calculate the forces in AB and BC

QUESTION 2 Question 2 A cross-section of a beam is shown in Figure Q2. If the shear force in this section is V = 125 KN, determine the value and the location of the maximum shear stress in the section. In Figure Q2, a = 30 mm and the origin of the coordinate system is at centroid of the cross section. 7 y= Z= A a AY S= 20 4a mm; mm; O Figure Q2 Answer The vertical coordinate (y-coordinate; the y-axis serves as the axis of symmetry of the cross- section.) and horizontal coordinate (z-coordinate) of the location where the maximum shear stress occurs in the section are ← a The vertical distance from the location where the maximum shear stress occurs in the section to the bottom side (AB cross section can be calculated as Distance = mm (units: mm) 3a Second moment of area The second moment of area employed in the equation to calculate maximum shear stress can be calculated as I₂ = a (units: mm²) Shear stress The second moment of area employed in the equation to calculate maximum shear…

/Find the two forces Fbc & Fac as shown :in the figure below Fab=30 N Fbc 50 30 Fac (Fbc=55 N and Fac= 70 N) O (Fbc=40 N and Fac= 50 N) O (Fbc=46 N and Fac= 59 N) O

The beam is supported by a fixed support at point A. There are 2 distributed loads applied to the beam. Neglect the weight and thickness of the beam. Take the origin for all functions to be at A. , i.e. start at the left and go right. Must use positive sign convention for V and M. W1 A W2 d1 Values for the figure are given in the following table. Note the figure may not be to scale. Variable Value di 3 m W1 6 (kN)/m W2 3 (kN)/m a. For the interval 0 < x < 3 m, determine the equation for the Shear Force as a function of x, V(x). b. For the interval 0 < x < 3 m, Use integrals to determine the equation for the Moment as a function of X, M(x). o Enter the coefficient of the x term as a fraction and the rest as decimals. for example, x4 1.4x c. Determine the magnitude of the max shear on the beam,Vmax d. Determine the magnitude of the max bending moment on the beam, Mn max

Part A: Q1 to Q4 Vc 0.9 m Figure 1 VB 2.4 m 1.2 m B 12 VA 15 m 1.2 m A

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3) A point mass is attached to one end of a beam levered against a spike which acts as the pivot. The point mass weighs 250N and the beam has a uniform mass of 10kg and is 8m long. Parts a and b Part c Fapplied 3 of 6mov 1m 8m Krist 0 = 60° Scale 250 N Scale a. If a scale placed beneath the point mass reads 250N, what is the magnitude and direction for the force at the pivot point? Direction: North, 90° from x-axis Magnitude to k* 250N = 2500 N b. For the same scenario, how far must the pivot point be located relative to the left edge of the beam? (Hint, both the boxes weight and the force from the scale act at the farthest right edge of the beam.) www.acta 0672nos onl c. The pivot is then moved to be 6m from the left edge of the beam which now makes a 60 angle with the floor. How much downward force would be needed at the left-most edge of the beam to reduce the reading on the scale to zero? (Hint: the beam is still in static eq. when the scale reads zero.)

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1. Note: for part a), you can do this without a calculator; it might be easier to work out the problem symbolically, and then substitute the numbers at the end of the problem. В 2L The axial rod is fixed to the two walls, as shown above. Portion AB is composed of one material (EAB = E), while portion BD is composed of a different material (EBD = 2E). Portion AB has cross sectional area AAB = A, while portion BD has cross sectional area ABD = 2A. EAB = E = 15×106 psi AAB = A = 1 in² EBD = 2E = 30×10° psi ABD = 2A = 2 in² LAB = L = 10 in. LBC = 2L = 20 in. LcD = L = 10 in. An external force, P = 100 kips, is applied at point C as shown below. a) Determine the reaction forces at wall A and D, Ra and Rp. Specify the direction, (→) or (+). b) Deflection at point C. Specify the direction, (→) or (+). (Ans: RA = 100/7 kips →, Rp = 600/7 kips →, & = 1.429x10² in –)