Curious Civil Engineer

  Binding wires are used to tie steel bars together. These wires are playing a significant role in maintaining the reinforcement stability and rigidity. Wires are used to tie the steel bars at intersections points. By tying the steel bars together, we ensure that the steel bars will not move from their locations during construction or during the time of concreting. In slabs, binding wires are used to tie longitudinal and transverse bars together. In columns, it is used to tie vertical bars with stirrups.   Figure 1 There are various types of binding wires such as  Black annealed baling wire, Stainless Steel Binding Wire, and PVC coated binding wires.  Black annealed baling wire used to tie black steel. The popular size for black wire ranges from 16 to 22 gauges. Stainless steel binding wires are used to bind stainless steel reinforcement. Stainless steel reinforcement is used in a harsh environment where the black steel gets rusted quickly. Therefore, stainless steel reinforcement is u

  Determine the distance "a" so the displacement at the edge of the beam equals the displacement at the center of the beam? Figure 1 The conjugate beam will be, as shown in figure no:2. The pin and roller at A and B will become internal pin. The free ends will become fixed, as shown in figure no:2. We need to determine the M/EI diagram for the real beam, which will be the load in the conjugate beam, so we can determine the distance “a” that will cause an equal displacement at the end and center of the beam.  Figure 2 Using the equilibrium equations, we will determine the reaction at point A and point B. Figure 3 After determining the reactions at support A and B, we can now draw the shear and moment diagram. The M/EI will be the load on the conjugate beam as shown in figure no:4 and 5. Figure 4 Figure 5 Now we need to equal the moment at the end beam and the center of the beam. we assume the distance between A and B equals L1 and L1=L-2*a. To simplify the solution, we will ta

  Glass fiber-reinforced polymer (GFRP) reinforcement has emerged as a new opponent to conventional reinforcement. The GFRP owns a higher tensile strength, non-corrosive reinforcement, lighter weight, and higher strength-weight ratio. The corrosion of reinforcement can severely damage the structure by causing the concrete to cracks or spall. The Maintainance of damaged structures can significantly be costly. The GFRP is a non-corrosive reinforcement. Therefore, the use of GFRP can eliminate the deterioration of structure due to reinforcement corrosion. The use of GFRP has increased recently for many applications such as bridge deck, pavement, walls, and other applications. However, the use of GFRP is limited due to the lack of information about the long-term performance of GFRP. Figure 1 GRFP bars have many advantages, such as tensile strength. The tensile strength of GRFP bars is higher than the conventional reinforcement. Table no:1 shows the mechanical properties of GRFP bars. We ca

  TMT bars are thermally mechanically treated steel bars with high strength and good mechanical properties compared to mild steel bars. TMT bar is a new enhanced version of mild steel bars. The treatment process of TMT bars includes 3 stages: quenching, self tempering, and atmospheric cooling. The quality of produced TMT bars depends on the quality of raw materials, quality of rolling mill, and quality of quenching and tempering process.  Figure 1 The process of manufacturing steel is divided into two stages. The first stage is molten iron production. In this stage, iron-sand and coal are heated in a multi-hearth furnace—heating the raw materials done in stages. First, the furnace pre-heat the material to reduce the volatile matter present in coal from 44% to 9%. There are 12 hearths in each furnace. The mixture temperature will be increased gradually through these hearths. Finally, the temperature will be adjusted to 620 C.  Figure 1 The output after removal of the impurities introduc

 Problem no:1, Determine the slope and displacement for the following beam at point C? Figure 1 In this example, the pin support will stay the same in the conjugate beam. While the roller will change to hinge and the free end to a fixed support. So the conjugate beam will be the same as shown in figure no:2. Figure 2 Now we need to determine the moment diagram for the real beam.  First step is to determine the force at support A and B by the use of equilibrium equations: Figure 3 Now we can draw the moment diagram for the real beam. To draw the moment diagram, we can draw the shear first, and the moment diagram will be the area of shear. The moment diagram for the real beam will be, as shown in figure no:4. Figure 4 After drawing the shear and moment diagram, our conjugate beam will be loaded with the M/EI diagram of the real beam, as shown in figure no:5. Figure 5 now to determine slope and displacement at c we need to calculate the shear moment at c by the use of equilibrium equatio

 Ductility is an important property of steel reinforcement. Ductility is the ability of the material to undergo plastic deformation before failure. The ductility of reinforcement is related to the elongation property. The ductility of reinforcement ensures safer and durable structures. Material that undergoes little or no plastic deformation is known as a brittle material. Brittle reinforcement can cause sudden structure failure because it doesn't undergo any plastic deformation before failure.  Figure 1 The ductility factor for reinforcement can be computed using the following equation: µ=ϵu/ϵy where, µ is the ductility factor ϵu is the ultimate strain ϵy is the yield strain In figure no:1-a, we can see the stress-strain curve for mild steel. The mild steel has well-defined yield stress and strain, as shown in the stress-strain curve. The reinforcing bar can recover all the elongation if the applied stress is lesser than the yield stress. This portion of the curve is known as the

 The Conjugate-Beam method can be used to determine the deflection and slope for beams. In this method, the shear in the conjugate beam at any point will be the real beam slope at this point. While the moment in the conjugate beam will be the displacement of the real beam at the same point. The conjugate-beam will be loaded with M/(EI), where M is the moment derived from the real beam, E is the modulus of elasticity of the beam material. I is the moment of inertia of the beam. When we draw the conjugate beam, we need to pay attention to supports. The support can be different in the conjugate beam compared to the real beam. Table no:1, show the support in the real beam and the corresponding support in the conjugate beam. Table 1 In table 1, we can notice that the pin and roller support stay the same for the conjugate beam. The reason behind this is that the displacement at pin supports equal zero. The corresponding displacement in the conjugate beam is the moment, and the moment at pin

There are many types of steel structures, and it ranges from single story to multi-stories structures. The simplest form of steel structure is composed of the roof truss or open web joist supported on concrete columns or masonry walls, as shown in figure no:1. The single-bay frame is an alternative to roof trust and open web joist. Figure no:2 shows a single-bay frame structure.   Figure 1 Figure 2 Another type of steel structure is framed structures that consist of beams, girders, columns, and diaphragms. This type of structure can be single or multistory. Figure no:2 shows the second story of a steel building. We can see the floor diaphragm spanning east-west above the supporting beams. The floor diaphragm consists of concrete fill and steel deck, as shown in figures no:4 and 5. The beams span north-south and supported on girders, as shown in figure no:4. Figure 3   Figure 4 Figure 5 These structures should be designed to resist lateral loads produced by winds or eart