2. A simply supported beam is carrying the following loads: Values . Pm (refer to the table at the next page) KN 100 KN w = (# of letters of your first name times 4) kN/m = 16 kN/m The beam has a rectangular cross-section with the following dimensions: b = 125 mm . h = (refer to table at the next page) mm = 350 mm g = (refer to table at the next page) m = 8 m (a) What are the reactions at support D? (b) Sketch the shear force and bending moment diagrams. (c) Locate all points of zero shear. (d) Determine the maximum positive and negative (if possible) shear force and bending moment values. (e) Calculate the moment of inertia of the section. (f) Solve for the maximum positive and negative (if possible) flexural and shearing stresses. A W g B g FEP2 Pm C D

2. A simply supported beam is carrying the following loads: Values . Pm (refer to the table at the next page) KN 100 KN w = (# of letters of your first name times 4) kN/m = 16 kN/m The beam has a rectangular cross-section with the following dimensions: b = 125 mm . h = (refer to table at the next page) mm = 350 mm g = (refer to table at the next page) m = 8 m (a) What are the reactions at support D? (b) Sketch the shear force and bending moment diagrams. (c) Locate all points of zero shear. (d) Determine the maximum positive and negative (if possible) shear force and bending moment values. (e) Calculate the moment of inertia of the section. (f) Solve for the maximum positive and negative (if possible) flexural and shearing stresses. A W g B g FEP2 Pm C D

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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Concept explainers

System of Forces

A force is an external influence or agent that changes the state of a system. According to Newton's second law, the force applied to a body is proportional to the change in momentum of a body. Hence, whenever a force is applied to a body, its momentum changes. If the body is at rest, under the action of force, its state of rest changes to motion. In other words, the magnitude of the momentum of the body changes from zero to a measurable value. Similarly, if the body possesses some initial velocity, it has some associated momentum, the effect of force can cause the body to halt or come to rest, that is, the momentum reduces to zero.

Question

2. A simply supported beam is carrying the following loads: Values
. Pm (refer to the table at the next page) KN 100 KN
w = (# of letters of your first name times 4) kN/m = 16 kN/m
The beam has a rectangular cross-section with the following dimensions:
b = 125 mm
.
•
h = (refer to table at the next page) mm = 350 mm
g = (refer to table at the next page) m = 8 m
(a) What are the reactions at support D?
(b) Sketch the shear force and bending moment diagrams.
(c) Locate all points of zero shear.
(d) Determine the maximum positive and negative (if possible) shear force and bending
moment values.
(e) Calculate the moment of inertia of the section.
(f) Solve for the maximum positive and negative (if possible) flexural and shearing stresses.
W
mim
A
g
B
g
FEP2
Pm
C
D

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