Jawaban Soal Statika Struktur 3-44 Dan 3-47

Jawaban Soal Statika Struktur

3/44. Determine the external reactions at A and F for the roof truss loaded as shown. The vertical loads represent the effect of the supported roofing materials, while the 400-N force represents a wind load.

Jawab:

∆ACF AC = AF Sin 30o = 10. (0.5) = 5 m AB = BC = 2,5 m (Segitiga ACG sama sisi dan BG = garis tingginya) +

Σ M A = 0 …………………………………………….. 2

= FB.dB + 500.dC + 500.dD + 500.dE + 250.dF - FF.dF = 400.(2.5) + 500.(5 Cos 60o) + 500.(5) + 500.(7.5) + 250.(10) - FF.(10). = 1000 + 1250 + 2500 + 3750 +2500 – 10 FF FF

= 1100

N

+

↑ ΣFy = 0 (1)

= FAy - FA - FBy - Fc - FD - FE - FF + FF = 0 FAy = FA + FBy + FC + FD + FE - FF Dari (1) FAy = 250 + 400 Sin 300 + 500 + 500 + 500 + 250 – 1100 = 250 + 200 + 500 + 500 + 500 + 250 - 1100 = 1100 N

+ →

Σ Fx = 0 = FBX + FAX FAX = - FBX = - 400 Cos 300 = - 346,41 N Arah ←  Terbalik dengan yang di DBB

3/47. The man pushes the lawn mower at a steady speed with a force P that is parallel to the incline. The mass of the mower with attached grass bag is 50 kg with mass center at G. If θ = 150, determine the normal forces NB and NC under each pair of wheels B and C. Neglect friction. Compare with the normal forces for the conditions of θ = 0 and P = 0 Ans NB = 214 N, NC = 260 N With θ = P = 0; NB = 350 N, NC = 140,1 N

y'

y

x' 150

∝ ∝

W

W=m.g = 50 . 9,81 = 490,5 N

Gunakan sumbu X’ – Y’ yang mempunyai kemiringan 150 terhadap sumbu X-Y →

Σ F' =0 x

P = W sin 15o = 129.4095 N +

ΣMB = 0 = W cos15o.(200) + P.(900) − N C − W sin 15o.(215) N C = 259,598 N +

↑ ΣFy ' = 0 …………………………………………………… 1 = N B + N C − W cos15o

Dari 1 N B = W cos15o − N C = 214,189 N

Untuk θ = 0 ‘ P = 0 +

Σ MB = 0 ; = W .200 − N C .(700) = 0

N C = 140,1 N

+

↑ Σ Fy = 0 ; −W + N B + N C = 0

N B = 490,5 − 140 = 350 N

x