0.00 1.25 2.50 3.75 5. 00 6.25 7. 50−30. 00−20. 00−10. 00- 00
- 00
- 00
- 00
0.00 1.25 2.50 3.75 5.00 6.25 7. 50Shearforce(kN )Coalface distance (m)L 1 L 2 L 3 L 4 L 5 L 6Type 1(a)0.00 1.25 2.50 3.75 5. (^00) 6. 25
−30. 00
−20. 00
−10. 00
- 00
- 00
- 00
- 00
0.00 1.25 2.50 3.75 5.00 6. 25Shearforce(kN )Coalface distance (m)L 1 L 2 L 3 L 4 L 5Type 2(b)Figure 10: Shear force.(^0 1 2 3 4 5 6)
−10. 00
−8. 00
−6. 00
−4. 00
−2. 00
- 00
- 00
0 150 300 450 600 750 900De ection(mm)Points of control along the panel- 75 m
- 25 m
- 50 m
L 1 L 2 L 3 L 4 L 5 L 6Type 1Figure 11: Deflection versus length of spans.Appendix
The not null elements of the matrices푀(24×24)={푚푖푗}and
퐵(24×1)={푏푖}of the problem type 1 are as follows.
Row 1.
푚1,1=푚1,3=1,푏 1 =−푦푝1.(A.1)
Row 2.
푚2,1=푚2,2=푚2,4=1,푚2,3=−1.(A.2)
0.00 1.25 2.50 3.75 5. 00 6.25 7. 50−22. 00−17. 00−12. 00−7. 00−2. 00- 00
0.00 1.25 2.50 3.75 5.00 6.25 7. 50De ection(mm)Coalface distance (m)500 m
1500 m
2500 mL 1 L 2 L 3 L 4 L 5 L 6Type 1Figure 12: Deflection versus depth of the panel.Row 3.푚3,1=푒퐾^1 푥^1 ⋅cos퐾 1 푥 1 ,푚3,2=푒퐾^1 푥^1 ⋅sin퐾 1 푥 1 ,푚3,3=푒−퐾^1 푥^1 ⋅cos퐾 1 푥 1 ,푚3,4=푒−퐾^1 푥^1 ⋅sin퐾 1 푥 1 ,푚3,5=−푒퐾^2 푥^1 ⋅cos퐾 2 푥 1 ,푚3,6=−푒퐾^2 푥^1 ⋅sin퐾 2 푥 1 ,푚3,7=−푒−퐾^2 푥^1 ⋅cos퐾 2 푥 1 ,