Effects of Wind on Billboard Structures

By

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August 7, 2020

Billboards are large outdoor advertising structures that are found along highways and city centres. Due to their large size and flexibility for creative output, billboards are usually eye-catching and offer a viable solution to advertisers for maximum exposure to vehicular traffic along highways. Billboard advertisements can be digital, painted, or printed. Most billboards on our highways are usually one-sided, two-sided, or three-sided, and supported on a single column. This type of cantilever structure can have a large windward area which can lead to high internal stresses in the members. Wind force is the most critical action on billboards, and failure due to wind and hurricane has often been reported.

Other effects such as imperfection and p-delta will also need to be checked for the optimal performance of such structures under wind load.
The types of failure that billboards can experience are;

Damage of the plate cladding
Failure of supporting structure
Overturning of the entire structure from the foundation

collapsed billboard — Failure of billboard from foundation

failure of billboard members — Failure of supporting structure

In this article, we are going to investigate the effect of wind on a three-sided highway billboard using Staad Pro software.

billboard model — 3D model of a 3-sided billboard

The members of the billboard are as follows

Circular column – CHS – 500mm diameter, 25 mm thick
Main cantilever beams (projecting from column) – 127 x 76 x 13 UK
Diagonals – UA 50 x 50 x 8
Member supporting billboard plates – UA 70 X 70 X 8

A rendered 3D view of the billboard is shown below.

ANOTHER VIEW OF BILLBOARD 3D MODEL — *3D rendering of a 3-sided billboard*

Wind Load Analysis

The wind load analysis on the billboard has been carried out in accordance with ASCE-7-2002 using Staad Pro v8i software. The wind load information is as follows;

The wind pressure on the structure is shown below.

wind load on billboard — *Wind load on a billboard*

Analysis Results

deflected shape of billboard — Deflected shape of the billboard structure

Maximum lateral deflection of billboard = 5.665 mm

Maximum bending moment on column = 31.552 kNm
Maximum axial force on column = 24.5 kN

A little consideration of other analysis results showed that the sections provided are a bit uneconomical for the structure. The design and modelling can be readjusted accordingly.

Design of Moment-Resisting Column Base Plates

By

Ubani Obinna

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August 6, 2020

In the design of portal frames and other steel structures, moment-resisting bases can be provided if large lateral forces and bending moments are anticipated. This can lead to the economical design of the frame members, but the base will have to be carefully designed as a moment-resisting connection. Therefore such bases are expected to transfer bending moment and axial forces between the steel members and the concrete substructure. Depending on the direction of the moment and the magnitude, base plates can be symmetric or asymmetric.

For the design of moment-resisting base plates, the smallest diameter of bolts to be considered should be M20, even though M30 bolts are deemed the most appropriate. Larger bolts in smaller numbers are usually preferable. All holding-down bolts should be provided with an embedded anchor plate for the head of the bolt to bear against.

The sizes of anchor plates are chosen so as not to apply stress of more than 30 N/mm² at the concrete interface, assuming 50% of the plate is embedded in concrete. When the moments and forces are high, the holding-down system may need to be designed in conjunction with the reinforcement in the base.

Design Methodology

The design methodology for a moment-resisting base plate connection involves using an iterative approach/experience to select the best base dimensions and bolt configuration. This arrangement is then evaluated for its resistance against the anticipated actions from the superstructure.

In EN 1993-1-1:2005, the resistance of a base plate subjected to bending moment and axial force is assumed to be provided by two T-stubs, one in tension, and the other in compression. The resistance of the stub in tension is assumed to be provided by the holding down bolts, while the resistance of the stub in compression is assumed to be provided by a compression zone in the concrete, which is concentric with the column flange as shown below.

The design verification of a moment-resisting base plate, therefore, involves the following steps;

STEP 1
Evaluate the design forces in the equivalent T-stubs for both flanges. For a flange in compression, the force may be assumed to be concentric with the flange. For a flange in tension, the force is assumed to be along the line of the holding down bolts.

STEP 2
Evaluate the resistance of the equivalent T-stub in compression.

STEP 3
Evaluate the resistance of the equivalent T-stub in tension

STEP 4
Verify the adequacy of the shear resistance of the connection.

STEP 5
Verify the adequacy of the welds in the connection.

STEP 6
Verify the anchorage of the holding down bolts.

Design Example

Verify the capacity of the base plate arrangement in grade S275 steel shown below to resist the following actions;
M_Ed= 125 kNm
N_Ed = -870 kN (compression)
V_Ed= 100 kN

Column
From data tables for 254 x 254 x 132 UKC in S355
Depth h_c = 276.3 mm
Width b_c = 261.3 mm
Flange thickness t_f,c = 25.3 mm
Web thickness t_w,c = 15.3 mm
Root radius r_c = 12.7 mm
Elastic modulus (y-y axis) W_el,y,c = 1630000 mm³
Plastic modulus (y-y axis) W_pl,y,c = 1870000 mm³
Area of cross section A_c= 16800 mm²
Depth between flanges h_w,c = h_c – 2 t_f,c = 276.3 – (2 x 25.3) = 200.4 mm

Base plate
Steel grade S275
Depth h_bp = 500 mm
Gross width b_g,bp = 500 mm
Thickness t_bp = 50 mm

Concrete
The concrete grade used for the base is C30/37

Bolts
M24 8.8 bolts
Diameter of bolt shank d = 24 mm
Diameter of hole d₀ = 26 mm
Shear area (per bolt) A_s = 353 mm²
Number of bolts on either side n = 3

MATERIAL STRENGTHS

Column and base plate
The National Annex to BS EN 1993-1-1 refers to BS EN 10025-2 for values of nominal yield and ultimate strength. When ranges are given the lowest value should be adopted.

For S355 steel and 16 < t_f,c< 40 mm
Column yield strength f_y,c= R_eH = 345 N/mm²
Column ultimate strength f_u,c = R_m = 470 N/mm²

For S275 steel and t_bp > 40 mm
Base plate yield strength f_y,bp = R_eH = 255 N/mm²
Base plate ultimate strength f_u,bp = R_m= 410 N/mm²

Concrete
For concrete grade C30/37
Characteristic cylinder strength f_ck = 30 MPa = 30 N/mm²

The design compressive strength of the concrete is determined from:
f_cd = α_ccf_ck/γ_c
f_cd = (0.85 x 30)/1.5 = 17 N/mm²

For typical proportions of foundations, conservatively assume:
f_jd = f_cd = 17 N/mm²

Bolts
For 8.8 bolts
Nominal yield strength f_yb = 640 N/mm²
Nominal ultimate strength f_ub = 800 N/mm²

PARTIAL FACTORS FOR RESISTANCE

Structural steel
γ_M0 = 1.0
γ_M1 = 1.0
γ_M2 = 1.1

Parts in connections
γ_M2 = 1.25 (bolts, welds, plates in bearing)

DISTRIBUTION OF FORCES AT THE COLUMN BASE
The design moment resistance of the base plate depends on the resistances of two T-stubs, one for each flange of the column. Whether each T-stub is in tension or compression depends on the magnitudes of the axial force and bending moment. The design forces for each situation are therefore determined first.

Forces in column flanges
Forces at flange centroids, due to moment:
N_L,f = M_Ed/(h – t_f) = [125/(276.3 – 25.3)] x 10³ = 498 kN (tension)
N_R,f = -M_Ed/(h – t_f) = -[125/(276.3 – 25.3)] x 10³ = -498 kN (compression)

Forces due to axial force:
N_L,f = N_R,f = N_Ed/2 = -870/2 = -435 kN

Total force:
N_L,f = 498 – 435 = 63 kN (tension)
N_R,f = – 498 – 435 = -933 kN (compression)

In this case, the left side is in tension and the right side is in compression. The resistances of the two T-stubs will, therefore, be centred along the lines shown below:

Forces in T-stubs of the base plate
Assuming that tension is resisted on the line of the bolts and that compression is resisted concentrically under the flange in compression, the lever arms from the column centre can be calculated as follows:

z_t= 380/2 = 190 mm
z_c = (276.3 – 25.3)/2 = 125.5 mm
In this design situation, the left flange is in tension and the right in compression.
Therefore, z_L = z_t and z_R = z_c

N_LT= [M_Ed/(z_L + z_R)] + [(N_Ed x z_R)/(z_L + z_R)] = [125/(190 + 125.5)] x 10³ + [-(870 x 125.5)/(190 + 125.5)] = 50.127 kN
N_RT= -[M_Ed/(z_L + z_R)] + [(N_Ed x z_L)/(z_L + z_R)] = -[125/(190 + 125.5)] x 10³ + [-(870 x 125.5)/(190 + 125.5)] = -742.265 kN

RESISTANCE OF EQUIVALENT T-STUBS

Resistance of compression T-stub
The resistance of a T-stub in compression is the lesser of:

The resistance of concrete in compression under the flange (F_c,pl,Rd)
The resistance of the column flange and web in compression (F_c,fc,Rd)

Compressive resistance of concrete below the column flange. The effective bearing area of the joint depends on the additional bearing width, as shown below.

Determine the available additional bearing width (c), which depends on the plate thickness, plate strength and joint strength.

c = t_bp x [√f_ybp/(3f_jdγ_M0)] = 50 x [√255/(3 x 17 x 1.0)] = 112 mm

Thus the dimensions of the bearing area are,
b_eff = 2c + t_fc = (2 x 112) + 25.3 = 249.3 mm
l_eff = 2c + b_c = (2 x 112) + 261.3 = 485.3 mm

The area of bearing is,
A_eff= 249.3 x 485.3= 120985 mm²

Thus, the compression resistance of the foundation is,
F_c,pl,Rd= A_efff_jd
F_c,pl,Rd = 120985 x 17 x 10^-3 = 2057 kN, > N_R,T = 742.265 kN (maximum value).
This is therefore satisfactory

Resistance of the column flange and web in compression
The resistance of the column flange and web in compression is determined from:
F_c,fcRd = M_c,Rd/(h_c – t_fc)

M_c,Rd is the design bending resistance of the column cross-section = 645 kNm (see page 402 of P363)

If V_Ed > V_c,Rd/2, the effect of shear should be allowed for

V_c,Rd= 705 kN (see page 234 of P363)
V_Ed = 100 kN
By inspection:
V_Ed< V_c,Rd/2

Therefore, the effects of shear may be neglected and hence
M_c,Rd= 645 kNm

F_c,fcRd = [(645 x 10⁶)/(276.3 – 25.3)] x 10-3 = 2569.4 kN

As, F_c,pl,Rd < F_c,fc,Rd the compression resistance of the right hand T-stub is:
F_c,pl,Rd = 2057 kN
F_c,pl,Rd> N_R,T = 742.265 kN Satisfactory

RESISTANCE OF TENSION T-STUB
The resistance of the T-stub in tension is the lesser of:

The base plate in bending under the left column flange, and
The column flange/web in tension

Resistance of base plate in bending
The design resistance of the tension T-stub is given by:
F_t,pl,Rd = F_t,Rd= min{F_T,1-2,Rd; F_T,3,Rd}

Where F_T,1-2,Rd is the ‘Mode 1 / Mode 2’ resistance in the absence of prying and F_T,3,Rd is the Mode 3 resistance (bolt failure).

F_T,1-2,Rd = 2M_Pl,1,Rd/m

M_Pl,1,Rd = (0.25∑l_eff,1t_bp²f_ybp)/γ_M0

∑l_eff,1 = is the effective length of the T-stub, which is determined from the equations below;

b_p = 500 mm
p = 190 mm
e = 60 mm
e_x = 60 mm
m_x = 52 mm
n = 3 (number of rows of bolt)

Non-circular patterns

Single curvature; l_eff,nc = b_p/2 = 500/2 = 250 mm
Individual end yielding; l_eff,nc = 0.2n(4m_x + 1.25e_x) = 1.5(4 x 52 + 1.25 x 60) = 424.5 mm
Corner yielding of outer bolts, individual yielding between; l_eff = 2m_x + 0.625e_x + e + (n – 2)(2m_x + 0.625e_x) = 2(52) + 0.625(60) + 60 + 1 x (2 x 52 + 0.625 x 60) = 324.25 mm
Group end yielding; l_eff = 2m_x + 0.625e_x + 0.5(n – 1)p = 2(52) + 0.625(60) + (0.5 x 2 x 190) = 331.5 mm

Circular patterns

Individual circular yielding; l_eff,cp = nπm_x= 3 x π x 52 = 490 mm
Individual end yielding; l_eff,cp = 0.5n(πm_x + 2e) = 1.5(π x 52 + 2 x 60) = 425 mm

The minimum is l_eff,1 = 250 mm

M_Pl,1,Rd = (0.25∑l_eff,1t_bp²f_ybp)/γ_M0 = [(0.25 x 250 x 50² x 255)/1.0] x 10^-6 = 39.84 kNm
F_T,1-2,Rd = 2M_Pl,1,Rd/m = (2 x 39.84)/0.052 = 1532.3 kN

F_t,3,Rd = ∑F_t,Rd
For class 8.8. M24 bolts; F_t,Rd = 203 kN
F_t,3,Rd = 3 x 203 = 609 kN

F_t,pl,Rd = F_t,Rd= min{F_T,1-2,Rd; F_T,3,Rd} = min{1532.3; 609} = 609 kN

N_LT = 50.127 kN < F_t,pl,Rd = 609 kN Okay.

WELD DESIGN
Welds to the tension flange
The maximum tensile design force is significantly less than the resistance of the flange, so a full-strength weld is not required. The design force for the weld may be taken as that determined between column and base plate in STEP 1, i.e. 498 kN (N_L,f)

For a fillet weld with s = 12 mm, a = 8.4 mm
The design resistance due to transverse force is:

F_nw,Rd = (Kaf_u/√3)/β_wγ_M2

where K = 1.225, f_u = 410 N/mm² and β_w = 0.85 (using the properties of the material with the lower strength grade – the base plate)

F_nw,Rd = [(1.225 x 8.4 x 410) / √3]/(0.85 x 1.25) = 2.29 kN/mm

Length of weld, assuming a fillet weld all around the flange:
For simplicity, two weld runs will be assumed, along each face of the column flange. Conservatively, the thickness of the web will be deducted from the weld inside the flange. An allowance equal to the leg length will be deducted from each end of each weld run.

L = (261.3 – 2 x 12) + (261.3 – 12 – 4 x 12) = 438.6 mm
F_t,weld,Rd = 2.29 x 438.6 = 1004 kN, > 498 kN – Satisfactory

Welds to the compression flange
With a sawn end to the column, the compression force may be assumed to be transferred in bearing. There is no design situation with moment in the opposite direction, so there should be no tension in the right-hand flange. Only a nominal weld is required. Commonly, both flanges would have the same size weld.

Welds to the web
Although the web weld could be smaller and sufficient to resist the design shear, it would generally be convenient to continue the flange welds around the entire perimeter of the column.

Structural Design of Signposts and Billboards

By

Ubani Obinna

-

August 3, 2020

As the name implies, signposts are post bearing structures that offer information or guidance to people. They are a prominent feature of our highways, streets, city centers, villages, and areas of public gathering. Signposts are usually placed strategically away from obstruction as they are intended to show information like route direction, warnings, route assurance, traffic signs, commercial advertisements, etc.

EN 12899-1:2007 requires that signposts made of steel structures should conform with EN 1993-1-1:2005 (Eurocode 3). One of the major concerns in the design of billboards and signposts is the risk of failure under wind load, which has serious economic and safety consequences. A failed highway sign structure can cause injury to pedestrians, damage vehicles, and obstruct traffic. As a result, such structures that are exposed to the public must satisfy all needed safety considerations. Additional risks of vehicles colliding with sign structures should also be checked, with passive protection provided for such structures.

Actions on Sign Structures

Wind action on signposts and billboards can be evaluated according to EN 1991-1-4:2005 (Eurocode 1 Part 4). ASCE 7-10 code of practice can also be used for the evaluation of wind load on sign structures. The National Annex to BS EN 12899-1:2007 recommends suitable wind loads for the majority of signs in the UK. Whilst is it more conservative than performing a full analysis, it is simpler and quicker.

Other forces that may need to be taken into account when designing sign structures are point loads and dynamic snow load (not applicable in Nigeria). The UK National Annex recommends that signs should be able to withstand a force of 0.5 kN applied at any point. This represents the load that might be exerted by, for example, a glancing blow from a vehicle mirror, a falling branch or malicious interference with the sign. This point load is the critical factor only for very small signs, but for signs mounted on a single support, it causes torsional forces that need to be considered.

For large billboards, live loads and the weight of services should be accounted for the in the design.

Design example
Provide adequate sections for a sign structure with the configuration shown below. The sign post is located in an area that is 76 m above sea level with a wind speed of 35 m/s.

Mounting height above ground h_m = 2.0 m
Width of sign face l = 3.0 m
Height of sign face b = 2.0 m
Total height H = h_m + b = 4.0 m
Height to centroid of sign area z = h_m + b/2 = 3.0 m
Depth of post buried above foundation h_b = 200 mm

Basic wind velocity
Let the basic wind velocity from wind map = v_b,map = 35 m/s
The altitude of site above sea level A = 76 m
Altitude factor c_alt = 1 + 0.001A = 1 + (0.001 × 76) = 1.076
v_b,0 = v_b,map ⋅ c_alt= 35 × 1.076 = 37.66 m/s

Assess Terrain Orography
Site is not very exposed site on cliff/escarpment or in a site subject to local wind funnelling. Therefore c_o = 1.0

Determine Design Life Requirement

p = design annual probability of exceedence
p = 1/design life = 1/25 = 0.04 (for signs design life is 25 years)
K = Shape parameter = 0.2
n = exponent = 0.5

Basic Wind Velocity
v_b = c_dir ⋅ c_season ⋅ v_b,0
c_dir = directional factor = 1.0
c_season = season factor = 1.0
v_b = 1.0 × 1.0 × 37.66 = 37.66 m/s

10 minute mean wind velocity having probability P for an annual exceedence is determined by:

v_{b,25 years}= v_b ⋅ c_prob
v_{b,25 years}= 37.66 × 0.96 = 36.15 m/s

Mean Wind
The mean wind velocity V_m(z) at a height z above the terrain depends on the terrain roughness and orography, and on the basic wind velocity, V_b, and should be determined using the expression below;

V_m(z) = c_r(z). c_o(z).V_b

Where;
c_r(z) is the roughness factor (defined below)
c_o(z) is the orography factor often taken as 1.0

c_r(z) = k_r. In (z/z₀) for z_min ≤ z ≤ z_max
c_r(z) = c_r.(z_min) for z ≤ z_min

Where:
Z₀ is the roughness length
k_r is the terrain factor depending on the roughness length Z₀ calculated using;

k_r = 0.19 (Z₀/Z_0,II)^0.07

Where:
Z_0,II = 0.05m (terrain category II)
Z_min is the minimum height = 2 m
z = 3 m
Z_max is to be taken as 200 m
K_r = 0.19 (0.05/0.05)^0.07 = 0.19
c_r(3) = k_r.In (z/z₀) = 0.19 × In(3/0.05) = 0.78

Therefore;
V_m(3.0) = c_r(z). c_o(z).V_b = 0.78 × 1.0 × 36.15 = 28.197 m/s

Wind turbulence
The turbulence intensity I_v(z) at height z is defined as the standard deviation of the turbulence divided by the mean wind velocity. The recommended rules for the determination of I_v(z) are given in the expressions below;

I_v(z) = σ_v/V_m = k_l/(c₀(z).In (z/z₀)) for z_min ≤ z ≤ z_max
I_v(z) = I_v.(z_min) for z ≤ z_min

Where:
k_l is the turbulence factor of which the value is provided in the National Annex but the recommended value is 1.0
C_o is the orography factor described above
Z₀ is the roughness length described above.

For the structure that we are considering, the wind turbulence factor at 3 m above the ground level;

I_v(60) = σ_v/V_m = k₁/[c₀(z).In(z/z₀)] = 1/[1 × In(3/0.05)] = 0.244

Peak Velocity Pressure
The peak velocity pressure q_p(z) at height z is given by the expression below;

q_p(z) = [1 + 7.I_v(z)] 1/2.ρ.V_m²(z) = c_e(z).q_b

Where:
ρ is the air density, which depends on the altitude, temperature, and barometric pressure to be expected in the region during wind storms (recommended value is 1.25kg/m³)

c_e(z) is the exposure factor given by;
c_e(z) = q_p(z)/q_b
q_b is the basic velocity pressure given by; q_b = 0.5.ρ.V_b²

q_p(60m) = [1 + 7(0.244)] × 0.5 × 1.25 × 28.197² = 1345.66 N/m²

Therefore, q_p(3m) = 1.345 kN/m²

Determination of force coefficient (Table NA 2 BS EN 12899)
λ = effective slenderness ratio of sign or aspect ratio
λ = l/b = 3.0 / 2.0 = 1.5
Therefore c_f= 1.30

Calculation of the total wind force (Clause 5.3 of EN 1991-1-4)
F_w = c_sc_d ⋅ cf ⋅ q_p(z_e) ⋅ A_ref (A_ref = area of sign)
c_sc_d = 1.0 (for sign posts)

F_w = 1.0 × 1.30 × 1.345 × 3.0 × 2.0 = 10.5 kN

Partial Factor for Action γ_F (Table 6 EN 12899-1:2007 (E))
ULS (bending and shear) γ_F = 1.5
SLS (deflection) γ_F = 1.0
γ_f3 = 1.0

Design Wind Force on the sign
F_w,d = F_w ⋅ γ_F ⋅ γ_f3
F_w,d (ULS) = 10.5 × 1.5 × 1.0 = 15.75 kN
F_w,d (SLS) = 10.5 × 1.0 × 1.0 = 10.5 kN

Ultimate Action Effects
Ultimate design bending moment per post, M_Ed
M_Ed = Wind force × lever arm to foundation / number of posts
M_Ed = F_w,d (ULS) ⋅ (z + h_b) / n
M_Ed = 15.75 × (3.00 + 0.2) / 2 = 25.2 kNm

Ultimate design shear per post
V_Ed = Wind force / number of posts
V_Ed = F_w,d(ULS) /2 = 15.75 / 2 = 7.9 kN

The 0.5 kN point load on the sign is not critical, since it is less than the wind action and there are no torsional effects with 2 posts.

Try circular hollow section CHS 139.7 x 8 (S355)

A = 33.1 cm²; W_pl = 139 cm³; I_x = 720 cm⁴

Section classification
ε = √235/f_y = √(235/355) = 0.81
Tubular sections (Table 5.2, sheet 3 of EN 1993-1-1:2005):
d/t = 139.7/8 = 17.46
Limit for Class 1 section = 50ε² = 40.7
40.7 > 17.46; section is Class 1

Member resistance at ULS
According to Table 7 of EN 12899-1:2007 (E), the material factor of safety for steel is γ_m = 1.05
Moment Capacity M_Rd= f_y⋅W_pl/γ_m = [(355 x 10³ x 139 x 10^-6)/√3)]/1.05= 46.99 kNm
M_Ed/M_Rd = 25.2/46.99 = 0.536 < 1.0 Okay

Shear capacity V_Rd = A_v(f_y/√3)/γ_m
Av = 2A/π = (2 x 33.1)/π = 21.027 cm²
V_Rd = 21.027 × 10^-4× [(355 x 10³)/√3)]/1.05 = 430.96 kN
VEd/VRd = 7.9/430.96 = 0.018 < 1.0 Okay

Calculation of Temporary deflection
The wind velocity for calculating the temporary deflection (SLS) criterion is 75% of the reference wind velocity, as it is based upon a 1 year mean return period. The 0.96 factor below reverses the c_prob conversion from 50 to 25 year return period used above (Clause 5.4.1, note 1 EN 12899-1).

F_{wd(1 year)} = F_wd(SLS) x 0.75²/0.96² = 10.5 x 0.75²/0.96² = 6.41 kN

Uniformly distributed load along sign face = F_{w,d (1 year)} / b
F_{w,d (1 year)} / b = 6.41/2.0 = 3.2 kN/m
where b = height of the sign face

Maximum deflection at top of sign (bending), δ

E = 210000 N/mm²
I = 720 cm⁴
n = number of posts = 2
Other parameters are as defined above

δ = [3.2/(24 x 210000 x 720 x 10⁴ x 2)] x [3(4000 + 200)⁴– 4(2000 + 200)³ x (4000 + 200) + (2000 + 200)⁴] = 34.3 mm

Deflection per linear metre = δ’ = δ/(H + h_b) = 34.3/(4 + 0.2)= 8.16 mm/m
Maximum temporary deflection taken as class TDB4 = 25 mm/m
8.16 mm/m < 25 mm/m. Therefore deflection is okay.

References
The Institution of Highway Engineers (2010): SIGN STRUCTURES GUIDE Support design for permanent UK traffic signs to BS EN 12899-1:2007 and structural Eurocodes

Structural Design of Corbels

By

Ubani Obinna

-

July 31, 2020

A corbel is a very short structural cantilever member projecting from a wall or a column for the purpose of carrying loads. In reinforced concrete structures, corbels are cast monolithically with the walls or columns supporting them. They are found mainly in bridges, industrial buildings, and commercial buildings with precast construction.

corbel 2 — Corbels in precast construction

If a structure is subjected to in-plane forces, then the stress distribution will consist of normal stresses in the two planes (σ_x and σ_y) and an accompanying shear stress τ_xy. These stresses will lead to two principal stresses σ₁ and σ₂. If for instance σ₁ is tensile and σ₂ is compressive, then the external load on the structure can be idealized as being resisted by a combination of tensile and compressive stresses.

In reinforced concrete structures, the compressive stresses can be resisted by the concrete, while the tensile stresses can be resisted by the steel reinforcement. Generally, these are idealized by a series of concrete struts and steel ties, and forms the underlying principle behind the strut-and-tie method of design. Structures that can be designed using this method are pile caps, deep beams, corbels, half-joints, etc.

Under certain conditions, Eurocode 2 permits the use of strut and tie method for the analysis and design of corbels. According to clause 6.5.1 of EN 1992-1-1:2004, strut-and-tie method can be used where non-linear stress distribution exists (for example at supports, near concentrated loads or plain stress).

Strut-and-tie models utilize the lower-bound theorem of plasticity which can be summarized as follows: for a structure under a given system of external loads, if a stress distribution throughout the structure can be found such that;

(1) all conditions of equilibrium are satisfied and
(2) the yield condition is not violated anywhere,

Then the structure is safe under the given system of external loads.

When using the strut−and-tie method of design, because the given structure is replaced by a pin-jointed truss, different types of nodes occur where members meet. These nodal zones need to be carefully designed and detailed.

(a) CCC Node: If three compressive forces meet at a node, it is called a CCC node. According to Eurocode 2 equation (6.60), the compression stress in each strut is restricted to a maximum value of 1.0(1 – f_ck/250)f_cd.

(b) CCT node: If two compressive forces and a tie force anchored in the node through bond meet, it is called a CCT node. According to Eurocode 2 equation (6.61), the compression stress in each strut is restricted to a maximum value of 0.85(1 – f_ck/250) f_cd.

(c) CTT node: If two tensile forces at a compressive force meet at a node it is called a CTT node. According to Eurocode 2 equation (6.62),
the compression stress in each strut is restricted to a maximum value of 0.75(1 – f_ck/250)f_cd.

strut tie model for corbel — Strut−tie model for a corbel with different types of nodes.

The following conditions are to be applied in the design of corbels;

(1) Corbels with 0.4h_c < a_c < h_c may be designed using a simple strut and tie model.
(2) For deeper corbels (a_c < 0.4h_c), other adequate strut and tie models may be considered.
(3) Corbels for which a_c > h_cmay be designed as cantilever beams
(4) Unless special provision is made to limit horizontal forces on the support, or other justification is given, the corbel should be designed for the vertical force F_v, and a horizontal force H_c > 0.2F_v acting at the bearing area.
(5) The overall depth (h_c) of the corbel should be determined from considerations of shear.
(6) The local effects due to the assumed strut and tie system should be considered in the overall design of the supporting member.
(7) The detailing requirements shall be met.

Design Example of a thick short corbel (a_c < h_c/2)

Design the corbel shown below. The bearing area of the support is 200 x 300 mm. The width of the corbel is 150 mm.

Materials
Concerete C30/35, f_ck= 30 N/mm²; f_yk = 500 N/mm²

Design strength of concrete f_cd= α_ccf_ck/γ_c = (0.85 x 30)/1.5 = 17 N/mm²
Design strength of steel f_yd= f_yk/γ_s = 500/1.15 = 435 N/mm²

Compressive strength of nodes;

CCC nodes:
σ_1Rd,max = 1.0 x (1 – f_ck/250)f_cd = 1.0 x (1 – 30/250) x 17 = 14.96 N/mm²

CCT nodes:
σ_2Rd,max = 0.85 x (1 – f_ck/250)f_cd = 0.85 x (1 – 30/250) x 17 = 12.716 N/mm²

CTT nodes:
σ_3Rd,max = 0.75 x (1 – f_ck/250)f_cd = 0.75 x (1 – 30/250) x 17 = 11.22 N/mm²

Actions
F_Ed = 600 kN

Load eccentricity with respect to the column side: e = 125 mm
The beam vertical strut width is evaluated by setting the compressive stress equal to σ_1Rd,max:

x₁ = F_Ed/σ_1Rd,maxb = (600 x 10³)/(14.96 x 350) = 114.6 mm
Node 1 is therefore located at x₁/2 = 114.6/2 = 57.3 mm

Let the cover from the top of the cantilever to the reinforcements be 30 mm. Assuming 16 mm bars;

The effective depth d = 450 – 35 – (16/2) = 407 mm

The distance y₁ of the node 1 from the lower border is evaluated setting the internal drive arm z equal to 0.8⋅d (z = 0.8 x 407 = 325.6 mm):

y₁ = 0.2d = 0.2 x 407 = 81.5 mm

Rotational equilibrium: F_Eda = F_cz
a = ac + x₁/2 = 125 + 57.3 = 182.3 mm
600 x 10³ x 182.3 = F_c x 325.6
F_c = F_t = 335933.66 N = 335.93 kN

Node 1 verification;
σ = F_c / (b∙2y₁) = (335.93 x 10³)/(350 x 2 x 81.5) = 5.89 N/mm² < σ_1Rd,max (14.96 N/mm²)

Main Top Reinforcement Design
A_s1 = F_t/f_yd = (335.93 x 10³)/435 = 772 mm²
Provide 5H16 (A_s,prov = 1005 mm²)

Shear reinforcement design
The beam proposed in EC2 is indeterminate, then it is not possible to evaluate the stresses for each single bar by equilibrium equations only, but we need to know the stiffness of the two elementary beams shown below in order to make the partition of the diagonal stress [F_diag = F_c/cosθ = F_Ed/sinθ] between them.

Based on the trend of main compressive stresses resulting from linear elastic analysis at finite elements, some researchers from Stuttgart have determined the two rates in which F_diag is divided, and they have provided the following expression of stress in the secondary reinforcement.

truss models — Strut-and-tie model resolution in two elementary beams and partition of the diagonal stress Fdiag.

F_wd = [(2z/a – 1)/(3 + F_Ed/F_c)] x F_c = [(2(325.6/182.3) – 1)/(3 + 600/335.93)] x 335.93 = 180.525 kN

A_sw= F_wd/f_yd= (180.525 x 10³)/435 = 415 mm² > k₁A_s,prov = 0.25 x 1005 = 251.25 mm²

Trying 2 legs of H10 mm, Provide 4 number of H10 mm stirrups (A_sw,prov = 628 mm²)

Node 2 verification, below the load plate
The node 2 is a tied-compressed node, where the main reinforcement is anchored; the compressive stress below the load plate is;
σ = F_Ed /(200 x 300) = (600 x 10³)/(200 x 300) = 10 N/mm² < σ_2Rd,max (12.716 N/mm²) Okay

Detailing Requirements

(1) The reinforcement, corresponding to the ties considered in the design, should be fully anchored beyond the node under the bearing plate by using U-hoops or anchorage devices, unless a length l_b,net is available between the node and the front of the corbel, l_b,net should be measured from the point where the compression stresses change their direction.

(2) In corbels with h_c > 300 mm, when the area of the primary horizontal tie A_s is such that (where A_cdenotes the sectional area of the concrete in the corbel at the column), then closed stirrups, having a total area not less than |0.4|As, should be distributed over the effective depth d in order to cater for splitting stresses in the concrete strut. They can be placed either horizontally or inclined.

Detailing 1 — Typical detailing of a corbel when the diameter of the main bars is 20mm or greater

detailing 2 — *Typical detailing of a corbel when the diameter of the main bars is 16mm or smaller*

P-Delta Analysis

By

Ubani Obinna

-

July 29, 2020

P-delta is a geometric non-linear effect that occurs in structures that are subjected to compressive loads and lateral displacement. Under the action of compressive loads and lateral displacement, tall slender structures will experience additional stresses and deformations due to the change of position of the structure. First-order structural analysis will normally consider small displacements and will compute the equilibrium of the structure and internal stresses based on the undeformed geometry. However, second-order analysis considers the deformed geometry of the structure and may require an iterative approach for the computation of equilibrium.

During the simultaneous action of vertical and horizontal loads, the structure deflects due to the action of the horizontal load. As the structure deflects, the position of the vertical load P shifts by a distance ∆ such that the vertical load instead of acting axially along the column now induces a moment reaction at the base P∙∆. The interaction of the compressive force (P) and lateral displacement (∆) to produce additional secondary effects in a structure is handled under P-delta analysis.

Solved Example

Let us quickly investigate the effect of P-delta on the cantilever column loaded as shown below.

Let;
P = 150 kN
H = 20 kN

1st iteration
First-order moment at the base of the cantilever M = H x L = 20 kN x 7.5m = 150 kNm
So, the lateral displacement (Δ1) at the column tip = ML2/3EI = (150 x 7.5²)/(3 x 210 x 10⁶ x 4.09 x 10^-4) = 0.0327 m

Also, the vertical load P acting on displaced Δ₁ column tip generates additional moment M₁ at the base.
M₁ = P x Δ₁= 150 x 0.0327 = 4.905 KNm

The total moment at the base M_T1 = M + M₁ = 150 + 4.905 = 154.905 kNm

The modified horizontal displacement “Δ₂” undergone for the first modified moment M_T1
Δ₂= (M_T1L²)/ (3EI) =(154.905 x 7.5²)/(3 x 210 x 10⁶ x 4.09 x 10^-4) = 0.0338 m

2nd iteration
The vertical load P acting on the newly displaced column tip (Δ₂– Δ₁) results in additional moment M₂ at the base;

M₂ = P x (Δ₂– Δ₁) = 150 x (0.0338 – 0.0327) = 0.165 KNm
Now, the total moment at the base, M_T2= (M + M₁ + M₂) = (M_T1 + M₂) = 154.905 + 0.165 = 155.07 KNm

The second modified horizontal displacement Δ₃against the second modified Moment M_t2
Δ₃= (M_T2L²)/3EI = (155.07 x 7.5²)/(3 x 210 x 10⁶ x 4.09 x 10^-4) = 0.0338 m

For all practical purposes, the analysis above will be taken to have converged.

There are two types of P-Delta effects which are;

P-“large” delta (P-∆) – a structure effect
P-“small” delta (P-δ) – a member effect

While large delta is associated with the global displacements and effects on the structure, small delta takes into account the local effects and displacements of the members. The sensitivity of a structure to P-Delta effect is related to the magnitude of the axial load P, stiffness/slenderness of the structure as a whole, and slenderness of individual elements. Second-order effects are prominent in structures under high compressive load with low stiffness or high slenderness. The closer the compressive load is to the critical buckling load, the more sensitive the structure is to P-delta effects.

P-delta analysis combines two approaches to reach a solution which are;

Large-displacement theory, and
Stress stiffening

In large-displacement theory, the resulting forces and moments take full account of the effects due to the deformed shape of both the structure and its members. The large-displacement theory identifies the fact that there is a clear distinction between the deformed and undeformed shapes of the structure. In small displacement theory, the strains and the displacements are small, while large-displacement theory assumes small strains in the structure but large displacements.

Stress stiffening is concerned with the effect of stress state on a structure’s stiffness. For instance, tensile loads straighten the geometry of an element thereby stiffening it. On the other hand, compressive loads accentuate deformation thereby reducing the stiffness of the element. The effect of stress stiffening is accounted for by generating and then using an additional stiffness matrix (stress stiffness matrix) for the structure. The stress stiffness matrix is added to the regular stiffness matrix in order to give the total stiffness.

Second-order effects have been incorporated in many codes of practice for the design of concrete, timber, and steel structures. This has usually been included in the design equations and verifications to ensure that the behaviour of the structure stays within the prescribed limits. However, with the availability of advanced computational power of software, engineers are now encouraged to carry out full second-order analysis for design purposes.

P-Delta Analysis on Staad Pro

The procedure adopted by Staad Pro for P-delta analysis are as follows;

First, the primary deflections are calculated based on the provided external loading.
Primary deflections are then combined with the originally applied loading to create the secondary loadings. The load vector is then revised to include the secondary effects.
A new stiffness analysis is carried out based on the revised load vector to generate new deflections.
Element/Member forces and support reactions are calculated based on the new deflections.

For proper analysis of P-delta effect in Staad, it is important that the vertical and horizontal load cases be combined under one single load case. This can be achieved using the ‘REPEAT LOAD CASE’, which allows the user to combine previously defined primary load cases to create a new primary load case. A REPEAT LOAD is treated as a new primary load which enables the P-Delta analysis to reflect the correct secondary effects. It is important to note that ‘LOAD COMBINATIONS’ command will algebraically combine the results such as displacements, member forces, reactions, and stresses of previously defined primary loadings evaluated independently. Therefore, it is not suitable for P-Delta analysis.

Furthermore, one can perform the P-Delta Analysis in Staad directly by considering the Geometric Stiffness Matrix [K_G]. In this approach, the stress stiffening effect due to the axial stress is used directly to modify the actual Stiffness Matrix [K]. In the view of this approach, the compressive force (depending on its magnitude) reduces the lateral load-carrying capacity of the structure. This ultimately modifies the geometric stiffness of the member and is referred to as ‘stress softening effect’. In Staad, the change in the GEOMETRIC STIFFNESS from [K] to [K + K_G] can be achieved by using the K_G option as shown below.

So, by this approach the P-Delta stiffness equation is directly linearized by the [K + K_G] matrix and the solution can be obtained directly and exactly, without iteration.

According to Staad Pro technical manual, there are two options in carrying out P-Delta analysis on Staad;

(1) When the CONVERGE command is not specified: The member end forces are evaluated by iterating “n” times. The default value of “n” is 1 (one).

(2) When the CONVERGE command is included: The member end forces are evaluated by performing a convergence check on the joint displacements. In each step, the displacements are compared with those of the previous iteration in order to check whether convergence is attained. In case “m” is specified, the analysis will stop after that iteration even if convergence has not been achieved. If convergence is achieved in less than “m” iterations, the analysis is terminated. (DO NOT ENTER “n” when CONVERGE is provided).

(3) To set convergence displacement tolerance, enter SET DISPLACEMENT f command. Default is maximum span of the structure divided by 120.

Solved Example

Let us consider the portal frame loaded as shown shown below.

Action Effect	1st order linear analysis	P-delta (1 iteration)	P-delta (2 iterations)	P-delta (5 iterations)	P-delta (10 iterations)	P-delta (15 iterations)
Displacement (mm)	126.204	143.634	146.602	147.235	147.239	147.239
Maximum bending Moment (kNm)	406.704	425.321	427.837	428.322	428.325	428.325
V_B (kN)	128.343	128.343	128.343	128.343	128.343	128.343
H_B (kN)	58.101	58.101	58.101	58.101	58.101	58.101

As can be seen from the table above, 5 to 10 iterations is sufficient for second-order analysis of portal frames for practical purposes. While internal forces and displacements varied with second-order analysis, the equilibrium of the structure was maintained.

Plates on Elastic Foundation

By

Ubani Obinna

-

July 27, 2020

Many problems of practical importance can be related to the solution of plates resting on an elastic foundation. Reinforced concrete pavements of highways and airport runways, raft foundation slabs of buildings, bases of water tanks and culverts etc., are well-known direct applications. Just like beams on elastic foundation, it is also based on the assumption that the foundation’s reaction q(x, y) can be described by the following relationship;

q(x, y) = kw

Where k is a constant termed the modulus of subgrade reaction, which has the unit (kN/m²/m), and q(x, y) is the resisting pressure of the foundation, and w is the deflection of the plate.

rigid pavement — Rigid pavement can be idealised as plate on elastic foundation

When the plate is supported by a continuous elastic foundation, the external load acting in the lateral direction consists of the surface load p(x, y) and of the reaction of the elastic foundation q(x, y). Thus, the differential equation of the plate becomes the following:

(∂⁴w)/∂x⁴+ 2(∂⁴w)/(∂x²∂y²) + (∂⁴w)/∂y⁴ = (1/D) x [p(x,y) – q(x,y)]

In this differential equation, the reactive force, q(x, y) exerted by the elastic foundation is also unknown, because it depends on the deflection, w(x, y) of the plate.

D∇² ∇²w + kw = p

This equation can be solved using the classical methods developed by Navier and Levy. Note that D is the flexural rigidity of the plate, and it is given by;

D = Eh³/[12(1 – μ²)]

The different methods that can be employed in the analysis of plates on elastic foundation are;

Classical methods
Finite difference methods
Finite element method

Cylindrical bending of plates on elastic foundation

Let us consider the cylindrical bending of a thin plate on an elastic foundation, and rigidly supported at the edges as given in Timoshenko.

plate on elastic foundation 2 — Cylindrical plate on elastic foundation

Cutting out an elemental strip, we may consider it as a beam on an elastic foundation. Assuming that the reaction of the foundation at any point in time is proportional to the deflection w at that point, we can obtain the equation given below;

D(∂⁴w)/∂x⁴ = q – kw

Introducing the notation;
β =L/2 ∜(k/4D)

The general solution to the equation above can be written as;

w = q/k + C₁sin2βx/L sinh2βx/L + C₂sin2βx/L cosh2βx/L + C₃cos2βx/L sinh2βx/L + C₄cos2βx/L cosh2βx/L —- (1)

Where the four constants of integration must be determined from the end conditions of the strip. With the case under consideration, we can assume that the deflection is symmetrical with respect to the middle strip. Taking the coordinate axis as shown in the figure above, we can conclude that C₂ = C₃ = 0. The constants C₁ and C₄ are found from the conditions that the deflection and bending moment of the strip are zero at the end (x = L/2). Hence;

w_(x=l/2) = 0

(d²w)/dx²)_{(x = l/2)} = 0

Substituting expressions (a) for w and observing that C₂ = C₃ = 0, we obtain;

q/k + C₁sinβsinhβ + C₄cosβcoshβ = 0
C₁cosβcoshβ – C₄sinβsinhβ = 0

From which we find;
C₁ = -q/k (2sinβsinhβ)/(cos2β + cosh 2β)
C₂ = -q/k (2cosβcoshβ)/(cos2β + cosh 2β)

On substituting into equation (1);

w = qL⁴/64Dβ⁴ ∙{1 – [(2sinβsinhβ)/(cos2β + cosh2β) ∙sin2βx/L∙sinh2βx/L] – [(2cosβcoshβ )/(cos2β + cosh2β)∙cos2βx/L∙cosh2βx/L)]} — (2)

The deflection at the middle is obtained by substituting x = 0, which gives;

w_(x=0) = (5qL⁴/384D)∙φ(β)

Where;
φ(β) = 6/5β⁴∙[1 – (2 cosβcoshβ )/(cos2β + cosh 2β)]

To obtain the angles of rotation of the edges of the plate, we differentiate expression (2) with respect to x and put x = -L/2.

In this way we obtain;

(dw/dx)_{(x= -l/2)}= (5qL⁴/384D)∙φ₁(β)

Where φ₁(β) = 3/4β³ ∙ [1 – (sinh⁡2β – sin2β)/(cosh2β + cos2β)]

The bending moment at any cross-section of the strip is obtained from the equation;

M = -D(d²w/dx²)

Substituting expression (D) for w, we find for the middle strip;

M_(x=0) = (qL²/8)∙ φ₂(β)
Where φ₂(β) = 2/β²[1 – (sinhβ – sinβ)/(cosh2β + cos 2β)]

To simplify the calculation of deflection and stresses, numerical values of functions of φ, φ₁, and φ₂ are presented in the Table below;

Plates on Elastic Foundation Using Finite Element Analysis

Software like Staad Pro can be used in the analysis of plates on elastic foundation. The general approach to solving such problems is to sub-divide the slab into several plate elements. Each node of the meshed slab will then have an influence area or a contributory area, which is to say that soil within the area surrounding that node acts as a spring. The influence area is then multiplied by the subgrade modulus to arrive at the spring constant. Subgrade modulus has units of force per length³. So, the spring will have units of force/length.

plate on elastic foundation model — *Typical model of a plate on grade supported on soil springs*

The influence area is calculated automatically in Staad Pro using the ‘Foundation’ type of support. The two options available for doing this are of doing this are;

ELASTIC MAT OPTION, and
PLATE MAT OPTION

The elastic mat method calculates the influence area of the various nodes using the Delaunay triangle method. The distinguishing aspect of this method is that it uses the joint-list that accompanies the ELASTIC MAT command to form a closed surface. The area within this closed surface is then determined and the share of this area for each node in the list is then calculated. Without a properly closed surface, the area calculated for the region may be indeterminate and the spring constant values may be erroneous.

If the foundation slab is modeled using plate elements, the influence area can be calculated using the principles used in determining the tributary area of the nodes from the finite element modeling standpoint. In other words, the rules used by the program in converting a uniform pressure load on an element into fixed end actions at the nodes are used in calculating the influence area of the node, which is then
multiplied by the subgrade modulus to obtain the spring constant.

Solved Example

A 150 mm thick rectangular slab of dimensions 6m x 5m is resting on a soil of modulus of subgrade reaction k_s= 30000 kN/m²/m. Evaluate the response of the slab when subjected to a full pressure load of 35 kN/m² all over the surface and a concentrated load of 200 kN at the centre.

Solution

The meshing of the plate has been carried out as follows;
In the longer direction: 24 divisions
In the shorter span: 20 divisions
Size of each finite element = 250 mm x 250 mm

The modelling of the foundation is shown below;

The loading of the plate is shown below.

Analysis Result

(a) Settlement

Deformed profile under concentrated load

Maximum settlement under concentrated load = 1.912 mm

Deformed profile under uniformly distributed pressure load

Maximum settlement under uniform pressure load = 1.167 mm

(b) Base Pressure

BASE PRESSURE UNDER CONCENTRATED LOAD — *Base pressure under concentrated load*

base pressure under UDL — *Base pressure under UDL*

(c) Bending Moment

MX P — *Transverse bending moment under concentrated load*

mx q — *Transverse bending moment under UDL*

my p — *Longitudinal bending moment under concentrated load*

my q — *Longitudinal bending moment under UDL*

Modulus of Subgrade Reaction of Soils

By

Peace Ikpotokin

-

July 27, 2020

The modulus of subgrade reaction k_s (also called the coefficient of subgrade reaction of soil) is the ratio of the pressure against a flat surface on soil and the settlement at that point. Mathematically, this is expressed as;

k_s = q/δ ——- (1)

Where;
k_s = Coefficient of subgrade reaction expressed in force/length²/length
q = pressure on the surface at the given point
δ = settlement at the same point

Against many popular opinions, the modulus of subgrade reaction is not an exclusive property of soil but depends mainly on the loaded area (size of the footing or mat). Other factors that affect the modulus of subgrade reaction are the shape of footing, depth of foundation, type of soil, and type of foundation. Different approaches have been suggested for the evaluation of the modulus of subgrade reaction, but the most practical approach is to carry out an in-situ plate load test on the soil.

Plate bearing test is an in-situ load-bearing test that is used to evaluate the ultimate bearing capacity and likely settlement of soil under a given load. The test can be carried out in accordance with BS 1377 Part 9: 1990. It basically consists of loading a steel plate of known diameter and recording the settlements corresponding to each load increment.

The test load is gradually increased till the plate starts to settle at a rapid rate. The total value of the load on the plate divided by the area of the steel plate gives the value of the ultimate bearing capacity of the soil. A factor of safety is applied to give the safe bearing capacity of the soil.

experiment for determination of modulus of subgrade reaction — **Fig 1:** Typical plate load test set-up

By implication, the coefficient of subgrade reaction is the unit pressure required to produce a unit settlement in soil. In saturated clay soils, the settlement under load will take time due to consolidation, so the coefficient of subgrade reaction should be determined on the basis of the final settlement.

On granular soil deposits, the settlement should take place immediately after the application of load. Therefore, the modulus of subgrade reaction is premised on two simplified assumptions which are;

The value of k_s is independent on the magnitude of the pressure
The value of k_s has the same value for every point on the surface of the footing

These two assumptions are however not strictly correct.

Terzaghi in 1955 presented empirical relationships for determining the coefficient of subgrade reactions (k_sf) for full-scale foundations, based on results from plate load tests. This is based on a rigid 1 ft x 1 ft (0.305 m x 0.305m) slab placed on a soil medium.

(1) For a square footing on cohesionless soil (B x B);

k_sf = k_s[(B + 0.305)/2B]² ——- (2)

(2) For a rectangular footing on cohesionless soil (B x L)

k_sf,r = [k_s_f(1 + 0.5B/L)]/1.5 ——- (3)

(3) For a long strip footing of width B, the coefficient of subgrade reaction is approximately 0.67k_sf

(4) For clay soils, it has been observed that the value of k_s varies with the length of the footing. Therefore, for clays;

k_sc = k_s[(L + 0.152)/1.5L)] ——- (4)

Where;

k_s = plate-load test value of modulus of subgrade reaction kN/m²/m, using square plate (1 × 1) ft or circular plate with diameter = 0.305 m
k_sc = corrected plate-load test value of modulus of subgrade reaction kN/m²/m, using square plate (1 × 1) ft or circular plate with diameter = 0.305 m for clays
k_sf = desired value of modulus of subgrade reaction for full-sized square footings B × B, kN/m²/m
k_sf,r = desired value of modulus of subgrade reaction for rectangular full-sized footings B × L, kN/m²/m
B = footing width, or least dimension of rectangular or strip.
L = Length of footing

Some formulas suggested by different authors for evaluation of the modulus of subgrade reaction are given in the Table below;

Author	Year	Suggested formula
Terzaghi	1955	k_sf = k_s[(B + B’)/2B]²
Vesic	1961	k_s = [0.65E_s/B(1 – μ²)] x (E_sB⁴/EI)^1/12
Meyerhof and Baike	1965	k_s = E_s/[B(1 – μ²)]
Selvadurai	1985	k_s = 0.65/B x [E_s/(1 – μ²)]
Bowles	1998	k_s = E_s/[B'(1 – μ²) x mI_sI_f]

k_s = the coefficient of subgrade reaction. B’ = side dimension of square base used in the plate load test. B = width of footing. ks = the value of subgrade reaction for 0.3 × 0.3 (1 ft wide) bearing plate. k_sf= value of modulus of subgrade reaction for the full-size foundation. E_s = modulus of elasticity. μ = Poisson’s ratio. EI = flexural rigidity of footing, m = takes 1, 2 and 4 for edges, sides and center of footing, respectively. I_s and I_f = influence factors depend on the shape and depth of footing.

One of the most popular relationships between allowable bearing capacity and modulus of subgrade reaction is given in equation (5) according to Bowles (1996);

k_s = 40.(FS).(q_a) ——- (5)

Where q_a is the allowable bearing capacity of the soil, and FS is the factor of safety that was used in converting the ultimate pressure (q_ult) to allowable pressure (q_a). It is important to note that in equation (5), the author assumed a 25 mm settlement value for the soil.

Ping-Sien Lin, Li-Wen Yang, and C. Hsein Juang (1998) made a series of plate-load tests to investigate the load settlement characteristics of a gravelly cobble deposit and estimated the value of modulus of subgrade reaction k_s as follows:

k_s = q_a/δ_a ——- (6)

Where;
k_s = Coefficient of subgrade reaction
q_a= Allowable bearing capacity = (ultimate bearing capacity/factor of safety)
δ_a = Allowable settlement

Typical values of modulus of subgrade reaction

The typical values of modulus of subgrade reaction of different types of soils are given below;

Soil description	Modulus of subgrade reaction (kN/m²/m)
Humus soil or peat	5000 – 15000
Recent embankment	10000 – 20000
Fine or slightly compacted soil	15000 – 30000
Well compacted sand	50000 – 100000
Very well-compacted sand	100000 – 150000
Loam or clay (moist)	30000 – 60000
Loam or clay (dry)	80000 – 100000
Clay with sand	80000 – 100000
Crushed stone with sand	100000 – 150000
Coarse crushed stone	200000 – 250000
Well compacted crushed stone	200000 – 300000

Types of Floor Systems for Steel Framed Buildings

By

Ivy Ugorji

-

July 27, 2020

Table of Contents

Floors in buildings have a primary function of carrying loads and supporting the activities of the occupants. In addition to carrying loads, floors in buildings also provide the needed rigid diaphragm action for transmitting horizontal loads to the stabilising vertical components. Furthermore, floors also support additional superimposed loads such as ceilings, building services, and finishes such as screeds and tiles.

composite slab supporting building services — Composite slab supporting building services

The types of floor systems that used in steel-framed buildings are;

Short-span composite beams and composite slabs with metal decking.
Slimdek^®.
Cellular composite beams with composite slabs and steel decking.
Slimflor^® beams with precast concrete units.
Long-span composite beams and composite slabs with metal decking.
Composite beams with precast concrete units.
Non-composite beams with precast concrete units.

Short span composite beam and composite slab with metal decking

In this floor system, shear connectors are welded through the metal decking to the top flange of downstand beams to enable it act compositely with an in-situ composite slab. For short span floor systems, the secondary beams are typically spaced between 3 – 4m and are supported by the primary beams. The primary and secondary beams act compositely with the composite slab, but the edge beams are usually non-composite. At 3-4m spacing, secondary beams will span about 6 – 7.5m when positioned orthogonally to the slab, while the primary beams will span about 6-9m (positioned parallel to the slab).

short span composite slab section — Short span composite decking

The floor slab consists of composite profiled metal decking with a typical depth of about 130 mm thick with in-situ concrete topping. The profiles may be re-entrant decking or trapezoidal. Re-entrant decking uses more concrete than trapezoidal decking, but has increased fire resistance for a given slab depth. Trapezoidal decking generally spans further than re-entrant decking, but the shear stud resistance is less with trapezoidal decking than with re-entrant decking. The profiles are usually between 0.5 to 1.2mm thick.

rentrant decking — Re-entrant profile composite decking

trapezoidal composite decking — Trapezoidal profile composite decking

Mesh reinforcement is provided at the top of the slab to help reduce cracking, spread localised loads, enhance fire resistance, and act as shear reinforcement around the shear connectors. The decking is normally designed to support the wet weight of the concrete and construction loading as a continuous member over at least two spans, but the composite slab is normally designed as simply supported between beams (but some continuity reinforcement is required). The design of the decking is usually picked from the manufacturer’s technical data sheet.

Advantages
(1) Shallower beams than non-composite floors
(2) More economical
(3) Light weight

Disadvantages
(1) More columns needed than with long-span systems.
(2) Deeper overall floor zone than shallow floor systems.
(3) Generally, beams require fire protection.

Slimdek

Slimdek is a shallow floor system comprising asymmetric floor beams (ASBs) supporting heavily ribbed composite slabs with 225 mm deep decking. ASBs are proprietary beams with a wider bottom flange than top. The section has embossments rolled into the top flange and acts compositely with the floor slab without the need for additional shear connectors.

The decking spans between the bottom flanges of the beams and acts as permanent formwork to support the slab and other loads during construction. The in-situ concrete acts compositely with the decking and encases the beams so that they lie within the slab depth – apart from the exposed bottom flange. The floor normally spans between 6 – 9m grid with floor depth of about 280 – 350 mm. Reinforcements of 16mm to 25mm bars are placed in the ribs of the slab to improve strength in the fire condition, while mesh reinforcement is paced above ASB.

Advantages
(1) Shallow floor zone – reduction in overall building height and cladding.
(2) Virtually flat soffit allows easy service installation and offers flexibility of internal wall positions.

Disadvantages
(1) Steel weight is often greater than other floor systems.
(2) Connections require careful detailing due to the width of the bottom flange.

Cellular composite beams with composite slabs and steel decking

Cellular beams are beams with openings at short regular intervals along their length. The beams are either fabricated from 3 plates or made from rolled sections. Openings, or ‘cells’, are normally circular, which are ideally suited to circular ducts, but can be elongated, rectangular or hexagonal. Cells may have to be filled in to create a solid web at positions of high shear, such as at supports or either side of point loads along the beam.

cellular beam composite slab — Cellular beam composite slab construction

Cellular beams can be arranged as long-span secondary beams, supporting the floor slab directly, or as long-span primary beams which are aligned parallel to the span of the slab supporting other cellular beams or conventional rolled section secondary beams. The secondary beams are typically placed at 3 – 4 m spacing, supported by primary beams on a 6 m, 7.5 m or 9 m column grid. The decking and slab can be designed using decking manufacturer’s design tables or software.

Advantages
Long, column-free floor spans.
Relatively lightweight beams compared with other long-span systems.
Economic long-span solution.
Precamber can be accommodated during the fabrication of the members.
Regular openings in the web allow ducts and other services to pass through the beams.

Disadvantages
Increased fabrication costs compared with plain sections.

Slimflor^® beams with precast concrete units

In slim floor system where the beams are contained within the structural floor depth. A steel plate (typically 15 mm thick) is welded to the underside of a UC section to make the Slimflor beam. This plate extends beyond the bottom flange by 100 mm either side, and supports the precast floor units. A structural concrete topping with reinforcement is recommended to tie the units together and the topping thickness should cover the units by at least 30 mm.

slimfloor section — Typical slimflor arrangement

A composite Slimflor beam can be achieved by welding shear connectors (normally 19 mm diameter by 70 mm long) to the top flange of the UC. Reinforcement is then placed across the flange into slots prepared in the precast units, or on top of shallow precast units. If the steel beams are to be designed compositely, the topping should cover the shear connectors by at least 15 mm, and the precast units by 50 mm.

Only 152 UCs and 203 UCs are normally suitable as composite beams because the overall depth of the floor slab becomes impractical for larger serial sizes. Precast units are usually cambered to cancel out dead load deflections between beams, and the floor spans are typically between 4.5m to 7.5m even though spans of 10m can be achieved.

Advantages
(1) Beams normally require no fire protection for up to 60 minutes fire protection.
(2) Shallow floor zone – reduction in overall building height and cladding. Virtually flat soffit allows easy service installation and offers flexibility of internal wall positions.
(3) Shear connectors can be welded off-site, enabling larger stud diameters to be used and reducing site operations.

Disadvantages
(1) The steelwork is relatively heavy.
(2) Extra fabrication is involved in welding the plate to the UC. Connections require more detailing as the plate is wider than the column.
(3) Precast units involve more individual lifting operations than decking, which is delivered and erected in bundles. The erection sequence requires access for installation of the concrete units.

Long-span composite beams and composite slabs with metal decking

This system consists of composite beams using rolled steel sections supporting a composite slab in a long-span arrangement of, typically, 10 to 15 m. Grids are either arranged with long-span secondary beams at 3 m to 4 m spacing supporting the slab, supported by short-span primary beams, or with short- span secondary beams (6-9 m span) supported by long-span primary beams.

long span cellular beam — Long span composite cellular beam floor

The depth of the long-span beams means that service openings, if required, are provided within the web of the beam. Openings can be circular, elongated or rectangular in shape, and can be up to 70% of the beam depth. They can have a length/depth ratio of up to 2.5. Web stiffeners may be required around holes. Shear studs are normally positioned in pairs, with reinforcing bars placed transversely across the beams to act as longitudinal shear reinforcement.

Advantages
(1) Large column-free areas.
(2) Service ducts pass through openings in the web of the beams

Disadvantages
(1) Deeper floor zones.
(2) Heavier steelwork than some short-span solutions.
(3) Fire protection required for 60 minutes fire resistance and above.

Composite beams with precast units

This system consists of rolled steel beams with shear studs welded to the top flange. The beams support precast concrete units with a structural concrete infill over the beam between the ends of the units, and often with an additional topping covering the units. The precast units are either hollow core, normally 150 – 260 mm deep, or they are solid planks of 75 mm to 100 mm depth.

composite beam with precast units — Different configurations if composite floors with precast units

The shear studs and transverse reinforcement allow the transfer of the longitudinal shear force from the steel section into the precast units and the concrete topping, so that they can act together compositely. Composite design is not permitted unless the shear connectors are situated in an end gap (between the concrete units) of at least 50 mm. Minimum flange widths are crucial for providing a safe bearing for the precast units and room for the shear studs.

Advantages
(1) Fewer secondary beams, due to long-span precast units.
(2) Shear connectors for most beams can be welded off site, enabling larger stud diameters to be used and fewer site operations. It is usually convenient to weld studs to edge beams on site.

Disadvantages
(1) The beams are subject to torsion and may need stabilising during the construction stage.
(2) The precast units need careful detailing for adequate concrete encasement of shear connectors and installation of transverse reinforcement.
(3) More individual lifting operations compared to the erection of decking, and the erection sequence requires access for installation of the concrete units.

Non-composite beams with precast units

This system consists of rolled steel beams supporting precast concrete units. The precast units may be supported on the top flange of the steel beams, or, to reduce construction depth, supported on ‘shelf’ angles. Shelf angles are bolted or welded to the beam web, with an outstand leg long enough to provide adequate bearing of the precast units and to aid positioning of the units during erection. Precast concrete units are generally grouted in position. The units may have a screed (which may be structural), or may have a raised floor. The precast units are either hollow core, normally 150-260 mm deep, or they are solid planks of 75 mm to 100 mm depth.

precast beam and slab in non composite construction — Floor construction with precast concrete units in non-composite construction

In order to meet robustness requirements, mesh and a structural topping may be required, or reinforcement concreted into hollow cores and passed through holes in the steel beam web. Tying may also be required between the concrete units and the edge beams.

Advantages
(1) Fewer secondary beams, due to long-span precast units.
(2) A simple solution involving basic member design.

Disadvantages
The beams are subject to torsion and may need stabilising during the construction stage.
More individual lifting operations compared with the erection of decking, and the erection sequence requires access for installation of the concrete units.

Source:
Brown D.G., Iles D.C., Yandzio E. (2009): Steel building design: Medium rise braced frames (In accordance with Eurocodes and the UK National Annexes). The Steel Construction Institute, UK

Second-Order Effects in Steel Structures

By

Ubani Obinna

-

July 25, 2020

Second-order effects involve the analysis of a structure based on the deformed geometry. In other words, second-order analysis recognizes the deflection in a structure due to an externally applied load, and determines its effect on the internal forces generated thereof. The magnitude at which the internal forces in a structure increase due to second-order effects depend on the geometry, stiffness, and support conditions of the structure. This is usually employed in the verification of the stability of steel structures against phenomena such as buckling.

The sensitivity of a frame to second order effects may be illustrated simply by considering one ‘bay’ of a multi-storey building in simple construction (i.e. with pinned connections between beams and columns); the bay is restrained laterally by a spring representing the bracing system. First and second order displacements are illustrated below.

first and second order effects in a pinned braced frame 1

For first order effects;
kδ₁ = H₁

For second order effects;
kδ₂= H₁ +V(δ₂ / h) = H₂

On rearranging the equation for second order effects, the equilibrium condition can be expressed as;

H₂ = H₁[1 / (1 – V/kh)]

Hence, it can be observed that if the stiffness of the structure k is large, there will be little amplification of the horizontal force, and first order analysis will be adequate for the structure. On the other hand, if the value of vertical force tends toward a critical value V_cr (= kh) then displacements and forces in the restraint tend toward infinity. The ratio V_cr/V, which may be expressed as a parameter α_cr, is thus an indication of the second order amplification of displacements and forces in the bracing system due to second order effects.

The amplifier is given by:

[1/(1 – 1/α_cr)

This amplification factor applies to horizontal forces such as wind loads and imperfection loads.

Criteria for considering second order effects

According to BS EN 1993-1-1, 5.2.1(2), the effects of the deformed geometry of the structure (second order effects) need to be considered if the deformation significantly increase the forces in the structure or if the deformations significantly modify structural behaviour. For elastic global analysis, clause 5.2.1 says that the second order effects are significant if the parameter α_cr < 10, where α_cr is determined by first order analysis and for a braced frame is defined by the approximate expression:

α_cr = (H_Ed/V_Ed) x (h/δ_h,Ed)

where:
H_Ed is the design value of the horizontal reaction at the bottom of the storey to the horizontal loads and the equivalent horizontal forces1
V_Ed is the total design vertical force on the structure on the bottom of the storey
δ_h,Ed is the horizontal displacement at the top of the storey, relative to the bottom of the storey, when the frame is loaded with horizontal loads (e.g. wind) and equivalent horizontal forces which are applied at each floor level
h is the storey height.

Methods for determining second order effects

Where second order effects need to be evaluated, BS EN 1993-1-1, 5.2.2 says that they may be allowed for by:

An appropriate second-order analysis, taking into account the influence of the deformation of the structure.
Using appropriate (increased) buckling lengths of members
Amplification of an elastic first order analysis using the initial geometry of the structure.

Second Order Analysis

A range of second order analysis software is available. Use of any software will give results that are to some extent approximate, depending on the solution method employed, the types of second-order effects considered and the modelling assumptions. Generally, second-order software will automatically allow for frame imperfections, so the designer will not need to calculate and apply the equivalent horizontal forces. The effects of deformed geometry (second-order effects) will be allowed for in the analysis. The effect of member imperfections and such things as residual stresses are allowed for if verifying members in accordance with the rules in Section 6 of BS EN 1993-1-1.

Use of increased column buckling length

The use of increased column buckling effective lengths is generally not recommended, simply because of the manual effort involved in calculating the effective length factors. However, if this option is chosen, effective length factors can be determined using a source of non-conflicting complementary information (NCCI), such as BS 5950 Annex E or DD ENV 1993-1-1 Annex E.

Amplification of first order effects

Use of amplified first order effects is subject to the limitation that α_cr ≥ 3 (if α_cr is less than 3, second order analysis must be used). The amplifier is given by:

[1/(1 – 1/α_cr)

Only the effects due to the horizontal forces (including the equivalent horizontal forces) need to be amplified. In a braced frame, where the beam to column connections are pinned and thus do not contribute to lateral stiffness, the only effects to be amplified are the axial forces in the bracing members and the forces in columns that are due to their function as part of the bracing system.

Vibration Serviceability of Composite Slabs

By

Ubani Obinna

-

July 20, 2020

Vibration analysis of floors is not entirely new in the field of structural engineering. Numerous studies and researches have been devoted to human-structure interaction, with emphasis on human perception of vibration in civil engineering structures. Recently, the quest for construction of slender structures with large unobstructed areas (with flexible partitioning) has made the idea of vibration serviceability very important in the design of structures. Once a building is constructed, it is usually challenging to ameliorate vibration issues since it involves modifying the mass and stiffness of the structure.

In the UK, the vibration sensitivity of composite slabs has been traditionally checked by ensuring that the vertical natural frequency of secondary and primary beams is greater than 4Hz. However, a new approach has been recommended in the publication SCI P354 (2009) for checking the vibration serviceability of composite slabs. An example using the approach has been presented in this article.

In P354, two modes of vibration have been recommended for checking acceptability. In mode A, alternate secondary spans may be deflecting up and down (assuming simply supported conditions) with the participation of the slabs (as fixed ended) but not the primary beams. The primary beams are assumed to form nodal lines with zero deflection. In mode B, the primary beams may be deflecting in the same manner, but the secondary beams and slabs which are effectively fixed ended contribute to extra deflection. Therefore for mode B, the deflection is a sum of three contributions. The lower frequency of the two modes is taken as the fundamental frequency and should be at least 3Hz to ensure that walking activities will be outside the frequency zone that will cause resonance.

The design procedures for determining the dynamic performance of composite floors is as follows;

Determine the natural frequency
Determine the modal mass of the floor
Evaluate the response of the floor
Verify the response of the floor against the requirements

Solved Example

The figure below shows a part plan of a composite floor. The slab is to be constructed using profiled metal decking and normal weight, grade 30 concrete. The longitudinal beams are of grade S275 steel with a span of 7.5 m and spaced 3 m apart. Check the acceptance of the floor for vibration. 406 x 178 x 67 UKB as the internal beams. T

Member Geometry

Primary beams 610 x 229 x 140 UKB
Secondary beams 406 x 178 x 67 UKB

Beam span = 7.5 m
Beam spacing = 3.0 m
Total depth of slab h_s = 130 mm
Depth of profile h_p = 60 mm
Overall height of profile h_d= 72 mm
Depth of concrete above profile = 58 mm
Profile: SMD TR60⁺ (1.2 mm gauge)
Gauge = 1.2 mm
Mesh: A142

Floor loading
130 mm deep concrete slab = 2.21 kN/m² (manufacturer’s data)
Self weight of profile decking = 0.131 kN/m²
Ceiling and services = 1 kN/m²
Finishes = 1.2 kN/m²
10% Imposed = (0.1 x 5 kN/m²)= 0.5 kN/m²
Sub total = 5.041 kN/m²

Primary beams = (140 kg/m x 9.81/7.5) x 10^-3 = 0.183 kN/m²
Scondary beams = (67 kg/m x 9.81/3) x 10^-3 = 0.22 kN/m²
Total = 5.44 kN/m² = (5.44 x 10³)/9.81 = 554.54 kg/m²

Typical SMD TR60+ profile

Calculation of composite slab properties

Profile neutral axis = 33.7 mm
Profile area/unit width = 1633 mm²/m
Profile moment of inertia = 119.8 cm⁴/m
Height of re-entrant rib = 60 mm

Concrete area/unit width for 130 mm thick slab= 0.096 m²/m
Therefore effective slab thickness = 96 mm

For dynamic properties, take the gross uncracked moment of inertia. Dynamic modulus of elasticity of concrete E_cm = 38 kN/mm².

Modular ratio α = E_s/E_cm = 210/38 = 5.526

We can calculate the elastic neutral axis of the composite slab from the table below.

Section	Area A (cm²)	Neutral axis y (cm)	Area x y (cm³)
Concrete	960/α = 173.724	9.6/2 = 4.8	816.5
Profile	16.33	13 – 3.37 = 9.63	157.257
Total	∑A = 190.054		∑Ay = 973.757

Elastic neutral axis of the composite slab NA = ∑Ay/∑A = 973.757/190.054 = 5.123 cm below the top of the slab

The moment of inertia of the composite slab can be calculated from the table below;

Section	Distance from NA (cm)	Area x Distance² (cm⁴)	I_y,local (cm⁴)
Concrete	0.323	18.124	3139
Profile	4.507	331.712	119.8
		349.836	3258.8

Second moment of area of composite slab = 349.836 + 3258.8 = 3608.636 cm⁴ (36.0863 x 10^-6 m⁴)

Moment of inertia of the composite secondary beam

406 x 178 x 67 UKB
Span = 7.5m; Weight = 67.1 kg/m
Depth = 409.4 mm; Area = 85.5 cm²
I_y = 24300 cm⁴; Effective breadth b_eff = 1875 mm

Composite section properties

Section	Width (cm)	Depth (cm)	y (cm)	A (cm²)	Ay (cm³)	Ay² (cm⁴)	I_local
Slab	187.5	7.0	3.50	1312.5/5.526 = 237.513	831.297	2909.532	969.847
Beam			33.47	85.5	2861.68	95780.596	24300
				∑A = 323.013	∑Ay = 3692.977	∑Ay² = 98690.128	∑I_local = 25269.847

Position of elastic neutral axis = ∑Ay/∑A = 3692.977/323.013 = 11.432 cm from the top of the slab

Gross moment of area = 98690.128 + 25269.847 – (11.432²x 323.013) = 81745.204 cm⁴ (8.1745 x 10^-4 m⁴)

Moment of inertia of the composite primary beam

610 x 229 x 140 UKB
Span = 6.0m; Weight = 139.9 kg/m
Depth = 617.2 mm; Area = 178 cm²
I_y = 112000 cm⁴; Effective breadth b_eff = 1500 mm

Composite section properties

Section	Width (cm)	Depth (cm)	y (cm)	A (cm²)	Ay (cm³)	Ay² (cm⁴)	I_local
Slab	150	9.6	4.8	1440/5.526 = 260.586	1250.81	6003.9	2001.3
Beam			43.86	178	7807.08	342418.53	112000
				∑A = 438.586	∑Ay = 9057.89	∑Ay² = 348422.43	∑I_local = 114001.3

Position of elastic neutral axis = ∑Ay/∑A = 9057.89/438.586 = 20.652 cm from the top of the slab

Gross moment of area = 114001.3 + 348422.43 – (20.652²x 438.586) = 275364.5625 cm⁴ (27.5364 x 10^-4 m⁴)

Fundamental natural frequency

Mode A

(i) Slab – Fixed Ended
w = 5.041 kN/m² x 3 m = 15.123 kN/m
δ₁ = wL³/384EI = (15.123 x 3³)/(384 x 210 x 10⁶ x 36.0863 x 10^-6) = 1.403 x 10^-4m = 0.14 mm

(ii) Secondary beam – Simply Supported
w = (5.041 kN/m² + 0.22 kN/m²) x 3 m = 15.783 kN/m
δ₂ = 5wL⁴/384EI = (5 x 15.123 x 7.5⁴)/(384 x 210 x 10⁶ x 8.1745 x 10^-4) = 3.629 x 10^-3m = 3.629 mm

Total deflection for mode A δ = 0.14 + 3.629 = 3.760 mm

Natural frequency for mode A f_A = 18/√δ = 18/√3.760 = 9.282 Hz

Mode B

(i) Slab – As above

(ii) Secondary beam (assume fixed ended)
w = (5.041 kN/m² + 0.22 kN/m²) x 3 m = 15.783 kN/m
δ₂ = wL⁴/384EI = (15.123 x 7.5⁴)/(384 x 210 x 10⁶ x 8.1745 x 10^-4) = 7.2589 x 10^-4m = 0.725 mm

(iii) Primary beam (assume simply supported)
Reactive force from secondary beam = (15.783 x 7.5)/2 = 59.186 kN
w = (5.041 kN/m² + 0.183 kN/m²) x 3.75 m = 19.59 kN/m

δ₃ = 5wL⁴/384EI + PL³/48EI = [(5 x 19.59 x 6⁴)/(384 x 210 x 10⁶ x 27.5364 x 10^-4)] + [(59.186 x 6³)/(48 x 210 x 10⁶ x 27.5364 x 10^-4)] = (5.716 x 10^-4) + (4.605 x 10^-4) = 1.032 x 10-3 m = 1.032 mm

Therefore total deflection = 0.14 + 0.725 + 1.032 = 1.897 mm

Natural frequency for mode B f_B = 18/√δ = 18/√1.879 = 13.13Hz

Therefore the fundamental frequency of the floor is found in Mode A f₀ = 9.292 Hz > 3.0 Hz (Okay)

Calculation of Modal Mass

The effective floor length can be calculated from;

L_eff = 1.09(1.10)^{n_y – 1}(EI_b/mbf₀²)^1/4 L_eff≤ n_yL_y

Where;
n_y is the number of bays in the secondary beam direction (n_y ≤ 4)
EI_b is the flexural rigidity of the composite secondary beam (expressed in Nm² when m is expressed in kg/m²)
b is the spacing of the secondary beam = 3.0 m
f₀ is the fundamental natural frequency
L_y is the span of the secondary beams

L_eff = 1.09(1.10)^{1 – 1}(210 x 10⁹ x 8.1745 x 10^-4/554.54 x 3 x 9.292²)^1/4 = 7.049 m < (1 x 7.5 = 7.5 m)

The effective width of the slab can be calculated from;

S = η(1.15)^{n_x – 1}(EI_s/mf₀²)^1/4

η = 0.71 for f₀ > 6.0
n_x = 2.0

S = 0.75(1.15)^{2 – 1} x (210 x 10⁹ x 36.0863 x 10^-6/554.54 x 9.292²)^1/4 = 3.05 m < (2 x 6 = 12)

Modal mass M = mL_effS = 554.54 x 7.04 x 3.05 = 11907 kg

Calculation of Floor Response

As f₀ < 10 Hz, the response is to be assessed according to the low frequency floor recommendations.

a_w,rms = μ_eμ_r (0.1Q/2√2Mζ) x Wρ

Q = 746 N based on an average weight of 76 kg
Take critical damping ratio ζ as 4.68%
Since f₀ > 8Hz, take the weighting factor W = 8/f₀ = 8/9.292 = 0.860
Conservatively assume that μ_e = μ_r = 1.0

Assuming a walking frequency f_p of 2Hz in a maximum corridor length L_p of 15 m;
v = 1.67f_p² – 4.83f_p + 4.5 = 1.52 m/s

ρ = 1 – e^{(-2πζL_pf_p/v)} = 1.0

a_w,rms = 1.0 x 1.0 x [(0.1 x 746)/(2√2 x 11907 x 0.0468)] x 0.86 x 1.0 = 0.0407 m/s²

Response Factor

R = a_w,rms/0.005 = 0.0407/0.005 = 8.14

For an office building, the floor is is acceptable for vibration since the response factor is greater than 8.0.

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Long-span composite beams and composite slabs with metal decking

Composite beams with precast units

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Second-Order Effects in Steel Structures

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Methods for determining second order effects

Vibration Serviceability of Composite Slabs

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