E 90 С X D 31000 d 3 FB 4 B MA C A b a

Statics-Two Force Members 1. Determine the force P required to maintain equilibrium of the pulley system, and the reactions at the ceiling. 2. Assume W=1.25kN 3. Show and clearly label all Free Body Diagram to determine solution.

Solve correctly

50 kN 30 kN 4 m 3 m t. In the truss shown , AE for all members is constant A = 200 mm2 , E = 200 GPa %3D If member BD is removed, which of the following most nearly gives the displacement (mm) of B relative to D

The beam is supported by a pin at B and a roller at D. The applied loads shown in the figure have 180 lb/ft, and M = 2200 ft-lb. Draw the shear force and bending moment diagrams (in your homework documentation) and determine the equations for V(x) and M(x). 0 at point A at the left magnitudes F1 = 630 lb, F2 = 800 lb, w = %3D Also enter the support reactions as an intermediate check on your work. Take x edge of the beam. 2013 Michael Swanbom 1F, BY NC SA F2 M a Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value а 5.80 ft 12.2 ft 4.64 ft Support Reactions The reaction at B is lb. The reaction at D is lb. Shear Force and Bending Moment Equations In section AB: V(x)= lb and M(x)=| lb-ft. In section BC: V(x)= lb and M(x)=| lb-ft. In section CD: V(x)= lb and M(x)= lb-ft.

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Determine the reactions at the supports A and B. EI is constant

Q1. (а) Figure Q1(a) shows a beam loaded with a point load and a partially UTM uniform distributed load. respectively. support at A and B are roller 3 UTM The M 5 UTM (i) Draw the free body diagram of the structure. * UTM UTM Determine the support reactions at A and B by taking the (ii) structure in equilibrium condition. ITM O UTM - UTM 5 UTM O UTM 30kN 30° 10KN/m F = 840 N 2m 24 m 2m 4m A \3 m G UTM 15 m 2 m STMUTM 5 UTM FIGURE Q1(a) 10 ni FIGURE Q1(b) 一

80 mm, A E -170 mm B D C 125 mm 75 mm Problem 06.085.a - Frame with 2-force member Determine the components of the reactions at A and E if a 930-N force directed vertically downward is applied at B. The components of the reaction at A are: Ax = lb- Ay = lb t The components of the reaction at E are: E₁= lb→

At step 3, at joint F, how does the angles 16.26 and 36.67 gets from?