The beam can be endorsed with the load applied at the mid-span. To cast the tested beam, group one is exploited on 07.02.2011 and its dimensions are observed as 1400x250x100. Upon laying the beam in the loading machine, the load is applied gradually, while the applied is kept as upgrading as long as the first crack is found. This test continues until the crack prototype is built at the both corner of the mid-span. The table below demonstrates the results of the deflection observed at all reading of load applied. Get uk essay help.
The following graph depicts the load deflection curve, in which the curve shape is clearly depicting the first crack, appearing as strength line with the slop of 13.529 KN/mm. Best dissertation writer uk.
- Prior to conducting the beam test, cylinders testing for the beam samples have been carried out, whose results are as follows:
Upon conducting the compressive test of concrete cylinder, it is found that average concrete strength is f_c^'=19.2 MPa and so f_cf^'=0.6×√(f_c^' )=2.63 MPa
- Moreover, the critical loading of the tested beam has been observed as Pcr=24.7KN. Upon calculating the theoretical value, the following is observed:
The cracking moment is M_cr=f_cf^'×Z=f_cf^'×(B×D^2)/6=2.63×(100×〖250〗^2)/6=2.74KN.mm
Additionally, the theoretical critical load is P_cr=〖4M〗_cr/L=(4×2.74×〖10〗^3)/1200=9.1KN and this value is out-of-the-way from the value observed in the test (24.7KN). The reason of this difference in the values can be ignoring the steel effect on finding the beam strength f’c. Professional assignment writing online.
- The following drawing depicts the advance expansion of cracks in the tested beam:
In the drawing above, the dots demonstrate the position of the first crack emerged when the load is estimated as 25KN.
- In order to identify the young modulus E for the concrete, E=30GPa is used as theoretical value, as shown below:
The deflection for supported beam can be observed asδ=(P×L^3)/(48×E×I)→P/δ=(48×E×I)/L^3
The first right equation in the above equation is similar to the slope of the load deflection curve. Therefore, the slope is=13.53KN/mm
When using the above equation E=slopexL3/(48xI)E=(slope×L^3)/(48×I)=(13.53×〖1200〗^3)/(48×(100×〖220〗^3)/12)=5.5GPa
It is observed that the experiment value of E is less than the theoretical value and this difference is because of ignoring the steel contribution in strength calculation. Therefore, there is need to identify the strength of the beam. buy thesis paper.
There is need to find k_u=1/0.85γ×f_sy/(f_c^' )×A_s/bd=1/(0.85×0.788)×500/19.2×314/(100×220)=0.56
The column is found as another structure component, which must be taken into account in the design. Buckling has been determined as main concern in the column design. The experiment made for column testing; load is used to the column as long as it buckles. For avoiding the breaking of column, crooked column is used where, column has the following: online coursework writer.
The bottom dimension: B=20.06mm and D=50.86mm
Top: B=20.01mm and D=50.47mm
Therefore, the standard dimension of the column B=20.045mm and D=50.665mm
Where, the crookedness is found with the dimensions as shown below:
The length of the column L=1000mm
It is observed that the column buckles are to the shortest side, hence, the axis of buckling is equal to the longest dimension. Moreover, D=20mm and B=50mm. Here, D is the shortest side.
The drawing below depicts the set-up of the experiment.
Once the column is laid according to the above-mention requirement, the load is applied in the specific upgrading as well as the corresponding lateral deformation is also documented. In addition, the calculation for deflection over the applied is undertaken, which shows the following result: Get academic writer service.
The graph shown below demonstrates the curve of the deflection against the ration of deflection to the applied load.
For having the line of best fit from the curve shown above, the slope is 0.0185 KN-1 which=1/Pcr
Therefore, from the experiment Pcr= 1/0.0185= 54.05 KN and theoretically P_cr=(π^2×E×I)/L^2=(π^2×200×(50×〖20〗^3)/12)/〖1000〗^2=67KN
Since the two values are in close proximity to each other except the little inconsistency is owing to neglecting of steel contribution.