Engineering Science: Behavioural Characteristics of Static Engineering Systems
TASK 1
Q1.
Distribution of shear forces (Dragoni, 2011).
Total length is 3.2+4.8+1.6 = 9.6 m U.D.L = 4.2kN/M
First calculate the reactions at the right and at the left of the beam supports
RA +RB = 90+ 35+(9.6*4.2)= 165.32 kN
Taking moments at RA we find that (90*3.2) + (35*8.0) + (40.32*4.8) =( RB *9.6) = 761.536 kN
Therefore RB = 761.536 /9.6 = 79.33 kN, hence RA =165.32-79.33 = 85.99 kN
At 0M shear force = RA = 85.99 kN
At 1.6M shear force = RA -wx= 85.99 kN-6.72=79.27 kN
At 3.2M shear force = RA-wx= 85.99 -13.44= 72.55 kN
At 4.8M shear force = RA-90-wx= 85.99 – 90-20.16= -24.17 kN
At 6.4M shear force = RA-90-wx= 85.99 – 90-26.88= -30.89 kN
At 8.0M shear force = RA-90-35-wx= 85.99 – 90-33.6= -37.61 kN
At 9.6 M shear force = RA-90-35-wx= 85.99 – 90-35-40.32= -79.33 kN
| Length (M) | 0 | 1.6 | 3.2 | 4.8 | 6.4 | 8.0 | 9.6 |
| Shear force(kN) | 85.99 | 79.27 | 72.55 | -24.17 | -30.89 | -37.61 | -79.33 |
This means that the reaction force at the left is 79.33 kN to balance a shear force of -79.33 kN
Moments of the forces
At RA& RB=M0= o, just before x1.6= M= +85.99 kN-M, just before x9.6 =M= 79.33 kN-M
TASK 2
Q1.
According to (Yu, 2010) the appropriate factor of safety is 6 and the selected ultimate tensile strength of steel to be used is normally 500 MPa.
Taking that this beam is loaded symmetrically and that its weight is negligible in calculating the maximum moment, then reactions at both ends are equal. Therefore, RA=RB= 60/2= 30 kN each reaction.
The maximum bending moment (M) = RA*2.8 at the centre of the beam = 30*103*2.8 = 84 kNm
Maximum allowable stress σ = ultimate tensile strength/factor of safety = 500*106/6 = 83.3 * 106 Pa
Elastic modulus section, S = M/ σ = 84*103 /83.3 *106 = 1008 * 10-6 m3. Hence converting into cm3 we get = 1008 cm3
From the standard table
| 914 x 419 | 388 | 920.5 | 420.5 | 21.5 | 36.6 | 24.1 | 791.5 | 493.9 |
We get the dimensions for the steel as 920.5*420.5
TASK 3
Q1
τ min = Mx (τ) r/J and τ max = Mx (τ) R/J where R= 54/2mm is external radius and r=48/2mm is internal radius and applied torque is 128Nm
J = π (D4 –d2 ) /32 = π (0.0544 -0.0484 ) /32= 313.6 *10-9 m4
But Mx τ = T and
Therefore τ max = TR/J where T=128Nm, R= 27mm=0.027m and J=value calculated above.
τ max = (128*0.027)/ 313.6 *10-9 m4 = 11.02 *106 Pa = 11.02 MPa
(Dragoni, 2011).
Q2
POWER
P= Tω where T is the torque and ω is the angular velocity
But ω = 2πN where N is number of revolutions per minute.
Given shear stress τ = 54MPa, N=3600rpm=3600/60=60rev/s. And D=54mm, d=48mm
Hence R=27mm = 0.027m and r= 24mm = 0.024 m
But Torque T, = J τ max /R = (54 * 10 6 N/m2 * 313.6 *10-9 m4)/0.027m.
Therefore, T= 627.2 Nm
Therefore, Power = 2πNT = 2π * 60 * 627.2 = 236,448 W= 236.5 kW
(Zienkiewicz, 2005).
Q3.
- = τ L/GR
Where L= 0.5 m τ = 25MPa G = 70GPa and R= 6.25mm = 0.0625m
- = 25*106 * 0.5/(0.0625*70*109 ) = 2.86*10-3 radians
Or (180*2.86*10-3 )/π = 0.16 degrees
(Sharpe, 2008).
REFERENCES
Dragoni, E., & Castagnetti, D. (2011). Concentration of shear stresses in shallow periodic notches. The Journal of Strain Analysis for Engineering Design, 46(6), 397-404.
Gorenc, B., Gorenc, B. E., Tinyou, R., & Syam, A. A. (2005). Steel Desingners’ Handbook. NewSouth Publishing.
Sharpe Jr, W. N., Sharpe, J., & William, N. (Eds.). (2008). Springer handbook of experimental solid mechanics. Springer.
Yu, W. W., & LaBoube, R. A. (2010). Cold-formed steel design. Wiley. com.
Zienkiewicz, O. C., & Taylor, R. L. (2005). The finite element method for solid and structural mechanics. Butterworth-Heinemann.
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