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Why Concrete Cracks?

Generally, it is assumed that cracks are due to some problems in the foundation, whereas it is not always correct and should not be considered failure of structure or improper design or bad quality work. Generally, 1/16 to 1/4-inch-wide cracks is acceptable limits.

The American Concrete Institute as per ACI 302.1-04 addresses this issue, even the best construction & concreting cannot prevent cracking in concrete, and 0% cracks is an unrealistic thing.

Causes of Cracks in Concrete

Causes of cracks in concrete can be many summarized as:

• Concrete expands and shrinks due to temperature differences
• Settlement of structure
• Due to heavy load applied or
• Due to loss of water from concrete surface shrinkage occurs
• Insufficient vibration at the time of laying the concrete
• Improper cover provided during concreting
• High water cement ratio to make the concrete workable
• Due to corrosion of reinforcement steel
• Many mixtures with rapid setting and strength gain performance have an increased shrinkage potential.

Types of Cracks in Concrete

The following figure shows types of cracks in concrete:

How to Prevent Cracks in Concrete Structures?

Preventive measures to avoid creation of cracks:

Preventive measures must be taken at the time of concreting and later to reduce cracks after concrete formation. Main factors are:

Reduce Water Content in Concrete:

A low water cement ratio will affect the quality of concrete. W/C ratio is weight of water to the weight of cement used. A lower w/c ratio leads to high strength in concrete and lesser cracks.

W/C ratio shall not exceed 0.5 in concreting, which reduces the workability of concrete which can be covered by use of plasticizer or superplasticizer. Less water content increases the durability of concrete.

Concrete expands and shrinks with changes in moisture and temperature. The overall tendency is to shrink. Shrinkage is the main cause of cracks, when concrete hardens it evaporates the excess water and thus shrinks, so lesser the water content, lesser is the shrinkage.

Cracking shrinkage in slabs is ½ inch per 100 ft. The shrinkage of concrete pulls the slab apart showing it as cracks on surface.

Proper Concrete Mix Design and use of Quality Materials

The concrete itself must be properly proportioned, and properly mixed. If you use too little cement, you can almost guarantee cracks. Using too much water will make the concrete weak, leading to cracking.

Use good quality aggregates so will produce lower shrinkage concrete. Hard, dense aggregate, using a large top size aggregate and optimizing the gradation of the aggregate is able to reduce the shrinkage of the concrete.

If the aggregate is of poor quality, maximizing the size, gradation, and content may have little effect on the concrete shrinkage. Mixing large aggregate with poor qualities to a mid-size aggregate with good properties may increase the shrinkage of the concrete.

Avoid the use of shrinkage-promoting admixtures (such as accelerators, dirty aggregate which increases water demand and using a cement with high shrinkage characteristics.

Finishing of Concrete Surface

Use proper finishing techniques and proper timing during and between finishing operations. Flat floating and flat troweling are often recommended.

Avoid overworking the concrete, especially with vibrating screeds. Overworking causes aggregate to settle and bleed water and excess fines to rise.

Don’t finish the concrete when there is bleed water on the surface, finishing leads the water back to concrete instead of evaporating thus leading to cracks.

Proper Curing of Concrete

Stop rapid loss of water from surface or drying of concrete due to hydration (liquid concrete converts to plastic and then to solid state) causes drying of the slab, so it’s recommended to cure the slab for several days.

As soon as the concrete on slab sets its general practice to make boundary with mortar on the slab and keep it filled with water. Cover slab with cotton mats soaked with water or spray on a curing compound also prevents loss of water.

The concrete should not be subjected to load during the curing period, which can last up to one month.

Proper Placement and Vibration of Concrete

Properly placed, vibrated, finished concrete reduces the chances of producing cracks. Properly vibrate to release entrapped air which later leads to cracks.

Proper Compaction of Soil to Prevent Settlement Cracks in Concrete

The area below the concrete slab has to be compacted properly and in layers so as to ensure against settlement of soil later. If the soil is left loose it will settle over time and create cracks on surface. This applies in the home as well as constructions on highways.

Providing Control Joints in Concrete

Control joints should be located at regular intervals so as to adjust the shrinkage of concrete. Generally, for 4-inch depth of slab joints are provided 8 to 12 ft. apart. Control joints are pre-planted cracks. An engineer should have an idea that concrete will crack at control joints instead of cracking any other location.

Some Other Preventive Control Measures for Cracks in Concrete:

• Applying good acrylic silicone sealer yearly to concrete works
• Avoid calcium chloride admixtures
• Prevent extreme changes in temperature
• Consider using a shrinkage-reducing admixture
• Warm the subgrade before placing concrete on it during cold weather
• Consider using synthetic fibers to help control plastic shrinkage cracks.
• Repairing Methods of Cracks in Concrete

Various types of Concrete Crack Repair Methodologies:

• Stitching
• Muting and sealing
• Resin injection
• Dry packing
• Polymer impregnation
• Vacuum impregnation
• Autogenously healing
• Flexible sealing
• Drilling and plugging
• Bandaging

To summarize, always prevention is better than cure. Prevention of concrete cracks give good quality, saves time, money and peace of mind to the owner.

Following articles might also be of interest to you:

☞ How to Prevent Plastic Shrinkage Cracking of Concrete?
☞ Cracks in Concrete Floors
☞ How to Repair Cracks in Concrete Floors?
☞ Cracks in Concrete Construction

The longest ocean-crossing bridge in the world, the Hangzhou Bay Bridge is an S-shaped stayed-cable bridge with six lanes in both directions that shortens the distance between Shanghai and Ningbo by 120 kilometers. The 36 kilometer long bridge required a great number of new techniques, new materials, new equipment and new theories due to the large scale and design of the project. It took close to 600 experts and a total of nine years to design the bridge. The Hangzhou Bay Bridge is expected to boost the economic development of the Yangtze River Delta, also called the Golden Industrial Triangle. Work on the bridge began in June 2003 and was completed in June 2007. The bridge was opened to the public in May 2008 and carried about 50,000 vehicles per day in its first year of operation. The total project cost was approximately $1.5 billion.

Construction

Hangzhou Bay is located in an area that has complicated geological conditions, is prone to typhoons and the Qiantang River Tide creates fast water and large waves. The tides frequently change direction making it difficult for construction vessels to maintain a position. Operations are only possible for half of the year and the seawater and strong currents have a corrosive effect on the steel infrastructure. Additionally, beneath the seabed lies large pockets of natural gas which would make any construction work in the area hazardous. Construction feasibility was a major concern and over 120 technical studies were carried out before any actual work started.

Hangzhou Bay Bridge (Image: Stuffpoint)

Engineers finally agreed on a cable-stayed bridge design because it could withstand the adverse conditions, multi-directional currents, high waves, and geologic conditions at the site. To ensure the bridge looks exquisite and grand, designers ensured that the aesthetics were prominent in every aspect from the shaping and color of each component, transitions between structures, choice of bridge deck systems, and structural lighting. From a bird’s eye view, the bridge is meant to look like a dragon nestled in the Hangzhou Bay.

To ward off the difficulty of onshore construction, engineers built some parts on land such as bridge foundations, piers and box girders. The steel piles were 71 – 88 meters in height and 1.5 – 1.6 meters in diameter. Numerous studies and tests were conducted by scientists before they finally settled on a spiral welding method as a suitable construction method for the steel piles. When they were completed, they were transported to the desired location and fitted in. Giant floating cranes with accurate anchoring devices and launching gantries were used to ship and erect these prefabricated components. Severe marine conditions also made it difficult for engineers to anchor barges and construction vessels. The floating cranes made it possible to transport the girder from the shore to the site and then anchor it stably to erect and install the precast concrete box girder.

In order to resist the corrosive effects of the seawater, engineers mixed a large quantity of coal ash and slag to the concrete to produce a special high-density variety. Silicone water repellents were also applied over the concrete and steel to protect the structure from salt water corrosion. The silicone water repellents are able to penetrate the pores of concrete and make the surface more resistant to water penetration. Engineers decided to tackle the natural gas issue by carrying out an exploration to investigate the distribution of the gas and the property of the soil during and after releasing the gas. The gas was then released before pile driving to avoid any disturbance to the soil, collapsing of ground or eruption and flaming of gas.

The entire structure of the Hangzhou Bay Bridge comprises of nine sections:

• The first section is the bank lead road to the north approach.
• The second section is the north approach that leads to the north navigable bridge; a cable-stayed bridge with twin diamond-shaped towers, double cable and steel box-girders.
• The third section has north piers with continuous 70 meter, post-tensioned, concrete box-girder spans with a total length of 1,470 meters.
• The fourth section is the middle bridge approach, laid on low piers with 70 meter, post-tension, concrete box-girder spans with a total length of 9,380 meters.
• The fifth section is the south navigable bridge – a cable-stayed bridge with an A-shaped single tower, double-cable and steel box-girders.
• The sixth section is the main span which is 318 meters.
• The seventh section features the south high piers that has continuous 70 meter, post-tension, concrete box-girder spans with a total length of 1,400 meters.
• The eighth section measures to a total of 19,373 meters, and is composed of three parts: an in-water section with girders and steel piles, a mud-flat section with girders and drill-shafts; and a land section with drill shaft foundations
• The final section is the Bank Lead Road at the south approach.

A Global Positioning System (GPS) was used to monitor the progress of the construction which required precise positioning and accurate placing of piles and pre-fabricated sections of the bridge.

The middle of the bridge has a service island targeted towards drivers who can enjoy a full range of services, including hotels, restaurants, petrol stations and a viewing tower. It also acts as a tourist site for the Qiantang River Tide. During the construction, this offshore platform acted as a living and working base for the offshore constructors. It also served as a relay station for offshore survey and communication as well as an offshore location for emergency rescue and maritime administration.

The Hangzhou Bay Bridge was built to serve for 100 years and can stand up to earthquakes that measure seven on the Richter scale. The bridge has a height of 62 meters, which allows fourth and fifth generation container ships to pass through in all conditions. The Hangzhou Bay Bridge is a stellar example of the possibilities allowed by new innovations and technology in the world today.

HEAVILY LOADED NIBS

Figure 1 Primary strut-and-tie model.

It is possible that a heavily loaded nib may require more than one load path to transfer a load safely. Strut-and-tie models can demonstrate alternative methods for reinforcing. However, it should be realized that the least direct paths will cause the most distortion and cracking, and should not be used for the serviceability state.

Figure 1 shows the most direct strut-and-tie (primary) model. The force paths are closest to that of an elastic model and will create the least internal distortion to achieve equilibrium. Figure 2 shows a secondary strut and tie model. This may be accompanied by distortion and cracking of the concrete before it can achieve equilibrium.

Figure 2 Secondary strut-and-tie model.

If the forces on the nib are too great for the primary model, it is reasonable to superimpose the secondary model to provide sufficient resistance for the total ultimate loads. However, it is important to ensure that the primary model is sufficient to resist the serviceability loads and provide crack control.

Comment: This approach was used for a major viaduct. The combination of the two models (see Figure 3) enabled all the reinforcement to fit—just!

SHEAR WALL WITH HOLES AND CORNER SUPPORTS

A multi-storey shear wall required so many openings (windows, doors, etc.) that the load path became very complicated. The designer assumed that the load would flow to the corners and then track vertically down the edge of the wall (see Figure 4a). In fact, since the wall was built in situ as a homogeneous structure, strain compatibility caused the load to flow back into the full width of the wall. The result was that several storeys of load were supported by a deep beam that transferred the load to its end supports (see Figure 4b).

Figure 3 Example of use of the two strut-and-tie models. (Courtesy of Gill Brazier.)

The limiting height of the natural arch of a deep beam (0.6 × span) was not considered (see Figure 4b) and this resulted in the omission in the design of much of the reinforcement needed for the bottom tie. Construction had reached several floors up by the time the mistake was recognised and this led to a redesign of the wall during construction and heavy remedial work. Each part of the wall required careful re-appraisal.

Figure 4a Incorrect simple modelling | Figure 4b Correct simple modelling Figure 4 Multi-storey shear wall

This led to the requirement of much more reinforcement at each floor level. The bottom corner reinforcement details required special attention to ensure that the junction between the tie and compression struts was adequately designed.

Figure 5 shows in a simple diagrammatic form how the force paths automatically flow out and back again. The assumed force path down the edges would not require ties at top and bottom, but without these the actual force path would cause large cracks to open up from the top and bottom surfaces. Even after cracking, the angle struts would still exist and so would the consequential horizontal component. Without sufficient tie force to resist, the support joint would move outward and eventually failure would follow.

Comment: The consequence of missing this simple principle of deep beam behavior before construction reached such an advanced state meant that it required the redesign of the structure and reprogramming of construction which were extremely costly.

DESIGN OF BOOT NIBS

Where nibs are attached to the bottom of a beam it is important to understand the load path of the forces. Figure 5.6 shows a typical section of such a nib.

Figure 5 Modelling deem beams

The conventional assumption for a short cantilever of dc and zc (shown in red in Figure 6) is unsafe for such a nib. The design compression zone for such a model would be close to the bottom face of the beam and likely to fall outside the beam reinforcement (both the links and main reinforcement). Strut-and-tie modelling is helpful to explain why this is so. The strut (shown in red in Figure 6) would just cause the cover to the reinforcement to spall off. The strut must be supported mechanically by the reinforcement of the supporting beam (shown in black in Figure 6). The effective lever arm becomes much smaller and the tension force in the nib top reinforcement much larger than assumed by the short cantilever approach.

It should also be noted that the force in the supporting links of the beam, F_t2d, is likely to be much greater than the applied load on the nib, F_Ed, to satisfy equilibrium. For the situation shown in Figure 6, it is conservative to assume the compression acts at the centroid of a triangular compression stress block. Hence the force in the link, in addition to any shear, may be calculated as follows:

Figure 6 Nib attached to bottom of a beam

Comment: There are probably many nibs of this type that have been designed incorrectly and survive because of built-in safety factors and the fact that the load assumed in the design has not occurred. The error described in this case study was found in the design of a nib for a very prestigious project. It was very fortunate that it was discovered before construction started.

The concept of force can be taken from our daily experience. Although forces cannot be seen or directly observed, we are familiar with their effects. For example, a helical spring stretches when a weight is hung on it or it is pulled. Our muscle tension conveys a qualitative feeling of the force in the spring. Similarly, a stone is accelerated by gravitational force during free fall, or by muscle force when it is thrown. Also, we feel the pressure of a body on our hand when we lift it. Assuming that gravity and its effects are known to us from experience, we can characterize a force as a quantity that is comparable to gravity.

In statics, bodies at rest are investigated. From experience we know that a body subject solely to the effect of gravity falls. To prevent a stone from falling, to keep it in equilibrium, we need to exert a force on it, for example our muscle force. In other words:

Characteristics and Representation of a Force

Figure 1

A single force is characterized by three properties: magnitude, direction, and point of application. The quantitative effect of a force is given by its magnitude. A qualitative feeling for the magnitude is conveyed by different muscle tensions when we lift different bodies or when we press against a wall with varying intensities. The magnitude F of a force can be measured by comparing it with gravity, i.e., with calibrated or standardized weights. If the body of weight G in Fig. 1 is in equilibrium, then F = G. The “Newton”, abbreviated N, is used as the unit of force.

The truss shown in Fig. 1 is loaded by an external force F. Determine the forces at the supports and in the members of the truss.

Figure. 2

A truss is a structure composed of slender members that are connected at their ends by joints. The truss is  one of the most important structures in engineering applications. After studying this chapter, students should be able to recognise if a given truss is statically and kinematically determinate. In addition, they will become familiar with methods to determine the internal forces in a statically determinate truss.

A structure that is composed of straight slender members is called a truss. To be able to determine the internal forces in the individual members, the following assumptions are made:

1. The members are connected through smooth pins (frictionless joints).

2. External forces are applied at the pins only.

A truss that satisfies these assumptions is called an “ideal truss”. Its members are subjected to tension or to compression only. In real trusses, these ideal conditions are not exactly satisfied. For example, the joints may not be frictionless, or the ends of the members may be welded to a gusset plate. Even then, the assumption of frictionless pin-jointed connections yields satisfactory results if the axes of the members are concurrent at the joints. Also, external forces may be applied along the axes of the members (e.g., the weights of the members). Such forces are either neglected (e.g., if the weights of the members are small in comparison with the loads) or their resultants are replaced by statically equivalent forces at the adjacent pins.

In this article we focus on plane trusses; space trusses can be treated using the same methods. As an example, consider the truss shown in Fig. 1. It consists of 11 members which are connected with 7 pins (the pins at the supports are also counted). The members are marked with Arabic numerals and the pins with Roman numerals.

To determine the internal forces in the members we may draw a free-body diagram for every joint of the truss. Since the forces at the pins are concurrent forces, there are two equilibrium conditions at each joint. In the present example, we thus have 7·2 = 14 equations for the 14 unknown forces (11 forces in the members and 3 forces at the supports).

A truss is called statically determinate if all the unknown forces, i.e., the forces in the members and the forces at the supports, can be determined from the equilibrium conditions. Let a plane truss be composed of m members connected through j joints, and let the number of support reactions be r. In order to be able to determine the m + r unknown forces from the 2j equilibrium conditions, the number of unknowns has to be equal to the number of equations:

2 j = m + r (Equation 1)

This is a necessary condition for the determinacy of a plane truss. As we shall discuss later, however, it will not be sufficient in cases of improper support or arrangement of the members. If the truss is rigid, the number of support reactions must be r = 3. In the case of a space truss, there exist three conditions of equilibrium at each joint, resulting in a total of 3 j equations.

Therefore,

3 j = m + r (Equation 2)

is the corresponding necessary condition for a space truss. If the truss is a rigid body, the support must be statically determinate: r = 6.

For the truss shown in Fig. 2a we have j = 7, m = 10 and r = 2 · 2 (two pin connections). Hence, since 2 · 7 = 10+4, the necessary condition (Equation 1) is satisfied. The truss is, in addition, completely constrained against motion. Therefore, it is statically determinate.

A truss that is completely constrained against motion is called a kinematically determinate truss. In contrast, a truss that is not a rigid structure and therefore able to move is called kinematically indeterminate. This is the case if there are fewer unknowns than independent equilibrium conditions. If there are more unknowns than equilibrium conditions, the system is called statically indeterminate. We shall only consider trusses that satisfy the necessary conditions stated in Equations (1) or (2), respectively. Even then, a truss will not be statically determinate if the members or the supports are improperly arranged. Consider, for example, the trusses shown in Figs. 2b and 2c. With j = 6, m = 9 and r = 3 the necessary condition (1) is satisfied. However, the members 7 and 8 of the truss in Fig. 2b may rotate about a finite angle ϕ, whereas the members 5 and 8 of the truss in Fig. 2c may rotate about an infinitesimally small angle dϕ. Each of the improperly constrained trusses is statically indeterminate.

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When a beam is subjected to applied forces, internal forces develop in the beam. Since the beam has supports and the system is in an equilibrium condition, the beam should resist these internal forces. This type of force is called shear force, or simply shear. The magnitude of the shear has a direct effect on the magnitude of the shear stress in the beam, and also on the design and analysis of the structural members.

Shear forces develop along the entire length of the beam, and their magnitude varies at each cross section of the beam. They actually act perpendicularly to the longitudinal axis of the beam. Our main objectives in finding the shear in beams are twofold. First, we have to know the maximum value of the shear in order to design the beam properly so that it can resist overall shear and will not fail when it is loaded.

The second objective is to calculate the values of shear along the length of the beam, so that we can find out where the beam fails under bending and where the shear value is zero.

As mentioned earlier, the designer must first know the magnitude of the shear force at any section of the beam to get the right picture of beam design. The magnitude of shear force at any section of the beam is equal to the sum of all the vertical forces either to the right or to the left of the section. To compute the shear force for any section of the beam, we simply call the upward forces (reactions) positive and downward forces (loads) negative. Then, the magnitude of the shear force at any section of the beam is equal to sum of the reactions minus the sum of the loads to the left of the section. This can be stated as:

Shear = Reactions - Loads

Example:

Find the shear force at sections C, D, and E for the simply supported beam shown (Fig. 1).

First, we calculate the reactions.

Using the moment equilibrium equation,

and
Point C

We usually represent a shear force by V, and we designate this shear force at C as V_c.
Vc = +6000 - 0 = 6000lb

Point D


Point E

The moment of inertia of an area is the capacity of a cross section to resist bending or buckling. It represents a mathematical concept that is dependent on the size and shape of the section of the member. The bending axis of a member is also the centroidal axis; therefore, the ability to locate the centroid of a shape is closely associated with moment of inertia. Engineers use the moment of inertia to determine the state of stress in a section, and determine the amount of deflection in a beam.

The definition of the moment of inertia of an area can be thought of as the sum of the products of all the small areas and the squares of their distances from the axis being considered. This gives

If we represent the moment of inertia by the letter I, then the moments of inertia with respect to the x and y axis axis are

Units

Moment of inertia is expressed in units of length to the fourth power. Although dimensionally speaking it seems unusual, it is just a mathematical abstract and is an important property in the design of beams and columns. We will see in the following examples the methods of calculating the moment of inertia for a given beam section subjected to bending. If we choose the unit of length as mm., then the unit of the moment of inertia is

mm² x mm² = mm⁴

While the elongation or contraction of axially loaded members along their longitudinal axes is usually of little consequence, beams may experience excessive deflection perpendicular to their longitudinal axes, making them unserviceable. Limits on deflection are based on several considerations, including minimizing vibrations, thereby improving occupant comfort; preventing cracking of ceiling materials, partitions, or cladding supported by the beams; and promoting positive drainage (for roof beams) in order to avoid ponding of water at midspan. These limits are generally expressed as a fraction of the span, L (Table 1).


Formulas for the calculation of maximum deflection are shown in Table 2, along with additional values for the recommended minimum depth of reinforced concrete spanning elements.


The maximum (mid span) deflection, Δ, of a uniformly loaded simple span can also be found from the equation:

where w distributed load (lb/in. or kips/in.), L span (in.), E modulus of elasticity (psi or ksi), and I moment of inertia (in 4 ). When using Equation 8.1 with L in feet, w in lb/ft or kips/ft, E in psi or ksi (compatible with load, w ), and I in in⁴, as is most commonly done, multiply the expression by 12³ to make the units consistent.
 

If a member is loaded beyond its ultimate stress, it will fail or rupture. In engineering structures, it is essential that the structure not fail. Thus, the design is based on some lower value called allowable stress or design stress. If, for example, a certain type of steel is known to have an ultimate strength of 110,000 psi, a lower allowable stress would be used for design, say 55,000 psi. This allowable stress would allow only half the load the ultimate stress would allow. Allowable stress values are different for different materials, and they are tabulated and recommended by the International Building Code Association.
The ratio of the ultimate stress to the allowable stress is known as the factor of safety.

factor of safety = ultimate strength / allowable stress

Image courtesy: http://www.smpes.com

Example

Determine the required size for a steel rod to support a tensile load of 50,000 lb if the allowable tensile stress of the steel is 25, 000 psi.

Solution

σ = F/A

or

A = F/σ = 50, 000lb/25, 000psi = 2in²
A = πD2/4 = 2in²
Solving for D, we have:
D = 1.6 in

When the ratio of the longer to the shorter side (L/S) of the slab is at least equal to 2.0, it is called one-way slab. Under the action of loads, it is deflected in the short direction only, in a cylindrical form. Therefore, main reinforcement is placed in the shorter direction, while the longer direction is provided with shrinkage reinforcement to limit cracking. When the slab is supported on two sides only, the load will be transferred to these sides regardless of its longer span to shorter span ratio, and it will be classified as one-way slab.

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