[cm^4]	[cm^4]
1	58	20	325187	650373	1160	877	892189640	1784379280
2	70	20	571667	1143333	1400	628	552137600	1104275200
3	181	20	9882902		3600	317	361760400	723520800
4	20	290	193333	386667	5800	360	751680000
5	110	20	2218333		2200	353	274139800
6	185	20	10552708		3700	0	0
Sum			22653943	2180373			1025819800	3612175280
							Sum of I1	4662829397

• I₂=I₆

nr	h	b	Iy	x2	P	z	Iy	x2
	[cm]	[cm]	[cm^4]	[cm^4]	[cm^2]	[cm]	[cm^4]	[cm^4]
1	278	20	35808253	71616507	5560	767	3270886840	6541773680
2	63	20	416745	833490	1260	507	323881740	647763480
3	20	333	222000		6660	360	863136000
4	15	20	5625		300	378	42865200
5	567	20	303807105		11340	67	50905260
6	20	290	193333		1360	351	167553360
7	20	23	15333		460	360	59616000
8	68	20	524053		5800	360	751680000
Sum			304767450	72449997			1935755820	7189537160
							Sum	9502510427

• I₃=I₅

nr	h	b	Iy	x2	P	z	Iy	x2
	[cm]	[cm]	[cm^4]	[cm^4]	[cm^2]	[cm]	[cm^4]	[cm^4]
1	356	20	75196693	150393387	10700	638	4355370800	8710741600
2	20	600	400000	800000	12000	360	1555200000	3110400000
3	374	20	87189373		7480	0	0
Sum			87189373	151193387			0	11821141600
							Sum	12059524360

• I₄

nr	h	b	Iy	x2	P	z	Iy	x2
	[cm]	[cm]	[cm^4]	[cm^4]	[cm^2]	[cm]	[cm^4]	[cm^4]
1	594	20	349307640	698615280	12000	607	4421388000	8842776000
2	414	600	3547897200		8280	0	0	0
Sum			3547897200	698615280			0	8842776000
							Sum	13089288480

I=65539016847 cm⁴=655,39016847m⁴

1. Wind action

The base wind speed:

It was assumed that the relevant building is located in the I zone of the wind load.

(Szczecin A=50 m.a.s.l.) A≤300 m.a.s.l.

Value of the primary base wind speed: v_b,0= 22 m/s

Assumed direction coefficient of the wind: c_dir=1,0

Season coefficient: c_season=1,0

Base wind speed:

v_b = v_b, 0 * C_dir * C_season = 22 m/s  * 1, 0 * 1, 0 = 22 m/s 

Pressure wind speed: q_d=0,30

1. Wind blowing on the long side of the building

h=85,5 m
b=36,5 m
d=18,36 m

Exposure coefficient:

$$c_{e}\left( z \right) = 3,0*\left( \frac{z}{10} \right)^{0,17}$$

$$c_{e}\left( z \right) = 3,0*\left( \frac{36,5}{10} \right)^{0,17} = 3,73$$

Value of exposure coefficient due to height

III category of terrain	Height	Ce(z)
	[m]	[-]
	36,5	3,738618
	39,6	3,790787
	42,7	3,839671
	45,8	3,885692
	48,9	3,929197
	85,5	4,320714

q_p(z) = q_b * c_e(z)

$${\rho = 1,25\frac{\text{kg}}{m^{3}}\text{\ \ \ air\ density}\backslash n}{q_{p}\left( 36,5 \right) = 3,74*0,3 = 1,12}$$

0 category of terrain	Height	Exposure coefficient	Pressure wind speed	Peak value of pressure wind speed [qp]
	[m]	ce(z)	qb, 0	$$\left\lbrack \frac{\text{kN}}{m^{2}} \right\rbrack$$
	36,5	3,74	0,3	1,12
	40	3,8	0,3	1,14
	45	3,87	0,3	1,16
	50	3,94	0,3	1,18
	55	4,01	0,3	1,20
	60	4,07	0,3	1,22
	85,5	4,32	0,3	1,30

Peak value of pressure wind speed

Wind pressure acting on the outer surfaces of the structure, expressed by the formula:

w_e = q_p(h) * c_pe, 10

∖n

The values ​​of the external pressure coefficients for vertical walls of buildings in a rectangular plan:

$$\frac{h}{d} = \frac{85,5}{18,36} = 4,65$$

c_{pe, 10, D} = +0, 8

c_{pe, 10, E} = −0, 6825

$$w_{e,D} = 1,12*0,8 = 0,896\frac{\text{kN}}{m^{2}}$$

$$w_{e,E} = 1,12*\left( - 0,6825 \right) = - 0,7654819\frac{\text{kN}}{m^{2}}$$

Height	Exposure coefficient	Pressure wind speed	Peak value of pressure wind speed	External preassure for vertical walls	Preassure on the surface	External preassure for vertical walls	Preassure on the surface	Summation
[m]				„D” area		„E” area
36,5	3,738618	0,3	1,1215853	0,8	0,897268	-0,6825	-0,7654819	1,66275016
39,6	3,790787	0,3	1,1372362	0,8	0,909789	-0,6825	-0,7761637	1,68595273
42,7	3,839671	0,3	1,1519012	0,8	0,921521	-0,6825	-0,7861726	1,7076936
45,8	3,885692	0,3	1,1657076	0,8	0,932566	-0,6825	-0,7955955	1,72816157
48,9	3,929197	0,3	1,178759	0,8	0,943007	-0,6825	-0,804503	1,74751019
85,5	4,320714	0,3	1,2962141	0,8	1,036971	-0,6825	-0,8846661	1,92163744

Summation of the wind preassure

Changing the load to evenly distributed:

$$\frac{146,7531\ }{h} = \frac{146,7531\ }{85,5} = \mathbf{1,716\ }\frac{\mathbf{\text{kN}}}{\mathbf{m}^{\mathbf{2}}}$$

1. Characteristic and design values

Longer side:

$$W_{e}^{*} = 1,716\frac{\text{kN}}{m^{2}}$$

$$\mathbf{W}_{\mathbf{e}}\mathbf{=}\mathbf{W}_{\mathbf{e}}^{\mathbf{*}}\mathbf{\bullet}\mathbf{\gamma}_{\mathbf{f}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{716}\mathbf{\bullet}\mathbf{1}\mathbf{,}\mathbf{5}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{574}\frac{\mathbf{\text{kN}}}{\mathbf{m}^{\mathbf{2}}}$$

We use two ways of calculating value of q:

-way of bending moment

-way of area of influences

Then we will use bigger value in further calculations.

Calculation the bending moment:

Bending moment of each section, width each section is 1m:

$$M_{1} = 1,609\frac{\text{kN}}{m^{2}}*36,5\ m*16,25\ m*1\ m = 1071,80\ kNm$$

M₂ = 1, 686 * 3, 1 * 38, 3 * 1 = 200, 18 kNm

M₃ = 1, 707 * 3, 1 * 41, 4 * 1 = 219, 08 kNm

M₄ = 1, 728 * 3, 1 * 44, 5 * 1 = 238, 37 kNm

M₅ = 1, 748 * 3, 1 * 47, 6 * 1 = 257, 94 kNm

M₆ = 1, 921 * 36, 5 * 67, 2 * 1 = 4711, 83 kNm

$$\sum_{}^{}M_{i} = 6699,2\ kNm$$

$$M = \frac{qh^{2}}{2} \rightarrow q = \frac{2M}{h^{2}} = \frac{2*6699,2}{{85,5}^{2}} = 1,83\frac{\text{kN}}{m^{2}}$$

∖n

Calculation the area of influences:

Rectangular area of each section:

$$A_{1} = 1,669\frac{\text{kN}}{m^{2}}*36,5m = 60,92\frac{\text{kN}}{m}$$

$$A_{2} = 1,686\frac{\text{kN}}{m^{2}}*3,1\ m = 5,23\frac{\text{kN}}{m}$$

$$A_{3} = 1,707\frac{\text{kN}}{m^{2}}*3,1\ \ m = 5,29\frac{\text{kN}}{m}$$

$$A_{4} = 1,728\frac{\text{kN}}{m^{2}}*3,1\ \ m = 5,36\frac{\text{kN}}{m}$$

$$A_{5} = 1,748\frac{\text{kN}}{m^{2}}*3,1\ m = 5,42\frac{\text{kN}}{m}$$

$$A_{6} = 1,9,21\frac{\text{kN}}{m^{2}}*36,5\ m = 70,12\frac{\text{kN}}{m}$$

$$\sum_{}^{}A_{i} = 152,34\frac{\text{kN}}{m}$$

$$q = \frac{\sum_{}^{}A_{i}}{H} = \frac{152,34\frac{\text{kN}}{m}}{85,5m} = 1,78\frac{\text{kN}}{m^{2}}$$

The bigger value of q is from the first method ($1,83\frac{\text{kN}}{m^{2}}$) so we will use this for further calculations.

1. Calculations of internal forces:

We consider wall 3 (axis 3) in our calculations.

Areas:

F_3A = F_3C = 5, 36 • 0, 20 + 6, 00 • 0, 2 = 2, 272m²

F_3B = 3, 74 • 0, 20 = 0, 478m²

Moments of inertia:

$$I_{3B} = \frac{{3,74}^{3}*0,20}{12} = 0,872\text{\ m}^{4}$$

$$I_{3A} = I_{3C} = \frac{{5,36}^{3}*0,20}{12} + 0,20*5,36*{1,52}^{2} + {1,26}^{2}*6*0,2 + \frac{{0,2}^{3}*6}{12} = 6,9523\ m^{4}$$

There is two lines of openings, symmetrical:

$$\alpha = \sqrt{\left( \frac{2l^{2}}{\sum_{}^{}I_{i}} + \frac{1}{F_{1}} \right)*\frac{12I_{p}}{h*b^{3}}}$$

$\sum_{}^{}I_{i} = 6,9523*2 + 0,872 = 14,7766\ m^{4}$ - total moment of inertia, sum for own central of gravity

F₁ = F_3A = 2, 272m² – area of cross-section no 1

l = 1, 36 + 1, 63 + 1, 87 = 4, 86m

h = 3,00 m – storey height

b = 1,63 m – width of opening

I_p – moment of inertia of header

$$I_{p} = \frac{d*h_{p}^{3}}{12*\left\lbrack 1 + 2,8{*\left( \frac{h_{p}}{b} \right)}^{2} \right\rbrack} = \frac{0,20*{0,8}^{3}}{12*\left\lbrack 1 + 2,8*\left( \frac{0,80}{1,63} \right)^{2} \right\rbrack} = 0,005096\ m^{4}$$

d = 0,20 m – thickness of wall

h_p = 0,80 m – height of header

$$\alpha = \sqrt{\left( \frac{2{*4,86}^{2}}{14,7766} + \frac{1}{2,272} \right)*\frac{12*0,005096}{3,00*{1,63}^{3}}} = 0,1308$$

We use the Rossman approach:

α * H = 0, 1308 * 85, 5 = 11, 18 m

Individual values for different strips:

x_i = ξ * H ξ − − thickness of slide η, η’ = f(α * H ; ξ)

x₀ = 0, 0 * H = 0, 0 * 85, 5 m = 0, 00 m η_o = 0, 000    and    η₀^′ = 0, 091

x₁ = 0, 1 * H = 0, 1 * 85, 5 m = 8, 55 m η₁ = 0, 021 and η₁^′ = 0, 130

x₂ = 0, 2 * 85, 5 m = 17, 10 m η₂ = 0, 055 and η₂^′ = 0, 216

x₃ = 0, 3 * 85, 5 m = 25, 65 m η₃ = 0, 106 and η₃^′ = 0, 303

x₄ = 0, 4 * 85, 5 m = 34, 2 m η₄ = 0, 176 and η₄^′ = 0, 400

x₅ = 0, 5 * 85, 5 m = 42, 75 m η₅ = 0, 266 and η₅^′ = 0, 496

x₆ = 0, 6 * 85, 5 m = 51, 3 m η₆ = 0, 374 and η₆^′ = 0, 588

x₇ = 0, 7 * 85, 5 m = 59, 85 m η₇ = 0, 500 and η₇^′ = 0, 663

x₈ = 0, 8 * 85, 5 m = 68, 4 m η₈ = 0, 636 and η₈^′ = 0, 680

x₉ = 0, 9 * 85, 5 m = 76, 95 m η₉ = 0, 766 and η₉^′ = 0, 567

x₁₀ = 1, 0 * 85, 5 m = 85, 5 m η₁₀ = 0, 835 and η₁₀^′ = 0, 000

Data for the calculations of internal forces:

$$w_{c} = q*L*\frac{I}{\sum_{}^{}I} = 1,83*36,5*\frac{120,59}{655,39} = 12,29\ kN/m$$

w_o = w_c * 1, 5 = 12, 29 * 1, 5 = 18, 435 kN/m

$$\psi = \frac{l}{\sum_{}^{}I_{i}}*\frac{12{*I}_{p}}{h_{k}*b^{3}} = \frac{4,86}{14,7766}*\frac{12*0,005096}{3,00*{1,63}^{3}} = 0,00155$$

$$M_{H}^{o} = \frac{w_{o}*H^{2}}{2} = \frac{18,435\ *{85,50}^{2}}{2} = 67382,23\text{\ kNm}$$

T_H^o = ω_o * H = 18, 435 * 85, 50 = 1576, 19 kN

Shearing forces:

$$T = \eta*\frac{\psi}{\alpha^{2}}*M_{H}^{o} = 0,021*\frac{0,00155}{{0,1308}^{2}}*67382,23 = 127,96\text{\ kN}$$

$$T^{'} = \eta^{'}*\frac{\psi}{\alpha^{2}}*T_{H}^{o} = 0,13*\frac{0,00155}{{0,1308}^{2}}*1576,19 = 18,53\text{\ kN}$$

$$Q = \eta^{'}*\frac{\psi}{\alpha^{2}}*T_{H}^{o}*h = 0,13*\frac{0,00155}{{0,1308}^{2}}*1576,19*3,00 = 55,59\text{\ kN}$$

M_x^o = M_H^o * ξ² = 67382, 23 * 0, 1² = 673, 8223 kNm

M_x = M_x^o − T * 2l = 673, 82 − 127, 96 * 2 * 4, 86 = −569, 96 kNm

Tabulated results:

ξ	η	η'	T (kN)	T' (kN)	Q (kN)	Mxo (kNm)	Mx (Knm)
0	0,000	0,091	0,00	12,97	38,91	0,00	0,00
0,1	0,021	0,130	127,96	18,53	55,59	673,82	-569,96
0,2	0,055	0,216	335,14	30,79	92,36	2695,29	-562,24
0,3	0,106	0,303	645,90	43,19	129,56	6064,40	-213,75
0,4	0,176	0,400	1072,44	57,01	171,04	10781,16	357,05
0,5	0,266	0,496	1620,85	70,70	212,09	16845,56	1090,94
0,6	0,374	0,588	2278,93	83,81	251,43	24257,60	2106,38
0,7	0,500	0,663	3046,70	94,50	283,50	33017,29	3403,36
0,8	0,636	0,680	3875,40	96,92	290,77	43124,63	5455,70
0,9	0,766	0,567	4667,55	80,82	242,45	54579,61	9211,06
1	0,835	0,000	5087,99	0,00	0,00	67382,23	17926,96

Calculating the stresses for 1,0 m area:

$$M_{1} = M_{x}*\frac{I_{1}}{\sum_{}^{}I_{i}} = 17926,96*\frac{6,952}{14,7766} = 8434,52\text{\ kNm}$$

$$\sigma_{1} = \frac{M_{1}*a}{2*I_{1}} = \frac{8434,52*5,56}{2*6,952} = 3372,69\ kN/m^{2}$$

$$\sigma_{2} = \sigma_{1}*\frac{\left( \frac{a}{2} - 1,0 \right)}{\frac{a}{2}} = 3073,31*\frac{\left( \frac{5,56}{2} - 1,0 \right)}{\frac{5,56}{2}} = 2159,49\ kN/m^{2}$$

Final total perpendicular force:

N = N₁ + N₂

$$N_{1} = \frac{T}{a} = \frac{5087,99}{5,56}*1mb = 1003,55\text{\ kN}$$

$$N_{2} = \frac{\sigma_{1} + \sigma_{2}}{2}*d*1mb = \frac{3372,69 + 2159,49}{2}*0,2*1,0 = 553,22\text{\ kN}$$

N = 553, 22 kN + 1003, 55 kN = 1556, 77 kN

Tabulated results:

ξ	M1 (kNm)	σ1 (kN/m2)	σ2 (kN/m2)	N2 (kN)	N1 (kN)	N (kN)
0	0,00	0,00	0,00	0,00	0,00	0,00
0,1	-268,16	-107,23	-68,66	-17,59	25,24	7,65
0,2	-264,53	-105,78	-67,73	-17,35	66,10	48,75
0,3	-100,57	-40,21	-25,75	-6,60	127,40	120,80
0,4	167,99	67,17	43,01	11,02	211,53	222,54
0,5	513,28	205,24	131,42	33,67	319,69	353,36
0,6	991,04	396,28	253,74	65,00	449,49	514,50
0,7	1601,26	640,29	409,97	105,03	600,93	705,95
0,8	2566,87	1026,41	657,20	168,36	764,38	932,74
0,9	4333,75	1732,92	1109,57	284,25	920,62	1204,87
1	8434,52	3372,69	2159,49	553,22	1003,55	1556,77

Deflection:

f = f_ultimate ≥ f_s + f_f

Additional 0,5 m for foundation in each direction.

B + 0,5 m + 0,5 m = 18,36 + 1 = 19,36 m

L + 0,5 m +0,5 m = 36,5 + 1 = 37,5 m

w = q * L = 1, 83 * 36, 5 = 66, 795 kN/m

$$W = w*H = 66,795\ \frac{\text{kN}}{m}*85,5\ m = 5710,97\ kN$$

$$z = \frac{H}{2} + H_{f} = \frac{85,5}{2} + 13,4 = 56,15\ m$$

$$I_{f} = \frac{bh^{3}}{12} = \frac{\left( L + 1 \right)\left( B + 1 \right)^{3}}{12} = \frac{37,5*{19,36}^{3}}{12} = 22675,9808\ m^{4}$$

Calculations Mx for characteristic values:

$$w_{c} = q*L*\frac{I}{\sum_{}^{}I} = 1,83*36,5*\frac{120,59}{655,39} = 12,29\ kN/m$$

$$\psi = \frac{l}{\sum_{}^{}I_{i}}*\frac{12{*I}_{p}}{h_{k}*b^{3}} = \frac{4,86}{14,7766}*\frac{12*0,005096}{3,00*{1,63}^{3}} = 0,00155$$

$$M_{\text{Hc}}^{o} = \frac{w_{c}*H^{2}}{2} = \frac{12,29*{85,50}^{2}}{2} = 44921,4863\text{\ kNm}$$

T_Hc^o = ω_c * H = 12, 29 * 85, 50 = 1050, 795 kN

Shearing forces:

$$T_{c} = \eta*\frac{\psi}{\alpha^{2}}*M_{\text{Hc}}^{o} = 0,021*\frac{0,00155}{{0,1308}^{2}}*44921,4863 = 85,31\text{\ kN}$$

$$T_{c}^{'} = \eta^{'}*\frac{\psi}{\alpha^{2}}*T_{\text{Hc}}^{o} = 0,130*\frac{0,00155}{{0,1308}^{2}}*1050,795 = 12,35\text{\ kN}$$

$$Q_{c} = \eta^{'}*\frac{\psi}{\alpha^{2}}*T_{\text{Hc}}^{o}*h = 0,130*\frac{0,00155}{{0,1308}^{2}}*1050,795*3,00 = 37,06\text{\ kN}$$

M_xc^o = M_Hc^o * ξ² = 44921, 4863 * 0, 1² = 449, 21 kNm

M_xc = M_xc^o − Tc * 2l = 449, 21 − 85, 31 * 2 * 4, 86 = −379, 98 kNm

ξ	η	η'	Tc (kN)	Tc' (kN)	Qc (kN)	Mxoc (kNm)	Mxc (kNm)
0	0,000	0,091	0,00	8,65	25,94	0,00	0,00
0,1	0,021	0,130	85,31	12,35	37,06	449,21	-379,98
0,2	0,055	0,216	223,42	20,53	61,58	1796,86	-374,83
0,3	0,106	0,303	430,60	28,79	86,38	4042,93	-142,50
0,4	0,176	0,400	714,96	38,01	114,03	7187,44	238,03
0,5	0,266	0,496	1080,56	47,13	141,40	11230,37	727,30
0,6	0,374	0,588	1519,29	55,87	167,62	16171,74	1404,25
0,7	0,500	0,663	2031,13	63,00	189,00	22011,53	2268,90
0,8	0,636	0,680	2583,60	64,62	193,85	28749,75	3637,13
0,9	0,766	0,567	3111,70	53,88	161,64	36386,40	6140,70
1	0,835	0,000	3391,99	0,00	0,00	44921,49	11951,30

Areas Mxc			Sum
Point	+	-
1	0,00	-1614	-1614,00
2	0,00	-3207	-3207,00
3	0,00	-2198	-2198,00
4	632,00	-226	406,00
5	4101,00	0	4101,00
6	9057,00	0	9057,00
7	15610,00	0	15610,00
8	25100,00	0	25100,00
9	41555,00	0	41555,00
10	76891,00	0	76891,00

Point	Mxc	Mc	Final area
0,1	-1614,00	50,50125	-81509,0175
0,2	-3207,00	151,5038	-485872,687
0,3	-2198,00	252,5063	-555008,847
0,4	406,00	353,5088	143524,5728
0,5	4101,00	454,5113	1863950,841
0,6	9057,00	555,5138	5031288,487
0,7	15610,00	656,5163	10248219,44
0,8	25100,00	757,5188	19013721,88
0,9	41555,00	858,5213	35675852,62
1	76891,00	959,5238	73778744,51
			144632912

C = 30,08 MN/m³ = 30080 kN/m³

E = 32 GPa Concrete C30/37

$$f_{f} = \frac{W*z*\left( H + H_{f} \right)}{I_{f}*c} = \frac{5710,9725*56,15*\left( 85,5 + 13,4 \right)}{22675,98*30080} = 0,0464\ m = 4,64\text{\ cm}$$

$$M_{\text{xc}}*{\overset{\overline{}}{M}}_{c} = 144\ 632\ 912$$

$$f_{s} = \int_{0}^{H}{\frac{M_{\text{xc}}*{\overset{\overline{}}{M}}_{c}}{E*I} = \frac{144\ 632\ 912}{32*10^{6}*120,59}} = \ 0,03748\ m = 3,75\text{\ cm}$$

$$f_{\text{ultimate}} = \frac{H}{2500} = \frac{85,5}{2500} = 0,0342\ m = 3,42\ cm$$

f = f_ultimate ≥ f_s + f_f

f = 3, 42 cm  ≥ 4, 64 cm + 3, 75 cm = 8, 39 cm condition not fulfilled

Unfulfilled condition. This should increase thickness of the wall or use a higher grade of concrete