4000 kg4=0.12BAnswer: 18.15°H=0.2Fig. P7.507-50 A small wedge is placed beneath corner B of the 4000-kg block of marble.Determine the largest angle e for which the wedge is self-locking; that is, thewedge will not slide out from under the block. Neglect the mass of the wedge andthe small angle between surface AB and the horizontal.

Answered: Image /qna-images/answer/1d99e260-8c8c-40e2-a3d9-ef5dc85b0ab2.jpg

Answered: 4000 kg 4=0.12 To B Answer: 18.15° "4=…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Q1: It is required that the resultant force acting on the evebolt in the Figure to be directed along the positive x axis and that F2 have a minimum magnitude. Determine this magnitude, the angle 0, and the corresponding resultant force. F, - 800 N 60 F2
1)Determine the horizontal and vertical components of the deflection at joint B B A 35 kN 4m 84 kN Plate #5 3m 4m E = 200 Gpa A 1200 mm²
Consider the element in with Txy = 30 MPa. 80 MPa Part A 01, 02 Determine the principal stresses. Enter your answers in megapascals to three significant figures separated by a comma. = T vec Submit VE ΑΣΦ | 41 Txy Request Answer
y = 125 EZL -(-x5 + 2L²x³ – L*x) Eq. (1) where L=6 m, E=5x10*kN/m², Z=3x10+m“, w,-250 kN/m, and x is distance of deflection at any point from A to B. (a) Use bisection method to determine point x at maximum deflection (i.e, the value of x where dy/dx = 0). Use initial guess of x= 0 m and x4= 6 m to achieve ɛs = 7%. Only show the 1* and 2nd iterations in full calculation and use 4 decimal places for all your answers. Summarize your answers in Table 1. (a) (x = L, y = 0) (r = 0, y = 0) B (b) Figure 1(a) The original beam, 1(b) The deflected beam lteration l: xi, Хa, Х, fx), fx.), { х) х fx) Iteration 2: x, Xu, Xr, f(x1), f(x-), f(x1) x f(x;), Eal% Table 1 i f(x) f(x,) f(x)xf(x,) 1 6. 3 4 (b) Find the value of maximum deflection, y by substituting the x value from part (a) into Eq. (1)
Situation #2 El is Constant (Assume El = 1 kN-m²) B kN/m All units must be in kN or m 7 m B Determine the location of the maximum upward deflection from B. Determine the maximum upward deflection. Determine the maximum downward deflection A= 8 B= 6
= For the section shown, if a = 15 mm, b 30 mm, h = 120 mm, 1 = 7646787 mm^4 and M = 30 kN.m, then the bending stress at the bottom of the hole is b hole h a 218.6 MPa zero -75.63 MPa O 75.63 MPa -218.6 MPa
The 75-lb block has a downward velocity of 2 ft/s when it is 3 ft from the top of the beam. Determine the maximum bending stress in the beam due to the impact, and calculate the maximum deflection of point B. Ew = 1.9(103) ksi.
Use the moment-area method to determine the lateral deflection at point D for the structure shown in Figure 1. 59 kN 100 kN 1m - 1m B D E = 200 GPa I = 1x10-4 m* 4 m A TTTT t
: Check the slab thickness for deflection control if slab thickness = 170 mm and the values of flexural stiffness (a) as shown on the beams in the figure below. fc = 21 MPa, fy = 420 MPa and Cols. = (375*375) mm. 7.6m 7.6m 18:57 1257 B 1. 3792 3.79 89 89 3 379 +379 89 89 7.6m - 6m 6m 6m
3. Using Superposition method for the beam below calculate: (Given E= =29 x 10³ ksi) a) the deflection and Slope at C b) the deflection and Slope at B 15 lb/in. -30 in.- B 125 lb 10 in. с 1.75 in. H Answer: a) Deflection at C = -0.364 in; Slope (rotation) at C = - 12.55 x 10 -³ rad b) Deflection at B = -0.24 in; Slope (rotation) at B = - 12.08 x 10 -³ rad
4:43 Determine the largest angle B for which the uniform box can be in equilibrium. 0.5 m -Hs= 0.45 O 22.6 degrees O 22.9 degrees O 22.2 degrees O 22.0 degrees
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