METHOD OF JOINTS

Getting the reactions at supports:

∑ 𝑴𝑷 = 0

RA (9.5) = -2.1175kN(1.1875m + 2.375m + 3.5625m) – 5.101kN(4.75m) –

7.1465kN(5.9375m + 7.125m + 8.3125m) - 3.8kN(9.5m) + 1.487kN (0.625m + 1.25m +
1.875m) + (0.744kN)(2.5m) + 0.478kN(2.5m) + 0.956 kN (1.875m + 1.25m +0.625m) +
3.279 kN (9.5m)
RA𝒚 = 19. 45332895 kN
∑ 𝑭𝒚 = 0
𝑅𝑃𝑌 = 3.8kN + 7.14653kN(3m) + 5.101kN + 2.1175 kN(3m) + 1.301kN –19.453kN –
3.279 kN
𝐑𝐏𝐘 = 11.98267105 kN
∑ 𝑭𝒙 = 0
𝑅𝐴𝑋 = 0.744 kN + 1.487kN(3) + 0.744kN + 0.478kN + (0.956kN)(3m) + (0.478kN)
𝑹𝑨𝑿 = 9.773 kN

METHOD OF JOINTS

At Joint A
ΣFy = 0
-1.721kN + 12.273071kN + AB sin(19.596) = 0
AB = 31.463 kN (C)

ΣFx = 0
(31.463kN) cos(19.596) + AC – 1.166 = 0
AC = 30. 807 kN (T)
METHOD OF SECTIONS

∑ 𝑴𝑨 = 0
-0.71kN(0.445m) + 3.158kN(1.25m) + BE cos(19.596)(0.445m) + BE
sin(19.596)(1.25m)=0
BE = 4.331 Kn (C)

METHOD OF JOINTS

At Joint B

ΣFy = 0
3.158 + 31.463 sin( 19.596) + BDsin(19.596) +
4.331 sin(19.596) = 0
BD = 26.378 kN (C)

4.33
-31.463 kN - 4.331 kN
kN kN

At Joint C

ΣFx = 0
-AC-CE = 0
-30.807 kN = CE
CE = 30.807 kN (T)
-30.807kN
CB = 0
∑ 𝑴𝑨 = 0
-0.71kN(0.445m) + 3.158kN(1.25m) – 0.71Kn(0.89m) + 3.158Kn(2.5m) +
DGcos(35.4508)(0.89) +DGsin(35.40((2.5m) =0
DG = 5.009 kN (C)

∑ 𝑴𝑬 = 0
12.273kN(2.5m) – 1.72kN(2.5m) – 3.158kN(1.25m) – (0.71kN)(0.445m) –
(0.71kN)(0.89m) –DGcos(35.451)(0.89m) +DFcos(19.596)(0.89m) = 0
DF = 21.293kN (C)

At Joint E
ΣFx = 0
-4.331 kN -EB cos (19.596) + EC + EG = 0
-4.33kN cos (19.596) + 30.807kN + EG = 0
EG = 26.726 kN (T)
ΣFy = 0
DE + EBsin19.596 =0
-30.807 kN
DE + 4.33kNsin(19.596) = 0
-30.807 kN DE = 1.453 kN (T)
∑ 𝑴𝑨 = 0
(-0.71kN)(0.445m) +3.158 kN(1.25m + 2.5 m +3.75m) – 0.71 kN(0.89 + 1.335) +
FIcos(46.883)(1.335m) + FIsin(46.833)(3.75m) = 0

FI = 5.970 kN (C)

At joint F

ΣFy = 0
-FHcos19.596 +5.967 kN cos(46.883) -
21.293kNcos(19.596) + 0.71 =0
FH= 16.209 kN (C)

-21.293kN -5.970 kN
 At joint G

ΣFy = 0
5. 009 kN -GF + 5.009kNsin(35.4509) = 0
L GF = 2.905 kN (T)

ΣFx = 0
-GI + 26.726 - 5. 009cos35.450
=0

GI=22.646 kN (T)
26.726 kN

At joint I
5.970 kN 4.648 kN
ΣFy = 0

-IH + 5.970kNsin(46.883) +
4.648sin(46.883)=0
22.646 21.742 kN
kN
IH = 7.751 kN