Compute for the Shear and Moment Equation each segment (kindly include the FBD). Compute the Reactions and Draw the detailed Shear and Moment Diagram indicating values. 3,000 lby800 lb/ftA8,000 lb-ft | CDE5 ft4 ft7 ft4 ft If the shear-force diagram is positive and looks like this ...VEVDLinearLinear(first order)(first order)Constant(zero order)VBVFVc.then the bending-moment diagram looks like this:..MFMBMDQuadratic(second order)Linear(first order)Quadratic(second order)MEMcMASlope becomesmore positiveSlope becomesless positiveConstant slope(an arch)(a hill thatgets steeper)(an upward ramp)V.FVcVAVBLinearConstantLinear(zero order)(first order)(first order)VD... then the bending-moment diagram looks like this:MEMcQuadratic(second order)MAQuadratic(second order)MDLinear(first order)MBSlope becomesmore negativeSlope becomesless negativeConstant slope(a valley)(a waterfall)|(a downward ramp)Shear-Force Diagram VBending-Moment Diagram MLoad Diagram wEquationRule 1: Concentrated loads create discontinuities in the shear-force diagram. [Equation (7.5)]Slope = VB%3DPositive jumpin shear force VMBSlope = VAMBMAVBPoAV = PoVMAXAXBXBXAXBXARule 2: The change in shear force is equal to the area under the distributed-load curve. [Equation (7.3)]AV= V½ – V¼ = J",w(x) dxw(x) dxMBw(x)XAXAMMBBMΔνVg - VA = [" w(x) dxXAXBXAXBXBXARule 3: The slope of the V diagram is equal to the intensity of the distributed load w. [Equation (7.1)]WBSlope = WBMBWBSlope = w"AMMBMAdV= w(x)dxXAXBXBXAXBRule 4: The change in bending moment is equal to the area under the shear-force diagram. [Equation (7.4)]AM=M;-M= | "V(x) dxV(x) dxXAXAVAMBAMMB - MA = |" V dxMXAXAB VBXBXAXBXARule 5: The slope of the M diagram is equal to the intensity of the shear force V. [Equation (7.2)]Slope = VSlope =w-Slope = VAMBMB%3DdM= VdxXBXAXBXAXBRule 6: Concentrated moments create discontinuities in the bending-moment diagram. [Equation (7.6)]Negative jump inNo effect onshear force Vbending momentMMBVBMAAM = -MoXBMoXBXAXBAXA

Answered: Image /qna-images/answer/dbb204da-b2a8-48e6-84d9-db8537c4a1fb.jpg

Answered: Compute for the Shear and Moment…

Compute for the Shear and Moment Equation each segment (kindly include the FBD). Compute the Reactions and Draw the detailed Shear and Moment Diagram indicating values

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: P. = 30KN Pz = 20KN 4m a = 1 L = 6m R2

A: Let R1=RAR2=RBRA+RB=50KN∑MA=030×1+20×5-6RB=0RB=21.666 KNRA=28.334 KN

Q: a) Find the reactions at the supports. (Report here in Moodle the reaction at B) b) Derive equations…

Q: B. The bent beam ABC is attached to pin at C and rests against a roller support at B. Neglecting the…

Q: a structures. All members have the same EI and are inextensible. P P 工作上有 + L L 4 LOL YLE

Q: PROBLEM 2: Refer to Example 4.9 page 131, and Determine the force in HI using section 1-1. Having…

Q: Use 3 Decimal places

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

Q: Determine the reactions and member forces of the truss shown below using Method of Joints.

A: Question is based on overhanging truss supported on hinge and roller support . Loading condition on…

Q: framework ABC consists of two rigid bars AB and BC, each having length b (Figure 4a). The bars have…

A: given figure span of AC is L1 cosα=L12bL1=2bcosαα=45° and b=100mmL1=2×100×cos45=141.42mm

Q: Consider the following truss system in figure 4: (a) Calculate the reactions at joints A and B. (b)…

Q: Determine the reactions in the supports using three-moment equation. 50 k 50 k E B D 8 ft 8 ft 8 ft…

A: Given,. The beam is A continuous beam Having two spans. Each length of the span = 16ft There are…

Q: Problem 5. Find member forces and reactions for the truss with loads shown. 800 N 600 N D 500 N B 2m…

A: Given Problem :

Q: Q.2) For the truss shown in figure below, determine the forces in lo00 4. 2000 N LE 4. in 3m 3m 3m…

Q: Calculate step by step Use 3 decimal places !!! Draw FBD

Q: -3 m - 4 m- -3 m C. D В 3 m E 3 m H 30° F L L L

Q: 3) Compute the reactions and bar forces in all members for the truss shown below. The area of all…

A: Reaction at Supports and Member forces of truss using castigliano's theorem

Q: Using Three Moment Equation Method, Determine the reactions in the supports and draw the shear…

A: For the given beam, the three moment equation can be written as:…

Q: compound beam ABCDE shown determine the reactions at the supports A Show necessary Free body…

A: degree of static indeterminacy, Ds = number of unknown support reactions - number of equilibrium…

Q: Problem 1. Solve for the reactions at the supports and Determine the stress or the internal force…

Q: Situation 1. For truss shown, use method of section. Compute the reaction at B (Bv and Bh) 10 N C E…

Q: 10 ft A В 14 3 20 k 25 ft 0.5 k/ft

Q: For the truss shown in the below figure. Determine the reactions and the force member in BE & BF. D…

Q: YOKNIM A telf 100mm 28 mm 60mm 28XN zm + + B 20mm 80mm 20mm 111 ¥ Im

Q: given the image, find the support reaction components at the pin joint A and the roller B 2 m 5 KN E…

Q: 20 kN 45° All vertical members 1.5 m long All horizontal members 3.0 m long

Q: for the given truss, calculate the support reaction at A in kips. express your answer in 2 decimal…

Q: D. Support Reactions

Q: 1. Determine the reactions at supports A and E for the truss shown below. (Must include a free body…

A: Given: To find: 1. The reactions at support A and E. 2. Free body diagram needed to calculate the…

Q: 30 B kips 3 kips 15

Q: Example 4: A beam with a hinge. Identify and find the supports reactions., And draw the moment and…

A: Given:- To find:- Support reactions Shear force and bending moment diagrams Concept:- Shear force…

Q: QUESTION 1: [Method of Sections] For the simple truss in figure Q1 below, Figure QI: Simple tras…

A: Given:- To find: b To draw free Body Diagram; To find reactions at support…

Q: Calculate the reactions at A and E, and draw the free body diagram of the sectioned-end members.…

Q: P. = 30KN P2= 20KN a = 1 L = 6m R: R2

Q: 2. Determine the reactions at the supports and forces at hinge B acting on member CBA and the…

A: To find support reactions

Q: REACTIONS

Q: Please solve the following probem step by step. Solve eveything!

A: Step 1: RAx = 25k, no other forces against it. Step 2: REy-RAy = 0 Step 3: At Point A, = F x 12 -REy…

Q: Determine the moments at all supports. 2 KN 2 KN 4 kN/m 1 m/ R₁ 4 m Th₁ R₂ m 2 m 4 m 1 m R3 3 m 6…

Question

Compute for the Shear and Moment Equation each segment (kindly include the FBD). Compute the Reactions and Draw the detailed Shear and Moment Diagram indicating values.

3,000 lb
y
800 lb/ft
A
8,000 lb-ft | C
D
E
5 ft
4 ft
7 ft
4 ft

If the shear-force diagram is positive and looks like this ...
VE
VD
Linear
Linear
(first order)
(first order)
Constant
(zero order)
VB
VF
Vc.
then the bending-moment diagram looks like this:
..
MF
MB
MD
Quadratic
(second order)
Linear
(first order)
Quadratic
(second order)
ME
Mc
MA
Slope becomes
more positive
Slope becomes
less positive
Constant slope
(an arch)
(a hill that
gets steeper)
(an upward ramp)
V.
F
Vc
VA
VB
Linear
Constant
Linear
(zero order)
(first order)
(first order)
VD
... then the bending-moment diagram looks like this:
ME
Mc
Quadratic
(second order)
MA
Quadratic
(second order)
MD
Linear
(first order)
MB
Slope becomes
more negative
Slope becomes
less negative
Constant slope
(a valley)
(a waterfall)
|(a downward ramp)
Shear-Force Diagram V
Bending-Moment Diagram M
Load Diagram w
Equation
Rule 1: Concentrated loads create discontinuities in the shear-force diagram. [Equation (7.5)]
Slope = VB
%3D
Positive jump
in shear force V
MB
Slope = VA
MB
MA
VB
Po
AV = Po
V
MA
XA
XB
XB
XA
XB
XA
Rule 2: The change in shear force is equal to the area under the distributed-load curve. [Equation (7.3)]
AV= V½ – V¼ = J",
w(x) dx
w(x) dx
MB
w(x)
XA
XA
M
MB
B
M
Δν
Vg - VA = [" w(x) dx
XA
XB
XA
XB
XB
XA
Rule 3: The slope of the V diagram is equal to the intensity of the distributed load w. [Equation (7.1)]
WB
Slope = WB
MB
WB
Slope = w
"A
M
MB
MA
dV
= w(x)
dx
XA
XB
XB
XA
XB
Rule 4: The change in bending moment is equal to the area under the shear-force diagram. [Equation (7.4)]
AM=M;-M= | "V(x) dx
V(x) dx
XA
XA
VA
MB
AM
MB - MA = |" V dx
M
XA
XA
B VB
XB
XA
XB
XA
Rule 5: The slope of the M diagram is equal to the intensity of the shear force V. [Equation (7.2)]
Slope = V
Slope =w
-Slope = VA
MB
MB
%3D
dM
= V
dx
XB
XA
XB
XA
XB
Rule 6: Concentrated moments create discontinuities in the bending-moment diagram. [Equation (7.6)]
Negative jump in
No effect on
shear force V
bending moment
M
MB
VB
MA
AM = -Mo
XB
Mo
XB
XA
XB
A
XA

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1ter

Step 2

Step 3

Step 4

Step 5

Step 6

Step by step

Solved in 6 steps with 6 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

Determine the reactions at the supports for the beam shown. Please type answer note write by hend.
2 ● Find all reactions (Moment & forces) at A and B through method of consistent deformations Note: you must replace fixed support with MUST pin connection a Draw moment diagram
The structure is subjected to loadings as shown, Draw the shear and moment curves label the maximum value of the shear and moment. Joint A is fixed support.
Stcr1
Using the method of joints, determine the force in each member of the truss shown below (solve for FAB, FAD, FBD, FDE, FBE, FBC, and FCE. Using Inverse Method) 2000 lb 1000 lb B -12 ft D -12 ft- E cl 8 ft 12 ft 6 ft 6 ft Hints: 1. The reaction at the roller support at E is 10,000 lbf upwards. 2. The reaction of the support at C is 7,000 lbf downwards.
Ex-6. A steel tape with a cross-sectional area of 0.039 cm²and weighing 0.909 kg has a length of 30.009 m. when supported throughout 20°C and subject to a tension of 5.5 kg. This tape is used to measure a distance along a uniform 4% grade and is supported throughout during the measurement. The tension applied is 12 kg. The temperature of the tape distance is recorded 381.615 m. What is the corrected horizontal distance? 2°C. The measured slope in Alwa
Show calculations for the reactions at A and E, and the free body diagram of thesectioned-end members. Sketch the normal force, shear force, bending moment diagrams, and the deformed shape for the frameshown below.
65 ll docs.google.com/forms/d/e, :D determine the force in member AB for.5 the truss show in fig, if the reactions at supportes A and B is Ax= 25kN to the left, Ay= 5kN up ,and By= 35kN up 40kN G. 5KN 4m 10kN C 10KN F D!E 4m A B 2m Im,Im, Im!lm 2m 22.5 kN (T) O 32.7 KN (C) 32.7 kN (T) 27.9 kN (C) O أخرى Classifv the fram show in in Fia. as .2
Example 2 Determine the force in members bd, cd & ce as shown in truss by section method? Find reactions at b & a → Σ Fx = 0 +→ 4m 4 4m C 300 kN 4m
1. Determine the reactions at the supports of the beam loaded as shown below: 4 A 10cm 12N 5600 B 10cm C 5cm 6N D
Please show complete solution with the formulas used and illustration
1. Given: Truss shown below with P = 100 N Find: (1) Reactions at A and E and (2) forces in all the members 4 m- B. 60° Ap 60 D E 4 m 4 m