PRINCIPLES OF STRUCTURAL DESIGN BY RAMS. GUPTA

# PRINCIPLES OF STRUCTURAL DESIGN BY RAMS. GUPTA

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**INTRODUCTION**

Snow is a controlling roof load in about half of all the states in the United States. It is a cause of frequent and costly structural problems. Snow loads are assumed to act on the horizontal projection of the roof surface.

Snow loads have the following components:

1. Balanced snow load

2. rain-on-snow surcharge

3. Partial loading of the balanced snow load

4. Unbalanced snow load due to drift on one roof

5. Unbalanced load due to drift from an upper roof to a lower roof

6. Sliding snow load For low-slope roofs, ASCE 7-10 prescribes a minimum load that acts by itself and not combined with other snow loads.

The following snow loading combinations are considered:

1. Balanced snow load plus rain-on-snow when applicable, or the minimum snow load

2. Partial loading (of balanced snow load without rain-on-snow)

3. Unbalanced snow load (without rain-on-snow)

4. Balanced snow load (without rain-on-snow) plus drift snow load

5. Balanced snow load (without rain-on-snow) plus sliding snow load

### DEAD LOADS

Dead loads are due to the weight of all materials that constitute a structural member. This also includes the weight of fixed equipment that is built into the structure, such as piping, ducts, air conditioning, and heating equipment. The specific or unit weights of materials are available from different sources. Dead loads are, however, expressed in terms of uniform loads on a unit area (e.g., pounds per square foot).

The weights are converted to dead loads taking into account the tributary area of a member. For example, a beam section weighting 4.5 lb/ft. when spaced 16 in. (1.33 ft.) on the centre will have a uniform dead load of 4.5/1.33 = 3.38 psf. If the same beam section is spaced 18 in. (1.5 ft.) on centre, the uniform dead load will be 4.5/1.5 = 3.5 psf.

The spacing of a beam section may not be known to begin with, as this might be an objective of the design. Moreover, the estimation of a dead load of a member requires knowledge as to what items and materials constitute that member. For example, a wood roof comprises roof covering, sheathing, framing, insulation, and ceiling. It is expeditious to assume a reasonable dead load for a structural member, only to be revised when found grossly out of order. A dead load of a building of light-frame construction is about 10 lb/ft. for a flooring or roofing system without plastered ceilings and 20 lb/ft. 2 2 with plastered ceilings.

For a concrete flooring system, each 1 in. the thick slab has a uniform load of about 12 psf; this is 36 psf for a 3 in. slab. To this, at least 10 psf should be added for the supporting system. Dead loads are gravity forces that act vertically downward. On a sloped roof, the dead load acts over the entire inclined length of the member.

**STRESS DISTRIBUTION IN BEAM **

The transverse loads on a beam segment cause a bending moment and a shear force that vary across the beam cross-section and along the beam length. At point (1) in a beam shown in Figure 16.1, these contribute to the bending (flexure) stress and the shear stress, respectively, expressed as follows: f and f b v My I = (16.1a) VQ Ib = (16.1b) where M is bending moment at a horizontal distance x from the support y is the vertical distance of the point (1) from the neutral axis I is a moment of inertia of the section V is the shear force at x Q is a moment taken at the neutral axis of the cross-sectional area of the beam above point

(1) b is the width of the section at (1) The distribution of these stresses is shown in Figure 16.2. At any point (2) on the neutral axis, the bending stress is zero and the shear stress is maximum (for a rectangular section).

On a small element at point (2), the vertical shear stresses act on the two faces balancing each other, as shown in Figure 16.2. According to the laws of mechanics, the complementary shear stresses of equal magnitude and opposite sign act on the horizontal faces as shown, so as not to cause any rotation to the element.

If we consider a free-body diagram along the diagonal a–b, as shown in Figure 16.3, and resolve the forces (shear stress times area) parallel and perpendicular to the plane a–b, the parallel force will cancel and the total perpendicular force acting in tension will be 1.414f A. Dividing by the area 1.41A along a–b, the tensile stress acting on a plane a–b will be f v v. Similarly, if we consider a free-body diagram along with the diagonal c–d, as shown in Figure 16.4, the total compression stress on the plane c–d will be f.

Thus, the planes a–b and c–d are subjected to tensile stress and compression stress, respectively, which has a magnitude equal to the shear stress on the horizontal and vertical faces. These stresses on the planes a–b and c–d are the principal stresses (since they are v not accompanied by any shear stress). The concrete is strong in compression but weak in tension. Thus, the stress on the plane a–b, known as the diagonal tension, is of great significance. It is not the direct shear strength of concrete but the shear-induced diagonal tension that is considered in the analysis and design of concrete beams.

**COMBINATION OF COMPRESSION AND FLEXURE FORCES:**

THE BEAM-COLUMN MEMBERS Instead of axial tension, when an axial compression acts together with a bending moment, which is a more frequent case, a secondary effect sets in. The member bends due to the moment. This causes the axial compression force to act off-centre, resulting in an additional moment equal to axial force times lateral displacement. This additional moment causes further deflection, which in turn produces more moment, and so on until an equilibrium is reached. This additional moment, known as the P–Î” effect, or the second-order moment, is not as much of a problem with axial tension, which tends to reduce the deflection.

There are two kinds of second-order moments, as discussed in the following sections. M EMBErs without s sideways Consider an isolated beam-column member AB of a frame with no sway in Figure 12.2. Due to load w, u on the member itself, a moment M results assuming that the top joint B does not deflect with respect to the bottom joint A (i.e., there is no way). This causes the member to bend, as shown in Figure 12.3. The total moment consists of the primary (first-order) moment, M u1 , and the second-order moment, P Î´. Thus, u =+Î´ (12.3) where M u1 MMP 1 noswayuu is the first-order moment in a member assuming no lateral movement (no translation). M EMBErs with s sideways Now consider that the frame is subject to a sideway where the ends of the column can move with respect to each other, as shown in Figure 12.4. M u2 u1 is the primary (first-order) moment caused by the lateral translation only of the frame. Since the end B is moved by Î” with respect to A, the second-order moment is P u Î”.

**REINFORCED CONCRETE BEAMS**

A concrete beam is a composite structure where a group of steel bars are embedded into the tension zone of the section to support the tensile component of the flexural stress. The areas of the group of bars are given in Appendix D, Table D.2. The minimum widths of the beam that can accommodate a specified number of bars in a single layer are indicated in Appendix D, Table D.3. These tables are very helpful in designs.

Equation 14.1 in the case of beams takes the following form similar to wood and steel structures: MM, where M u ≤Ï† (14.2) un is maximum moment due to the application of the factored loads M is nominal or theoretical capacity of the member Ï†, is strength reduction (resistance) factor for flexure n According to the flexure theory, M n = F bS, where F is the ultimate bending stress and S is the section modulus of the section. The application of this formula is straightforward for a homogeneous section for which the section modulus or the moment of inertia could be directly found. However, for a composite concrete–steel section and nonlinear stress distribution, the flexure formula presents a problem.

A different approach termed the internal couple method is followed for concrete beams. b In the internal couple method, two forces act on the beam cross-section represented by a compressive force, C, acting on one side of the neutral axis (above the neutral axis in a simply supported beam) and a tensile force, T, acting on the other side. Since the forces acting on any cross-section of the beam must be in equilibrium, C must be equal and opposite of T, thus representing a couple. The magnitude of this internal couple is the force (C or T) times the distance Z between the two forces called the moment arm.

This internal couple must be equal and opposite to the bending moment acting at the section due to the external loads. This is a very general and convenient method for determining the nominal moment, M n, in concrete structures.

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