Question

P1 P2 B C D E F 20 10 5 20 20 What is the max NEGATIVE moment at E for the following truck to the nearest -01 k-ft? Note use the negative sign. P1 - 38k P2 = 6k L = 4ft

Answer 1

to A 8 E F os f gives the We will draw the Influence Line Diagrams according Muller - Breslau] Principle of proceed further To draw the ILD for BM at E; introduce 20 ** -20 -20" a hinge at point & and rotate it shown alongside. The deeagram this formed ILD for BM 8 at Alsos e x2=17042109 Nocoj moxm negative moment: at E Then; Y = y will >> H-ISXD0 occur if one of two loads 15 out of P L P P2 B and or; s = 7.5 the other lies either to the left or right of the other. Case : Pj lies at B, P2 to the right of it Then; moment 6, Mez48sx Px + Yzx P2] =- (765x38 +S•SX 6) Here; 142 ys 318 K-ft) » 422 18z 5.5 20 20- 20 20 lies at 11 15 11x7.5 15 Case 2 → Pl lies Here; 8. 16 20 20 M to the left of P₂ at B Y 16x7.5 Moment cet E, ME2 =-(Y₃ x P q + Y q X P2) OY) 2 Y 3= 6 €2-- (6x38+7.5X6) or; M 2² -273 K-ifth maximum Negative Moment at 62 -318 K-ftl XOX X Hence; X 8 E -G H -50 35' K -20 1.75 --- f i above, 35 20 The 'ILD for Shear Force at drawn as according the Muller Breslau principle Then: 5 we mest have length IF=1 1.e. JH IF » JH-Tox )= = = 1.75 >> (IHE 1075) also; 13x1.75 = 0.65 2 y = 0.6s Now; inspection; we By general dedure that the manimcem Shear at F will be Negative s with Ps load acting at K and P2 load the free end 943 = 1-75 35 35 can acting at H: Man negative shear , Thus, Velmares at f -(P x y + P₂X HJ) (32 x 0:65 + 15% 1.75) 47:05 K NEGATIVE f its - at F į Hence; monimum Shear volue is 647.1 kips