Road traffic signs are signs that are erected by the side or above the road to give information or instructions to road users. We are all road users, and a thorough understanding of these signs will make our usage and interaction with highway facilities safer and more convenient. On the contrary, lack of knowledge of traffic signs can mean incompetency and disobedience to traffic regulations. This may result in your vehicle being impounded, fines/penalties, and in the worst case, a highway crash.
In Nigeria, whenever Federal Road Safety Corps (FRSC) Marshals are on highway patrol, penalty tickets often issued that are related to lack of road signs knowledge, and their penalties are briefly highlighted below;
1. Construction Area Speed Limit Violation (CAV) – Fine = N3,000 – 3 Penalty points 2. Light/Sign Violation (LSV) – Fine = N2,000 – 2 penalty points 3. Road Marking Violation (RMV) – Fine = N5,000 – 5 penalty points 4. Route Violation (RTV) – Fine = N10,000 – 10 penalty points 5. Speed Limit Violation (SLV) – Fine = N5,000 – 5 penalty points 6. Unauthorized Tampering / Removal of Road Signs (UTS) – Fine = N5,000 – 5 penalty points 7. Wrongful Overtaking (WOV) – Fine = N3,000 – 3 penalty points 8. Other offenses / Violations (OFV) – Fine = N3,000 – 2 penalty points
Penalty points are points allotted to traffic offenses accumulated in a driver’s record. The implications of such penalty points are given below;
Warning (10 – 14 penalty points)
Suspension of license (15 – 20 penalty points)
Withdrawal of license (21 and above penalty points)
Nigeria road traffic signs are divided into the following categories;
A. Warning Signs – Warning signs are usually triangular in shape, with yellow or white background, black inscription, and red border. Another variety is the ‘Give Way’ or ‘Yield’ sign, which is an upside-down triangle. As the name implies, they are signs that warn you about what is ahead. They can also give instructional warning about a certain situation.
B. Regulatory Signs – Regulatory signs can be prohibitive, or mandatory. (i) Regulatory Signs (Prohibitive): These signs are usually circular in shape with yellow or white background, black inscription, and red border. Another variety is the STOP sign, which is octagonal in shape. The STOP sign requires that you stop your vehicle completely whenever you see it. Generally, prohibitive signs give instructions that are prohibitive in nature. (ii) Regulatory Signs (Mandatory): Regulatory signs are usually circular with a blue background, white inscription, and sometimes with a white border. They give positive instructions.
C. Informative Signs – These signs are usually rectangular in shape with a green background, and they provide guidance or information while on the highway.
These signs are shown in full below. All traffic signs are labelled at the bottom;
(a) Warning Signs
FIG A1: YIELD EVERY RIGHT OF WAY. GIVE WAY TO TRAFFIC ON YOUR RIGHT OR TRAFFIC ON YOUR LEFT
These are electronic light signaling devices used to control traffic flow at intersections to avoid conflict of vehicles. The basic signals are shown in the pictures below.
TO DOWNLOAD ALL NIGERIAN ROAD SIGNS (A FULL PAPER) IN A PRINTABLE FORMAT, CLICK HERE This post is dedicated to members of NYSC Federal Road Safety Club (Batch ‘B’ 2015 and Batch ‘A’ 2016), Ikot Ekpene L.G.A., Akwa Ibom State, and the FRSC Marshals at RS 6.32 Ikot Ekpene Unit Command. God bless you all for your efforts in preaching the message of safety.
In multi-storey buildings, ramps, elevators, escalators, and stairs are often used to facilitate vertical circulation. Circulation refers to the movement of people and goods between interior spaces in buildings, and to entrances and exits. Stairs are important building elements that are used to provide vertical circulation and access across different floor levels in a building. It is also recommended that when an access height exceeds 600mm, a staircase should be provided. A staircase can be made of reinforced concrete, steel, timber, and other composite construction materials.
In modern architecture, stairs are designed to be aesthetically pleasing, and they contribute immensely to the interior beauty of a building. There are different types of stairs with different configurations. For stairs in a building, the recommended slope for comfort is 27°, but for practical purposes, this can sometimes be extended to 35°.
Types of staircase
Generally, stairs are usually of the following types:
Straight stairs are stairs along which there is no curvature or change in direction on any flight between two successive floors or levels. There are several possible arrangements of straight stairs. For example, they may be arranged in a straight run with a single flight between floors, or a series of flights without a change in direction.
Also, straight stairs may permit a change in direction at an immediate landing. When the stairs require a complete reversal of direction, they are called parallel stairs or half-landing stairs (turning through 180°). When successive flights are at an angle to each other, (usually 90°), they are called angle stairs or quarter-turn stairs. In addition, straight stairs may be classified as scissors stairs when they comprise a pair of straight runs in opposite directions and are placed on opposite sides of a wall.
Straight staircase with half landing
Circular stairs when viewed from above appear to follow a circle with a single centre of curvature and large radius.
Curved stairs when viewed from above appear to follow a curve with two or more centres of curvature, such as an ellipse.
Spiral stairs are similar to circular stairs except that the radius of curvature is small and the stairs may be supported by a column.
Terminologies in staircase design
Functional parts of a staircase
Flight: A series of steps extending from floor to floor, or from a floor to an intermediate landing or platform.
Guard: Protective vertical barrier along edges of stairways, balconies, and floor openings.
Landings (platforms): Used where turns are necessary or to break up long climbs. Landings should be level, as wide as the stairs, and at least 1000mm long in the direction of travel.
Step: Combination of a riser and the tread immediately above.
Rise: Distance from floor to floor.
Run: Total length of stairs in a horizontal plane, including landings.
Riser: Vertical face of a step. Its height is generally taken as the vertical distance between treads.
Tread: Horizontal face of a step. Its width is usually taken as the horizontal distance between risers.
Nosing: Projection of a tread beyond the riser below.
Soffit: Underside of a stair.
Railing: Framework or enclosure supporting a handrail and serving as a safety barrier.
Baluster: Vertical member supporting the handrail in a railing.
Balustrade: A railing composed of balusters capped by a handrail.
Handrail: Protective bar placed at a convenient distance above the stairs for a handhold.
Ideas on the selection of staircase dimensions
Headroom Ample headroom should be provided not only to prevent tall people from injuring their heads but to give a feeling of spaciousness. A person of average height should be able to extend his hand forward and upward without touching the ceiling above the stairs. The minimum vertical distance from the nosing of a tread to overhead construction should preferably never be less than 2100 mm.
Stairway Width The width of a stairway depends on its purpose and the number of persons to be accommodated in peak hours or emergencies. Also, there are building codes that regulate the geometric design of stairways. The following can be used as guidelines;
For residential flats between two to four storeys, use a minimum width of 900 mm, for flats more than 4 storeys, use a width of 1000 mm.
For public buildings of under 200 persons per floor, use a width of 1000 mm, for buildings between 200 – 400 persons per floor, use a width of 1500 mm. For over 400 persons, use a width between 1500 – 3000 mm. However, when the width of a stairway exceeds 1800 mm, it is necessary to divide it using handrails.
Step Sizes Risers and treads generally are proportioned for comfort and to meet accessibility standards for the handicapped, although sometimes space considerations control or the desire to achieve a monumental effect, particularly for outside stairs of public buildings. Treads should be at least 250 mm, exclusive of nosing.
The most comfortable height of riser is 175 mm. Risers less than 100 mm and more than 200 mm high should not be used. The steeper the slope of the stairs, the greater the ratio of the riser to tread. In the design of stairs, account should be taken of the fact that there is always one less tread than riser per flight of stairs. No flight of stairs should contain less than three risers.
Structural design of a staircase
The theoretical procedures employed in the structural analysis of stairs is the concept of an idealised line structure and when detailing the reinforcement for the resulting stairs, additional bars should be included to limit the formation of cracks at the points of high-stress concentration that inevitably occur. Typical detailing of corners (edge between flight and landing) of a staircase is shown below;
Detailing of end support of staircase with landing
Detailing of end support of staircase with landing
The ‘three-dimensional’ nature of the actual structure and the stiffening effect of the triangular tread areas, both of which are usually ignored when analysing the structure, will result in actual stress distributions that differ from those calculated, and this must be remembered when detailing (Reynolds et al, 2008). The typical nature of internal stresses induced in a simply supported straight flight stair and reinforcement pattern is as shown in the picture below.
Bending behaviour of a single flight staircase
Simple straight flights of stairs can span either transversely (i.e. across the flight) or longitudinally (i.e.along the flight). When spanning transversely, supports must be provided on both sides of the flight by either walls or stringer beams. In this case, the waist or thinnest part of the stair construction need be no more than 60 mm thick say, the effective lever arm for resisting the bending moment being about half of the maximum thickness from the nose to the soffit, measured at right angles to the soffit. When the stair spans longitudinally, deflection considerations can determine the waist thickness.
In principle, the design requirements for beams and slabs apply also to staircases, but designers cannot be expected to determine the deflections likely to occur in the more complex stair types. BS 8110 deals only with simple types and allows a modified span/effective depth ratio to be used. The bending moments should be calculated from the ultimate load due to the total weight of the stairs and imposed load, measured on plan, combined with the horizontal span. Stresses produced by the longitudinal thrust are small and generally neglected in the design of simple systems.
Sample design of reinforced concrete staircase
A section of a staircase is shown above. The width of the staircase is 1160mm. We are expected to carry out a full structural analysis and design of the staircase according to EC2 using the following data; Density of concrete = 25 kN/m3; Compressive strength of concrete (fck) = 30 N/mm2; Yield strength of steel (fyk) = 460 N/mm2; Concrete cover = 25mm; Imposed load on staircase (qk) = 4 kN/m2 (category C3).
The structural idealisation of the staircase is shown below;
Loadingof the staircase
Thickness of waist and landing = 200 mm Depth of riser = 150mm Load actions on the stairs Concrete self weight (waist area) = 0.2 × 25 = 5 kN/m2 (normal to the inclination) Stepped area = 0.5 × 0.15 × 25 = 1.875 kN/m2 (global vertical direction) Finishes (say) = 1.2 kN/m2
We intend to apply all gravity loads purely in the global y-direction, therefore we convert the load at the waist of the stair from local to global direction by considering the angle of inclination of the flight area to the horizontal;
? = tan−1(1.2/1.75) = 34.438989°
Therefore the UDL from waist of the stair in the global direction is given by = (5 × cos 34.438989) = 4.124 kN/m2
Total dead load on flight area (gk) = 4.124 + 1.875 + 1.2 = 7.199 kN/m Variable load on staircase (qk) = 4 kN/m2 The load on the flight area at ultimate limit state = 1.35gk + 1.5qk Ed = 1.35(7.199) + 1.5(4) = 15.719 kN/m2
On the landing; gk = 5 + 1.2 = 6.2 kN/m2; qk = 4 kN/m2
The load on the landing at ultimate limit state = 1.35gk + 1.5qk n = 1.35(6.2) + 1.5(4) = 14.370 kN/m2
The loading of the structure for dead and live loads at ultimate limit state is shown below;
The ultimate bending bending moment diagram due to ultimate loads is shown below;
The ultimate shear force diagram is shown below;
Flexural design of the staircase span
A little consideration will show that it is best to use the design moment MEd = 41.119 kNm to design the entire stairs.
MEd = 41.119 kNm d = h – Cc – ϕ/2 Assuming ϕ12mm bars will be used for the construction d = 200 – 25 – 6 = 169mm; b = 1000mm (designing per unit width)
βs = 310/242.358 = 1.2789 Taking the distance between supports as the effective span, L = 4.35m The allowable span/depth ratio = βs × 23.0258 = 1.2789 × 23.0358 = 29.460 Actual deflection L/d = 4350/169 = 25.739 Since 25.739 < 29.460, deflection is ok.
Shear design
Ultimate shear force VEd = 35.358 kN v = VEd/bd = (35.358 × 1000)/(1000 × 169) = 0.209 N/mm2 VRd,c = [CRd,ck(100ρ1fck)1/3] ≥ (Vmin)
CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/169) = 2.087 > 2.0, therefore, k = 2
Bridges are structures designed to span over obstacles, and as such provide access routes for humans, animals, vehicles, fluid, and other objects as may be desired. EN 1990 (Basis of Structural Design) classified bridges under consequence class 3 (CC3) which implies that the consequences are high if failure should occur. As a result, the design of bridges is taken very seriously and all possible load regimes that the structure will be subjected to in its design life are taken into account.
Bridges are important for adequate development, safety, and efficiency of road transportation system in any country including Nigeria. Nigerian engineers have stepped up to solve the engineering problems facing them internally and independently.
Loads from vehicles are one of the the most prominent loads that bridges are subjected to in their design life. Eurocode 1 Part 2 (EN 1991-2) handles the actions of traffic on bridges.
The code specifies four load models for vertical loads on highway bridges. Among these four load models, Load Model 1 (LM1) is the one used to represent the effect of normal traffic on bridges and it is used for both local and global verifications. LM1 consists of tandem axle loads (TS) which are concentrated loads applied in the longitudinal direction of the bridge, and uniformly distributed loads (UDL’s) which are applied both transversely and longitudinally. The values of the loads depends on the notional lanes on which they are applied unlike BS 5400 which uses lane factors. (See more on the paper uploaded). It is however pertinent to point out that LM1 is an artificial model and does not represent the weight of real vehicles. The models are created and calibrated based on traffic data collected in France between 1980 and 1994, and the effects they induce on bridge decks are similar to that induced by real traffic. See the form of LM1 below.
In the paper downloadable in this post, a 15.0m single span bridge simply supported between two abutments (the first picture on this post) is subjected to LM1 and analysed as static load using Staad Pro software. The bridge has a deck that is 10.1m wide and a carriageway that is 7.2m wide. The reinforced concrete slab is supported by five precast beams (girders) and the aim of the analysis is to find out the effects of the traffic load on the girders. The load was arranged to induce the worst effect on the first and second girders. The finite element analysis results obtained were very comparable to the ones obtained using other rational and computational methods.
Relevant Pictures from Analysis
(a) Another 3D view of the bridge (Modelled using Autodesk Revit 2009)
(b) A 2D elevation view of the bridge
(c) Section through the bridge deck
(d) Division of the carriageway into notional lanes
(e) Sectional view of the notional lanes, remaining area, and the pedestrian areas
(f) Finite element modelling of the bridge deck on StaadPro V8i
(g) Loading of the bridge with LM1 (including crowd load)
(h) Full 3D Model of the bridge deck on Staad Pro
(i) Bending moment diagram of the girders due to LM1
(j) Shear force diagram of the girders due to LM1
To download the full loading, analysis, and results paper, click HERE.
The most common construction concept of sports stadiums today is a composite type where usually precast concrete terrace units (seating decks) span between inclined (raker) steel or reinforced concrete raker beams and rest on each other, thereby forming a grandstand.
The raker beams are usually formed in-situ with the columns of the structure, or sometimes may be preferably precast depending on site/construction constraints. This arrangement usually forms the skeletal frame of a stadium structure.
A typical section through the grandstand and the L-shaped seating unit are shown in Figures 1 and 2 respectively.
Figure 1: Section through the grandstand
Figure 2: Section through the L-shaped seating terrace unit
This process was used in the construction of Cape Town Stadium South Africa, for the 2010 World cup (see Figure 3);
Figure 3: Installation of precast seating units on RC raker beam, Cape Town Stadium, South Africa
In this article, a raker beam isolated from a double-tiered reinforced concrete grandstand (see Figure 1) has been presented for the purpose of structural analysis and design. Each grandstand frame has precast L-shaped seating terrace units that span in between reinforced concrete raker beams inclined at angles between 20° – 22° with the horizontal. The raker beams are spaced at 7m centre to centre.
Crowd live loads and other loads are transferred from the seating units to the raker beams, which then transfers them to the columns and then to the foundations. Loads from the service and concourse areas are also transferred using the same load path.
The details of the structure and the loading on the first tier are shown below;
Variable Actions Imposed load for structural class C5 (qK) = 5 kN/m2 Total imposed load for seating unit (QK) = (5 × 0.95 × 7) = 33.25 kN
Total action on L-shaped seating terrace unit at ultimate limit state by Eurocode 2 (PEd) = 1.35Σ(GKi) + 1.5QK = 1.35(44.8) + 1.5(33.25) = 110.355 kN
Loading on the raker beams
Height of beam = 1200mm Width of beam = 400mm
Self weight of raker beam Concrete own weight (waist area) = 1.2m × 0.4m × 25 kN/m3 = 12.00 kN/m (normal to the inclination i.e. in the local direction) Height of riser in the raker beam = 0.4m; Width of tread in the raker beam = 0.8m; Angle of inclination (𝛼) = 20.556° Stepped area (risers) = ½ × 0.4 × 25 = 5 kN/m (in the global direction)
For purely vertical load in the global y-direction, we convert the load from the waist of the beam by; UDL from waist of the beam = (12.00 × cos 20.556°) = 11.236 kN/m Total self weight (Gk) = 11.236 + 5 = 16.235 kN/m Self weight of raker beam at ultimate limit state; n = 1.35Σ(GKi) = 1.35 × 16.235 = 21.917 KN/m
Load from precast seating units End shear from precast seating unit = PEd/2 = 110.355/2 = 55.1775 kN Total number of the precast seating units on the beam = 24/0.8 = 30 units For intermediate beam supporting seating units on both sides; Total number of precast seating units = 2 × 30 = 60 units Therefore, the total shear force transferred from the seating units to the raker beam = 55.1775 × 60 = 3310.65 kN Equivalent uniformly distributed load in the global direction at ultimate limit state = 3310.65/24 = 137.94 kN/m Total load on intermediate raker beams at ultimate limit state in the global direction = 137.94 + 21.917 = 159.857 kN/m
Structural Analysis
A full 3D Staad Pro Model of the grandstand is shown in Figure 5;
Figure 5: 3D Modelling of the grandstand on Staad Pro
The raker beam was analysed for the loading obtained as shown above. Due to its inclination, it was subjected to significant bending, axial, and shear forces. The internal stresses are shown below;
Figure 6: Bending moment diagram
Figure 7: Shear force diagram
Figure 8: Axial force diagram
The summary of the internal stresses in the raker beams is shown in the Table below;
Structural Design
The structural design of the raker beam using EN 1992-1-1 has been carried out and all the parameters used in the, and steps followed are shown below in the subsequent sections.
Design compressive of concrete fck = 35 N/mm2 Yield strength of steel fyk = 460 N/mm2 bw = 400mm; h = 1200mm; Cc = 40mm
Flexural Design of span AB (MABspan) MEd = 948.078 kNm d = h – Cc – ϕ/2 – ϕlink d = 1200 – 40 – 16 – 10 = 1134 mm k = 𝑀𝐸𝑑/𝑓𝑐𝑘 𝑏𝑑2 = (948.078 × 106)/35 × 400 × 11342 = 0.0527 Since k < 0.167 No compression reinforcement required z = d[0.5+ √(0.25−0.882𝐾)] = z = d[0.5 + √(0.25 − 0.882(0.0527)] = 0.95d
To calculate the minimum area of steel required; (Table 3.1 EC2) fctm =3.2099 N/mm2 As,min = 0.26 × fctm/fyk × bw × d = 0.26 × 3.2099460 ×400 × 1134 = 822.962 mm2 Check if ASmin < 0.0013 × bw × d (589.68 mm2) Therefore, As,min = 822.962 mm2
Check for Deflection Since the span is greater than 7m, allowable span/depth ratio = 𝛽𝑠 × 31.842 × 7000𝐿 = 1.11 × 31.842 × (7000/12816) = 19.374 Actual deflection L/d = 12816/1134 = 11.301 Since 11.301 < 19.374, deflection is ok.
Flexural Design of support A (MA); MEd = 1967.54 kNm k = 0.1093; la = 0.8919; AS1 = 4861 mm2; ASmin = 822.9785 mm2 Provide 4Y32mm + 4Y25mm TOP (ASprov = 5180 mm2)
Flexural Design of support B (MB); MEd = 2283.18 kNm k = 0.1268; la = 0.8717; AS1 = 5772 mm2; ASmin = 822.9785 mm2 Provide 6Y32mm + 4Y20mm TOP (ASprov = 6080 mm2)
Flexural Design of span BC (MBCspan) MEd = 1249.787 kNm k = 0.0694; la = 0.9345; AS1 = 2947mm2; ASmin = 822.9785 mm2 Provide 5Y25mm + 2Y20mm BOT (ASprov = 3083 mm2)
Shear Design
Let us consider support B VEd = 983.88 kN; NEd = 339.376 kN (Tension) Note that due to the tensile axial force in the section, the second term of VRd equation assumes a negative value. 𝜌1 = 0.0134; 𝜎𝑐𝑝 = − 0.7070 N/mm2; vmin = 0.3504 N/mm2; VRd = 230.6532 kN Since VRd,c < VEd, shear reinforcement is required
Assuming that the strut angle 𝜃 = 21.8° v1 = 0.5160; fcd = 19.8450 N/mm2; z = 0.9d = 1020.6 mm; VRD,max = 1440.64 kN Since VRD,max > VEd 𝐴𝑠𝑤/S = 0.9635 Trying 3Y10mm @ 200mm c/c (235/200 = 1.175) 1.175 > 0.9153 Hence shear reinforcement is ok.
The design results show that the effect of axial force was not very significant in the quantity of shear reinforcement required. Asv/Sv ratio of 1.175mm (3Y10mm @ 200 c/c) was found to satisfy shear requirements at all sections. The greatest quantity of longitudinal reinforcement was provided at the intermediate support with a reinforcement ratio of 1.3404%. The provided reinforcement was found adequate to satisfy ultimate and serviceability requirements.
It is worthwhile to observe the variation in the nature of axial forces on the beams (i.e from compressive to tensile). The effect can be more significant when using BS 8110-1:1997 for design than Eurocode 2 (EN 1992-1-1).
You can view some sections of the reinforced concrete raker beam below;
To download the full calculation sheet (PDF), click HERE
A structure undergoes free vibration when it is brought out of static equilibrium and can then oscillate without any external dynamic excitation. Free vibration of structures occurs with some frequencies which depend only on the parameters of the structures such as the boundary conditions, distribution of masses, stiffnesses within the members etc, and not on the reason for the vibration.
At each natural frequency of free vibration, the structure vibrates in simple harmonic motion where the displaced shape (mode shape) of the structure is constant but the amplitude of the displacement is varying in a sinusoidal manner with time. The number of natural frequencies in a structure coincides with the number of degrees of freedom in the structure. These frequencies are inherent to the given structure and are often referred to as eigenfrequencies. Each mode shape of vibration shows the form of an elastic curve which corresponds to a specific frequency.
A method of obtaining the natural frequencies and mode shapes of vibration is modal analysis. This is a technique by which the equations of motion, which are originally expressed in physical coordinates, are transformed to modal coordinates using the eigenvalues and eigenvectors gotten by solving the undamped frequency eigenproblem. The transformed equations are called modal equations.
For an undamped free vibration, the equation of motion is;
Mü + ku = 0 ———-(1)
From Figure 1.1, the following equations can be developed; M1Ü1 + K1U1 + K2 (U1 − U2) = 0 M2Ü2 + K2(U2 − U1) = 0
This can be expanded to give; M1Ü1 + U1(K1 + K2) + U2(−K2) = 0 M2Ü2 + U1(-K2) + U2(K2) = 0
Arranging it in matrix form we obtain;
The characteristic polynomial equation is thus;
The solution of the equation is; U1 = A1sin(𝜔𝑡+ 𝜑), U2 = A2sin(𝜔𝑡+ 𝜑)
Where Ai is the amplitude of the displacement of mass Mi, and 𝜑 is the initial phase of vibration. [(K1 + K2) − M1ω2]A1 − K2A2=0 −K2A1 + (K2−M2ω2) A2 = 0
To obtain the frequency equation, a non-trivial solution exists (non-zero amplitudes Ai), if the determinant of the coefficients to the amplitude is zero. This is also called the characteristic polynomial equation and is thus;
The solution of equation (3) presents the eigenfrequencies of the system. The system does not allow us to determine amplitudes directly, but we can find the ratio between these amplitudes. Hence for equation 3 above;
If we assume A1 = 1.0, then entries [1 𝐴2]𝑇 defines for each eigenfrequency, the column 𝜑 of the modal matrix Φ. The shape of each mode of free vibration is unique but the amplitude of the mode shape is undefined. The mode shapes are usually normalised such that the largest term in the vector is 1.0. The mode shapes have the important property of being orthogonal with respect to the mass and stiffness matrix of the structure.
Worked Example
A frame with an infinitely rigid floor is supported by 300 x 300 mm columns. If it is loaded as shown below, carry out the full modal dynamic analysis of the structure. (Take EI = 2.1 × 107 kN/m2)
In the paper downloadable in this post, modal dynamic analysis was carried out on the three-storey frame shown above. The results obtained regarding the three assumed degrees of freedom are shown in the pictures below;
(a) Mode 1 vibration parameters and displaced shape
(b) Mode 2 vibration parameters and displaced shape
(c) Mode 3 vibration parameters and displaced shape
The displacement time history response is therefore;
If velocity (v) = du/dt; The velocity time history response is therefore; The velocity time history response is obtained by direct differentiation of the displacement time history equations;
Internal stresses are induced in statically indeterminate frames when one support settles relatively to another. This is one of the major detrimental effects of differential settlement in civil engineering structures, and it is capable of inducing cracks and structural failures in buildings. Therefore, wherever it matters or where it is anticipated, the effects of the differential settlement must be taken into account during the design of civil engineering structures.
In this article, an indeterminate frame whose support has been subjected to an indirect action of differential settlement has been presented and fully solved. As stated above, if any statically indeterminate structure experiences differential settlement of supports, internal stresses will be developed in the members of the structure.
Analysis of such structures may be effectively performed by the force (flexibility) or stiffness (displacement) method in canonical form. In the paper downloadable in this article, the frame loaded as shown below is subjected to a differential settlement of 25mm at fixed support A. The internal stresses induced were calculated using the force method, stiffness method, and slope deflection method.
Canonical Equations
Consider a structure that is unable to be solved solely based on equilibrium equations (statically indeterminate). This structure has additional constraints (redundant constraints) that number “n.” Additionally, certain supports of this structure can undergo linear or angular movements (displacements) denoted by “di.”
The following set of equations are known as the Canonical equations and will be used to analyze this structure:
where free terms δk∆ (k = 1, 2,…, n) represent displacements of the primary system in the direction of primary unknowns Xk due to settlements of the supports. For the calculation of these terms, we need to use the theorem of reciprocal unit displacements and reactions (Rayleigh’s second theorem). This is given by;
δi∆ = -EIC (∆.R)
Where: EIC = Flexural rigidity of the column ∆ = Settlement at the point under consideration R = Support reaction at the point under consideration
Steps in the Analysis of Structures Due to Sinking of Supports
This section outlines the step-by-step procedure for analyzing redundant structures experiencing support settlements:
1. Kinematics and Redundancy Determination:
Carry out a kinematic analysis of the structure.
Identify the degree of redundancy (number of redundant constraints, n).
Select a primary system based on the force method.
Formulate the canonical equations using the chosen primary unknowns (Xi).
2. Unit Load Analysis:
Construct unit bending moment diagrams for each redundant member.
Calculate the unit displacements (δi) corresponding to each redundant constraint (di).
3. Free Term Calculations:
Determine the free terms within the canonical equations, considering the external loading and support settlements.
4. Solving for Primary Unknowns:
Solve the system of canonical equations with respect to the primary unknowns (Xi).
5. Internal Force Determination:
Construct the internal force diagrams (axial forces, shear forces, bending moments) for all members using the obtained primary unknowns (Xi) and the principle of superposition.
6. Support Reaction Verification:
Calculate the reactions at all supports using equilibrium equations and the internal forces.
Verify the calculated support reactions through equilibrium checks.
Worked Example
The portal frame shown below is fixed at column base A, and hinged at point C as shown below. The column of the frame has a square cross-section of 30cm x 30cm while the beam has a depth of 60cm and a width of 30cm. The foundation of support A settles vertically by 25mm. Draw the bending moment, shearing force, and axial force diagram due to the differential settlement action on the frame (E = 2.17 × 107 kN/m2).
SOLUTION Geometrical properties Moment of inertia of column IC = (bh3) / 12 = (0.3 × 0.33)/12 = 6.75 × 10-4 m4 Moment of inertia of beam IB = (bh3) / 12 = (0.3 × 0.63)/12 = 5.4 × 10-3 m4 We desire to work in terms of IC such that IC/IB = 0.125
Hence flexural rigidity of the column, EIC = (2.17 × 107) × (6.75 × 10-4) = 14647.5 KN.m2
The deformation of a structure at a point due to support settlement is given by;
δi∆ = -EIC (∆.R) ———— (1) Where; EIC = Flexural rigidity of the column ∆ = Settlement at the point under consideration R = Support reaction at the point under consideration
Basic system The next step in the analysis is to reduce the structure to a basic system, which is a system that must be statically determinate and stable. The frame is statically indeterminate to the second order, which means that we are going to remove two redundant supports.
The degree of static indeterminacy (RD) is given by;
RD = (3m + r) – 3n – s ————————— (2)
So that RD = (3 × 2) + 5 – (3 × 3) – 0 = 2
By choice, I am deciding to remove the two reactive forces at support C of the frame to obtain the basic system as shown in Figure 1.2 below;
Analysis of Case 1; X1 = 1.0, X2 = 0 The simple static analysis of the structure with case 1 loading is shown below in terms of the internal stresses diagram. Realise that there is a unit value support reaction acting at support A (pointed downwards) and a bending moment of 5.0 units.
Analysis of Case 2; X1 = 0, X2 = 1.0 The simple static analysis of the structure loaded with case 1 loading is shown below in terms of the internal stresses diagram. It will be so important to realise that there is no vertical support reaction at point A. A unit horizontal reaction is developed at the support to counter the applied unit load for case 2.
The appropriate canonical equation for this structure is therefore;
δ11X1 + δ12X2 + δ1∆ = 0 δ21X1 + δ22X2 + δ2∆ = 0
Computation of the influence coefficients
Influence coefficients are based on Mohr’s integral such that δi = 1/EI∫Mmds.
We can compute this by using the graphical method (making use of bending moment diagrams), we directly employ Vereschagin’s rule which simply states that when we are combining two diagrams of which one must be of a linear form (due to the unit load) and the other of any other form, the equivalent of Mohr’s integral is given by the area of the principal combiner (diagram of arbitrary shape) multiplied by the ordinate which its centroid makes with the linear diagram. The rule can also work vice versa. This process has been adopted in this work.
Where; EI = Flexural Rigidity of the section M = Bending moment due to externally applied load ̅M = Bending moment due to unit load at the point where the deflection is sought
δ21 = δ12 (Deformation at point 2 due to unit load at point 1 which is equal to deformation at point 1 due to unit load at point 2 based on Maxwell-Betti’s law). This is obtained by the bending moment diagram of case 1 combined with the bending moment diagram of case 2. This is shown below.
EICδ21 = (1/2 × 5 × 4 × 4) = 40
δ22(Deformation at point 2 due to unit load at point 2) This is obtained by the bending moment diagram of case 2 combining with itself. This is shown below.
EICδ11 = (1/3 × 4 × 4 × 4) = 21.333
Influence coefficients due to support settlement
Case 1 (Take a good look at the support reactions) δi∆ = -EIC(∆.R) R = Support reaction at point A = -1.0 ∆ = Support settlement in the direction of support reaction = -25 mm = -0.025m Hence δ1 ∆ = – 14647.5 × (-1 × -0.025) = -366.1875
Case 2 (Take a good look at the support reactions again) Horizontal support reaction at point A = R = 1.0 ∆ = Support settlement/movement in the direction of support reaction = 0 Hence δ2∆ = 0
The appropriate canonical equation now becomes;
105.208X1 + 40X2 = 366.187 40X1 + 21.333X2 = 0
On solving the simultaneous equations; X1 = 12.123 kN X2 = -22.730 kN
The final value of the internal stresses is given by the equations below;
Final Axial Force (Ndef) NA – NBB = (12.123 × 1) = 12.123 kN NBR – NCL = (-22.73 × -1) = 22.730 kN
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1.1 Introduction
A structure is stable when it maintains balance in force and moment. As a result, we know that from statics, if a structure is to be at equilibrium;
∑Fy = 0; ∑Fx = 0; ∑Mi = 0; ——————- (1.1)
Where;
∑Fy = Summation of the vertical forces
∑Fx = Summation of the horizontal forces
∑Mi = Summation of the moment of force components acting in the x-y plane passing through point i.
When the number of constraints in a structure permits the use of equation of statics (equation 1.1) to analyse the structure, the structure is said to be statically determinate. Otherwise, it is statically indeterminate, and additional equation which is derived from load-deformation relationship is used for analysis. For the records, there are two well known approaches to the analysis of indeterminate structures and they are;
1. Flexibility Methods – When the structure is analysed with respect to unknown forces
2. Stiffness Methods – when the structure is analysed with respect to unknown displacements
A structure may be indeterminate due to redundant components of reaction and/or redundant members. Note that a redundant reaction or member is one which is not really necessary to satisfy the minimum requirements of stability and static equilibrium. However, redundancy is desirable in structures because they are cheaper alternatives to determinate structures. The degree-of-indeterminacy (referred to as RD in this post) is equal to the number of unknown member forces/external reactions which are in excess of the equations of equilibrium available to solve for them.
1.2 Determinacy of rigid frames
In rigid frames, the applied load system is transferred to the supports by inducing axial loads, shear forces and bending moments in the members. Since three components of reaction are required for static equilibrium the total number of unknowns is equal to;
U = (3 × m) + r ————— (1.2)
Since we have three equations of equilibrium, we have (3 × n) equations, hence;
RD = (3m + r) – 3n – s ————— (1.3)
Where;
m = number of members
r = Number of support reactions
n = Number of nodes
S = Number of special conditions (e.g. internal hinge)
Another equation that can be used for calculation of degree of indeterminacy in frames is;
RD = R – e – S ———— (1.4)
Where;
R = Number of support reactions
e = Number of equations of equilibrium (i.e 3)
S = Number of special conditions (e.g. internal hinge)
Whenever;
RD = 0 (structure is statically determinate and stable
RD < 0 (structure is unstable)
RD > 0 (structure is statically indeterminate)
1.3 Solved examples
In the frames shown below, classify the following frames as statically determinate or indeterminate. All internal hinges are denoted by G.
Solution
Using equation 1.3
RD = (3m + r) – 3n – s
(a) m = 6, r = 6, n = 7, s = 1
RD = [3(6) + 6] – 3(7) – 1 = 2, Hence the frame is indeterminate to the 2nd order
(b) m = 5, r = 5, n = 6, s = 2
RD = [3(5) + 5] – 3(6) – 2 = 0, Hence the frame is statically determinate
(c) m = 4, r = 4, n = 5, s = 1
RD = [3(4) + 5] – 3(6) – 1 = 0, Hence the frame is statically determinate
(d) m = 3, r = 5, n = 4, s = 0
RD = [3(5) + 5] – 3(4) – 0 = 2, Hence the frame is statically indeterminate to the 2nd order
(e) m = 4, r = 5, n = 5, s = 1
RD = [3(4) + 5] – 3(5) – 1 = 1, Hence the frame is statically indeterminate to the 1st order
(f) m = 3, r = 3, n = 4, s = 1
RD = [3(3) + 3] – 3(4) – 1 = -1, Hence the frame is unstable
Alternatively using equation 1.4;
RD = r – e – s
(a) r = 6, e = 3, s = 1
RD = 6 – 3 – 1 = 2, Hence, the frame is indeterminate to the 2nd order
(b) r = 5, e = 3, s = 2
RD = 5 – 3 – 2 = 0, Hence, the frame is statically determinate
(c) r = 4, e = 3, s = 1
RD = 4 – 3 – 1 = 0, Hence, the frame is statically determinate
(d) r = 5, e = 3, s = 0
RD = 5 – 3 – 0 = 0, Hence, the frame is statically indeterminate to the 2nd order
(e) r = 5, e = 3, s = 1
RD = 5 – 3 – 1 = 0, Hence, the frame is statically indeterminate to the 1st order
(f) r = 3, e = 3, s = 1
RD = 3 – 3 – 1 = -1, Hence, the frame is unstable
A manhole (alternatively called utility hole, cable chamber, maintenance hole, inspection chamber, access chamber or confined space) is the top opening to an underground utility vault used as an access point for making connections or performing maintenance on underground and buried utility and other services including sewers, telephone, electricity, storm drains, and gas. The types of manholes that would be discussed in this article include sanitary sewer manholes and stormwater drain manholes.
Location of sewer manholes
Storm sewer manholes are usually provided at;
(a) Intersections of stormwater drains (b) Junctions between different sizes of stormwater drains (c) Where a stormwater drain changes direction/gradient (d) On long straight lengths at the following intervals; – For pipes with a diameter less than 600mm, provide manholes at a maximum interval of 40m – For pipes with a diameter between 600mm – 1050mm, provide manholes at a maximum interval of 80m – For pipes with a diameter greater than 1050mm, provide manholes at a maximum interval of 120m
In addition, manholes should wherever possible be positioned such that the disruption to the traffic will be at a minimum when their covers are lifted under normal maintenance operations.
Access opening of manholes
A manhole opening for man access should not be smaller than 550mm by 550mm. If cat ladders are installed in a manhole, the minimum clear opening should be 675mm by 675mm. A man access opening should be placed off the centreline of the stormwater drain for deep manholes, and along the centreline of the stormwater drain for shallow manholes with depths less than 1.2m.
Covers to manholes
Manhole covers should be sufficiently strong to take the live load of the heaviest vehicle likely to pass over, and should be durable under different weather conditions. Manhole covers should not rock when initially placed in position, or develop a rock with wear.
Step irons and cat ladders
Step-irons should be securely fixed in position in manholes, and should be equally spaced and staggered about a vertical line at 300mm centres. Cat ladders should be used in manholes deeper than 4.25m or where manholes are frequently entered. It is safer and easier to go down a ladder when carrying tools or equipment.
Forces acting on round manhole shafts
The forces acting on circular manhole shafts are;
– Lateral earth pressure – Hydrostatic pressure
Because both loads are uniformly distributed around the periphery of the manhole, no bending moment is experienced by the manhole section. The following equation may be used to calculate the total lateral pressure at a given depth (H).
p = (Ws × H × Ks × Cosi) + (Ww × H)
Where; p = total earth and hydrostatic pressure Ws = Unit weight of backfill material H = depth of manhole Ks = coefficient of earth pressure i = angle of internal friction of soil Ww = unit weight of water
In theory, the pressure (P) acts equally around the periphery of the manhole section, placing the ring in pure compression without introducing bending moments into the concrete section in the horizontal plane. The compressive stress in any section of the round manhole riser is given by the equation;
S = pD/2t
Where; S = compressive stress in the ring p = total lateral earth and hydrostatic pressure D = diameter of manhole t = thickness of the manhole
The frame shown above is supported with rollers at A, C and D, and pinned at point B. It is loaded as shown , and all columns have a cross-section of 30cm x 30cm, while the beams have a cross-section of 45cm x 30cm. Draw the bending moment diagram due to the externally applied load. Where necessary take E = 21.5 KN/mm2. In this case we are utilizing the force method, and we will apply Vereshchagin’s rule.
(1) First we reduce the structure to a basic system. A basic/primary system is a system that is statically determinate and stable. This is done by removing the vertical redundant supports at points C and D.
(2) We draw the bending moment for case 1 (Unit load applied at support C). X1 = 1.0
(3) We draw the bending moment for case 2 (Unit load applied at support D). X2 = 1.0
(4) We load the basic system with externally applied load and analyse it.
(5) We draw the bending moment diagram due to the externally applied load on the basic system.
(7) We write our appropriate canonical equation, and compute the influence coefficients using Vereshchagin’s rule. Solve the canonical equation and obtain X1 and X2.
(8) Substitute your final values into the Mdef equation.
(9) Plot your final moment diagram.
To download the full calculation sheet, click HERE.
The famous Maxwell-Mohr’s integral forms the backbone of the force method of structural analysis. It is also an important and universal method of computing displacements in beams and frames. It is more ideal for hand calculation purposes than the direct stiffness method which may involve large matrices.
The Mohr’s integral may be solved by direct multiplication and integration of bending moment equations (one linearly and the other of any arbitrary form), or by using the bending moment diagram multiplication/graphical method which is based on Vereschagin’s rule. Vereschagin’s rule is the graphical solution of Maxwell-Mohr’s integral.
In the paper downloadable in this post, a lot of formulas were derived for combining different shapes of bending moment diagrams for use in the graphical method than can be found in many structural engineering textbooks.
The diagram multiplication method presents the most effective way for computation of any displacement (linear, angular, mutual, etc.) of bending structures, particularly for framed structures. The advantage of this method is that the integration procedure according to Maxwell–Mohr integral is replaced by an elementary algebraic procedure on two bending moment diagrams in the actual and unit states.
This method was developed by Russian engineer Vereschagin in 1925 and is often referred to as the Vereschagin’s rule, in which the area of the bending moment diagram in the actual state multiplies the ordinate that its centroid makes with the unit state diagram in order to obtain the deformation.
In the article, the following examples can be found;
Example 1
The procedure for the combination of the two shapes shown above is presented in the screenshot shown below;
Example 2
In this case, we have to split the shapes at the point of contraflexure, so we have two shapes with areas A1 and A2 combining with the triangle below them. So as usual, we have;
To download the paper the full paper on equation formulation where series of shapes encountered in analysis have been combined, click HERE
To know how Vereschagin’s rule is applied in analysis, click HERE
