Contoh Soal dan penyelesaian Statis Tak Tentu Teknik Sipil

 Contoh Soal Statis Tak Tentu

Conroh Soal

Penyelesaian:

A. Angka kekakuan dan ditribusi faktor

Joint A= KAB = 0.75 EI/6.5= 0.115 ; Df AB= 0 (Karena tumpuan jepit)

Joint B= KBA = 0.75 EI/60.5 = 0.115 ; Df BA = 0.115/0.329 = 0.35

Joint B= KBC = 1.5 EI/7 = 0.214 ; Df BC = 0.214/0.329 = 0.65

(karena joint B ada dua arah kanan dan kiri maka dijumlahkan)

Joint B Total = KBA+KBC = 0.329 ; Df Total = 0.35+0.65 = 1.00 (OK)

Joint C= KCB = 1.5EI/7 = 0.214 : Df CB = 1

*Tulisan tangan*

B. Momen Primer

*Tulisan tangan*

C. Tabel Cross

D. Free Body

E. Analisis Pengaruh Bidang D

Analisis penggambaran Bidang Momen

MA = 214.956 kg

(2m) = RVA (2)-214.956=168.606(2)-214.956=122.256 kg.m

(3.5m) = RVA(3.5)-214.956-150(1.5)=168.606(3.5)-214.956-150(1.5)=150.165 kg.m

(6.5m) = RVA (6.5)-214.956-150(4.5)-200(3)=168.606(6.5)-214.956-150(4.5)-200(3)= -394.017 kg.m

MB = -394.015 kg

(1m) = 406.288(1)-394.015-100(1)(0.5) = -37.727 kg.m

(2m) = 406.288(2)-394.015-100(2)(1) = 218.561 kg.m

(3m) = 406.288(3)-394.015-100(3)(1.5) = 374.849 kg.m

(4m) = 406.288(4)-394.015-100(4)(2) = 431.137 kg.m

(5m) = 406.288(5)-394.015-100(5)(2.5) = 387.425 kg.m

(6m) = 406.288(6)-394.015-100(6)(3) = 243.713 kg.m

(7m) = 406.288(7)-394.015-100(7)(3.5) = 0.001 kg.m

Sekian semoga bermanfaat dan terima kasih.