Question

PROBLEM 3 (25%) For the problem shown below, the free body diagram (FBD) is superimposed showing the reactions forces (the reaction forces are given). Ax = 0; Ay = 9 kip, Cy = 33 kip 1. Draw the shear force diagram. Identify values at kink points (no equations required). 2. Draw the bending moment diagram. Identify values at kink points (no equations required). 3. Determine the value of the maximum positive moment in the beam (Mmax). 4. Compute the vertical distance from the base of the cross section to its centroid, 7. 5. Compute the moment of inertia of the cross-section with respect to horizontal axis that passes through its centroid, Icg. o Ibase = 785 in. (DO NOT RE-COMPUTE I base). 6. Using the maximum positive moment in the beam that you have previously obtained, Mmax, o Compute the normal bending stress at the top of the cross section, Otop. o Compute the normal bending stress at the bottom of the cross section, Obot. 7. Determine the maximum shear stress (absolute value) at section a-a. 2 kip 4 kip/ft a 6 in A, 9 in 10 in 1 in B D a Cross Section (a-a) 10 ft 10 ft 10 ft A, C,

Answer 1

First we have to find the Sheor Force and Bending Moments at various segments: Consider ; ;) Section AB M A 9 kip Shear Force, V = 9 kip Bending Moment, M = 9x » at A, iex=0 v=q kips M = 0 kip.ft at B, ie x= 10ft v=q kip M = 90 kip. ft

11) Section BC A 4 kiplet M WS 2 Sheas Force, V- 9 - 4(x-10) V = -4x + 49 Bending Moment , a kip 2 M= -2?+ 498 – 200 => atB, x = loft V= q kip M 90 kip. ft => at c, n=2oft V= .31 kip M = 20 kip.ft Shear Force 0 at O= - 4x + 49 n = 12.25 ft M at n = 12.25 ff Mr 100.125 kip.ft man

3 Segment De ,2kip ER V= 2 kip M = -22 at c a = 10ft 2 V= 2 kip M = -20 kip. ft at D, x=0 V = 2 kip M-o

Akiplex Bermum c a kip 33 kip Part I - 'a kip q kip 12 kip 2 kip Shear Force Diagram A Bi d ID + ! 12.25 ft -31 kip Part 2 90 kipuft 100.125 kip.ft Bending Moment Diagrang B ! - 20 kip.ft t 19-325 ft

5 Past 3 Maximum Positive Bending Moment (Mmax) = 100.125 kipit Part 4 Distance of Cendrold from base -610 * lin = (qx1x4-5) + (6x1x9.5 (9x1)+(6x1) a in ý 6.5 in, AK lin Part 5 Moment of Inertia about Centoid. (1xq² t Ica 1x9x2) Ce a benet) + (6x13 '+ 6x1x3² 3in 12 2 in T Ica 151.25 in 4

Part 6 My Bending stress, ob T M may = 100.125 kip ft 3:15. T 6.5 in Normal Bending stress at top is 160.125 X 3.5 X 12 Top 151.25 Top 27.803 kiplin2 or ksi Normal Bending stress at bottoms is bottom 100-125 X 6.5 X12 151.25 o bothm 51-635 kgi

7 Part 7 Shear Force at a-a q kip. Area of cross Section, A - 9x1 + 6x1 15 in? a Shear stress 15 Shear stress 0.6 ksi

The problem is solved for shear force diagram, bending moment diagram, bending stress and Shear stress