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)

0=Σ+

AM …………………………………………….. 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

0=Σ↑+

yF (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

W = m . g

= 50 . 9,81

= 490,5 N

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

0'=Σ

→

xF

oWP 15sin= = 129.4095 N

0=Σ+

BM

)215.(15sin)900.()200.(15cos oC

o WNPW −−+=

598,259=CN N

'yFΣ↑+

= 0 …………………………………………………… 1

oCB WNN 15cos−+=

Dari 1

189,21415cos =−= Co

B NWN N

Untuk θ = 0 ‘ P = 0

0=Σ+

MB ;

0)700.(200. =−= CNW 1,140=CN N

Σ↑+

Fy = 0 ; 0=++− CB NNW 3501405,490 =−=BN N

y y'

x'

x 150

W ∝

∝