Lab Experiment on the Two-Hinged Arches and Elaborating the Distribution of Their Internal Forces - Article Example

Name Institution Subject Instructor Date Abstract Ideally, there three main types of the arches that used in engineering practice. These include the hinge less arches, two and the three hinged arches. In the 19th century, the three hinged arch was mainly used for construction in performing the structural analysis n the structures that exhibited long spans. . However, with the development of engineering practice and technology, the three hinged arches have been transformed to a two hinged arch as well as the hinge less arches. In practice, the two hinge arch exhibit at an indeterminate structural structure that is related to the degree of one. The reaction that exhibits horizontally is related to be redundant as t is assumed that it would perform little work. This paper discusses the lab experiment performed on the two hinged arches and elaborating the distribution of their internal forces. 1 Introduction Arches have and an advantage over a beam in that they are capable of carrying many loads as compared to the beams. These have been utilized in the construction of the bridges, and the aqueducts (Vaidyanathan and Perumal 127). It functions by the virtue that, the pressure would act in a downward position that would make the arch to compress to together hence making to stay intact rather than pulling apart. In the event that it has not been restrained properly, the horizontal component which is the thrust would make it to collapse. 1.1 Two-pin arches Arches are continuous- curved, structural element that support loads in a plane. Arches can carry much more load than beams thus being the more advantageous to use. Historically their construction has been much more easily using small easily carried blocks or bricks rather than using massive stone beam or rental. Two-pin arches are commonly used as this form of structure has the pinned supports that allow the rotation of the arch at the ends under load. Temperature changes and horizontal support a settlement also makes them relatively flexible and less prone to developing high bending stresses. Due to their four reaction forces and three equilibrium equations these structures are indeterminate to the first degree of statically indeterminacy. Figure 1 Two-pinned Arch (Vaidyanathan and Perumal 129) A fourth equation may be written considering deformation of the arch. HB, the unknown redundant reaction is calculated by noting that the displacement of pin B is zero. Generally the horizontal reaction in the two pinned arches evaluated using straightforward application of least work theorem, which states that the partial derivative of the strain energy of a statically indeterminate structure with respect to statically indeterminate action should vanish. Thus to find, horizontal reaction, one should develop an expression for strain energy. Essentially, any part of the arch is subjected to shear force V, ending and the axial forces as shown Figure 2 force distribution of an arch (Vaidyanathan and Perumal 120) The strain energy due to the bending moment is calculate by the use of the below expression This expression is same as that used in the case of straight beams. Though here, the integration needs to be evaluated along the curve arch length. The parameters above are: s- Length of the center line of the arch I-moment of inertia of the of the arch cross section E-Young’s modulus of the arch material The strain energy due to shear is very small as compared to that due to bending and hence is always neglected in the analysis. The strain energy due to axial compression can be appreciated in cases of flat arches. This is done by the use of the expression The total strain energy of the arch is given by According to the principle of least work . Solving this equation, the horizontal reaction H is evaluated. 1.2 Temperature Effect When an arch undergoes a uniform temperature change of Tic, its span will increase by lit allowed to expand freely. Α is the coefficient of thermal expansion of the arch material. A horizontal force is induced at the support as the temperature increases since the arch is restrained from horizontal movement. Figure 3 horzontal movement of arches Applying Castiglione’s first theorem, Solving for H, H= The second term may be neglected because the axial rigidity is quite high. The equation becomes H= In the analysis of the two hinged arch, there are four reactions that have unknown reactions. The equations of equilibrium required to solve these reactions are only three. This makes the statically indeterminacy to be one from the two arches that are hinged. Figure 4 statically indeterminacy arch (Vaidyanathan and Perumal 79) The fourth equation is obtained by raking the arch to be deformed. The horizontal displacement would aid in deterring the redundant reaction. The reaction that emanates horizontally, is evaluated from the taking the least work done theorem. It highlights that, the partial derivative of the strain obtained would be statically indeterminate as related to e the structure that exhibits an indeterminate statically action that is disappearing. This calls for establishment of the strain energy so as to obtain the reaction that takes place horizontally. N a given arc, the section would be subjected to a bending moment M, shear force V and the axial compression N. The bending strain energy UBS would be obtained from the expression (33.1) Where s- centerline arch I -moment of inertia that occurs across the arch E -the material’s young modulus This equation may also be used in straight beams. Contrary, in this case, the integration process is conducted on a curved beam. The sheer force exhibits less strain energy as compared to the strain energy hat would be produced due to bending. The axial compressions would usually provide the strain energy where there are arches that are flat. 1.3 Two hinged arch that are symmetrical The arch responds to the loading so as the lateral displacement is avoided. In a scenario where one of the supports is equated to a roller, the arch would turn into a curved beam. This ensures that the anchorage system that is lad has to overcome the lateral thrust that is exhibited. The force of gravity also determines the spread of the forces across a given span of a full arch. This would give the wind, dead and the live loads. Exp 1 Graph of horizontal reaction against distance Ha (calculation of theoretical reaction for each position.) Figure 5 graph indicating displayed horizontal reaction against distance 1.3.1 Where load would be placed to attain maximum horizontal reaction. Do the formulae predict accurately the behavior of the arch? It is evident that, the horizontal displacement is almost the same. The maximum deflection would be exhibited the loads at the free ends with the maximum deflection taking at the centre taking in an upward directions. There is a slight variation in the theoretical values as relates to the experimental values. This may be explained by errors that may have occurred during the experiment. Figure 6 Graph of experimental value and calculated influence value against fraction span The challenge that may occur while the two pin arch occurs in a resisting load is the aspect of buckling. This requires the analysis to take into considerations the effect if buckling, stress and deflection. To determine the reaction of the forces, it would require that more reactions that are unknown be established as compared to the reactions of the equilibrium that are established. Figure 7 A graph representing the reaction forces, moments against distance The theoretical moment presumes a sine wave when plotted. . 2 Exp 3 Lab 2 2.1Redundant Truss A truss consists of the members which connect to each other by pin joints. At h end, they are supported by rollers of the hinged joints. This makes them to be statically determinate. The forces applying at a node and at each point applies the Newton laws of forces. The sum of the forces and moments that ct on a given node that have been subjected to the external forces have to be zero. A truss is a structure made up of two force members all pin-connected to each other. The members support each other although some don’t carry the force and they should be eliminated in calculating the forces. These members are called redundant or indeterminate. We can say that a truss is statically redundant when the total number of reactions and member axial forces exceed the total number of static equilibrium equation. The degree of redundancy can be determined from inspection in simple planar truss. The following formula to calculate indeterminacy of static planar whenever things become tedious. In this case, m, j and r are number of members, joints and unknown reaction components respectively. The redundancy in a truss can be either external or internal or both. Externally redundant planar truss have number of reactions exceeding the number static equilibrium equations available and has (2j-3) members. Internally redundant truss has exactly three reaction components and more than (2j-3) members. Finally a truss that is both internally and externally redundant has more than three reaction components and also has more than (2j-3) members. The simple method of analyzing redundant truss by force method is similar to the redundant beam analysis. You start by determining the degree of static redundancy of the structure. Identify the number redundant reactions equal to the degree of indeterminacy. The redundant should be so selected that when the restraint corresponding to the redundant is removed, automatically the resulting truss is statically determinate and stable. Redundant is selected as the reaction component in excess of three and the rest from the member forces. But one could choose redundant actions completely from member forces. Loading a beam with external load makes it subject to shear and moment force in each given strain. The bending moment would usually, tend to deflect the beam together with the stresses that are found internally so as to make its resistant to the bend. In analyzing this scenario, it is assumed that the materials take an aisotropic nature as it has a homogenous structure with elastic properties that are distributed in all directions. The beam would also be assumed to obey the hooks law, as it would be stressed till the elastic limit is attained. The regions that experiences bending would remain a plane even after the bending has taken place. And lastly, it would have the Young modulus of elasticity of the force of compression being the same as those of the tension forces. The stress due to bending may be obtained from the equation F=(M/I)y in which case, M – moment f- The stress exhibited as the sections I- Moment of inertia II- Y- Fibre distance considering the neutral axis. The digital stain indicator would operate by considering the balancing of the bridge network as well as the internal dummy arms that are involved. His gives the final strain value that is displayed at the screen. Figure 8. Graph of experiment strain against load. Figure 9. Graph of experimental force against load Figure 10 Graph of experimental force against load. From the experiment, when the wire has been stretched in a given length and diameter, wire stretches elastically. This measures the change in resistance that occurs due to the change in the load that is applied. Ideally, when a compressive load is applied on a given column, the member s may react to it depending on the nature of the material. It may crush, buckle or crush. 3 Results The deflection exhibited n the node is obtained by the equation [2] Where T represents the forces of the members U- The member under the he load that is at the point of location L-is the length of a given member, a area E-young modulus 3.1 Calculation of the theoretical members with the load of 250N adds values and compare with the theoretical values    -HA(1)+250(2)=0 HA=500  HA +HB=0 This means that HB=-500N  -VB-250=0 VB=-250N 3.2 Experimental force Member 1 F= =2.10105N/mm2 10-6 F=37407N 3.3 Explain the reading of members are the strain gauges effective transducers for the measurement of the forces in the framework The fifth member expresses a redundant nature as it does not offer compressive or tensile forces. 3.4 Does the frame work comply with the pin joint theory even though the points are not truly pin jointed? The results indicate that the framework complies with the pin joint theory. This is indicated by the way the forces comply with the Newton’s law in which the summation of forces adds up to zero. 3.5 Comparison of the results which theoretical values The experimental forces differences with the theoretical value. This was exhibited in the members 1, 2, 5, 6 and 8 due to the effect of parallax. This also may relate to the fact that the equipment may not be functioning correctly or not properly maintained. Additionally, the environmental condition may also alter the results as they may respond differently to the wind or vibration. Members 3, 4, and 7 exhibit almost similar results as the theoretical values. The experimental strain is directly proportionally to he load exerted making it o be prone to failure in case the load exceeds the strength of the structure. 3.6 Comparison of member’s forces and comment on their safety and structure From the experiment data, they exhibit negative forces that indicate the compression while the positive forces indicate the tensile at the members. Structures that are built considering trusses that are more than the minimum required would withstand the structure even if some of the adjacent members fail or are deflected. This is so as the forces that act on the members depend on the relative stiffness of the adjacent members as regards to the equilibrium condition. This makes it to be economical. The failure would occur in the scenario where, the load is more than the ability of a structure to withstand it. This may be estimated by taking the probability density function in conjunction with the ability of structure and the load to be high in the whole design. 4 Conclusion The graph of the load against the deflection gives the comparison in which the results are related to the linear. While in construction, a prototype maybe constructed so as to ensure that the structure is stable. In the case where there is redundant forces, the structure would fail in the scenario here he load is more than the capability of the structure to withstand. The external energy has to be similar as the internal energy. The loads under the given frames are calculated with the load taking place in the redundant numbers. References Vaidyanathan, R, and P Perumal. Comprehensive Structural Analysis: Volume Ii. New Delhi: Laxmi Publications, 2004. Print. Read More

Though here, the integration needs to be evaluated along the curve arch length. The parameters above are: s- Length of the center line of the arch I-moment of inertia of the of the arch cross section E-Young’s modulus of the arch material The strain energy due to shear is very small as compared to that due to bending and hence is always neglected in the analysis. The strain energy due to axial compression can be appreciated in cases of flat arches. This is done by the use of the expression The total strain energy of the arch is given by According to the principle of least work .

Solving this equation, the horizontal reaction H is evaluated. 1.2 Temperature Effect When an arch undergoes a uniform temperature change of Tic, its span will increase by lit allowed to expand freely. Α is the coefficient of thermal expansion of the arch material. A horizontal force is induced at the support as the temperature increases since the arch is restrained from horizontal movement. Figure 3 horzontal movement of arches Applying Castiglione’s first theorem, Solving for H, H= The second term may be neglected because the axial rigidity is quite high.

The equation becomes H= In the analysis of the two hinged arch, there are four reactions that have unknown reactions. The equations of equilibrium required to solve these reactions are only three. This makes the statically indeterminacy to be one from the two arches that are hinged. Figure 4 statically indeterminacy arch (Vaidyanathan and Perumal 79) The fourth equation is obtained by raking the arch to be deformed. The horizontal displacement would aid in deterring the redundant reaction.

The reaction that emanates horizontally, is evaluated from the taking the least work done theorem. It highlights that, the partial derivative of the strain obtained would be statically indeterminate as related to e the structure that exhibits an indeterminate statically action that is disappearing. This calls for establishment of the strain energy so as to obtain the reaction that takes place horizontally. N a given arc, the section would be subjected to a bending moment M, shear force V and the axial compression N.

The bending strain energy UBS would be obtained from the expression (33.1) Where s- centerline arch I -moment of inertia that occurs across the arch E -the material’s young modulus This equation may also be used in straight beams. Contrary, in this case, the integration process is conducted on a curved beam. The sheer force exhibits less strain energy as compared to the strain energy hat would be produced due to bending. The axial compressions would usually provide the strain energy where there are arches that are flat. 1.3 Two hinged arch that are symmetrical The arch responds to the loading so as the lateral displacement is avoided.

In a scenario where one of the supports is equated to a roller, the arch would turn into a curved beam. This ensures that the anchorage system that is lad has to overcome the lateral thrust that is exhibited. The force of gravity also determines the spread of the forces across a given span of a full arch. This would give the wind, dead and the live loads. Exp 1 Graph of horizontal reaction against distance Ha (calculation of theoretical reaction for each position.) Figure 5 graph indicating displayed horizontal reaction against distance 1.3.1 Where load would be placed to attain maximum horizontal reaction.

Do the formulae predict accurately the behavior of the arch? It is evident that, the horizontal displacement is almost the same. The maximum deflection would be exhibited the loads at the free ends with the maximum deflection taking at the centre taking in an upward directions. There is a slight variation in the theoretical values as relates to the experimental values. This may be explained by errors that may have occurred during the experiment. Figure 6 Graph of experimental value and calculated influence value against fraction span The challenge that may occur while the two pin arch occurs in a resisting load is the aspect of buckling.

Read More