1. Plot the moment (M), axial (N), and shear (S) diagrams as functions of z.a)b)F₁ = 1250 NF₁ = 600 NM₁ = 350 000 N mmF2 = 500 N200 Na = 600 mmb=1000 mma=750 mmb = 1000 mmd)M₁ = 350 000 N mmF₁ = 600 NF₂ =200 Na = 600 mmb = 1000 mmM₁ 175 000 NmmF = 900 Na-250 mmb-1000 mm-250 mm.Figure 1: Schematics problem 1.

Step 1: Step 2: Step 3: Step 4:

Answered: 1. Plot the moment (M), axial (N), and…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Consider the bracket shown in Fig-1. If: F1 = 10 N@ 120° • F2 = 75 N@ -90° then, what is the magnitude of the resultant force on the bracket? 8.66 N 66.5 N -75 N 75 N O ON 66.3 N*m 100 N*m O 10 N
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.)
L=9m W=5kN/m 360 UB 50.7
15 kN/m 50 kN 150 kN m A 3 m E B 6 m 3 m -3 m a) Determine the reactions at supports A (pin) and D (roller). b) Determine the internal loading at x= 3 m (point E) measured from point A. c) Write functions for shear force (V) in terms of x d) Write functions for bending moment (M) in terms of x.
F
A bar with varying cross section is subjected to various forces as shown below. Compute: a) Stresses in each section b) strain in each section c) total extension of the bar. Take E = 2.1x105 N/mm² 400 mm² P1=10kN A 1000 mm B 800 mm mm² P₂ P3=45KN+ 1500 mm 600 mm² 800 mm → P4-35kN
Question 7 Find the value of The reaction force at the support A .
5. Determine the value of the distributed load w and the concentrated loads x, y and z in the load diagram. [x=5.75kN, y=15kN, w= 5kN/m, z=34.25kN] 80 kN-m 5.75 X V (kN) 50N A 5m V (N) M (Nm) B 5 -9.25 6. Determine the value of A, B in the shear force diagram and C in the bending moment diagram. [A =50N; B = -50N; C = 125Nm] 10N/m 10m -5m- C W Z -.x(m) -34.25 50N -x (m) B -x (m)
Find the tension in each segment. 7 ft B 4 ft 50lb 4 ft 5 ft 3 ft 100 lb D Fbc = 47 #, Fba = 83 #, Fcd = 88 # Fbc = 40 #, Fba = 50 #, Fcd = 33 # Fbc = 27 #, Fba = 68 #, Fcd = 18 # ☐ none of the above
A wooden beam is loaded by a distributed load w = 5 N/mm as shown in Figure Q4(a). The cross-section of the beam is shown in Figure Q4(b). Assume that LAB = 1600 mm, LBC = 3400 mm, W=190 mm, Bo = 20 mm, and H = 200 mm, ho=40 mm. Units: use N, mm, MPa as consistent units for all calculations. Decimal points: any decimal points are fine and all the numerical questions have a 10% error tolerance. (a) (b) hol Во H B LAB LBC Cross-section W Figure Q4: (a) A wooden beam is loaded by a distributed load, (b) cross-section of the beam.
QUESTION 1 For the force system shown in the figure, the following additional data are given: M N.m 20 F1 N Z 5 N F2 N 15 7 xr is equal to (in m) F1 N (2.3m, 5.0m) (4.5m, 2.5m) (1.0m, 0m) X₁ M N.m R (7.0m, 4.5m) 10 N F2 N (7.0m, 1.2m) X You have been asked to replace the force system with a single force R acting at a distance xr as shown.
4.) A 20 kg block is suspended by 2 very thin ropes that meet at a 40° angle. What is the tension on the two ropes? AM ThN < Q口 Formula EQA D Untitled Notebook (2) - Untitled Notebook (2)- Satple Praloom s HI UNSM Ntbok2 Sanple Frirm 3 LnlT Sheet O O & C OBE 63 Atuoods Machine Block Suspended by a Strings 92=-a, (m. - mi). 9% = (mz +m.) Untitled Notebook (2) Tz = sine Block Sliding on a Romp Т. mg w/out friction: 9= gsine w/ friction:

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