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Simplification of forces & Couple Moments.

\( (x-y) \)

\( \sum F_{y} \ , \ \sum F_{x}. \ ① \)

\( F_{R}=\sqrt{\left(\sum F_{x}\right)^{2}+\left(\sum F_{y}\right)^{2}} \)  المحصلة   \(②\)

\( \theta=\tan ^{-1} \frac{\sum F_{y}}{\sum F_{x}} \ ③\)

\( (E) \quad \sum M \ ④\)

\( \sum M_{E}=F_{R y} d \leftarrow d \ ⑤\)

\( (3.D) \)

\( \sum F_{z} \quad, \sum F_{y}, \quad \sum F_{x}. \quad ①\)

\( F_{R}=\sqrt{\left(\sum F_{x}\right)^{2}+\left(\sum F_{y}\right)^{2}+\left(\sum F_{z}\right)^{2}} \) المحصلة  \(②\)

\( \alpha=\cos ^{-1}\left(\frac{\sum F_{x}}{F_{R}}\right), \beta =\cos ^{-1}\left(\frac{\Sigma F y}{F_{R}}\right) \quad ③\)

\( \gamma =\cos ^{-1}(\frac{\sum F_{z}}{F_{R}})\)

\( \sum M \quad ④\)

\( d \) الـ \(\longleftarrow \sum M \quad ⑤\)

Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point  O .

 

 

\( \stackrel{+}\longrightarrow \quad \Sigma F_{x}=-450 \sin 30=-225 N \)

\( +\uparrow \sum F_{y}=200-450 \cos 30=-190 N \)

\( F_{R}=\sqrt{(-225)^{2}+(-190)^{2}}=294 N \)

\( \theta=\tan ^{-1}\left(\frac{-190}{-225}\right)=40 \cdot 2^{\circ} \)

\( \sum M_{O}=-200+(200)(3 \cdot 5)-450 \cos 30(1.5)+450 \sin 30(0.2). \)

\(=-39.6 N.m\)

\(=39.6 N.m\)

 Replace the loading on the frame by a single  resultant force. Specify where its line of action intersects  member  \(C D,\) measured from  \(C .\)

\( \stackrel{+}\longrightarrow \quad \sum F_{x}= -500 \cos 60-250\left(\frac{4}{5}\right)=-450 N \)

\( +\uparrow \sum F_{ y}=-250(\frac{3}{5})-300-500 \sin 60=-883 N \)

\( F_{R}=\sqrt{(-450)^{2}+(-883)^{2}}=991 N \)

\( \theta=\tan ^{-1}\left(\frac{-883}{-450}\right)=63^{\circ} \)

\( \sum M_c=400-300(3)-250\left(\frac{3}{5}\right)(6)-500 cos 60 (2)-500 sin 60 (1)=2333 N.m \)

\( M_{c}=F_{y} * d \)

\(-2333=-883*d\longrightarrow d=\frac{2333}{883}=2.64 m\)

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