Loaded Beams, Columns and Struts - Assignment Example

Table of Contents SUMMARY 1 REPORT AND ANALYSIS 3 Task 1 3 Task 2a): Determine the moment of resistance of the beam 6 Task 2b): New depth of beam and tension area 7 Task Three 9 Task 4a): Buckling Load of the Strut 11 Task 4b): New buckling Load of the Strut 11 Task 4c): Types of end conditions 11 Task 5a) Strut experiment (pinned at both ends) 13 Objectives 13 Introduction 13 Apparatus 14 Procedure 15 Results 15 Discussion 15 Conclusion 16 Task 5b) Comparing the experimental and theoretical values of critical loads 16 Task 5c): Organized and detailed work showing the way the experiment was conducted. 16 Discussion 17 Conclusion 18 REFERENCES 18 Appendices 19 SUMMARY Beams, columns and struts have been used for a wide period of time in the construction sector. The beams fail by tension and they have to be assessed for their suitability for use in a project. For example the material to be used, the dimensions of the beam and the condition of the end supports all affect the strength of the beam. Deflection of the beam at various loadings is calculated as well as the correspondent slope. In the tasks given, the beams slope and deflection has been calculated. Beams can be constructed of reinforced concrete which provides a higher stability. In beams moments develop along the length. The maximum bending moment is important because it is at this point that the beam will fail. A bending moment diagram is drawn. A shear force diagram is also drawn below the beams sketch to show the point of zero deflection and maximum deflection. This information assist the engineer in determining the number and type of the steel bars to use when constructing the reinforced concrete beam. Columns and struts are members that fail in compression. A strut is slender and thin compared to a column. It fails by buckling. When subjected to a load the column is observed to shorten. However, after the critical load is attained the strut buckles. A behavior that is sudden and can result in many losses and should be avoided by carrying a proper and accurate design. The general formula for the Critical Load is where N is the effective length factor, E is modulus of elasticity, I is moment of inertia and L is the actual length of the strut. The engineer is required to accurately estimate the critical load as well as the expected loading of the strut so as to determine whether the strut can carry the load. From the formula above it can be said that the material used affects the stability of the strut. The length has an inverse relation with the critical load while the conditions at the end support conditions of the strut. In the tasks we have assessed columns with different end conditions and found out the differences in their critical load REPORT AND ANALYSIS Task 1 Moment about R1 to find R2 Moment about R2 to find R1 From Where: To find C and D, set conditions Use deflection equation to solve for C and D Then, 0=0+0+0+0+D Calculate slope (use slope equation) Slope at LHS of beam, x = 0 Find ’I’ Find ‘EI’ Where: Slope at RHS of beam, x = 1 Calculate deflection at the middle of the beam (deflection equation) Task 2a): Determine the moment of resistance of the beam First find “h”: Where: b = breadth of section = 250mm h = distance to neutral axis m = modular ratio = 16 A = total area of steel reinforcement = 0.0013m2 d = depth of section = 460mm Simplify this to: We can use aquod ratio equation to find the valve for “h” using: Therefore, h = we can now use the following equation to find the moment of resistance (based on compressive forces) Therefore, the moment of resistance (compressive) =. We can now use the following equation to find the moment of resistance (based on tensile forces) Moment of resistance (tensile) = Therefore, the safe moment of resistance which the beam can carry within both limiting stress values is Task 2b): New depth of beam and tension area Transpose for Therefore: To find d: Using: Now, the service loads have been up rated by 25%, therefore: To find m, we need to multiply the moment of resistance of the concrete reinforcement () calculated previously by 1.25 therefore: Substitute To solve for “d” Therefore: To find “A”: Therefore, the new area of steel for tension reinforcement required is Task Three On the tensile edge: This can also be rewritten as: Where: Force = 400kN Therefore: Therefore, the stress on the tensile edge = 1.416MN/m2 (tensile) On the compressive edge: Where: Force = 400kN Therefore: Therefore, the stress on the compressive edge = -4.243MN/m2 (compressive) The neutral axis mathematically A + B = 0.6 A = 0.6 – B Therefore, the distance from the bottom edge of the circle (column) =0.4498m Task 4a): Buckling Load of the Strut First find “I” Now find buckling load Because the bottom of the Strat is built into a ground socket and the top is unrestrained, the value for n = 0.25 If the end conditions of the same strut were pinned at both the top and bottom, calculate the new buckling load. “I” is the same = It will be stronger. Task 4b): New buckling Load of the Strut Since the end conditions are the same, The buckling load for this case, the ends are both pinned thus n=1. For such a strut there is no bending moments at the ends. Therefore; The pinned strut at both ends is much stronger than the strut that has one end fixed and the other unrestrained. Task 4c): Types of end conditions i. Struts with both ends pinned For a pinned strut, the effective length. Due to the ends being pinned, the column does not bend at all as the moments are shed off. The buckling load is calculated by the formula ii. One end fixed and other end free. The struts deflects in the unrestrained end through a distance. The effective length and therefore the buckling load is calculated as follows; iii. Both ends fixed The strut has the highest deflection at the center of the strut. These moments act at the opposite direction from the direction of the slope at the ends. Effective length . The buckling load is calculated as follows; iv. One end fixed and the other pinned The struts deflect such that the greater deflection is closer to the fixed end. Fixed end moments develop at the fixed end when a horizontal force is applied in the pinned end. The effective length and therefore buckling load can be calculated as Task 5a) Strut experiment (pinned at both ends) Objectives 1. To determine the critical load that would result to buckling 2. To find out the differences between the theoretical value and practical value for the critical load. Introduction Struts are slender columns which fail by buckling. The buckling is sudden and occurs only when the critical load is reached. At the point of buckling the strut undergoes permanent deformation which reduces the stability of the strut and can result in a lot of destruction. For this reason, it should always be avoided. The strut should be designed for a load less than the critical load. The strut has a variety of uses as in the columns in steel structure and other constructions. The strut are used as pistons in hydraulic systems and in lifting gears. Resistance of a strut to buckling is dependent on the diameter of strut, slenderness of strut and the material used. The fixity of the ends also determine the resistance to buckling. The ends either be 1 Fixed at both ends 2 Pinned at both ends 3 Fixed at one end and free at the other end 4 Fixed at one end and pinned at the other In this experiment we shall focus on a strut that has both ends pinned or hinged. The effective length Le is equal to the actual length L which shows that the deflection of the strut occurs with the entire length of the strut. There are no end moments at the ends as they are shed off. The critical load is found theoretically by the formula below which is referred to as Euler’s formula Where E is the rigidity modulus, I the moment of inertia and L the actual length of the strut. Apparatus 1 Force measurement device: - The principle of hydraulics is applied when measuring the force applied on the specimen. Pressure produced by the load is measured by a gauge that is attached to the device and the values displayed for recording. Any friction that might disturb the values is avoided by direct contact of rod to the device. 2 Specimen holders: - The specimen rod is held between the clamps in this device both at the top and at the bottom using screws. The specimen in this case is ensured to have pinned supports at the ends. 3 Deformation measurement device: - There is a gauge that is attached to the specimen to measure the lateral deflection after the specimen is loaded. 4 Load device: - The device is used to load the force in the specimen. After each increment, the value of deformation is recorded beside the corresponding load. 5 Specimen: - The test device consists of the devices above interconnected together to form the test device. The height of the specimen can vary and the cross bars have been designed to be adjustable for this reason. The arrangement can be horizontal or vertical which subject to the person carrying out the test. Procedure 1 The test device was set up vertically, the force gauge placed perpendicularly. 2 The specimen inserted to the notches and clamped lightly. The specimen is 0.8m long with a cross section width of 12mm and a depth of 12mm. The elastic modulus of the specimen is 200GN/m² and pinned on both ends. 3 Cross bar fastened along the guide column and a small distance left between the two for measurement and observation of deformation. 4 Measuring gauge now placed at 90° to the buckling direction in the middle portion of the specimen where deformation will be maximum thus buckling 5 The specimen is loaded with force at intervals of 1000N initially after which the value reduces to 5N incrementally past 5000N. 6 The loading is stopped when failure is observed. Results After the experiment, the critical load was recorded in the table below as shown. strut Critical load Pinned at both ends 5275 N Discussion Critical load is the load beyond which the strut will fail. The struts fail by buckling. The experimental critical load has been found to be 5275N which is less compared to the theoretical value of 5379N. The difference is 104 N. the environment of performing the experiment could be different from the one that the theoretical value was based on. In addition, there could be human error, apparatus error or reading error. The deviation from the theoretical value is not as large and is within acceptable limits. Conclusion The experiment was successful because all the objectives of the test were achieved. The experimental critical load was found to be 5275N and the difference from theoretical found to be acceptable. Task 5b) Comparing the experimental and theoretical values of critical loads The experimental critical load is 5275N which is less than theoretical value of 5329N. The difference is 104N which expressed as a percentage is 1.95% error. The specimen was a bit uneven thus not straight leading to some error. The error incurred was also in the level of bubble which on top level of test machine was not perfectly horizontal. There was observed misalignment of the load when being added incrementally on the strut resulting to error. Human error in recording data was another source of error in the strut experiment. The environment conditions within which the experiment was done also affect the results. For example a high temperature would result to expansion of specimen which could consequently result in the critical load changes. Task 5c): Organized and detailed work showing the way the experiment was conducted. time Task title Description 0-30minutes Setting up of the apparatus The testing machine and all the components were fixed together and set ready for the experiment. Specimen dimensions verified and recorded 30-45 minutes Trial and demonstration test A trial test was carried out to demonstrate how the experiment is to be conducted and how to make proper observations 45-55minutes Clamping of specimen to machine This procedure was keenly carried out to ensure that the specimen is free to rotate at the ends so it can be a pinned support 55-100 minutes Loading incrementally The loads were imposed on the specimen to produce compression forces on the specimen. Loads were added in multiples of 1000N up to 5000N after which the force was added in multiples of 25N. 100-120 minutes Buckling observed and recorded At 5275N, the strut failed by buckling. The deformation and the load were recorded 120-140minutes Dismantling of devices After the successful experiment the specimen was removed from the testing machines and machine cleaned. Discussion From the tasks conducted the behavioral properties of beams, columns and struts have been assessed. For the beam in task 1 conditions were set for purposes of calculation and the deflection at mid span was found to be 0.4565mm and the slope at D was 0.001316 and the engineer can now use the information for design of the beam and in loading. Task 2 found the safe moment of resistance as 76KN/m. this value can be increased if the number of steel bars is increased. Task 3 and 4 involved the analysis of a column and a strut respectively. The critical load of these compression members is determined as it is crucial during load analysis. The pinned strut has a higher critical load and is applicable where the design load is high. To increase the critical load the engineer can decide to reduce the length of the strut which in most cases is inapplicable as the height of a structure is fixed. All this analysis carried out are important in construction and should be carried out accurately. Conclusion The assignment has been a success and the purpose achieved. The behavioral characteristics of the beam, column and struts have been properly determined as illustrated in the tasks completed REFERENCES Christensen, R.M., 1986. Mechanics of low density materials. Journal of the Mechanics and Physics of Solids, 34(6), pp.563-578. Schwartz, V.V., Alperovich, T.A., Palej, S.M., Petrov, E.A. and Vilkovyskaja, G.B., 1984. Strength of Materials. In Illustrated Dictionary of Mechanical Engineering (pp. 21-33). Springer Netherlands. Timoshenko, S., Timoshenko, S.P., Timoshenko, S.P. and Timoshenko, S.P., 1956. Strength of materials (Vol. 210). New York: van Nostrand. CE IIT, Kharagpur Appendices E- Elastic modulus I-Moment of inertia KN/m – kilo Newtons per meter N/m- newton per meter GN/m²- Giga Newtons per meter squared b- Breadth of section h- Height of section d- Depth of section m- Modular ratio π- Pi F- Force A-area R1- Reaction 1 R2- Reaction 2 Read More

0013m2 d = depth of section = 460mm Simplify this to: We can use aquod ratio equation to find the valve for “h” using: Therefore, h = we can now use the following equation to find the moment of resistance (based on compressive forces) Therefore, the moment of resistance (compressive) =. We can now use the following equation to find the moment of resistance (based on tensile forces) Moment of resistance (tensile) = Therefore, the safe moment of resistance which the beam can carry within both limiting stress values is Task 2b): New depth of beam and tension area Transpose for Therefore: To find d: Using: Now, the service loads have been up rated by 25%, therefore: To find m, we need to multiply the moment of resistance of the concrete reinforcement () calculated previously by 1.

25 therefore: Substitute To solve for “d” Therefore: To find “A”: Therefore, the new area of steel for tension reinforcement required is Task Three On the tensile edge: This can also be rewritten as: Where: Force = 400kN Therefore: Therefore, the stress on the tensile edge = 1.416MN/m2 (tensile) On the compressive edge: Where: Force = 400kN Therefore: Therefore, the stress on the compressive edge = -4.243MN/m2 (compressive) The neutral axis mathematically A + B = 0.6 A = 0.

6 – B Therefore, the distance from the bottom edge of the circle (column) =0.4498m Task 4a): Buckling Load of the Strut First find “I” Now find buckling load Because the bottom of the Strat is built into a ground socket and the top is unrestrained, the value for n = 0.25 If the end conditions of the same strut were pinned at both the top and bottom, calculate the new buckling load. “I” is the same = It will be stronger. Task 4b): New buckling Load of the Strut Since the end conditions are the same, The buckling load for this case, the ends are both pinned thus n=1.

For such a strut there is no bending moments at the ends. Therefore; The pinned strut at both ends is much stronger than the strut that has one end fixed and the other unrestrained. Task 4c): Types of end conditions i. Struts with both ends pinned For a pinned strut, the effective length. Due to the ends being pinned, the column does not bend at all as the moments are shed off. The buckling load is calculated by the formula ii. One end fixed and other end free. The struts deflects in the unrestrained end through a distance.

The effective length and therefore the buckling load is calculated as follows; iii. Both ends fixed The strut has the highest deflection at the center of the strut. These moments act at the opposite direction from the direction of the slope at the ends. Effective length . The buckling load is calculated as follows; iv. One end fixed and the other pinned The struts deflect such that the greater deflection is closer to the fixed end. Fixed end moments develop at the fixed end when a horizontal force is applied in the pinned end.

The effective length and therefore buckling load can be calculated as Task 5a) Strut experiment (pinned at both ends) Objectives 1. To determine the critical load that would result to buckling 2. To find out the differences between the theoretical value and practical value for the critical load. Introduction Struts are slender columns which fail by buckling. The buckling is sudden and occurs only when the critical load is reached. At the point of buckling the strut undergoes permanent deformation which reduces the stability of the strut and can result in a lot of destruction.

For this reason, it should always be avoided. The strut should be designed for a load less than the critical load. The strut has a variety of uses as in the columns in steel structure and other constructions. The strut are used as pistons in hydraulic systems and in lifting gears. Resistance of a strut to buckling is dependent on the diameter of strut, slenderness of strut and the material used. The fixity of the ends also determine the resistance to buckling.

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