here for joint c, it\'s a frame not a truess so why there\'s no pin forces(horiz

Question

here for joint c, it's a frame not a truess so why there's no pin forces(horizntal and vertical pin reactions) included in FBD of pin c?

7.1 INTERNAL LOADINGS DEVELOPED IN MPLE 7.4 th e normal force, of the frame e.shear force, and bending moment acting mine the e loaded as shown in Fig. 7-1a. force 1 m 0.5 m 0.5 m 600 N 600 N ortes UTION port Reactions. By inspection, members AC and CD are two- members, Fig. 7-7b. In order to determine the inter nal loadings at ly ? mem determine the force R acting at the end of member AC. e will analyze the equilibrium of the pin at C rees in the vertical direction on the pin, Fig. 7-7b ut eceh Dther to ave mis se R = 848.5N sin 45°-600 N = 0 45° inoram of segment CE is

Explanation / Answer

The answer to your question lies in the fact that as mentioned in the text AC and CD are two force members and they can have force only along their axis .Therefore at C ,we do not have horizontal and vertical forces like in a pin joint. However the force at C can be resolved into components like R cos(theta ), R sin(theta), like in the given problem for Sigma Fy=0 , RSin (theta) ie RSin45 has been used.So a frame can have Two Force members which are identified by inspection and treated accordingly.By definition , a two force member is a member that has forces (and only forces, no moments) acting on it in only in two locations at its ends. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear.

Hope this will help in understanding.

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